Every lesson, checkpoint, FRQ practice set and test of this unit, in course order. Answer keys are marked ✓ KEY. Feedback, model responses and rubrics are shown in full.
11 lessons
Outcome: State what intermolecular forces are and compare their strength with the bonding within or throughout a substance.
Water boils in a pot at 100 °C, and bubbles of steam rise and escape.
Every bubble carries away H₂O molecules — so something that held those molecules in the liquid must have been overcome.
The figure shows the same water before and after: in the liquid the particles crowd together, and in the steam the same particles fly far apart.
Each particle in both boxes is one whole H₂O molecule — nothing inside any molecule changed.
You've already seen that ice melts at 0 °C because the attractions between molecules are much weaker than the covalent bonds within them.
Boiling tells the same story: what breaks is the attraction between separate molecules, never the O–H bonds inside them.
The figure shows two neighboring water molecules: solid lines are the covalent O–H bonds within each molecule, and the dashed line is the attraction between the two molecules.
Attractions like the dashed one — attractions between separate particles — are called 'intermolecular forces', or IMFs.
The separate particles can be atoms, molecules, or ions.
IMFs are much weaker than the covalent, ionic, or metallic bonding within or throughout a substance.
That is why boiling water at 100 °C pulls whole molecules away from one another while every O–H bond stays intact.
State that intermolecular forces (IMFs) are attractions between separate particles, and that they are much weaker than the covalent, ionic, or metallic bonding within or throughout a substance.
Check — multiple choice
Liquid nitrogen boils away at −196 °C. What is overcome as the liquid becomes a gas?
Feedback if chosen: Correct. Boiling pulls whole N₂ molecules away from one another; the intermolecular forces are overcome, and each N≡N bond stays intact.
Feedback if chosen: This option is wrong — you broke the molecules themselves. Boiling separates whole N₂ molecules; the covalent bond within each molecule stays intact.
Feedback if chosen: This option is wrong — you overcame both kinds of attraction. Only the much weaker intermolecular forces are overcome; the covalent bonds survive boiling.
Feedback if chosen: This option is wrong — you left the molecules with no attractions between them. Attractions do act between separate N₂ molecules — they are what holds the liquid together — and boiling overcomes them.
Check — multiple choice
Which of these is an intermolecular force?
Feedback if chosen: This option is wrong — you picked an attraction within one molecule. Intermolecular forces act between separate particles, not inside them.
Feedback if chosen: Correct. The two CO₂ molecules are separate particles, so the attraction between them is an intermolecular force.
Feedback if chosen: This option is wrong — you picked the bonding that extends throughout an ionic solid, not an attraction between separate particles.
Feedback if chosen: This option is wrong — you picked the bonding that extends throughout a metal, not an attraction between separate particles.
Check — multiple choice
How do intermolecular forces compare in strength with the bonding within or throughout a substance?
Feedback if chosen: This option is wrong — you flipped the comparison. Boiling water overcomes the attractions between molecules long before any O–H bond breaks, so IMFs are much weaker.
Feedback if chosen: This option is wrong — you made the two comparable. Water boils at 100 °C without breaking a single O–H bond, so the attractions between molecules must be much weaker than the bonds.
Feedback if chosen: Correct. IMFs are much weaker than the covalent, ionic, or metallic bonding within or throughout a substance.
Feedback if chosen: This option is wrong — you removed the attraction entirely. Neutral molecules do attract one another — that is what holds liquid water together — the attraction is just much weaker than a bond.
Outcome: State what London dispersion forces are and explain how temporary, fluctuating dipoles create them.
Chill argon to −186 °C and it turns into a liquid — its atoms must attract one another.
The figure shows why that is puzzling: an argon atom's electron cloud is, on average, perfectly even around its nucleus.
You've already seen that opposite charges attract and like charges repel (Coulomb's law).
You've already seen that partial charges — δ⁺ and δ⁻ — sit on the two ends of a polar bond.
An argon atom has no bonds, so it has no permanent partial charges anywhere — what could attract it to its neighbor?
The electrons in the cloud are in constant motion around the nucleus.
The figure shows the same atom at two moments: on average the cloud is even, but at any single instant the moving electrons can bunch on one side.
For that instant, the bunched side of the atom is slightly negative (δ⁻) and the exposed side is slightly positive (δ⁺) — a temporary dipole.
An instant later the electrons have moved on, so the dipole fades, reforms, and points a different way — the dipoles are fluctuating, never permanent.
When a temporary dipole appears in one atom, its momentary partial charges shift the neighbor's cloud too, producing an opposite momentary dipole in the neighbor.
The figure shows the result: two momentarily lopsided clouds with opposite partial charges facing each other, and an attraction between the atoms.
The atoms attract because temporary, fluctuating dipoles create Coulombic attractions between particles.
Attractions that arise this way are called 'London dispersion forces', or LDFs.
State that London dispersion forces (LDFs) are attractions that arise because temporary, fluctuating dipoles create Coulombic attractions between particles.
Check — multiple choice
Two neighboring Ne atoms attract each other. What creates the attraction?
Feedback if chosen: This option is wrong — you gave neon a polar bond. A Ne atom has no bonds, so it has no permanent partial charges; the dipoles involved are temporary.
Feedback if chosen: Correct. The moving electrons briefly bunch to one side of each atom, and the temporary, fluctuating dipoles create Coulombic attractions between the atoms.
Feedback if chosen: This option is wrong — you turned the atoms into ions. Neither Ne atom gains or loses an electron; each cloud only shifts to one side for an instant.
Feedback if chosen: This option is wrong — you bonded the atoms covalently. No electron pair is shared; the attraction is an intermolecular force between separate atoms.
Check — multiple choice
The figure shows a Kr atom whose electron cloud has slipped to one side for an instant. What produced the lopsidedness?
Feedback if chosen: Correct. The electrons never stop moving, and at any instant they can happen to crowd one side — that momentary crowding is the temporary dipole.
Feedback if chosen: This option is wrong — you moved the nucleus. The nucleus stays put; it is the electron cloud around it that shifts for an instant.
Feedback if chosen: This option is wrong — you changed the electron count. The atom keeps all of its electrons; they are only unevenly placed for a moment.
Feedback if chosen: This option is wrong — you gave Kr a bond. A krypton atom has no bonds; the shift comes from the electrons' own motion.
Check — multiple choice
A temporary dipole appears in one of two neighboring Ar atoms. What happens to the other atom's electron cloud?
Feedback if chosen: This option is wrong — you kept the neighbor's cloud fixed. The first atom's momentary partial charges shift the neighbor's cloud, producing an opposite momentary dipole in it.
Feedback if chosen: This option is wrong — you paired like charges. Like charges repel; the neighbor's cloud shifts so that opposite partial charges face each other and the atoms attract.
Feedback if chosen: Correct. The momentary dipole in one atom produces an opposite momentary dipole in its neighbor, and the facing opposite partial charges attract.
Feedback if chosen: This option is wrong — you made the dipole permanent. The electrons keep moving, so the dipoles fade and reform — they are temporary and fluctuating.
Outcome: State that London dispersion forces are present between all particles — atoms, molecules, and ions.
London dispersion forces need no bonds, no permanent partial charges, and no special structure.
They arise because temporary, fluctuating dipoles create Coulombic attractions between particles — and any electron cloud can form a temporary dipole.
Every particle — every atom, every molecule, every ion — has an electron cloud.
So London dispersion forces are present between all particles: atoms, molecules, and ions alike.
Even helium, with only two electrons, shows the effect: the figure shows a He atom's cloud even on average and slipped to one side an instant later.
The attraction between two He atoms is weak, but it is real — chilled below −269 °C, helium holds together as a liquid.
Polar molecules and ions also attract in other ways, but London dispersion forces are always present in addition.
State that London dispersion forces are present between all particles, whether atoms, molecules, or ions.
Check — multiple choice
Between which particles do London dispersion forces act?
Feedback if chosen: This option is wrong — you limited LDFs to nonpolar molecules. Any electron cloud can form a temporary dipole, so LDFs act between all particles.
Feedback if chosen: This option is wrong — you excluded ions. Ions have electron clouds too, so temporary dipoles arise in them as well.
Feedback if chosen: Correct. Every particle has an electron cloud, so London dispersion forces are present between all particles.
Feedback if chosen: This option is wrong — you required polar bonds. LDFs come from temporary, fluctuating dipoles, which need no bonds at all — two Ar atoms attract without any bond.
Check — multiple choice
Why are London dispersion forces present between all particles?
Feedback if chosen: This option is wrong — you made every dipole permanent. Most particles have no permanent dipole; the dipoles behind LDFs are temporary and fluctuating.
Feedback if chosen: This option is wrong — you required a polar bond. Single atoms like He have no bonds at all, yet LDFs still act between them.
Feedback if chosen: This option is wrong — you described covalent bonding. LDFs are attractions between separate particles, with no electron pair shared.
Feedback if chosen: Correct. Every particle has an electron cloud, the cloud's electrons are always moving, and their momentary bunching creates the temporary dipoles behind LDFs.
Outcome: State what polarizability is and pick the more polarizable of two particles from how their clouds respond.
Bring the same neighboring particle — carrying the same momentary dipole — up to two different particles, A and B.
The figure shows the two responses at the same instant: A's cloud has slipped far to one side, while B's cloud is still essentially even.
A's cloud distorts easily; B's cloud resists.
Chemists give that ease a name: 'polarizability' is the ease with which a particle's electron cloud is distorted to form a temporary dipole.
A particle whose cloud shifts easily when a charge approaches is highly polarizable — like A.
A particle whose cloud barely shifts is weakly polarizable — like B.
Every particle has some polarizability, high or low, because every electron cloud can be distorted at least a little.
State that polarizability is the ease with which a particle's electron cloud is distorted to form a temporary dipole.
Check — multiple choice
What does a particle's polarizability describe?
Feedback if chosen: This option is wrong — you reached for a permanent dipole. Polarizability is about temporary dipoles: how easily the cloud distorts to form one.
Feedback if chosen: This option is wrong — you counted electrons. Polarizability is not a count; it is the ease with which the cloud is distorted.
Feedback if chosen: This option is wrong — you put the nuclei in charge. Polarizability describes the electron cloud's response, not an attraction between nuclei.
Feedback if chosen: Correct. Polarizability is the ease with which a particle's electron cloud is distorted to form a temporary dipole.
Check — multiple choice
A charge approaches particles X and Y. X's electron cloud barely shifts; Y's cloud slips far to one side. Which particle is more polarizable?
Feedback if chosen: This option is wrong — you picked the cloud that resisted. The more polarizable particle is the one whose cloud distorts more easily, and that is Y.
Feedback if chosen: Correct. Y's cloud shifts easily when the charge approaches, so Y is the more polarizable particle.
Feedback if chosen: This option is wrong — you ignored the different responses. The clouds distorted by different amounts under the same approaching charge, so the polarizabilities differ.
Feedback if chosen: This option is wrong — you reached for a permanent dipole. Polarizability is how easily a cloud distorts under an approaching charge, and the sketch shows that directly: Y's cloud shifted far more.
Outcome: State how polarizability changes with electron count and cloud size, and pick the more polarizable of two particles.
You've already seen that polarizability is the ease with which a particle's electron cloud is distorted to form a temporary dipole.
The figure compares the clouds of F₂ (18 electrons) and I₂ (106 electrons): I₂'s cloud is far larger.
The nuclei hold a large cloud's distant outer electrons only weakly, because interaction strength increases as the distance between charge centers decreases (Coulomb's law).
Weakly held electrons shift easily, so the big cloud distorts easily.
Polarizability therefore increases with the number of electrons in a particle and with the size of its electron cloud.
More electrons and a bigger cloud almost always come together: I₂, with 106 electrons, is far more polarizable than F₂, with 18.
State that polarizability increases with the number of electrons in a particle and the size of its electron cloud.
Check — multiple choice
Which particle is the most polarizable?
Feedback if chosen: This option is wrong — you picked the smallest cloud. Polarizability increases with electron count and cloud size, so tiny He is the least polarizable of the four.
Feedback if chosen: This option is wrong — you stopped at a small cloud. Ne has more electrons than He but far fewer than Kr, and its tightly held cloud still resists distortion.
Feedback if chosen: This option is wrong — you stopped one cloud short. Ar's cloud outgrows Ne's, but Kr's is larger still, and the largest cloud distorts most easily.
Feedback if chosen: Correct. Kr has the most electrons and the largest cloud of the four, so its outer electrons are held least tightly and its cloud distorts most easily.
Check — multiple choice
Which change makes a particle more polarizable?
Feedback if chosen: Correct. With more electrons and a bigger cloud, the outermost electrons sit far from the nuclei, are held weakly, and shift easily.
Feedback if chosen: This option is wrong — you shrank the cloud. Electrons held close to the nucleus are held tightly, so a small cloud resists distortion.
Feedback if chosen: This option is wrong — you reached for a permanent dipole. Polarizability is about how easily a temporary dipole forms, not about any permanent one.
Feedback if chosen: This option is wrong — you strengthened the nucleus's grip. A stronger pull on the same small cloud makes the electrons harder to shift, not easier.
Check — multiple choice
Which molecule is more polarizable: CH₄ (10 electrons) or CBr₄ (146 electrons)?
Feedback if chosen: This option is wrong — you picked the small cloud. CH₄ holds only 10 electrons close to its nuclei, so its cloud is the harder one to distort.
Feedback if chosen: Correct. CBr₄ has far more electrons in a far larger cloud, so its outer electrons are held less tightly and its cloud distorts more easily.
Feedback if chosen: This option is wrong — you leveled the two clouds. The four Br atoms add well over a hundred electrons and a much larger cloud, and that larger cloud distorts more easily.
Feedback if chosen: This option is wrong — you required a permanent dipole. Distortion needs none: any cloud shifts when a charge approaches, and CBr₄'s large cloud shifts most easily of all.
Outcome: State that pi bonding enhances a molecule's polarizability, and say why.
You've already seen the two overlap patterns in the figure: a sigma bond's electron density lies directly between the nuclei, while a pi bond's density sits in lobes above and below the internuclear axis.
That placement matters for polarizability.
Pi electrons sit off the internuclear axis, away from the direct pull of both nuclei, so they are held relatively loosely.
Loosely held electron density distorts easily, so a pi system adds easily shifted density to a molecule's cloud.
The presence of pi bonding therefore enhances a molecule's polarizability.
Benzene, with its pi system, is more polarizable than a comparably sized molecule without pi bonds.
State that the presence of pi bonding enhances a molecule's polarizability.
Check — multiple choice
Molecules X and Y are comparably sized. X contains only sigma bonds; Y also contains a pi system. Which molecule is more polarizable?
Feedback if chosen: This option is wrong — you favored the all-sigma molecule. Sigma density is held tightly between the nuclei; it is the pi system that adds easily shifted density.
Feedback if chosen: Correct. Y's pi system holds electron density off the internuclear axis, where it is loosely held and easily distorted, so pi bonding enhances Y's polarizability.
Feedback if chosen: This option is wrong — you used size alone. At comparable size, the molecule with the pi system is the more polarizable one.
Feedback if chosen: This option is wrong — you reached for bond polarity. Polar bonds set permanent partial charges; polarizability here turns on how loosely the pi electron density is held.
Check — multiple choice
Why does pi bonding enhance a molecule's polarizability?
Feedback if chosen: Correct. The lobes of a pi bond lie above and below the internuclear axis, away from the nuclei's direct pull, so that density shifts easily.
Feedback if chosen: This option is wrong — you turned a pi bond into a permanent dipole. A pi bond need not be polar at all; the enhancement comes from loosely held, off-axis density.
Feedback if chosen: This option is wrong — you described a sigma bond. It is sigma density that concentrates between the nuclei; pi density sits off the axis, which is why it distorts easily.
Feedback if chosen: This option is wrong — you added electrons. A pi bond is made from electrons the atoms already have; it changes where density sits, not how many electrons there are.
Outcome: State that London dispersion forces increase with contact area between molecules, and pick the shape that attracts more strongly.
Pentane and neopentane are both C₅H₁₂ — the same 5 carbons, the same 12 hydrogens, the same 42 electrons.
Electron count and cloud size match, yet pentane's molecules hold on to one another noticeably more strongly than neopentane's.
The difference is shape: pentane is a long, uncoiled chain, while neopentane packs the same atoms into a compact, nearly spherical ball.
The figure shows each shape meeting a twin of itself: the long chains lie side by side and touch along their whole length, while the compact balls touch only at one small patch.
LDFs act wherever two clouds come close, so more contact means more places where temporary-dipole attractions act at once.
London dispersion forces therefore increase with increasing contact area between molecules.
Straight-chain pentane, with its long side-by-side contact, has stronger LDFs than compact neopentane.
State that London dispersion forces increase with increasing contact area between molecules.
Check — multiple choice
Butane and isobutane are both C₄H₁₀; butane is a straight chain, and isobutane is branched and compact. Between which molecules are the LDFs stronger?
Feedback if chosen: Correct. The straight chains can lie side by side and touch along their whole length, so more of their clouds interact at once.
Feedback if chosen: This option is wrong — you picked the compact shape. Branched, compact molecules touch a neighbor over a smaller area, so fewer temporary-dipole attractions act at once.
Feedback if chosen: This option is wrong — you stopped at the formula. Same formula means matching electron counts, but the long chain makes far more contact with its neighbor — and contact area also sets LDF strength.
Feedback if chosen: This option is wrong — you required polarity. LDFs act between all particles; nonpolar molecules attract through their temporary, fluctuating dipoles.
Check — multiple choice
Two molecules have identical formulas and electron counts, but one is long and uncoiled while the other is compact. Why are the LDFs between the long molecules stronger?
Feedback if chosen: This option is wrong — you confused the covalent bonds within each molecule with the attractions between molecules. LDF strength is about how much of the two clouds comes close together, not about the bonds inside them.
Feedback if chosen: This option is wrong — you added permanent dipoles. Both shapes are nonpolar hydrocarbons; the extra attraction comes from contact, not polarity.
Feedback if chosen: Correct. LDFs act wherever the clouds come close, and long molecules lying side by side bring far more cloud area close together.
Feedback if chosen: This option is wrong — you made the compact molecules repel. They still attract — just over a smaller contact area, so the total attraction is smaller.
Check — multiple choice
Four arrangements of two identical nonpolar molecules are compared. Which arrangement gives the strongest total LDF attraction between the two molecules?
Feedback if chosen: This option is wrong — you chose point contact. Two balls touch over one small patch, so few temporary-dipole attractions act at once.
Feedback if chosen: This option is wrong — you still have point contact. However long the chain is, touching a ball at one spot keeps the interacting area small.
Feedback if chosen: This option is wrong — you crossed the chains. Crossed chains touch only where they overlap; side-by-side alignment brings the whole length of both clouds close.
Feedback if chosen: Correct. Side-by-side chains keep their clouds close along the entire molecular length — the largest contact area, and so the most simultaneous attractions.
Outcome: State that London dispersion forces increase with the polarizability of the molecules.
You've already seen that polarizability increases with electron count and cloud size.
The figure shows temporary dipoles in Cl₂ and in Br₂ at the same instant: the larger Br₂ cloud has shifted further, separating more charge.
The more polarizable the molecule, the more easily its cloud distorts and the larger the temporary dipoles that form.
Larger temporary dipoles put larger momentary partial charges face to face, and the attraction grows, because interaction strength is proportional to the charge on each particle (Coulomb's law).
So London dispersion forces increase with increasing polarizability of the molecules.
Between two Br₂ molecules the LDFs are stronger than between two Cl₂ molecules, because Br₂'s larger cloud is more polarizable.
State that London dispersion forces increase with increasing polarizability of the molecules.
Check — multiple choice
Between which pair of atoms are the London dispersion forces stronger: two Ne atoms or two Xe atoms?
Feedback if chosen: This option is wrong — you picked the small, tightly held cloud. Ne's 10 electrons distort less easily, so its temporary dipoles — and the attractions they create — stay smaller.
Feedback if chosen: This option is wrong — you leveled the two clouds. Both are single atoms, but Xe's cloud is far larger and more polarizable, so its LDFs are stronger.
Feedback if chosen: This option is wrong — you removed the attraction. LDFs act between all particles; two atoms attract through their temporary, fluctuating dipoles.
Feedback if chosen: Correct. Xe's much larger cloud (54 electrons to Ne's 10) is more polarizable, so larger temporary dipoles form and the Coulombic attractions are stronger.
Check — multiple choice
Why do more polarizable molecules experience stronger London dispersion forces?
Feedback if chosen: This option is wrong — you swapped in permanent dipoles. LDF strength turns on the temporary dipoles a cloud can form, not on any permanent one.
Feedback if chosen: This option is wrong — you moved electrons between molecules. No electrons change molecules; each cloud only shifts within its own molecule for an instant.
Feedback if chosen: Correct. Easier distortion means bigger momentary partial charges, and interaction strength is proportional to the charge on each particle (Coulomb's law).
Feedback if chosen: This option is wrong — you made mass the cause. Heavy molecules do tend to have big clouds, but it is the polarizable cloud — not the mass itself — that creates stronger attractions.
Outcome: State that London dispersion forces are often the strongest net intermolecular force between large molecules.
A single temporary-dipole attraction between two small clouds is one of the weakest attractions in chemistry.
But a large molecule is one enormous cloud: many electrons spread over a long frame.
Between two large molecules, temporary-dipole attractions act at many places along the molecules at once.
You've already seen that LDFs grow with polarizability and with contact area — and a large molecule supplies both.
The many weak attractions add up, and the total is often larger than any other attraction between the molecules.
So London dispersion forces are often the strongest net intermolecular force between large molecules.
Between two long C₂₀H₄₂ chains, the summed LDFs outweigh the molecules' other attractions.
State that London dispersion forces are often the strongest net intermolecular force between large molecules.
Check — multiple choice
Two long C₁₈H₃₈ chains lie side by side in a liquid. Which attraction between the two molecules is strongest overall?
Feedback if chosen: Correct. Along two long chains, temporary-dipole attractions act at many points at once, and their sum dwarfs any other attraction between these molecules.
Feedback if chosen: This option is wrong — you gave the chains permanent dipoles. C₁₈H₃₈ is a nonpolar hydrocarbon; its molecules carry no permanent dipole to attract with.
Feedback if chosen: This option is wrong — you charged the molecules. The chains are neutral molecules, not ions; no full charges are present.
Feedback if chosen: This option is wrong — you bonded the chains together. The two molecules are separate particles; no covalent bond joins them.
Check — multiple choice
Each temporary-dipole attraction is weak. How can London dispersion forces still be the strongest net force between two large molecules?
Feedback if chosen: This option is wrong — you strengthened each single attraction past a bond. Each temporary-dipole attraction stays weak; it is the sum over many contact points that grows large.
Feedback if chosen: Correct. A large molecule offers a big, polarizable cloud and lots of contact, so many weak attractions act simultaneously and their total is large.
Feedback if chosen: This option is wrong — you turned the molecules into ions. Large or small, these molecules stay neutral; no full charges appear.
Feedback if chosen: This option is wrong — you froze the dipoles. The electrons never stop moving, so the dipoles keep fluctuating — the strength comes from summing, not from permanence.
Outcome: Distinguish the term London dispersion forces from the family term van der Waals forces.
Alongside 'London dispersion forces' you will meet a broader term: 'van der Waals forces'.
'Van der Waals forces' names a whole family of weak attractions between molecules.
London dispersion forces are one member of that family.
The family has other members, which you will meet in the coming lessons.
So the two terms do not mean the same thing: every London dispersion force is a van der Waals force, but not every van der Waals force is a London dispersion force.
Saying 'London dispersion forces are a type of van der Waals force' uses the terms correctly.
Saying 'van der Waals forces' when you mean specifically London dispersion forces treats a family name as a member's name — do not use the two terms as synonyms.
State that the term London dispersion forces should not be used synonymously with the term van der Waals forces, which names a wider family of attractions.
Check — multiple choice
How are London dispersion forces related to van der Waals forces?
Feedback if chosen: This option is wrong — you made the terms synonyms. 'Van der Waals forces' names a family of weak attractions, and LDFs are just one member of it.
Feedback if chosen: This option is wrong — you nested the family inside the member. It is the other way around: the van der Waals family contains London dispersion forces.
Feedback if chosen: Correct. Every LDF is a van der Waals force, but the family also holds other members, so the terms are not interchangeable.
Feedback if chosen: This option is wrong — you split the terms by particle type. Both terms describe weak attractions between molecules; the difference is member versus family.
Check — multiple choice
In an answer about two Ar atoms attracting, a student replaces 'London dispersion forces' with 'van der Waals forces' and claims the meaning is unchanged. What is wrong with the claim?
Feedback if chosen: This option is wrong — you accepted the swap. The terms are not synonyms: one names a family, the other names a single member.
Feedback if chosen: Correct. 'London dispersion forces' picks out the specific member — attraction through temporary, fluctuating dipoles — while the family name leaves the mechanism unstated.
Feedback if chosen: This option is wrong — you shrank the family. Van der Waals forces include LDFs, so the Ar–Ar attraction does belong to the family; the problem is lost precision, not exclusion.
Feedback if chosen: This option is wrong — you nested the family inside the member. 'Van der Waals forces' is the wider family name and 'London dispersion forces' the specific member, so the swap loses precision rather than adding it.
Outcome: Rank nonpolar substances by London dispersion force strength using polarizability and contact area.
You've already seen the two levers that set LDF strength: polarizability and contact area.
To rank nonpolar substances by LDF strength, ask two questions in order.
First: which particles have more electrons in a larger cloud — that is, which are more polarizable?
Second: if the polarizabilities are close — as with molecules of the same formula — which shape allows more contact between neighbors?
The table gives the four halogen molecules, their electron counts, and the temperature at which each boils.
| Molecule | Electrons per molecule | Boils at (°C) |
|---|---|---|
| F₂ | 18 | −188 |
| Cl₂ | 34 | −34 |
| Br₂ | 70 | 59 |
| I₂ | 106 | 184 |
All four molecules are the same shape, so contact area does not separate them — polarizability decides.
Electron count and cloud size climb from F₂ to I₂, so polarizability climbs the same way.
The LDF ranking follows: F₂ < Cl₂ < Br₂ < I₂.
The last column agrees: the more strongly a substance's molecules attract one another, the higher the temperature it stays liquid up to — F₂ boils away at −188 °C while I₂ holds together up to 184 °C.
When polarizabilities are close, shape takes over: straight-chain pentane outranks compact neopentane, its C₅H₁₂ twin, because the long chains make more contact.
Rank nonpolar substances by London dispersion force strength using their polarizability and their contact area.
Check — multiple choice
Between which element's atoms are the LDFs strongest: helium (He), neon (Ne), or krypton (Kr)?
Feedback if chosen: This option is wrong — you picked the smallest cloud. Helium's 2-electron cloud is the least polarizable of the three, so the He–He attractions are the weakest.
Feedback if chosen: This option is wrong — you stopped at neon. Neon's 10-electron cloud outranks helium's, but krypton's 36-electron cloud is far larger and more polarizable still.
Feedback if chosen: This option is wrong — you leveled the three. Polarizability grows with electron count and cloud size, so the three LDF strengths differ.
Feedback if chosen: Correct. Krypton has the most electrons and the largest cloud of the three, so its cloud distorts most easily and the Kr–Kr LDFs are the strongest.
Check — multiple choice
Rank these three nonpolar molecules by LDF strength, from weakest to strongest: n-pentane (a C₅H₁₂ chain), neopentane (a compact C₅H₁₂ ball), and n-hexane (a C₆H₁₄ chain).
Feedback if chosen: This option is wrong — you ranked the chain below the ball. The two C₅H₁₂ twins tie on electrons, so shape decides — and the uncoiled chain makes more contact than the compact ball.
Feedback if chosen: This option is wrong — you ranked the largest cloud weakest. n-hexane has the most electrons of the three, so it is the most polarizable and its LDFs are the strongest.
Feedback if chosen: Correct. Neopentane and n-pentane tie on electrons, so the compact ball's smaller contact area puts it lowest; n-hexane's larger cloud puts it on top.
Feedback if chosen: This option is wrong — you slotted n-hexane between the C₅H₁₂ twins. n-hexane beats both on electron count and cloud size, so it belongs at the strong end.
Check — multiple choice
Why are the LDFs between two I₂ molecules stronger than the LDFs between two F₂ molecules?
Feedback if chosen: This option is wrong — you polarized I₂. Both halogens are nonpolar; their attractions come from temporary, fluctuating dipoles.
Feedback if chosen: This option is wrong — you changed the shapes. Both are two-atom molecules of the same shape; it is cloud size, not contact area, that separates them.
Feedback if chosen: This option is wrong — you compared the covalent bonds inside the molecules. LDF strength is about attractions between molecules, set here by the clouds' polarizability.
Feedback if chosen: Correct. More electrons in a larger cloud mean higher polarizability, bigger temporary dipoles, and stronger Coulombic attractions between the molecules.
9 lessons
Outcome: State what a dipole-dipole interaction is and describe how neighboring polar molecules line up, δ⁺ end toward δ⁻ end.
Butane and acetone are molecules with nearly the same size of electron cloud — the figure compares them.
| Substance | Electrons per molecule | State at 25 °C |
|---|---|---|
| butane, C₄H₁₀ | 34 | gas |
| acetone, CH₃COCH₃ | 32 | liquid |
You've already seen that particles with electron clouds of similar size experience London dispersion forces of similar strength.
So London dispersion forces alone cannot explain why acetone molecules hold together as a liquid at 25 °C while butane molecules fly apart as a gas.
Acetone has something butane lacks: a permanent dipole.
You've already seen that a polar molecule carries a permanent separation of charge: a partially positive end (δ⁺) and a partially negative end (δ⁻) — HCl is a simple example.
When two polar molecules drift near each other, their charged ends do not stay in random positions.
Opposite charges attract, so each molecule turns until its δ⁺ end sits near a neighbor's δ⁻ end.
The figure shows two HCl molecules settled into that lined-up arrangement, with the attraction drawn as a dashed line.
An attraction between polar molecules aligned this way is called a 'dipole-dipole interaction'.
In liquid acetone, the partially positive carbon end of one molecule attracts the partially negative oxygen end of the next — that extra grip is what butane's molecules lack.
Dipole-dipole interactions act in addition to the London dispersion forces that every pair of neighboring particles already has.
State that dipole-dipole interactions are attractions between polar molecules that align so the partially positive end of one molecule sits near the partially negative end of a neighbor.
Check — multiple choice
The figure shows two HCl molecules in liquid HCl. What does the dashed line between them represent?
Feedback if chosen: This option is wrong — you turned an attraction between molecules into a bond within one. No electrons are shared across the dashed line; each molecule keeps its own H–Cl covalent bond.
Feedback if chosen: Correct. The δ⁺ H of one molecule sits near the δ⁻ Cl of the other, and the opposite partial charges attract — a dipole-dipole interaction.
Feedback if chosen: This option is wrong — you read the facing ends as like charges. A δ⁺ end faces a δ⁻ end here, and opposite charges attract.
Feedback if chosen: This option is wrong — you moved electrons between the molecules. The partial charges are fixed features of each polar molecule; nothing is transferred or shared between them.
Check — multiple choice
In which of these substances do dipole-dipole interactions act between neighboring particles?
Feedback if chosen: Correct. Acetone is a polar molecule with a permanent dipole, so neighboring molecules line up δ⁺ end to δ⁻ end and attract.
Feedback if chosen: This option is wrong — you stopped at the polar C=O bonds. The two bond dipoles point in opposite directions and cancel, so a CO₂ molecule has no permanent dipole to line up.
Feedback if chosen: This option is wrong — you gave identical atoms a dipole. The two N atoms pull the shared electrons equally, so N₂ has no bond dipole and no permanent dipole.
Feedback if chosen: This option is wrong — you gave a lone atom a dipole. Ar has no bonds and no permanent separation of charge, so its atoms attract only through London dispersion forces.
Check — multiple choice
In liquid acetone, the central C of each molecule is δ⁺ and the O is δ⁻. Toward which part of a neighboring molecule does a given molecule's δ⁻ O end turn?
Feedback if chosen: This option is wrong — you lined up like charges. Two δ⁻ ends repel each other; the molecules turn away from that arrangement.
Feedback if chosen: This option is wrong — you dismissed partial charges as too weak to matter. They are smaller than full charges, but they still attract and set a preferred lined-up arrangement.
Feedback if chosen: This option is wrong — you aimed the attraction at the neutral middle of the molecule. The attraction acts between the charged ends, δ⁻ toward δ⁺.
Feedback if chosen: Correct. Opposite partial charges attract, so the δ⁻ O end of one molecule turns toward the δ⁺ C end of its neighbor.
Outcome: Explain how the strength and preferred orientation of a dipole-dipole interaction follow from the signs and positions of the interacting partial charges.
You've already seen that opposite charges attract and like charges repel.
You've already seen dipole-dipole interactions: two polar molecules align so a δ⁺ end sits near a δ⁻ end, as the HCl pair in the figure shows.
Whether a given arrangement attracts or repels is decided by the signs of the partial charges that face each other.
Flip one HCl over, and its δ⁻ Cl end faces the neighbor's δ⁻ Cl end.
Like charges repel, so the flipped arrangement pushes the molecules apart — and they turn back toward the lined-up one.
The positions of the partial charges set the strength: the lined-up arrangement brings the opposite partial charges as close together as they can get.
That closeness matters because interaction strength increases as the distance between charge centers decreases (Coulomb's law).
So two HCl molecules attract most strongly when a δ⁺ H points straight at a neighbor's δ⁻ Cl.
In the liquid, collisions constantly knock molecules out of this arrangement, but it is the one they keep returning to — the preferred orientation.
Explain, from the signs and positions of the interacting partial charges, which orientation of two polar molecules attracts most strongly and why.
Check — multiple choice
Two HCl molecules drift into the arrangement shown. Why do they not stay in this arrangement?
Feedback if chosen: This option is wrong — you took HCl to be nonpolar. Cl pulls the shared electrons harder than H, so every HCl molecule carries a permanent dipole, and some arrangements are preferred over others.
Feedback if chosen: This option is wrong — you required contact. Partial charges attract and repel across the gap between molecules; the trouble with this arrangement is which ends face each other.
Feedback if chosen: Correct. The facing ends are both δ⁻, and like charges repel — the molecules turn until a δ⁺ H faces a δ⁻ Cl instead.
Feedback if chosen: This option is wrong — you read the dashed line as a covalent bond. It marks an interaction between the two molecules; each molecule's own H–Cl bond is not involved.
Check — multiple choice
Two polar molecules can meet in several ways. Which arrangement gives the strongest attraction between them?
Feedback if chosen: Correct. Opposite partial charges attract, and putting them close together maximizes the attraction, because interaction strength increases as the distance between charge centers decreases (Coulomb's law).
Feedback if chosen: This option is wrong — you brought like charges together. Two δ⁺ ends repel, so this close arrangement pushes the molecules apart.
Feedback if chosen: This option is wrong — you kept the right ends facing but gave up the closeness. Interaction strength increases as the distance between charge centers decreases (Coulomb's law), so the far-off arrangement attracts only weakly.
Feedback if chosen: This option is wrong — you kept the charged ends apart. With neither pair of opposite charges close together, little attraction can build.
Check — multiple choice
Two pairs of acetone molecules are each aligned δ⁺ C end toward δ⁻ O end. In pair 1 the molecules sit far apart; in pair 2 they sit close together. Which pair attracts more strongly?
Feedback if chosen: This option is wrong — you let alignment alone set the strength. Alignment picks the attracting arrangement, but the attraction weakens as the molecules move apart.
Feedback if chosen: Correct. Both pairs are aligned, but the close pair attracts more strongly because interaction strength increases as the distance between charge centers decreases (Coulomb's law).
Feedback if chosen: This option is wrong — you ignored the distance between the charges. Two aligned pairs at different separations attract with different strengths: the closer pair grips harder.
Feedback if chosen: This option is wrong — you swapped attract and repel. A δ⁺ end facing a δ⁻ end is the attracting arrangement.
Outcome: State what an ion-dipole interaction is and identify which end of a polar molecule turns toward a cation or an anion.
Stir a spoonful of table salt, NaCl, into water: the crystals vanish.
The ions have not gone anywhere — each Na⁺ has been pulled out of the crystal and surrounded by water molecules, as the figure shows.
You've already seen that a positive ion such as Na⁺ is a cation and a negative ion such as Cl⁻ is an anion.
You've already seen that water is a polar molecule: its O end is δ⁻ and its H ends are δ⁺.
What attraction holds a charged ion in place among neutral molecules?
A cation's full positive charge attracts the nearest negative charge on offer — and each water molecule offers its δ⁻ O end.
So the water molecules around Na⁺ turn their O ends toward it, as the figure shows.
An attraction between an ion and the oppositely charged partial end of a polar molecule is called an 'ion-dipole interaction'.
Around an anion the orientation flips: each nearby water molecule points a δ⁺ H end at the Cl⁻.
Either way the rule is the same: the ion attracts the end of the dipole whose partial charge is opposite to the ion's own charge.
State that ion-dipole interactions are attractions between an ion and the oppositely charged partial end of a polar molecule.
Check — multiple choice
Solid KCl is stirred into water, and its K⁺ ions spread among the water molecules. Which part of each nearby water molecule turns toward a K⁺ ion?
Feedback if chosen: This option is wrong — you oriented the water as it would sit around an anion. K⁺ is a cation, so it attracts the water molecule's negative partial end, not its positive one.
Feedback if chosen: This option is wrong — you left the orientation to chance. The ion's full charge attracts one end and repels the other, so the water molecules take a preferred orientation.
Feedback if chosen: This option is wrong — you aimed the attraction at the neutral middle of the molecule. The ion attracts the charged end of the dipole, not the midpoint.
Feedback if chosen: Correct. K⁺ carries a full positive charge, so it attracts the oppositely charged δ⁻ O end of each nearby water molecule.
Check — multiple choice
A student drew a water molecule next to a Cl⁻ ion, as shown. Which one change brings the drawing in line with the ion-dipole interaction that actually forms?
Feedback if chosen: Correct. Cl⁻ is an anion, so it attracts the oppositely charged δ⁺ H end of the water molecule — the O end belongs facing a cation instead.
Feedback if chosen: This option is wrong — you turned the interaction into a covalent bond. An ion-dipole interaction is an attraction between separate particles; no electrons are shared.
Feedback if chosen: This option is wrong — you rewrote the water molecule's polarity. O pulls the shared electrons harder than H, so the O end is δ⁻ in every water molecule.
Feedback if chosen: This option is wrong — you memorized one picture for both kinds of ion. The O end faces cations; an anion like Cl⁻ attracts the δ⁺ H end.
Check — multiple choice
Which pair of neighboring particles experiences an ion-dipole interaction?
Feedback if chosen: This option is wrong — you paired two full charges. A cation next to an anion is ionic bonding, not an ion paired with a polar molecule's partial charge.
Feedback if chosen: Correct. K⁺ carries a full charge, and H₂O is a polar molecule — the ion attracts the water's oppositely charged δ⁻ O end.
Feedback if chosen: This option is wrong — you paired two polar molecules. Neither carries a full charge, so their attraction is a dipole-dipole interaction.
Feedback if chosen: This option is wrong — you paired two neutral, nonpolar atoms. With no ion and no permanent dipole, they attract only through London dispersion forces.
Outcome: Explain why ion-dipole interactions are typically stronger than dipole-dipole interactions, from the full charge on the ion.
Picture a single Na⁺ ion sitting among acetone molecules.
You've already seen that acetone is polar, with a δ⁺ central C and a δ⁻ O.
The Na⁺ attracts the δ⁻ O end of each nearby acetone molecule — an ion-dipole interaction.
Each acetone molecule also attracts its neighbors, δ⁺ C end to δ⁻ O end — a dipole-dipole interaction.
Both attractions pull on the very same δ⁻ O end, so which pulls harder?
Na⁺ carries a full 1+ charge: a whole electron's worth of charge is missing.
A dipole end carries only a partial charge — a fraction of an electron's worth.
That difference decides the contest, because interaction strength is proportional to the charge on each particle (Coulomb's law).
So Na⁺ attracts an acetone molecule more strongly than a second acetone molecule does.
The same comparison holds in general: ion-dipole interactions are typically stronger than dipole-dipole interactions.
'Typically', because particle sizes and dipole sizes still vary from case to case — but full charge against partial charge is the difference that usually dominates.
Explain that ion-dipole interactions are typically stronger than dipole-dipole interactions because interaction strength is proportional to the charge on each particle, and an ion carries a full charge while a dipole end carries only a partial charge.
Check — multiple choice
A single Cl⁻ ion sits among many acetone molecules. Which attraction in this mixture is typically the stronger one?
Feedback if chosen: This option is wrong — you picked the dipole-dipole interaction. Both of the attracting ends there carry only partial charges, so the pull is weaker than the ion's.
Feedback if chosen: This option is wrong — you treated a full charge and a partial charge as equivalent. Interaction strength is proportional to the charge on each particle, and the ion's full charge outpulls a partial charge.
Feedback if chosen: Correct. Cl⁻ carries a full charge and attracts the δ⁺ end of the polar acetone molecule — an ion-dipole interaction, typically stronger than the acetone-acetone dipole-dipole interaction.
Feedback if chosen: This option is wrong — you required both particles to be charged. A neutral polar molecule still has a δ⁺ end for the anion to attract.
Check — multiple choice
Why is an ion-dipole interaction typically stronger than a dipole-dipole interaction?
Feedback if chosen: This option is wrong — you made size the deciding factor. Many ions are smaller than polar molecules; what matters is that interaction strength is proportional to the charge on each particle.
Feedback if chosen: This option is wrong — you canceled the dipole away. The two partial charges sit on opposite ends of the molecule, so each end still attracts oppositely charged neighbors.
Feedback if chosen: This option is wrong — you restricted what a partial charge can attract. A partial charge attracts any opposite charge, full or partial — the ion's full charge just pulls harder.
Feedback if chosen: Correct. Interaction strength is proportional to the charge on each particle (Coulomb's law), and a full charge is larger than a partial charge — so the ion grips the dipole end harder.
Outcome: State how a polar molecule induces a temporary dipole in a nonpolar neighbor, and explain why the resulting interaction is always attractive.
An HCl molecule drifts toward an argon atom.
Ar is a lone nonpolar atom: no bonds, no permanent dipole, no charged ends to line up.
You've already seen HCl's permanent dipole: δ⁺ on the H end, δ⁻ on the Cl end.
You've already seen that polarizability is the ease with which a particle's electron cloud is distorted to form a temporary dipole.
A permanent dipole parked next to the Ar atom is exactly the kind of nearby charge that does the distorting.
Let the δ⁺ H end of the HCl approach from the left: the Ar electron cloud is pulled toward it, as the figure shows.
The shifted cloud gives the Ar a δ⁻ side facing the HCl and a δ⁺ side facing away — a temporary dipole.
A temporary dipole created by a neighbor's charge, rather than by a random fluctuation, is called an 'induced dipole'.
The attraction between the polar molecule and the dipole it induces is called a 'dipole-induced dipole interaction'.
This interaction is always attractive.
Near a δ⁺ end the neighbor's cloud shifts toward the polar molecule; near a δ⁻ end it shifts away — either way, the side facing the polar molecule carries the opposite partial charge.
State that a polar molecule induces a temporary dipole in a nearby nonpolar molecule, producing a dipole-induced dipole interaction that is always attractive.
Check — multiple choice
A xenon atom drifts next to the δ⁻ Cl end of an HCl molecule. What happens to the xenon atom's electron cloud?
Feedback if chosen: This option is wrong — you shifted the electrons toward a negative charge. The δ⁻ Cl end repels electrons, pushing the cloud away, so the side facing the HCl ends up δ⁺.
Feedback if chosen: Correct. The δ⁻ Cl end pushes the xenon electrons away, so the cloud shifts away from the HCl — the exposed δ⁺ side faces the δ⁻ Cl, and the pair attracts.
Feedback if chosen: This option is wrong — you made the cloud rigid. Every particle's electron cloud can be distorted by a nearby charge; that ease of distortion is its polarizability.
Feedback if chosen: This option is wrong — you made the induced dipole permanent. The distortion lasts only while the polar neighbor is close; drift the HCl away and the cloud relaxes back to symmetric.
Check — multiple choice
Why is a dipole-induced dipole interaction always attractive?
Feedback if chosen: Correct. Whichever end of the polar molecule approaches, the neighbor's cloud shifts so the side facing the polar molecule carries the opposite partial charge — so the pair always attracts.
Feedback if chosen: This option is wrong — you fixed the polar molecule's orientation. Either end can face the nonpolar particle; the induced dipole adjusts to attract in both cases.
Feedback if chosen: This option is wrong — you gave a neutral particle a net charge. A nonpolar particle is neutral overall; only its charge distribution shifts.
Feedback if chosen: This option is wrong — you turned the interaction into a covalent bond. No electrons are shared; the attraction is between the permanent dipole and the shifted cloud.
Check — multiple choice
Which neighboring pair experiences a dipole-induced dipole interaction?
Feedback if chosen: This option is wrong — you paired two permanent dipoles. Two polar molecules attract through a dipole-dipole interaction; nothing needs to be induced.
Feedback if chosen: This option is wrong — you paired two nonpolar atoms. With no permanent dipole on either side, they attract only through temporary, fluctuating dipoles — London dispersion forces.
Feedback if chosen: Correct. HF has a permanent dipole and Ne does not — HF's dipole distorts Ne's electron cloud, inducing the dipole it then attracts.
Feedback if chosen: This option is wrong — you paired a full charge with a permanent dipole. That is an ion-dipole interaction; nothing needs to be induced in a polar molecule.
Outcome: State that a dipole-induced dipole interaction strengthens as the polar molecule's dipole grows, and pick the stronger case from two polar molecules.
An induced dipole is a response: the nonpolar particle's cloud shifts only as far as the neighboring dipole pulls it.
You've already seen that, for comparable single bonds, a larger electronegativity difference produces a larger bond dipole.
F is more electronegative than Cl, so HF's dipole is larger than HCl's.
Parked beside an Ar atom, HF therefore pulls the Ar electron cloud further off center than HCl can.
A more distorted cloud is a larger induced dipole, and a larger induced dipole is attracted more strongly.
So a dipole-induced dipole interaction strengthens as the magnitude of the polar molecule's dipole increases.
State that a dipole-induced dipole interaction strengthens as the magnitude of the polar molecule's dipole increases.
Check — multiple choice
HF induces a larger temporary dipole in a neighboring Ar atom than HCl does. What difference between the two molecules is responsible?
Feedback if chosen: This option is wrong — you compared the molecules' own clouds, and backwards: HCl has more electrons than HF. What induces the dipole is the polar molecule's dipole, not its cloud size.
Feedback if chosen: This option is wrong — you promoted a partial charge to a full one. HF is a neutral molecule; its F end carries only δ⁻, just a larger δ⁻ than Cl's.
Feedback if chosen: This option is wrong — you bonded the pair together. No electrons are shared with Ar; the molecule only distorts Ar's cloud from nearby.
Feedback if chosen: Correct. F out-pulls Cl for the shared electrons by a wider margin, so HF's dipole is larger — and a larger dipole distorts the Ar cloud more.
Check — multiple choice
Argon gas is mixed once with HCl and once with HBr. Cl is more electronegative than Br. Which molecule induces the larger temporary dipole in a neighboring Ar atom?
Feedback if chosen: This option is wrong — you picked the molecule with the larger electron cloud. HBr's own polarizability is not what counts here; its dipole is smaller than HCl's, so it induces less.
Feedback if chosen: Correct. The larger electronegativity difference in H–Cl gives HCl the larger dipole, and the larger dipole distorts the Ar cloud more.
Feedback if chosen: This option is wrong — you ignored the sizes of the two dipoles. HCl's dipole is larger, so it induces the larger response in Ar.
Feedback if chosen: This option is wrong — you required a permanent dipole on Ar. Induction needs none: the polar molecule creates the dipole it attracts.
Outcome: State that a dipole-induced dipole interaction strengthens as the polarizability of the nonpolar partner increases, and pick the stronger case from two nonpolar particles.
The polar molecule supplies the pull, but how far the cloud shifts also depends on whose cloud is being pulled.
You've already seen that polarizability increases with the number of electrons in a particle and the size of its electron cloud.
Compare He, holding 2 electrons, with O₂, holding 16 — the figure shows how different the two clouds are.
The same water-molecule dipole barely budges He's small, tightly held cloud, but shifts O₂'s large cloud easily.
A larger shift is a larger induced dipole, so water molecules attract an O₂ molecule mixed into the water more strongly than they attract a He atom.
So a dipole-induced dipole interaction strengthens as the polarizability of the nonpolar molecule increases.
State that a dipole-induced dipole interaction strengthens as the polarizability of the nonpolar molecule increases.
Check — multiple choice
HCl gas is mixed once with helium and once with xenon. In which mixture is the dipole-induced dipole attraction stronger?
Feedback if chosen: Correct. Xe holds far more electrons than He, so its larger, more polarizable cloud distorts further — a larger induced dipole and a stronger attraction.
Feedback if chosen: This option is wrong — you picked the smaller cloud. He's 2 electrons sit close to the nucleus and barely shift, so the induced dipole in He is tiny.
Feedback if chosen: This option is wrong — you left the partner out of the strength. The same HCl dipole induces a larger response in the more polarizable atom, so the two mixtures differ.
Feedback if chosen: This option is wrong — you required a permanent dipole on both particles. A polar molecule attracts a nonpolar atom through the dipole it induces.
Check — multiple choice
Water molecules attract each of these atoms when it is mixed into water: He, Ne, or Kr. Which atom do the water molecules attract most strongly?
Feedback if chosen: This option is wrong — you picked the least polarizable atom. He's 2-electron cloud shifts the least, so its induced dipole is the smallest of the three.
Feedback if chosen: This option is wrong — you stopped in the middle. Ne's 10-electron cloud distorts more than He's, but Kr's much larger cloud distorts more still.
Feedback if chosen: Correct. Kr holds the most electrons and the largest electron cloud of the three, so water's dipole induces the largest dipole in Kr — the strongest attraction.
Feedback if chosen: This option is wrong — you gave every atom the same response. The three clouds differ in size and polarizability, so the induced dipoles — and the attractions — differ too.
Outcome: State what hydrogen bonding is: a strong attraction between an H atom bonded to N, O, or F and the negative end of a dipole formed by N, O, or F in another molecule or another region of the same molecule.
Methane and water are both tiny molecules holding 10 electrons each — the figure compares them.
| Substance | Electrons per molecule | State at 25 °C |
|---|---|---|
| methane, CH₄ | 10 | gas |
| water, H₂O | 10 | liquid |
You've already seen that particles with electron clouds of similar size experience London dispersion forces of similar strength.
Water is polar, so dipole-dipole attractions add their grip — but that cannot be the whole story either.
H₂S is also polar, with an even larger electron cloud than water's, yet at 25 °C H₂S is a gas.
Something holds water molecules together that ordinary dipole-dipole attractions do not explain.
You've already seen how to rank elements by electronegativity: N, O, and F are among the most electronegative of all.
When an H atom is covalently bonded to N, O, or F, the shared electrons are pulled strongly away from the H.
H is a tiny atom with no inner electrons, so the δ⁺ left behind is unusually exposed and concentrated.
That exposed δ⁺ H is strongly attracted to the negative end of a dipole formed by N, O, or F in a different molecule.
The figure shows the attraction in water: the δ⁺ H of one molecule reaches toward the δ⁻ O of its neighbor.
This strong intermolecular attraction is called 'hydrogen bonding'.
Despite the name, a hydrogen bond is not a covalent bond — you've already seen that attractions between molecules are much weaker than the bonds within them.
It is, though, notably stronger than a typical dipole-dipole interaction — strong enough to keep water liquid at 25 °C while 10-electron methane is a gas.
Liquid HF shows the same attraction: the H of one HF molecule reaches toward the δ⁻ F of a neighboring HF.
In a very large molecule, hydrogen bonding can also form between an H in one region and an N, O, or F in a different region of the same molecule.
State that hydrogen bonding is a strong intermolecular interaction between an H atom covalently bonded to N, O, or F and the negative end of a dipole formed by N, O, or F in a different molecule or a different region of the same molecule.
Check — multiple choice
In liquid HF, a hydrogen bond forms between which two atoms?
Feedback if chosen: Correct. The δ⁺ H of one HF is attracted to the δ⁻ F of its neighbor — an H bonded to F reaching toward the negative end of another molecule's dipole.
Feedback if chosen: This option is wrong — you paired the two δ⁻ ends. Like partial charges repel; the attraction needs the δ⁺ H facing the δ⁻ F.
Feedback if chosen: This option is wrong — you paired the two δ⁺ ends. Like partial charges repel; the H must face a δ⁻ F, not another H.
Feedback if chosen: This option is wrong — you pointed at the covalent H–F bond. A hydrogen bond is an attraction between different molecules, not the bond inside one.
Check — multiple choice
A student drew the attraction that holds liquid water together as the dashed line shown. Which one change brings the drawing in line with hydrogen bonding?
Feedback if chosen: This option is wrong — you kept the attraction inside one molecule and made it a bond. The O–H bond is already there; hydrogen bonding is a separate attraction between molecules.
Feedback if chosen: Correct. A hydrogen bond acts between molecules: from the δ⁺ H of one water molecule to the δ⁻ O of its neighbor — not alongside a covalent bond within one molecule.
Feedback if chosen: This option is wrong — you joined the two δ⁻ ends. Like partial charges repel; the attraction needs the δ⁺ H facing a δ⁻ O.
Feedback if chosen: This option is wrong — you joined two δ⁺ ends. Like partial charges repel; each H must reach toward a neighbor's δ⁻ O.
Check — multiple choice
How does the strength of a hydrogen bond compare with the other attractions in a liquid made of small molecules?
Feedback if chosen: This option is wrong — you read 'bond' in the name literally. A hydrogen bond is an attraction between molecules, far weaker than the covalent bonds within them.
Feedback if chosen: This option is wrong — you filed hydrogen bonding as ordinary. The concentrated δ⁺ on an H bonded to N, O, or F makes the attraction notably stronger than a typical dipole-dipole interaction.
Feedback if chosen: Correct. Hydrogen bonding is a strong intermolecular attraction — stronger than a typical dipole-dipole interaction, yet still much weaker than a covalent bond.
Feedback if chosen: This option is wrong — you put hydrogen bonding below the weakest attraction. Between small molecules, London dispersion forces are weaker than a hydrogen bond, not stronger.
Outcome: Decide whether hydrogen bonding occurs within a pure substance or between two species by checking for an H atom bonded to N, O, or F and an available N, O, or F acceptor.
Here is methanol, CH₃OH.
One of its H atoms is bonded to O, and every molecule offers a δ⁻ O — so liquid methanol has hydrogen bonding between its molecules.
Now here is fluoromethane, CH₃F — a molecule with the same number of electrons, built around F, an atom that forms strong dipoles.
Every one of its H atoms is bonded only to C.
An H on C carries no concentrated δ⁺, so pure CH₃F has no hydrogen bonding — even though F is right there.
Hydrogen bonding needs two things at once.
First, an H atom covalently bonded to N, O, or F — the source of the exposed δ⁺ H.
Second, an N, O, or F atom whose δ⁻ end is available for that H to reach — the acceptor.
For a pure substance, run the check on one kind of molecule: CH₃OH passes both parts by itself, while CH₃F fails the first part because its H atoms are bonded only to C.
For two different species, one can supply the H while the other supplies the N, O, or F acceptor.
Identify whether hydrogen bonding can occur within a pure substance or between two species by checking for an H atom bonded to N, O, or F and an available N, O, or F acceptor.
Check — multiple choice
Which pure substance has hydrogen bonding between its molecules?
Feedback if chosen: This option is wrong — you let an H on C count as a donor. Every H in CH₄ is bonded only to C, so no exposed δ⁺ H exists.
Feedback if chosen: This option is wrong — you let any polar H–X bond qualify. The H must be bonded to N, O, or F; Cl is not on that list, so HCl molecules attract only through dipole-dipole interactions and London dispersion forces.
Feedback if chosen: Correct. Each NH₃ molecule has H atoms bonded to N, and each offers a δ⁻ N for a neighbor's H to reach — both parts of the check pass.
Feedback if chosen: This option is wrong — you skipped the first part of the check. CO₂ has O atoms to accept, but no H atoms at all, so nothing can donate.
Check — multiple choice
The figure shows dimethyl ether, CH₃OCH₃. Pure dimethyl ether has no hydrogen bonding between its own molecules. When dimethyl ether mixes with water, can hydrogen bonds form between the two kinds of molecule?
Feedback if chosen: Correct. Water supplies the H bonded to O, and the ether's O accepts — between two species, one molecule can donate while the other accepts.
Feedback if chosen: This option is wrong — you let an H on C donate. The ether's H atoms are all bonded to C, so none carries the concentrated δ⁺ that hydrogen bonding needs.
Feedback if chosen: This option is wrong — you applied only the donor half of the check. The ether cannot supply the H, but its O can still accept an H from a molecule that does supply one.
Feedback if chosen: This option is wrong — you restricted hydrogen bonding to pure substances. The check works across two species: one supplies the H, the other the N, O, or F acceptor.
Check — multiple choice
Which pure substance has no hydrogen bonding between its molecules?
Feedback if chosen: This option is wrong — you passed over the O–H bonds. Each water molecule has H bonded to O and an O to accept, so both parts of the check pass.
Feedback if chosen: Correct. Both H atoms in CH₂F₂ are bonded only to C — the F atoms could accept, but no molecule supplies an H bonded to N, O, or F.
Feedback if chosen: This option is wrong — you missed the O–H group. Methanol's H on O donates, and its O accepts, so pure methanol hydrogen bonds with itself.
Feedback if chosen: This option is wrong — you passed over the H–F bond. Each HF has its H bonded to F and offers a δ⁻ F, so both parts of the check pass.
6 lessons
Outcome: Identify every intermolecular force present in a pure sample of a substance from its structure and polarity.
A bottle of methanol, CH₃OH, holds nothing but methanol molecules.
Naming one attraction between those molecules leaves the picture incomplete, because several act at the same time.
A complete answer lists every intermolecular force present in the sample.
You've already seen that London dispersion forces (LDFs) are present between all particles, because temporary, fluctuating dipoles create Coulombic attractions between particles.
So the first entry on the list is settled before you look at the structure: LDFs are always present.
Second, decide from the structure whether the molecules are polar.
You've already seen that dipole-dipole interactions act between polar molecules, so a polar substance adds dipole-dipole interactions to the list.
Third, look for hydrogen bonding: one molecule must have an H atom bonded to N, O, or F, and a neighbor must offer an N, O, or F atom to attract it.
You've already seen this last check decide between two similar molecules: CH₃OH hydrogen bonds with itself, while CH₃F does not, because its H atoms are bonded only to C.
Here is the structure of methanol.
LDFs are present, because they act between all particles.
The bond dipoles of the C–O and O–H bonds do not cancel, so the molecule is polar and dipole-dipole interactions are present.
Each molecule has an H atom bonded to O, and every neighbor offers an O atom to attract it, so hydrogen bonding is present as well.
The complete list for pure methanol is LDFs, dipole-dipole interactions, and hydrogen bonding.
Here is the structure of carbon dioxide, CO₂.
The two C=O bond dipoles point in opposite directions and cancel, so the molecule is nonpolar.
With no dipole, and with no H atom at all, neither of the last two questions adds anything.
LDFs are therefore the only intermolecular force present in pure CO₂.
A short list is a complete answer whenever the structure gives no reason to add more.
Identify every intermolecular force present in a pure sample of a substance, using its structure and polarity.
Check — multiple choice
The figure shows the structure of ammonia, NH₃. Which forces are present in a pure sample of ammonia?
Feedback if chosen: This option is wrong — you stopped at the force every substance has. Ammonia is polar and its H atoms are bonded to N, so two more forces belong on the list.
Feedback if chosen: This option is wrong — you left out LDFs. London dispersion forces are present between all particles, so they belong on every list.
Feedback if chosen: This option is wrong — you missed the hydrogen bonding. Each N–H hydrogen is attracted to the N atom of a neighboring molecule.
Feedback if chosen: Correct. LDFs act between all particles, the polar NH₃ molecules attract one another through dipole-dipole interactions, and each N–H hydrogen is attracted to a neighbor's N atom.
Check — multiple choice
The figure shows the structure of ethane, C₂H₆. Which forces are present in a pure sample of ethane?
Feedback if chosen: Correct. The C–H bonds are effectively nonpolar and no H atom is bonded to N, O, or F, so LDFs — present between all particles — are the only force left.
Feedback if chosen: This option is wrong — you treated ethane as polar. Its C–H bonds are effectively nonpolar, so its molecules have no dipoles to align.
Feedback if chosen: This option is wrong — you counted C–H hydrogens as hydrogen-bond donors. Hydrogen bonding needs an H atom bonded to N, O, or F.
Feedback if chosen: This option is wrong — you read 'nonpolar' as 'no attractions'. LDFs are present between all particles, ethane included.
Check — multiple choice
The figure shows the structure of hydrogen sulfide, H₂S. Which forces are present in a pure sample of hydrogen sulfide?
Feedback if chosen: This option is wrong — you counted the S–H hydrogens as hydrogen-bond donors and dropped the dipole-dipole interactions. Hydrogen bonding needs an H atom bonded to N, O, or F.
Feedback if chosen: Correct. LDFs act between all particles and the bent H₂S molecule is polar, so dipole-dipole interactions are present; its H atoms are bonded to S, so there is no hydrogen bonding.
Feedback if chosen: This option is wrong — you added hydrogen bonding on the strength of the H atoms alone. Those H atoms are bonded to S, which is not N, O, or F.
Feedback if chosen: This option is wrong — you named the one force that is absent and left out both that are present. H₂S has LDFs and, because it is polar, dipole-dipole interactions.
Outcome: Identify every intermolecular force present between particles of two different chemical species from both structures.
A pure substance holds one kind of particle, so its list of forces covers one pairing.
Mix in a second substance and a new pairing appears: a particle of one species meeting a particle of the other.
The question becomes which intermolecular forces act between a particle of the first species and a particle of the second.
London dispersion forces (LDFs) come first again, because they are present between all particles, whatever the two species are.
Next, if both species are polar molecules, dipole-dipole interactions act between them.
Then check hydrogen bonding across the pair: one species must supply an H atom bonded to N, O, or F, and the other must supply an N, O, or F atom to attract it.
Either species can supply that H atom, as long as the other supplies the N, O, or F.
In the figure, an N–H hydrogen of an ammonia molecule is attracted to the O atom of a water molecule.
You've already seen an ion attracting the oppositely charged end of a polar molecule — an ion-dipole interaction.
So a fourth question applies once two species are in play: if one is an ion and the other is polar, ion-dipole interactions are present.
You've also already seen a polar molecule induce a temporary dipole in a nearby nonpolar particle — a dipole-induced dipole interaction.
So a fifth question applies: if one species is polar and the other is nonpolar, dipole-induced dipole interactions are present.
Here is the structure of methanol, CH₃OH, the liquid the ions are mixed into.
A dissolved Na⁺ ion is a single atom carrying a full positive charge.
LDFs act between the ion and the molecule, because LDFs are present between all particles.
The Na⁺ ion attracts the partially negative O end of the methanol molecule, so ion-dipole interactions are present.
The ion has no dipole of its own and no H atom, so neither dipole-dipole interactions nor hydrogen bonding acts across this pair.
The complete list between Na⁺ and CH₃OH is LDFs and ion-dipole interactions.
The hydrogen bonding between methanol molecules has not gone away — it acts within one species rather than across the pair.
Identify every intermolecular force acting between particles of two different chemical species, using both structures.
Check — multiple choice
The figure shows a chloride ion, Cl⁻, next to a water molecule in a mixture of water and dissolved sodium chloride. Which forces act between the chloride ion and the water molecule?
Feedback if chosen: This option is wrong — you left out LDFs. London dispersion forces are present between all particles, ions included.
Feedback if chosen: This option is wrong — you treated the water molecule's partially charged ends as full ionic charges. Water is a neutral polar molecule, so its attraction to Cl⁻ is an ion-dipole interaction, not an ion-ion one.
Feedback if chosen: Correct. LDFs act between all particles, and the full negative charge of Cl⁻ attracts the partially positive H end of the polar water molecule — an ion-dipole interaction.
Feedback if chosen: This option is wrong — you treated the ion as the polar end of a molecule. Cl⁻ carries a full charge rather than a partial one, so its attraction to water is an ion-dipole interaction.
Check — multiple choice
The figure shows the structures of hydrogen chloride, HCl, and argon, Ar. Which forces act between an HCl molecule and an Ar atom?
Feedback if chosen: Correct. LDFs act between all particles, and the polar HCl molecule induces a temporary dipole in the Ar atom, giving a dipole-induced dipole interaction.
Feedback if chosen: This option is wrong — you gave argon a dipole of its own. A lone Ar atom has none, so the attraction across this pair is dipole-induced dipole.
Feedback if chosen: This option is wrong — you counted the H of HCl as a hydrogen-bond donor. That H is bonded to Cl, and Ar offers no N, O, or F atom to attract it.
Feedback if chosen: This option is wrong — you stopped at the universal force. The polar HCl molecule also induces a dipole in the Ar atom, which adds a dipole-induced dipole interaction.
Check — multiple choice
The figure shows the structure of dimethyl ether, CH₃OCH₃, which is mixed with water. Which forces act between a dimethyl ether molecule and a water molecule?
Feedback if chosen: This option is wrong — you ruled out hydrogen bonding because dimethyl ether has no O–H bond. The H atom comes from the water molecule, and the ether's O atom attracts it.
Feedback if chosen: This option is wrong — you dropped the dipole-dipole interactions. Both molecules are polar, so their dipoles attract each other too.
Feedback if chosen: This option is wrong — you left out LDFs, which are present between all particles.
Feedback if chosen: Correct. LDFs act between all particles, both molecules are polar, and an O–H hydrogen of water is attracted to the O atom of the ether, so hydrogen bonding acts across the pair as well.
Outcome: Rank pure substances by overall intermolecular force strength using hydrogen bonding, polarity, and size or polarizability evidence from their structures.
Two substances can both be built from small molecules, yet one holds its particles together far more tightly than the other.
Ranking substances by overall intermolecular force strength means ranking the total strength of the attractions between their particles.
The evidence for that ranking is the structures themselves.
Read three features from the structures, in this order: hydrogen bonding, then polarity, then size.
Among molecules of similar size, hydrogen bonding generally gives the strongest attractions, so a substance whose molecules hydrogen bond with each other ranks above one whose molecules do not.
Where neither substance hydrogen bonds, polarity decides: the polar substance adds dipole-dipole interactions to the London dispersion forces (LDFs) both substances have, so it generally ranks higher.
Where the substances match on both counts, size decides.
You've already seen that polarizability increases with the number of electrons and the size of the electron cloud, so the larger particle has the stronger LDFs.
Reading polarity is only fair between molecules of similar size, because a much larger nonpolar molecule can have stronger LDFs than a small polar molecule.
Here is the structure of methane, CH₄.
Its C–H bonds are effectively nonpolar, so LDFs are the only force between its molecules.
Here is the structure of chloromethane, CH₃Cl.
The C–Cl bond makes the molecule polar, so dipole-dipole interactions act between its molecules as well as LDFs.
Here is the structure of methanol, CH₃OH.
Methanol is polar, and its O–H hydrogen is attracted to the O atom of a neighbor, so hydrogen bonding acts between its molecules too.
Reading the evidence in order gives the ranking, weakest first: CH₄ < CH₃Cl < CH₃OH.
Chloromethane has more electrons than methanol, yet methanol still ranks higher, because hydrogen bonding generally outweighs a size difference this small.
Rank pure substances by overall intermolecular force strength, using hydrogen bonding, polarity, and size or polarizability evidence from their structures.
Check — multiple choice
Ethane (C₂H₆), fluoromethane (CH₃F), and methylamine (CH₃NH₂) each contain 18 electrons per molecule, so they are of similar size. Which ranking puts them in order of overall intermolecular force strength, weakest first?
Feedback if chosen: This option is wrong — you ranked the nonpolar molecule above the polar one. C₂H₆ has LDFs alone, while polar CH₃F adds dipole-dipole interactions.
Feedback if chosen: Correct. Ethane is nonpolar and has LDFs alone, fluoromethane is polar and adds dipole-dipole interactions, and methylamine's N–H hydrogens add hydrogen bonding on top.
Feedback if chosen: This option is wrong — you ranked hydrogen bonding below simple polarity. Among molecules of similar size, hydrogen bonding generally gives the stronger attractions, so CH₃NH₂ ranks above CH₃F.
Feedback if chosen: This option is wrong — you reversed the ranking. Hydrogen bonding makes CH₃NH₂ the strongest of the three, and nonpolar C₂H₆ the weakest.
Check — multiple choice
Neon (Ne), argon (Ar), and krypton (Kr) are all nonpolar atoms. Krypton has the most electrons and the largest electron cloud, and neon the fewest and smallest. Which ranking puts them in order of overall intermolecular force strength, weakest first?
Feedback if chosen: This option is wrong — you ranked the largest atom weakest. Polarizability increases with the number of electrons and the size of the electron cloud, so Kr has the strongest LDFs.
Feedback if chosen: This option is wrong — you ranked argon below neon. Argon has more electrons and a larger electron cloud, so its LDFs are the stronger ones.
Feedback if chosen: Correct. All three have LDFs alone, and polarizability increases with the number of electrons and the size of the electron cloud, so the ranking follows the atoms' sizes.
Feedback if chosen: This option is wrong — you read 'no dipole' as 'no difference'. Each of these atoms has LDFs, and their strength grows with polarizability, so the three do not tie.
Check — multiple choice
Methane (CH₄), phosphine (PH₃), and ammonia (NH₃) are compared. PH₃ is polar and has more electrons than NH₃. Which ranking puts the three in order of overall intermolecular force strength, weakest first?
Feedback if chosen: Correct. Methane is nonpolar and has LDFs alone, phosphine is polar and adds dipole-dipole interactions, and ammonia's N–H hydrogens add hydrogen bonding, which outweighs phosphine's extra electrons.
Feedback if chosen: This option is wrong — you let the larger electron count settle the top place. Ammonia's hydrogen bonding generally outweighs the extra polarizability of phosphine.
Feedback if chosen: This option is wrong — you ranked the hydrogen-bonding substance weakest. Read hydrogen bonding first: NH₃ has N–H hydrogens attracted to a neighbor's N atom, while CH₄ has nothing beyond LDFs.
Feedback if chosen: This option is wrong — you ranked polar PH₃ below nonpolar CH₄. A polar molecule adds dipole-dipole interactions to the LDFs every substance has.
Outcome: Rank the strength of the interactions between different pairs of chemical species from their structures.
Stir two different substances into water and each one is attracted to the water molecules around it.
One of those two attractions is the stronger, and the structures say which.
The item being ranked is now a pair of species, not a single substance.
For each pair, name the strongest force type that acts across it, then compare those types.
You've already seen that ion-dipole interactions are typically stronger than dipole-dipole interactions, because interaction strength is proportional to the charge on each particle (Coulomb's law).
Generally, the force types rank in this order: ion-dipole above hydrogen bonding, hydrogen bonding above dipole-dipole, and dipole-dipole above dipole-induced dipole.
So a pair whose strongest force type sits higher in that order generally interacts more strongly.
Which interacts more strongly with a water molecule: a dissolved K⁺ ion, or a dissolved CH₃OH molecule?
The figure shows the first pair: the K⁺ ion carries a full positive charge, and it attracts the partially negative O end of the water molecule.
That is an ion-dipole interaction, the strongest force type across that pair.
Across the second pair, an O–H hydrogen of one molecule is attracted to the O atom of the other, so the strongest force type is hydrogen bonding.
Ion-dipole sits above hydrogen bonding, so K⁺ interacts with water more strongly than CH₃OH does.
Two pairs can share the same strongest force type, and then the magnitudes decide.
Water induces a temporary dipole in a dissolved Kr atom and in a dissolved Ne atom in the same way.
Krypton has far more electrons than neon, so it is the more polarizable partner and water interacts with it more strongly.
The same reasoning covers a larger charge and a larger dipole: whichever pair offers more of what the force type depends on interacts more strongly.
Rank the strength of the interactions between different pairs of chemical species, using both structures in each pair.
Check — multiple choice
Three substances are dissolved in water: argon atoms (Ar), ammonia molecules (NH₃), and chloride ions (Cl⁻). Rank the interaction of each with a water molecule, weakest first.
Feedback if chosen: This option is wrong — you reversed the ranking. The full charge on Cl⁻ makes its ion-dipole interaction with water the strongest of the three, and argon's induced dipole the weakest.
Feedback if chosen: This option is wrong — you ranked hydrogen bonding weakest. Ammonia hydrogen bonds with water, which is generally stronger than the dipole-induced dipole interaction water has with an argon atom.
Feedback if chosen: This option is wrong — you ranked the ion below the hydrogen-bonding molecule. Cl⁻ carries a full charge while NH₃ offers only partial charges, so the ion-dipole interaction is the stronger, because interaction strength is proportional to the charge on each particle (Coulomb's law).
Feedback if chosen: Correct. Water induces a dipole in nonpolar Ar, hydrogen bonds with NH₃, and attracts the full charge of Cl⁻ through an ion-dipole interaction — the strongest of the three.
Check — multiple choice
Xenon atoms (Xe, 54 electrons) and nitrogen molecules (N₂, 14 electrons) are both dissolved in water. How do their interactions with a water molecule compare?
Feedback if chosen: This option is wrong — you ranked the smaller particle stronger. N₂ has far fewer electrons than Xe, so it is less polarizable and water induces a weaker dipole in it.
Feedback if chosen: Correct. Water induces a temporary dipole in each nonpolar particle, and Xe, with far more electrons, is the more polarizable of the two.
Feedback if chosen: This option is wrong — you treated the two nonpolar particles as interchangeable partners. Their electron counts differ greatly, so their polarizabilities differ too.
Feedback if chosen: This option is wrong — you read 'nonpolar' as 'no attraction'. A polar water molecule induces a temporary dipole in a nonpolar particle, and that interaction is always attractive.
Check — multiple choice
A mixture contains dissolved Na⁺ ions, which carry a 1+ charge, and dissolved Mg²⁺ ions, which carry a 2+ charge. How do their interactions with a water molecule compare?
Feedback if chosen: This option is wrong — you ranked the smaller charge higher. The 2+ ion attracts a water molecule more strongly, because interaction strength is proportional to the charge on each particle (Coulomb's law).
Feedback if chosen: This option is wrong — you treated the two ions as interchangeable. A 2+ ion and a 1+ ion do not attract a water molecule equally strongly.
Feedback if chosen: Correct. Both ions attract the partially negative O end of a water molecule, and the 2+ ion attracts it more strongly, because interaction strength is proportional to the charge on each particle (Coulomb's law).
Feedback if chosen: This option is wrong — you gave an ion a hydrogen bond. Hydrogen bonding needs an H atom bonded to N, O, or F on one partner, and Mg²⁺ is a bare ion, so its attraction to water is an ion-dipole interaction.
Outcome: State that in large biomolecules and polymers, noncovalent interactions occur between different molecules or between different regions of the same large molecule.
Every substance you have ranked so far is built from small molecules, a few atoms each.
Living systems run on molecules built from thousands of atoms joined into long chains.
A chain that long can fold back on itself, bringing one stretch of the chain right up against another stretch of the same chain.
What holds a folded chain in that shape, when both stretches belong to one molecule?
The large molecules of living systems are called 'biomolecules', and proteins are one example.
Very long chain molecules, whether from a living system or made industrially, are called 'polymers'.
Between two separate small molecules, the attractions at work are the intermolecular forces you have already identified.
Attractions of this kind — attractions that are not covalent bonds — are called 'noncovalent interactions' when they act within or between large molecules.
In a biomolecule or a polymer, noncovalent interactions occur in two places: between different large molecules, and between different regions of the same large molecule.
The figure shows one folded protein chain, with noncovalent interactions between two regions of that single molecule.
The covalent bonds run unbroken along the chain, while the noncovalent interactions act across the gap between the folded regions.
You've already seen this possibility named for one force: hydrogen bonding can occur between two different molecules or between two different regions of the same molecule.
The other forces behave the same way along a long chain, so London dispersion forces (LDFs), dipole-dipole interactions, and ion-dipole interactions can all act between regions of one folded molecule.
State where noncovalent interactions occur in biomolecules and polymers: between different large molecules, and between different regions of the same large molecule.
Check — multiple choice
A protein chain folds back on itself inside a cell, alongside other molecules. Where can noncovalent interactions act?
Feedback if chosen: Correct. Noncovalent interactions act between different large molecules and between different regions of the same large molecule.
Feedback if chosen: This option is wrong — you allowed only the attractions between separate molecules. A chain long enough to fold brings two of its own regions close together, and noncovalent interactions act there as well.
Feedback if chosen: This option is wrong — you allowed only the contacts inside one chain. Two separate large molecules attract each other through the same noncovalent interactions.
Feedback if chosen: This option is wrong — you counted only the bonds along the chain. Covalent bonds join the atoms of the chain, while noncovalent interactions hold its folded regions against one another.
Check — multiple choice
Two long polymer chains lie side by side. Each chain is built from C and H atoms only. Which noncovalent interaction acts between the two chains?
Feedback if chosen: This option is wrong — you counted C–H hydrogens as hydrogen-bond donors. Hydrogen bonding needs an H atom bonded to N, O, or F.
Feedback if chosen: This option is wrong — you placed an ion in a chain of C and H atoms only. With no ion present, no ion-dipole interaction is possible.
Feedback if chosen: This option is wrong — you ruled out every attraction because the chains are nonpolar. LDFs are present between all particles, and along two long chains they add up.
Feedback if chosen: Correct. The chains are nonpolar, so LDFs are the noncovalent interaction between them — and you've already seen that LDFs are often the strongest net force between large molecules.
Check — multiple choice
One region of a protein chain carries an N–H group, and a folded-back region of the same chain carries an O atom; the two are attracted to each other. How is that attraction described?
Feedback if chosen: This option is wrong — you turned a noncovalent contact into a bond. No electron pair is shared between the N–H hydrogen and the O atom; they attract each other across a gap.
Feedback if chosen: Correct. An H atom bonded to N is attracted to an O atom, which is hydrogen bonding, and here it acts between two regions of one large molecule.
Feedback if chosen: This option is wrong — you placed the two groups on separate molecules. Both regions belong to the same folded chain.
Feedback if chosen: This option is wrong — you named a force that needs an ion. Neither the N–H group nor the O atom carries a full charge, so the attraction is hydrogen bonding.
Outcome: Explain that the function and properties of a large biomolecule or polymer depend strongly on its shape, which is largely set by its noncovalent interactions.
You've already seen that noncovalent interactions act between different regions of one large molecule.
Those attractions are what hold a folded chain in its particular shape.
A large molecule's shape is therefore set largely by its noncovalent interactions, rather than by the covalent bonds along the chain alone.
What the molecule then does depends strongly on that shape.
Much of what a protein does depends on the precise shape of its folded surface.
Heating a protein disrupts many of the noncovalent interactions holding its folded shape together.
The covalent bonds along the chain stay intact, so the chain is still one molecule with the same atoms in the same order.
The shape, though, is gone — and with it goes what that shape allowed the protein to do.
Raw egg white is clear and runny, and a few minutes in a hot pan leaves it white and firm.
Its protein chains have not been cut into pieces; the noncovalent interactions holding each folded shape have been disrupted.
The properties of the egg white changed because the shapes of its molecules changed.
A polymer's properties also depend on the shape its chains take and on the noncovalent interactions between them.
Chains that lie closely against one another interact more strongly, and the material is stiffer as a result.
So two samples of the same kind of polymer can differ in stiffness without any difference in their chains' covalent bonds.
Explain how the function and properties of a large biomolecule or polymer depend on its shape, and how that shape depends on its noncovalent interactions.
Check — multiple choice
A protein is heated until its folded shape is lost, while every covalent bond along its chain stays intact. Which statement describes the protein after heating?
Feedback if chosen: This option is wrong — you broke the chain. Heating disrupted noncovalent interactions, which act across gaps between regions; the covalent bonds joining the chain's atoms are intact.
Feedback if chosen: This option is wrong — you tied the function to the covalent bonds alone. The function depends strongly on the folded shape, and that shape was held by the noncovalent interactions heating disrupted.
Feedback if chosen: Correct. What a large molecule does depends strongly on its shape, so a chain that no longer holds its folded shape no longer does what that shape allowed.
Feedback if chosen: This option is wrong — you counted a change of shape as a change of substance. The same atoms are joined in the same order, and only the folded shape has changed.
Check — multiple choice
Two samples of the same kind of polymer differ in stiffness. Which difference between the samples explains it?
Feedback if chosen: Correct. Chains lying closely against one another interact more strongly through their noncovalent interactions, and the material is stiffer as a result.
Feedback if chosen: This option is wrong — you changed the substance. Both samples are the same kind of polymer, so their repeating units are the same.
Feedback if chosen: This option is wrong — you looked inside the chains. Both samples have the same covalent bonds, and what differs is how their chains lie against one another.
Feedback if chosen: This option is wrong — you ruled out a difference the arrangement can produce. The same chains, packed more or less closely, interact differently and give different stiffness.
Check — multiple choice
A folded chain and a stretched-out chain contain exactly the same atoms, joined in the same order. Which feature can still differ between the two?
Feedback if chosen: This option is wrong — you let the formula follow the shape. The same atoms in the same order give the same formula, folded or not.
Feedback if chosen: This option is wrong — you moved the covalent bonds. Folding brings regions of one chain together without changing which atoms are bonded to which.
Feedback if chosen: This option is wrong — folding neither adds nor removes atoms. Both chains contain the same atoms in the same order.
Feedback if chosen: Correct. What a large molecule does depends strongly on its shape, so two chains with identical atoms in identical order can behave differently when their shapes differ.
Check — multiple choice
Chlorine (Cl₂) contains 34 electrons per molecule; iodine (I₂) contains 106. Which molecule has the more polarizable electron cloud?
Feedback if chosen: Correct. Polarizability increases with the number of electrons and the size of the cloud — I₂'s 106-electron cloud is larger and more loosely held, so it distorts more easily.
Feedback if chosen: This option is wrong — you picked the smaller cloud. A cloud with fewer electrons held closer to its nuclei is harder to distort, not easier.
Feedback if chosen: This option is wrong — you treated family membership as what matters. Polarizability tracks the number of electrons and the size of the cloud, and these two clouds differ greatly on both counts.
Feedback if chosen: This option is wrong — you tied polarizability to a permanent dipole. Polarizability measures how easily a cloud distorts, and any electron cloud can distort, polar or not.
Check — multiple choice
Between which pair of molecules are the London dispersion forces stronger: two N₂ molecules (14 electrons each) or two Br₂ molecules (70 electrons each)?
Feedback if chosen: This option is wrong — you picked the less polarizable molecule. N₂'s small, tightly held cloud forms only small temporary dipoles, so its LDFs are weaker.
Feedback if chosen: This option is wrong — you assumed nonpolar molecules all attract equally. LDF strength tracks polarizability, and Br₂'s much larger cloud is far more polarizable than N₂'s.
Feedback if chosen: This option is wrong — you required a permanent dipole. LDFs come from temporary, fluctuating dipoles, which form in every molecule, polar or not.
Feedback if chosen: Correct. Br₂'s larger, 70-electron cloud is more polarizable, so larger temporary dipoles form and the facing partial charges attract more strongly.
Check — multiple choice
In an ICl molecule, the I end is δ⁺ and the Cl end is δ⁻. Two ICl molecules in a liquid sample can meet in several arrangements. Which arrangement gives the strongest attraction between the two molecules?
Feedback if chosen: This option is wrong — the orientation is right but the distance is not. As the separation grows, the attraction between the partial charges weakens, so distant molecules attract only weakly.
Feedback if chosen: This option is wrong — you faced two δ⁻ ends toward each other. Like partial charges repel; attraction needs a δ⁺ end facing a δ⁻ end.
Feedback if chosen: Correct. Opposite partial charges attract, and as the particles get closer, the attraction between them gets stronger — so a nearby δ⁺ I end facing a δ⁻ Cl end attracts most strongly.
Feedback if chosen: This option is wrong — you kept every δ⁺ end away from every δ⁻ end. With no opposite partial charges close together, only a weak attraction remains.
Check — multiple choice
A single Li⁺ ion sits among many molecules of chloroform, CHCl₃, a polar molecule. The attraction between the Li⁺ ion and a neighboring chloroform molecule is typically stronger than the attraction between two neighboring chloroform molecules. Why?
Feedback if chosen: This option is wrong — you reasoned from mass. Interaction strength follows the charges on the particles, not how light the particles are.
Feedback if chosen: This option is wrong — you canceled the dipole away. A polar molecule's δ⁺ and δ⁻ ends sit at different positions, so two chloroform molecules do attract, δ⁺ end to δ⁻ end — just more weakly than the ion does.
Feedback if chosen: This option is wrong — you turned an attraction between separate particles into a covalent bond. The ion and the molecule share no electrons; they attract while staying separate particles.
Feedback if chosen: Correct. An ion carries a full charge and a dipole end only a partial charge, so the ion-dipole interaction between Li⁺ and chloroform outranks the dipole-dipole interaction between two chloroform molecules.
Check — multiple choice
Which pure substance has no hydrogen bonding between its molecules?
Feedback if chosen: This option is wrong — you missed H₂O₂'s O–H bonds. Each H is bonded to an O, and neighboring molecules offer O acceptors, so hydrogen bonds form.
Feedback if chosen: Correct. PH₃'s H atoms are bonded only to P. P sits in the same group as N, but it is not N, O, or F, so no hydrogen bonding forms.
Feedback if chosen: This option is wrong — you missed the O–H group. CH₃CH₂OH has an H bonded to O and an O acceptor on each neighboring molecule, so hydrogen bonds form.
Feedback if chosen: This option is wrong — you missed the N–H bonds. CH₃NH₂ has H bonded to N and an N acceptor on each neighboring molecule, so hydrogen bonds form.
Check — multiple choice
Fluorine (F₂), hydrogen chloride (HCl), and methanol (CH₃OH) each contain 18 electrons per molecule, so their electron clouds are similar in size. Which ranking puts the three substances in order of overall intermolecular force strength, weakest first?
Feedback if chosen: This option is wrong — you placed F₂ above HCl. F₂'s two identical atoms share their electrons equally, so it is nonpolar and has only LDFs, while polar HCl adds dipole-dipole attractions on top of LDFs of similar strength.
Feedback if chosen: This option is wrong — you put HCl's dipole-dipole attractions above CH₃OH's hydrogen bonding. Hydrogen bonding is the stronger of the two, so CH₃OH ranks highest.
Feedback if chosen: Correct. With similar-sized clouds the LDFs are comparable, so the force types decide: nonpolar F₂ has only LDFs, polar HCl adds dipole-dipole attractions, and CH₃OH's O–H group adds hydrogen bonding, the strongest of the three.
Feedback if chosen: This option is wrong — you listed the substances strongest first. The comparison itself is right, but the question asks for the ranking weakest first.
Check — multiple choice
O₂ molecules, SO₂ molecules, and Ca²⁺ ions are all dissolved in the same sample of water. O₂ is nonpolar; SO₂ is polar. Rank the interaction of each dissolved particle with a neighboring water molecule, weakest first.
Feedback if chosen: This option is wrong — you listed the interactions strongest first. Ca²⁺'s full charge gives the strongest attraction to water, so it belongs at the end of a weakest-first list.
Feedback if chosen: Correct. Nonpolar O₂ manages only LDF and dipole-induced dipole attractions with water; polar SO₂ adds dipole-dipole attractions; and Ca²⁺'s full 2+ charge gives an ion-dipole interaction, the strongest of the three.
Feedback if chosen: This option is wrong — you ranked SO₂'s dipole-dipole attraction above Ca²⁺'s ion-dipole interaction. An ion carries a full charge while a dipole end carries only a partial charge, so the ion-dipole interaction is stronger.
Feedback if chosen: This option is wrong — you gave O₂ no attraction at all. Water's dipole induces a temporary dipole in O₂'s cloud, and LDFs act between every pair of particles, so O₂ still interacts — just weakly.
6 lessons
Outcome: State what gas pressure is — the force per unit area exerted by gas particles colliding with the walls of their container.
Press a gauge onto a tire's valve and it reads a pressure.
Yet the air inside the tire is nothing but tiny particles separated by empty space — what is pressing on the gauge?
The figure shows a sample of gas in a closed container: particles in constant motion, striking the walls from every direction.
Each collision delivers a tiny push on the wall it strikes.
Enormous numbers of particles strike every patch of wall every second, so the tiny pushes add up to a steady force spread over the walls.
That force, per unit area of wall, is called 'gas pressure'.
The tire's gauge reads a number because gas pressure is produced by particles colliding with the container walls, so more frequent or more forceful collisions mean higher pressure.
Gas pressure is usually measured in atmospheres (atm): 1 atm is roughly the pressure the air around you exerts at sea level.
State what gas pressure is: the force per unit area exerted by gas particles colliding with the walls of their container.
Check — multiple choice
Which of these is gas pressure?
Feedback if chosen: Correct. Every wall collision delivers a tiny push, and the pushes add up to a steady force — gas pressure is that force per unit area of wall.
Feedback if chosen: This option is wrong — you described how concentrated the gas is, not how hard it pushes. Pressure is a force per unit area, delivered by particles striking the walls.
Feedback if chosen: This option is wrong — you named a property of the particles' motion, not of their effect on the walls. Speed matters only through the collisions it produces.
Feedback if chosen: This option is wrong — you counted the particles instead of their pushes. Pressure is the force per unit area their wall collisions deliver, not a particle count.
Check — multiple choice
A sealed flask holds neon gas. What produces the pressure the neon exerts on the flask walls?
Feedback if chosen: This option is wrong — you let the atoms rest. Gas particles are in constant motion, striking the walls and rebounding, never lying on the bottom.
Feedback if chosen: This option is wrong — you built the push out of steady particle-to-particle contact. Gas particles are separated by empty space; only their collisions with the walls deliver the push.
Feedback if chosen: Correct. Gas pressure is produced by particles colliding with the container walls — each collision is a tiny push, and together they make the steady pressure the flask feels.
Feedback if chosen: This option is wrong — you let the atoms change size. They stay tiny; the pressure comes from their collisions with the walls, not from swelling.
Check — multiple choice
Two rigid containers have walls of equal area. The gas particles in container B strike the walls more often, and harder, than those in container A. Which gas exerts the higher pressure?
Feedback if chosen: This option is wrong — you reversed the connection. More frequent and more forceful collisions mean higher pressure, and container B has both.
Feedback if chosen: This option is wrong — you ignored the collisions. Pressure comes entirely from particles striking the walls, and B's walls are being struck more often and harder.
Feedback if chosen: This option is wrong — you reached for the particle count. However many particles each container holds, the pressure is set by the collisions actually delivered to the walls — given here as more frequent and more forceful in B.
Feedback if chosen: Correct. Gas pressure is produced by particles colliding with the container walls, so more frequent or more forceful collisions mean higher pressure — container B on both counts.
Outcome: State what vapor pressure is — the pressure exerted by the vapor above a liquid or solid in a closed container once particles escape and return at equal rates.
Seal water in a flask at 25 °C, pump the air out of the space above the liquid, and watch the gauge on that space.
The water just sits there — yet the gauge climbs from zero, then holds steady at about 0.03 atm.
Something is striking that gauge — where did a gas come from?
The figure shows the flask up close, at the surface between the liquid and the empty space above it.
You've already seen that intermolecular forces are attractions between separate particles, much weaker than the bonds within them.
At the liquid's surface, some particles break away from their neighbors and fly into the space above.
A substance's gas, sitting above its own liquid or solid, is called that substance's 'vapor'.
A liquid or a solid — particles in contact — is called a 'condensed phase'.
Particles cross both ways: vapor particles that strike the liquid's surface are caught by its attractions and rejoin it.
Just after sealing there is almost no vapor, so escapes far outnumber returns.
As vapor builds up, returns become more and more frequent.
Eventually the rate at which particles escape the condensed phase equals the rate at which they return.
From then on the amount of vapor holds steady — particles still escape and return constantly, but in equal numbers.
The gauge records that story as it happens.
| Time after sealing (min) | Gauge reading (atm) |
|---|---|
| 0 | 0.00 |
| 10 | 0.01 |
| 20 | 0.02 |
| 30 | 0.03 |
| 45 | 0.03 |
| 60 | 0.03 |
You've already seen that gas pressure is the force per unit area exerted by gas particles colliding with the walls of their container.
The vapor above the water is a gas like any other: its particles strike the flask walls and the gauge.
The pressure exerted by the vapor above a liquid or solid in a closed container, once the escape and return rates are equal, is called the 'vapor pressure'.
Water's vapor pressure at 25 °C is about 0.03 atm — the steady value the gauge reached.
State what vapor pressure is: the pressure exerted by the vapor above a liquid or solid in a closed container once the rate at which particles escape the condensed phase equals the rate at which they return.
Check — multiple choice
Liquid ethanol is sealed in a flask at 25 °C. After an hour, the gauge reading above the liquid has stopped changing. What does the steady reading tell you about the particles?
Feedback if chosen: This option is wrong — you stopped the escapes. Particles never stop leaving the liquid's surface; the reading steadies because returns have become just as frequent.
Feedback if chosen: Correct. A steady reading means a steady amount of vapor — particles still escape and return constantly, but the two rates have become equal.
Feedback if chosen: This option is wrong — you stopped the returns. Vapor particles strike the liquid's surface and rejoin it the whole time; it is the balance of the two rates that holds the reading steady.
Feedback if chosen: This option is wrong — you froze the vapor. Its particles keep moving and keep striking the walls — that constant motion is exactly what the gauge is reading.
Check — numeric entry
| Time after sealing (min) | Gauge reading (atm) |
|---|---|
| 0 | 0.00 |
| 10 | 0.14 |
| 20 | 0.23 |
| 30 | 0.28 |
| 45 | 0.28 |
| 60 | 0.28 |
The table shows the gauge readings above liquid bromine, Br₂, sealed in a flask at 25 °C with the air pumped out. What is the vapor pressure of bromine at 25 °C? Give your answer in atm to 2 decimal places.
Feedback when correct: Correct. From 30 minutes on the reading stays steady at 0.28 atm — escape and return have reached equal rates, and that steady value is bromine's vapor pressure at 25 °C.
Anticipated wrong answers (feedback shown):
| 0.23 | You read a value taken while the vapor was still building up — at 20 minutes the reading was still climbing. The vapor pressure is the steady value, 0.28 atm, reached once escape and return rates are equal. |
| 0.14 | You read an early value, taken while escapes still far outnumbered returns. The vapor pressure is the steady value the gauge settles at: 0.28 atm. |
| 0.00 | You read the starting value, before any vapor had formed. The vapor pressure is the steady value the gauge reaches once escape and return rates are equal: 0.28 atm. |
Check — multiple choice
The gauge reading above a freshly sealed liquid is still climbing. At that moment, how does the rate at which particles escape the liquid compare with the rate at which they return?
Feedback if chosen: This option is wrong — you read the imbalance the wrong way round. If returns outpaced escapes, the amount of vapor — and the reading — would be falling, not rising.
Feedback if chosen: This option is wrong — equal rates are what hold a reading steady. A climbing reading means vapor is still building up, so the rates have not yet matched.
Feedback if chosen: This option is wrong — the changing reading is itself the comparison: more vapor is appearing than disappearing, so escape must be outrunning return.
Feedback if chosen: Correct. Vapor is still building up, so more particles are leaving the liquid each second than are returning to it.
Outcome: State what the boiling point is — the temperature at which a liquid's vapor pressure equals the external pressure above it — and read a boiling point from vapor-pressure data.
Heat a pan of water at sea level and, at 100 °C, bubbles of vapor form throughout the liquid — not just at the surface.
Why 100 °C, and why does the whole liquid join in?
You've already seen that vapor pressure is the pressure exerted by a liquid's vapor once the escape and return rates are equal.
Vapor pressure rises fast as a liquid warms — the table shows water's.
| Temperature (°C) | Vapor pressure of water (atm) |
|---|---|
| 25 | 0.03 |
| 50 | 0.12 |
| 75 | 0.38 |
| 100 | 1.00 |
The gas above a liquid presses down on it — for an open pan at sea level, that is the air, pressing at about 1 atm.
The pressure pressing on the liquid from above, whatever gas supplies it, is called the 'external pressure'.
A bubble of vapor forming inside the liquid gets squeezed by that external pressure.
The bubble survives only if the vapor inside it pushes outward as hard as the outside squeezes in — that is, only once the vapor pressure has reached the external pressure.
The temperature at which a liquid's vapor pressure equals the external pressure above it is called the 'boiling point'.
Water under 1 atm reaches a vapor pressure of 1 atm at exactly 100 °C — the table's last row.
Change the external pressure, and the same liquid boils at a different temperature.
State what the boiling point is: the temperature at which a liquid's vapor pressure equals the external pressure above it.
Check — multiple choice
A liquid boils when which condition is met?
Feedback if chosen: Correct. Boiling begins when the liquid's vapor pressure has risen to match the external pressure above it — only then can bubbles of vapor survive inside the liquid.
Feedback if chosen: This option is wrong — you promoted water's sea-level value into a rule for every liquid. 100 °C is just where water's vapor pressure reaches 1 atm; other liquids, and other external pressures, give other temperatures.
Feedback if chosen: This option is wrong — you sent the vapor pressure to a maximum. Boiling needs the vapor pressure to reach the external pressure, whatever that happens to be — not some highest possible value.
Feedback if chosen: This option is wrong — you removed the external pressure instead of matching it. Boiling happens when the vapor pressure rises to the pressure that is actually pressing on the liquid.
Check — numeric entry
| Temperature (°C) | Vapor pressure of water (atm) |
|---|---|
| 25 | 0.03 |
| 50 | 0.12 |
| 75 | 0.38 |
| 90 | 0.69 |
| 95 | 0.83 |
| 100 | 1.00 |
The table gives the vapor pressure of water at several temperatures. At a mountain campsite the external pressure is 0.83 atm. At what temperature does water boil there?
Feedback when correct: Correct. Water boils where its vapor pressure equals the external pressure — and the table shows water's vapor pressure reaching 0.83 atm at 95 °C.
Anticipated wrong answers (feedback shown):
| 100 | You gave water's sea-level boiling point. Here the external pressure is only 0.83 atm, and the table shows water's vapor pressure reaching 0.83 atm at 95 °C — so it boils at 95 °C. |
| 90 | You read the row where the vapor pressure is 0.69 atm — still below the external pressure, so bubbles cannot yet survive. The match with 0.83 atm comes at 95 °C. |
| 83 | You turned the pressure's digits into a temperature. 0.83 atm is the external pressure to match; the table shows water's vapor pressure reaching it at 95 °C. |
Check — multiple choice
Inside a sealed pressure cooker, the external pressure above the water is held at 2 atm. Compared with an open pan at sea level, at what temperature does the water in the cooker boil?
Feedback if chosen: This option is wrong — you moved the boiling point the wrong way. A higher external pressure demands a higher vapor pressure, and vapor pressure only reaches higher values at higher temperatures.
Feedback if chosen: This option is wrong — you kept water's sea-level boiling point. 100 °C is where water's vapor pressure reaches 1 atm; at 2 atm the match must come at a higher temperature.
Feedback if chosen: Correct. To boil under 2 atm, the water's vapor pressure must rise to 2 atm — beyond its 100 °C value of 1 atm — so the match comes above 100 °C. That is the point of a pressure cooker: hotter liquid water.
Feedback if chosen: This option is wrong — you capped the vapor pressure at 1 atm. Kept liquid and heated further, water's vapor pressure keeps rising past 1 atm and does reach 2 atm.
Outcome: Explain why a liquid with stronger intermolecular forces has a lower vapor pressure, and predict which of two liquids has the higher vapor pressure at one temperature.
Acetone, CH₃COCH₃, is the liquid in many nail-polish removers.
Seal water in one flask and acetone in another, pump out the air above each liquid, and hold both at 25 °C.
The two gauges settle at very different steady values.
At the same temperature, acetone's vapor reaches a steady pressure ten times water's — why?
You've already seen that vapor pressure is the steady pressure a liquid's vapor reaches once particles escape and return at equal rates.
To escape, a surface particle must break away from every neighbor that attracts it — completely.
Escaping from the condensed phase into the vapor is called 'vaporizing'.
You've already seen how to rank liquids by overall intermolecular force strength from their structures.
Water molecules hydrogen-bond to their neighbors; acetone, with no O-H bond, holds together through dipole-dipole and London dispersion forces alone.
Water's stronger attractions hold its surface molecules back, so fewer of them escape each second.
Fewer escapes mean the vapor levels off at a smaller steady amount — and a smaller amount of vapor exerts a lower pressure.
Water's vapor pressure sits far below acetone's because vaporizing requires completely overcoming intermolecular attractions, so stronger attractions mean fewer particles escape and more energy is needed.
The rule is general: at any one temperature, the stronger a liquid's intermolecular forces, the lower its vapor pressure.
Explain why vapor pressure decreases as intermolecular force strength increases: escaping particles must completely overcome their attractions, so stronger attractions mean fewer escapes.
Check — multiple choice
Ethanol, CH₃CH₂OH, and acetone, CH₃COCH₃, are both liquids at 25 °C. Which has the higher vapor pressure?
Feedback if chosen: This option is wrong — you gave the hydrogen-bonding liquid the higher vapor pressure. Ethanol's O-H group hydrogen-bonds, those stronger attractions let fewer molecules escape, and its vapor pressure is the lower one.
Feedback if chosen: Correct. Acetone has no O-H bond, so its molecules attract only through dipole-dipole and London dispersion forces — weaker overall than ethanol's hydrogen bonds — and more acetone molecules escape each second.
Feedback if chosen: This option is wrong — you let the shared temperature force one shared vapor pressure. Temperature is only half the story: at one temperature, the liquid with the weaker attractions reaches the higher steady vapor pressure.
Feedback if chosen: This option is wrong — the structures decide it: ethanol's O-H group hydrogen-bonds and acetone's molecules cannot, so acetone's weaker attractions give it the higher vapor pressure.
Check — multiple choice
| Liquid | Vapor pressure at 25 °C (atm) |
|---|---|
| A | 0.45 |
| B | 0.05 |
Two molecular liquids are held at the same temperature, and their vapor pressures are given in the table. Which liquid has the stronger intermolecular forces?
Feedback if chosen: This option is wrong — you tied the higher vapor pressure to the stronger attractions. It is the other way around: stronger attractions hold particles back, so fewer escape and the vapor pressure is lower.
Feedback if chosen: This option is wrong — you asked for information you don't need. Whichever liquids these are, the one with the lower vapor pressure at the shared temperature is the one whose attractions hold its particles back more strongly.
Feedback if chosen: This option is wrong — you let one shared temperature force equal attractions. The very different steady pressures show very different escape rates, and escape rates reflect attraction strength.
Feedback if chosen: Correct. Far fewer of liquid B's particles escape each second at the same temperature, so its attractions must be the stronger ones — vapor pressure falls as intermolecular force strength rises.
Check — multiple choice
At 25 °C, water's vapor pressure is far below acetone's. Why?
Feedback if chosen: Correct. Water's hydrogen bonds are stronger than acetone's dipole-dipole and London dispersion attractions, so fewer water molecules escape each second and the vapor levels off at a smaller amount — a lower pressure.
Feedback if chosen: This option is wrong — you reached for mass, and in the wrong direction: a water molecule (18 amu) is lighter than an acetone molecule (58 amu). It is the strength of the attractions, not the mass, that sets the escape rate.
Feedback if chosen: This option is wrong — you broke covalent bonds. A molecule escapes the liquid with every O-H bond intact; only the attractions between separate molecules must be overcome.
Feedback if chosen: This option is wrong — you stopped acetone's returns. In both flasks vapor particles constantly return to the liquid; the steady pressures differ because the escape rates differ.
Outcome: Explain why a substance with stronger intermolecular forces has a higher boiling point, and predict boiling-point order from structures.
Methane, CH₄, and methanol, CH₃OH, are molecules of similar size.
Their boiling points are not similar at all.
| Substance | Boiling point at 1 atm (°C) |
|---|---|
| CH₄ | −162 |
| CH₃OH | 65 |
A gap of 227 °C between such similar molecules needs explaining.
You've already seen that the boiling point is the temperature at which a liquid's vapor pressure equals the external pressure above it.
You've already seen that at any one temperature, stronger intermolecular forces mean a lower vapor pressure.
A liquid with stronger attractions therefore starts further from boiling at every temperature: its vapor pressure sits lower across the board.
Lifting that lower vapor pressure all the way up to the external pressure takes more heating.
So the stronger a substance's intermolecular forces, the higher its boiling point.
Real data show the race: by 56 °C acetone's vapor pressure has reached 1 atm — so acetone boils — while water's is still far behind.
| Temperature (°C) | Vapor pressure of acetone (atm) | Vapor pressure of water (atm) |
|---|---|---|
| 25 | 0.30 | 0.03 |
| 56 | 1.00 | 0.16 |
Under 1 atm, acetone boils at 56 °C; water must climb to 100 °C before its vapor pressure gets there.
Methanol molecules hydrogen-bond through their O-H groups; methane, nonpolar, holds together through London dispersion forces alone.
Methanol boils far above methane because vaporizing requires completely overcoming intermolecular attractions, so stronger attractions mean fewer particles escape and more energy is needed.
That is the whole 227 °C gap in the first table.
Explain why boiling point increases as intermolecular force strength increases, and predict which of two substances boils higher from their structures.
Check — multiple choice
Ammonia, NH₃, and methane, CH₄, are molecules of similar size. Which boils at the higher temperature?
Feedback if chosen: This option is wrong — you gave methane the stronger attractions. CH₄ is nonpolar, with only London dispersion forces; NH₃ hydrogen-bonds, and stronger attractions mean a higher boiling point.
Feedback if chosen: This option is wrong — you let similar sizes force similar boiling points. Size is only one factor: NH₃'s hydrogen bonds are far stronger than CH₄'s London dispersion forces.
Feedback if chosen: Correct. NH₃ hydrogen-bonds (H bonded to N); CH₄ cannot. Stronger attractions mean fewer particles escape, so NH₃ must be heated much further before its vapor pressure reaches the external pressure.
Feedback if chosen: This option is wrong — the structures give you exactly what you need: an N-H group that hydrogen-bonds against a nonpolar molecule that cannot.
Check — multiple choice
Methanol, CH₃OH, boils at 65 °C while methane, CH₄, boils at −162 °C. Why does methanol boil so much higher?
Feedback if chosen: This option is wrong — you broke covalent bonds. Boiling leaves every bond within the molecule intact; it is the attractions between separate molecules that must be completely overcome.
Feedback if chosen: Correct. Methanol's O-H groups hydrogen-bond, so far fewer methanol molecules escape at any temperature — its vapor pressure sits lower, and much more heating is needed to lift it to the external pressure.
Feedback if chosen: This option is wrong — you reversed the vapor-pressure comparison. Stronger attractions mean a lower vapor pressure at every temperature — methanol's sits lower, which is why it needs more heating.
Feedback if chosen: This option is wrong — you removed methane's attractions entirely. CH₄ molecules do attract one another through London dispersion forces — that is why methane can be a liquid at all near −162 °C.
Check — multiple choice
Rank Ne, HCl, and HF from highest boiling point to lowest.
Feedback if chosen: Correct. HF hydrogen-bonds (H bonded to F), HCl is polar with dipole-dipole attractions but cannot hydrogen-bond, and Ne has only London dispersion forces — so the boiling points fall in exactly that order.
Feedback if chosen: This option is wrong — you reversed the ranking. The strongest attractions, HF's hydrogen bonds, make for the highest boiling point, not the lowest.
Feedback if chosen: This option is wrong — you placed dipole-dipole attractions above hydrogen bonding. HF's H is bonded to F, so HF hydrogen-bonds — the stronger attraction — and boils above HCl.
Feedback if chosen: This option is wrong — you placed Ne above HCl. Ne atoms attract only through London dispersion forces, weaker here than the dipole-dipole attractions between polar HCl molecules.
Outcome: State that melting points tend to correlate with intermolecular force strength but can be more subtle, because melting only rearranges the interactions rather than overcoming them completely.
You've already seen that melting point is the temperature at which a solid changes to a liquid under the stated pressure.
Boiling points track intermolecular force strength faithfully — do melting points do the same?
You've already seen the two forms of C₅H₁₂: chain-shaped pentane, whose long contact area strengthens its London dispersion forces, and compact neopentane, with less contact area and weaker ones.
The table gives both properties for the pair.
| Substance | Boiling point (°C) | Melting point (°C) |
|---|---|---|
| pentane | 36 | −130 |
| neopentane | 10 | −17 |
The boiling points behave: pentane, with the stronger London dispersion forces, boils higher.
The melting points flip: neopentane — the weaker-attracting molecule — melts more than 110 °C above pentane.
You've already seen that vaporizing means completely overcoming the attractions between particles — so boiling reads attraction strength directly.
Melting asks for much less: the ordered arrangement of the solid loosens into a liquid, but the particles stay in contact.
Melting only rearranges the interactions rather than overcoming them completely.
So a melting point also depends on details beyond attraction strength — above all, how neatly the particles fit into the solid's ordered arrangement.
Compact, symmetric neopentane fits an ordered arrangement so well that its solid keeps its order up to a much higher temperature than pentane's does.
Melting points therefore tend to correlate with intermolecular force strength, but the correlation can be more subtle.
In particular, two substances with similar boiling points can have quite different melting points.
State that melting points tend to correlate with intermolecular force strength but can be more subtle, because melting only rearranges the interactions rather than overcoming them completely.
Check — multiple choice
| Substance | Boiling point (°C) | Melting point (°C) |
|---|---|---|
| benzene | 80 | 6 |
| hexane | 69 | −95 |
The table compares benzene and hexane. Which claim do these data support?
Feedback if chosen: This option is wrong — you pinned every melting point just under its boiling point. Hexane refutes it in this very table: it melts about 164 °C below the temperature at which it boils.
Feedback if chosen: This option is wrong — you overcorrected to no connection at all. Melting points do tend to follow attraction strength; these data show the correlation is loose, not absent.
Feedback if chosen: This option is wrong — you turned one surprise into a rule. Here the higher-boiling substance, benzene, also melts higher — the two properties simply do not move in step.
Feedback if chosen: Correct. The boiling points sit only 11 °C apart, yet the melting points differ by about 100 °C — melting reflects more than attraction strength alone.
Check — multiple choice
Why do melting points track intermolecular force strength less tightly than boiling points do?
Feedback if chosen: This option is wrong — you removed the attractions from the solid. The particles in a solid attract each other strongly; melting loosens their ordered arrangement without ending the attractions.
Feedback if chosen: Correct. The particles stay in contact in the liquid, so melting is only a rearrangement — and a melting point therefore also reflects how neatly the particles fit their ordered arrangement, not attraction strength alone.
Feedback if chosen: This option is wrong — you treated melting like boiling. Only boiling separates the particles completely; melting leaves them touching, which is exactly why it is the looser test of attraction strength.
Feedback if chosen: This option is wrong — you changed the attractions between phases. The same intermolecular forces act in the solid and in its liquid; what changes at melting is the arrangement, not the kind of attraction.
Check — multiple choice
Substances X and Y boil within a few degrees of each other. What is the safest conclusion about their melting points?
Feedback if chosen: This option is wrong — you assumed the tight correlation that boiling points have. Melting only rearranges the interactions, so similar boiling points leave room for quite different melting points.
Feedback if chosen: This option is wrong — you placed every melting point just below its boiling point. Many substances melt hundreds of degrees below the temperature at which they boil.
Feedback if chosen: Correct. Similar boiling points show the attraction strengths are similar, but melting also depends on how neatly the particles fit an ordered arrangement — so the melting points could still be quite different.
Feedback if chosen: This option is wrong — you flipped the lesson into a rule. Similar boiling points do not force the melting points apart; they simply leave them unpredicted — close or far, either can happen.
2 lessons
Outcome: Draw several interacting molecules, showing each intermolecular attraction as a dashed line between the right atoms of neighboring molecules.
You've already seen how to identify every intermolecular force in a pure substance from its structure and polarity.
A particulate drawing has to show those forces without confusing them with the covalent bonds inside each molecule.
The figure shows the convention for two H₂O molecules attracting through one hydrogen bond.
Inside each molecule, every covalent O–H bond is a solid line.
The hydrogen bond is a dashed line drawn between the molecules — from an H atom of one molecule to the O atom of its neighbor.
Because an intermolecular force acts between molecules, the dashed line never sits inside a single molecule.
δ⁺ at the H end and δ⁻ at the O end mark the opposite partial charges doing the attracting.
You've already seen that a hydrogen bond needs an H covalently bonded to N, O, or F, attracted to the N, O, or F of a neighboring molecule.
So in a drawing, a dashed hydrogen bond starts only at an N–H, O–H, or F–H hydrogen — never at a hydrogen bonded to carbon.
It ends at the N, O, or F atom of the neighbor, the site of the partial negative charge.
Forming the attraction breaks nothing: every covalent bond in both molecules stays exactly as drawn.
With several molecules the convention simply repeats: draw each molecule complete, then add one dashed line for each attracting pair.
Draw a particulate representation of multiple interacting molecules, showing each intermolecular attraction as a dashed line between the correct atoms of neighboring molecules.
Check — drawing task (pen and paper, self-marked)
Take a pen and paper and draw two H₂O molecules interacting through one hydrogen bond. Draw the covalent bonds as solid lines, draw the hydrogen bond as a dashed line, and mark δ⁺ and δ⁻ at the two ends of the dashed line. When you are done, select Check to compare your drawing with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
Check — drawing task (pen and paper, self-marked)
Take a pen and paper and draw an NH₃ molecule and an H₂O molecule interacting through one hydrogen bond in which the NH₃ molecule provides the hydrogen. Mark δ⁺ and δ⁻ at the two ends of the dashed line. When you are done, select Check to compare your drawing with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
Check — multiple choice
The figure shows the structure of CH₃OH. Two CH₃OH molecules sit side by side, and the hydrogen bond between them is to be drawn as a dashed line. Between which two atoms does the dashed line run?
Feedback if chosen: This option is wrong — you started the dashed line at a carbon-bonded H, and a hydrogen bond needs an H bonded to N, O, or F.
Feedback if chosen: This option is wrong — you connected the two partial negative O atoms, and an attraction needs opposite partial charges at its ends.
Feedback if chosen: Correct. The O–H hydrogen carries the partial positive charge, and the neighboring molecule's O carries the partial negative charge the dashed line ends on.
Feedback if chosen: This option is wrong — you kept the dashed line inside one molecule, and an intermolecular attraction acts between two molecules.
Outcome: Evaluate a drawing of interacting molecules, and name the one specific feature that is wrong when the drawing has a flaw.
You've already seen the drawing convention: solid lines for the covalent bonds inside each molecule, and a dashed line between molecules for each intermolecular attraction.
Evaluating someone else's drawing means applying that convention as a checklist and pointing at the one feature that breaks it.
Check one: the dashed line is drawn between two molecules — an attraction drawn inside a single molecule shows no intermolecular force at all.
This drawing fails check one: its dashed line sits alongside a covalent O–H bond inside the left molecule.
Check two: the two ends of the dashed line carry opposite partial charges — for a hydrogen bond, an N–H, O–H, or F–H hydrogen at one end and the neighbor's N, O, or F at the other.
A hydrogen bonded to carbon fails check two, because C and H share their electrons almost evenly, leaving that H with almost no partial positive charge.
Check three: every covalent bond in both molecules is still there, drawn solid and unchanged.
A drawing that passes all three checks is correct — reject a drawing only when you can name the specific check it fails.
Evaluate whether a supplied particulate drawing shows the species and their interactions correctly, and identify the one specific flaw when it does not.
Check — multiple choice
The figure shows a student's drawing of two HCl molecules and the dipole-dipole attraction between them. One feature of the drawing is wrong — which one?
Feedback if chosen: This option is wrong — the marks are placed correctly: Cl pulls the shared electrons toward itself, so Cl is δ⁻ and H is δ⁺.
Feedback if chosen: Correct. A dipole-dipole attraction acts between molecules, so the dashed line must run from the δ⁺ H of one HCl molecule to the δ⁻ Cl of the other — not alongside a bond inside one molecule.
Feedback if chosen: This option is wrong — solid lines are required here: a solid line is the convention for a covalent bond.
Feedback if chosen: This option is wrong — separate, complete molecules are required: a dipole-dipole attraction acts between distinct molecules that stay whole.
Check — multiple choice
The figure shows a student's drawing of the hydrogen bond between two H₂O molecules. One feature of the drawing is wrong — which one?
Feedback if chosen: Correct. A hydrogen bond points from a δ⁺ hydrogen to the neighbor's δ⁻ O — as drawn, the dashed line joins two δ⁺ hydrogens, and like partial charges push apart rather than attract.
Feedback if chosen: This option is wrong — starting at an O–H hydrogen is required: that hydrogen carries the partial positive charge that does the attracting.
Feedback if chosen: This option is wrong — δ⁻ belongs on O: the O pulls the shared electrons toward itself, so it carries the partial negative charge.
Feedback if chosen: This option is wrong — a line drawn between the molecules is required: that is exactly what makes the attraction intermolecular.
Check — multiple choice
The figure shows a student's drawing of the hydrogen bond between two CH₃OH molecules. One feature of the drawing is wrong — which one?
Feedback if chosen: This option is wrong — δ⁻ belongs on the O atoms: each O pulls shared electrons toward itself.
Feedback if chosen: This option is wrong — ending on the neighbor's O is required: the O carries the partial negative charge the dashed line must end on.
Feedback if chosen: Correct. A hydrogen bond starts only at an H bonded to N, O, or F — this dashed line starts at a carbon-bonded H, which carries almost no partial positive charge. It belongs on the O–H hydrogen instead.
Feedback if chosen: This option is wrong — solid covalent bonds are required: forming an intermolecular attraction changes nothing inside either molecule.
Check — multiple choice
The figure shows a student's drawing of the hydrogen bond between an HF molecule and an NH₃ molecule. Which feature of the drawing, if any, is wrong?
Feedback if chosen: This option is wrong — the dashed line starts on the H of HF, and F is one of the three atoms (N, O, F) whose bonded H can form a hydrogen bond.
Feedback if chosen: This option is wrong — the dashed line ends on the N of NH₃, and N carries the partial negative charge a hydrogen bond ends on.
Feedback if chosen: This option is wrong — δ⁺ sits on the F-bonded H and δ⁻ sits on the N, matching how the electrons are actually shared.
Feedback if chosen: Correct. The dashed line is drawn between the molecules, from an F-bonded H to the neighbor's N, with δ⁺ and δ⁻ on those atoms — every check passes.
20 lessons
Outcome: State that ionic solids tend to have very low vapor pressures.
Open a box of mothballs and the smell reaches you across the room — the solid is sending molecules into the air.
Open a box of table salt and there is nothing to smell — ever.
You've already seen vapor pressure: the pressure exerted by the vapor above a liquid or solid in a closed container, once particles escape and return at equal rates.
A sealed jar of mothballs develops a small but measurable vapor pressure, and so does ice — which is why old ice cubes slowly shrink in the freezer.
Seal solid NaCl in a jar at room temperature and no measurable vapor forms at all.
You've already seen the structure of solid NaCl: a crystal lattice, with Na⁺ and Cl⁻ ions repeating throughout the solid.
Solids built this way barely produce vapor: ionic solids tend to have very low vapor pressures.
At room temperature the vapor pressure of solid NaCl is too small to measure.
The pattern is general — seal KBr, CaCl₂, or MgO in a jar at room temperature and you find essentially no vapor.
State that ionic solids tend to have very low vapor pressures — solid NaCl at room temperature produces no measurable vapor.
Check — multiple choice
Four sealed jars sit at room temperature. Which jar develops essentially no vapor pressure?
Feedback if chosen: This option is wrong — you treated a molecular solid like an ionic one. Solid I₂ produces enough vapor to tint the jar visibly purple.
Feedback if chosen: This option is wrong — you picked the jar with the most vapor. Acetone is a molecular liquid with a vapor pressure of about 0.30 atm at 25 °C.
Feedback if chosen: Correct. KBr is an ionic solid, and ionic solids tend to have very low vapor pressures — no measurable vapor forms.
Feedback if chosen: This option is wrong — you overlooked water's measurable vapor pressure, about 0.03 atm at 25 °C. Liquid water steadily produces vapor in a sealed jar.
Check — multiple choice
CaF₂ is an ionic solid. At 25 °C, which vapor pressure is expected above a sealed sample of it?
Feedback if chosen: This option is wrong — you assigned the vapor pressure of a molecular liquid like acetone. An ionic solid produces essentially no vapor at room temperature.
Feedback if chosen: Correct. Ionic solids tend to have very low vapor pressures — for CaF₂ at 25 °C, the vapor is too small to measure.
Feedback if chosen: This option is wrong — you assigned the vapor pressure of liquid water at 25 °C. An ionic solid sits far below even that small value.
Feedback if chosen: This option is wrong — a vapor pressure of 1 atm is what a liquid shows at its boiling point, not what an ionic solid shows at room temperature.
Outcome: State that ionic substances tend to have high boiling points as well as high melting points.
You've already seen why ionic substances have relatively high melting points: melting requires enough energy to overcome the strong Coulombic attractions holding oppositely charged ions throughout the lattice.
For NaCl, melting alone takes 801 °C.
Boiling sits far higher still.
| Change | Temperature (°C) |
|---|---|
| Melts | 801 |
| Boils | about 1465 |
You've already seen the boiling point: the temperature at which a liquid's vapor pressure equals the external pressure above it.
For liquid NaCl under 1 atm, that temperature is near 1465 °C — far above the boiling point of any small molecular substance.
The pattern is general: ionic substances tend to have high boiling points as well as high melting points.
Individual values differ from one ionic substance to another, but they sit high — hundreds to thousands of degrees Celsius.
State that ionic substances tend to have high boiling points as well as high melting points — NaCl boils near 1465 °C.
Check — multiple choice
KBr is an ionic substance. Which boiling point is consistent with that?
Feedback if chosen: This option is wrong — you treated KBr like a small molecular substance; −34 °C is where molecular Cl₂ boils. Ionic substances tend to boil high.
Feedback if chosen: This option is wrong — you treated KBr like a small molecular substance; 59 °C is where molecular Br₂ boils. Ionic substances tend to boil high.
Feedback if chosen: This option is wrong — you treated KBr like a small molecular substance; 184 °C is where molecular I₂ boils. Ionic substances tend to boil high.
Feedback if chosen: Correct. Ionic substances tend to have high boiling points — KBr boils at about 1435 °C, in the same territory as NaCl's 1465 °C.
Check — multiple choice
Substances W, X, Y, and Z are each either a small molecular substance or an ionic substance. Which is most likely ionic?
Feedback if chosen: This option is wrong — you read 100 °C as high, but that is where molecular water boils. Ionic substances boil hundreds to thousands of degrees higher.
Feedback if chosen: Correct. A boiling point near 1600 °C fits the ionic pattern — molecular substances in this comparison boil far lower.
Feedback if chosen: This option is wrong — you picked the lowest boiler of the four. A boiling point of −162 °C fits a small molecular substance like CH₄.
Feedback if chosen: This option is wrong — a boiling point of 56 °C fits a small molecular substance like acetone, not an ionic one.
Outcome: Explain why ionic substances have very low vapor pressures and high boiling points.
You've already seen the two properties: solid NaCl produces no measurable vapor at room temperature, and NaCl does not boil until about 1465 °C.
Both trace back to the same question — what it costs a particle to leave for the vapor.
You've already seen the lattice: oppositely charged ions attracting one another throughout the solid.
For any particle to escape to the vapor, it must completely overcome the attractions holding it to its neighbors.
In an ionic substance those attractions are Coulombic attractions between full charges — every Na⁺ carries a full 1+ charge and every Cl⁻ a full 1− charge.
Full charges make these attractions very strong, because interaction strength is proportional to the charge on each particle (Coulomb's law).
In a polar molecular substance, by contrast, the attractions between molecules act between partial charges — δ⁺ and δ⁻ — far smaller than full charges.
Escaping from the lattice costs an enormous amount of energy, because the full ionic charges attract each particle in every direction.
At room temperature, essentially no particles have anywhere near that energy, so essentially nothing leaves the solid.
With essentially nothing leaving, almost no vapor forms — the vapor pressure of an ionic solid is very low.
What trace of vapor does form is electrically neutral overall — the solid does not shed separate free ions.
Boiling requires the vapor pressure to rise all the way up to the external pressure, and for an ionic substance that takes an enormous temperature — near 1465 °C for NaCl.
Explain that ionic substances have very low vapor pressures and high boiling points because escaping to the vapor requires completely overcoming the strong Coulombic attractions between full ionic charges, so essentially no particles leave the solid.
Check — multiple choice
A sealed sample of solid KBr at room temperature develops no measurable vapor. Why?
Feedback if chosen: Correct. Escape to the vapor means completely overcoming the Coulombic attractions between full ionic charges — at room temperature essentially no particles can, so essentially nothing leaves the solid.
Feedback if chosen: This option is wrong — you blamed the mass of the ions. Heavy particles can still enter a vapor; what stops these ions is the strength of the full-charge attractions holding them.
Feedback if chosen: This option is wrong — you assigned covalent bonding to an ionic solid. KBr is held by Coulombic attractions between K⁺ and Br⁻ ions, not by covalent bonds.
Feedback if chosen: This option is wrong — you blamed the geometry of the lattice. Particles at the surface are already in position to leave; what holds them is the strength of the attractions, not a blocked path.
Check — multiple choice
KCl boils near 1420 °C, while HCl boils at −85 °C. What explains the enormous difference?
Feedback if chosen: This option is wrong — you ranked by mass. I₂ is heavier than KCl per particle yet boils at only 184 °C; what matters is full charges against partial charges.
Feedback if chosen: Correct. K⁺ and Cl⁻ carry full charges, and interaction strength is proportional to the charge on each particle — so escape from KCl costs vastly more than escape from HCl, where only partial charges attract.
Feedback if chosen: This option is wrong — you turned HCl's intermolecular attractions into repulsions. HCl molecules attract one another; the attractions are simply between partial charges, so they are weak.
Feedback if chosen: This option is wrong — you reasoned from density. Density does not set the escape cost; the strength of the attractions between the particles does.
Check — multiple choice
Solid LiF has both a vapor pressure too small to measure at room temperature and a boiling point above 1600 °C. Which single fact explains both observations?
Feedback if chosen: This option is wrong — you cited the arrangement rather than the strength. Molecular solids also pack into repeating lattices, yet they produce measurable vapor.
Feedback if chosen: This option is wrong — you reasoned from ion size. Small size does not stop escape; the full-charge Coulombic attractions do.
Feedback if chosen: Correct. One escape cost explains both properties: essentially no particles overcome the full-charge attractions at room temperature (no measurable vapor), and driving the vapor pressure up to 1 atm takes an enormous temperature (very high boiling point).
Feedback if chosen: This option is wrong — you paired each ion with a single partner. In the lattice each ion attracts every nearby ion of opposite charge, and escape means overcoming those attractions, not splitting a pair.
Outcome: State that a covalent network solid consists of atoms covalently bonded into a continuous three-dimensional network or into stacked two-dimensional layers.
Diamond is pure carbon — no metal atoms, no ions, and, strangely, no molecules.
Grind a diamond as fine as you like and you never arrive at a 'diamond molecule' — there is no small repeating particle to name.
So what kind of solid is it?
The figure shows a flat section of the answer: carbon atoms covalently bonded to their neighbors, which are bonded to their neighbors, without end.
You've already seen the covalent bond — one or more shared pairs of valence electrons between two atoms — and here it simply never stops repeating.
In diamond this bonding continues in three dimensions: every C atom is bonded into one continuous three-dimensional network.
A solid built this way is called a 'covalent network solid'.
The whole crystal is, in effect, one connected particle: a single network of covalently bonded atoms.
Covalent network solids come in two arrangements.
Some, like diamond, are continuous three-dimensional networks.
Others, like graphite — also pure carbon — are two-dimensional layers stacked on top of one another, with the covalent bonding extending through each flat layer.
The figure shows what a single graphite layer looks like: every C atom covalently bonded to its neighbors within one flat sheet.
State that a covalent network solid consists of atoms covalently bonded into a continuous three-dimensional network (like diamond) or into stacked two-dimensional layers (like graphite).
Check — multiple choice
In solid silicon, each Si atom is covalently bonded to four neighboring Si atoms, and the bonding continues in three dimensions throughout the entire crystal. What class of solid is this?
Feedback if chosen: This option is wrong — you looked for discrete molecules, but the covalent bonding never ends at a molecule; it continues throughout the crystal.
Feedback if chosen: Correct. Atoms covalently bonded into one continuous three-dimensional network — that is exactly the definition of a covalent network solid.
Feedback if chosen: This option is wrong — you assigned ions to a solid that has none. Every particle in solid silicon is a neutral Si atom sharing electrons covalently.
Feedback if chosen: This option is wrong — you assigned a delocalized electron sea to a solid whose electrons sit in fixed covalent bonds between specific atoms.
Check — multiple choice
Which description fits a covalent network solid?
Feedback if chosen: This option is wrong — you described a molecular solid. In a covalent network solid there are no discrete molecules for intermolecular forces to act between.
Feedback if chosen: This option is wrong — you described an ionic solid. A covalent network solid contains neutral atoms sharing electrons, not charged ions.
Feedback if chosen: Correct. A covalent network solid is atoms covalently bonded without end — as a continuous three-dimensional network or as stacked two-dimensional layers.
Feedback if chosen: This option is wrong — you described a metallic solid. In a covalent network the valence electrons stay in fixed covalent bonds rather than moving freely.
Check — multiple choice
Graphite consists of flat sheets of covalently bonded C atoms stacked on top of one another. Which statement about graphite is correct?
Feedback if chosen: Correct. Stacked two-dimensional layers of covalently bonded atoms are the second form of covalent network solid — graphite is the standard example.
Feedback if chosen: This option is wrong — you treated each endless sheet as a discrete molecule. A molecular solid is built of separate small particles; a graphite sheet is continuous covalent bonding.
Feedback if chosen: This option is wrong — you required three-dimensional bonding, but the definition includes both continuous three-dimensional networks and stacked two-dimensional layers.
Feedback if chosen: This option is wrong — you assigned charges to neutral sheets. Graphite is pure carbon, with no ions anywhere in its structure.
Outcome: State that covalent network solids form only from nonmetals and metalloids, either as elements or as binary compounds.
You've already seen how to classify an element as a metal or nonmetal from its position in the periodic table.
Elements sitting along the stair-step, with properties between metals and nonmetals, are called 'metalloids' — silicon is one.
In this course's classification — the one the AP exam uses — covalent network solids form only from nonmetals and metalloids.
Some are elements: diamond and graphite are networks of C atoms, and solid silicon is a network of Si atoms.
Others are binary compounds — compounds of exactly two elements.
Quartz is SiO₂ and the cutting abrasive carborundum is SiC: binary covalent network solids, each pairing a metalloid with a nonmetal.
The figure is a simplified flat section of the SiO₂ network — read the bonding pattern from it, not the formula.
So in this classification, no covalent network solid contains a metal.
State that covalent network solids form only from nonmetals and metalloids, either as elements or as binary compounds — SiO₂ and SiC are binary examples.
Check — multiple choice
Which of these substances could be a covalent network solid?
Feedback if chosen: This option is wrong — you included a metal. Covalent network solids form only from nonmetals and metalloids, and Cu is a metal.
Feedback if chosen: This option is wrong — you included a metal. Ca is a metal, so CaCl₂ forms an ionic solid, not a covalent network.
Feedback if chosen: Correct. Ge is a metalloid element, and covalent network solids form from nonmetals and metalloids — solid germanium is in fact a network like silicon.
Feedback if chosen: This option is wrong — you included a metal. Na is a metal, so NaF forms an ionic solid, not a covalent network.
Check — multiple choice
Which pair of elements could combine to form a binary covalent network solid?
Feedback if chosen: Correct. Si is a metalloid and C is a nonmetal — their binary compound SiC is a covalent network solid.
Feedback if chosen: This option is wrong — you paired a metal with a nonmetal. Na and Cl form an ionic solid, not a covalent network.
Feedback if chosen: This option is wrong — you paired two metals. Cu and Zn mix as an alloy, held by metallic bonding, not a covalent network.
Feedback if chosen: This option is wrong — you paired a metal with a nonmetal. K and O form an ionic solid, not a covalent network.
Outcome: State that covalent network solids have very high melting points.
The table compares two measured melting points.
| Substance | Melting point (°C) |
|---|---|
| NaCl (ionic) | 801 |
| SiO₂ (covalent network) | above 1600 |
NaCl's 801 °C already counts as high — and the covalent network solid sits far above it.
SiO₂ does not melt until above 1600 °C.
The pattern holds across the class: covalent network solids have very high melting points.
Diamond and SiC likewise stay solid past 2000 °C.
State that covalent network solids have very high melting points — SiO₂ melts above 1600 °C.
Check — multiple choice
GeO₂ is a covalent network solid. Which melting point is consistent with that?
Feedback if chosen: This option is wrong — a melting point below room temperature fits a small molecular substance, not a covalent network solid.
Feedback if chosen: This option is wrong — a melting point of 45 °C fits a soft molecular solid like white phosphorus, not a covalent network solid.
Feedback if chosen: Correct. Covalent network solids have very high melting points — GeO₂ melts at 1115 °C.
Feedback if chosen: This option is wrong — 210 °C is still molecular territory. Covalent network solids melt far higher, typically above 1000 °C.
Check — multiple choice
Silicon nitride is a covalent network solid used in engine parts. What does its class predict about its melting point?
Feedback if chosen: Correct. Covalent network solids have very high melting points — silicon nitride stays solid well past the reach of an engine's heat.
Feedback if chosen: This option is wrong — you inverted the pattern. It is small molecular substances that can melt below room temperature; network solids melt very high.
Feedback if chosen: This option is wrong — you ranked by atomic mass. Class, not mass, sets the scale here: network solids melt far above molecular solids of any comparable mass.
Feedback if chosen: This option is wrong — you reached for density, which does not set melting behavior. The class alone predicts a very high melting point.
Outcome: Explain why covalent network solids have very high melting points.
You've already seen that intermolecular forces — attractions between separate particles — are much weaker than the covalent, ionic, or metallic bonding within or throughout a substance.
The figure shows solid iodine: discrete I₂ molecules packed in a lattice, with only intermolecular attractions acting across the gaps between molecules.
Melting solid iodine only overcomes some of those weak between-molecule attractions — every covalent I–I bond stays intact — so iodine melts at just 114 °C.
Now look at the SiO₂ network: no separate particles anywhere, only atoms covalently bonded straight through the structure.
For these atoms to flow past one another as a liquid, covalent Si–O bonds themselves must break, throughout the network.
Breaking covalent bonds takes far more energy than overcoming weak attractions between separate particles, so SiO₂ does not melt until above 1600 °C.
That is the general reason: covalent network solids have very high melting points because melting them requires breaking covalent bonds throughout the network rather than overcoming weaker attractions between separate particles.
Explain that covalent network solids have very high melting points because melting requires breaking covalent bonds throughout the network — melting SiO₂ requires breaking Si–O bonds.
Check — multiple choice
Melting SiC requires temperatures above 2000 °C. What must happen for solid SiC to melt?
Feedback if chosen: Correct. SiC is a covalent network — there are no separate particles to pull apart, so melting means breaking covalent bonds throughout the structure, which takes enormous energy.
Feedback if chosen: This option is wrong — you gave SiC discrete molecules. Its covalent bonding is continuous; there are no separate SiC molecules with attractions between them.
Feedback if chosen: This option is wrong — you treated SiC as ionic. Its Si and C atoms are neutral and share electrons covalently; no ions exist in the solid.
Feedback if chosen: This option is wrong — you treated SiC as metallic. Its valence electrons sit in fixed covalent bonds, not in a delocalized sea.
Check — multiple choice
Quartz (SiO₂) melts above 1600 °C, while white phosphorus (P₄) melts at 44 °C. Both are built from nonmetal atoms. Why the huge difference?
Feedback if chosen: This option is wrong — you ranked by mass, and P₄ (about 124 amu) is in fact heavier than an SiO₂ formula unit (about 60 amu). Structure, not mass, decides here.
Feedback if chosen: This option is wrong — you turned intermolecular attractions into repulsions. P₄ molecules attract one another; the attractions are simply weak, so little energy overcomes them.
Feedback if chosen: This option is wrong — you gave SiO₂ ions. Its atoms are neutral and covalently bonded; the resistance comes from the bonds, not from charges.
Feedback if chosen: Correct. P₄ is a molecular solid — melting separates whole molecules by overcoming weak attractions. SiO₂ is a network — melting cannot happen until covalent Si–O bonds break throughout.
Outcome: State that three-dimensional covalent network solids are rigid and hard.
Drag a diamond across a pane of glass and the glass scratches — the diamond is unmarked.
Diamond is the hardest natural substance: it scratches everything, and almost nothing scratches it.
Hard and rigid go together — a hard solid resists being scratched, dented, or forced out of shape.
This is the pattern for the class: three-dimensional covalent network solids are rigid and hard.
Quartz shares it — SiO₂ is hard enough to scratch steel.
'Three-dimensional' is part of the claim: this lesson is about networks whose covalent bonding extends in all three dimensions, like diamond, quartz, and SiC.
State that three-dimensional covalent network solids are rigid and hard — diamond scratches glass.
Check — multiple choice
SiC is a three-dimensional covalent network solid. Which use fits its expected properties?
Feedback if chosen: This option is wrong — you assigned slipperiness to a rigid solid. A three-dimensional network resists sliding and deformation; it does not lubricate.
Feedback if chosen: Correct. Three-dimensional covalent network solids are rigid and hard — hard enough that SiC grinds and cuts metal.
Feedback if chosen: This option is wrong — you assigned softness. A three-dimensional network solid resists being forced out of shape, so it cannot be cast into flexible sheets.
Feedback if chosen: This option is wrong — you assigned flexibility. Bending is exactly the deformation a rigid three-dimensional network resists.
Check — multiple choice
Which observation is consistent with a three-dimensional covalent network solid?
Feedback if chosen: This option is wrong — flaking under light pressure is softness, the opposite of what a three-dimensional network shows.
Feedback if chosen: This option is wrong — being hammered into sheets is malleability, a metallic behavior; a three-dimensional network is rigid.
Feedback if chosen: Correct. Rigid and hard is the pattern — a three-dimensional covalent network solid scratches glass and resists denting.
Feedback if chosen: This option is wrong — melting just above room temperature fits a molecular solid; three-dimensional network solids are hard, rigid, and very high melting.
Outcome: Explain why three-dimensional covalent network solids are rigid and hard.
Look again at the three-dimensional network: every atom is joined to its neighbors by covalent bonds.
You've already seen that a covalent bond is one or more shared pairs of valence electrons between two specific atoms.
In the network those bonds hold fixed angles — each atom's bonds point in set directions.
An atom cannot lean, slide, or trade places without bending those bonds out of their fixed angles.
So for any atom to shift position, covalent bonds would have to break.
Scratching or denting a solid means forcing its atoms to shift — and in a three-dimensional network that means breaking covalent bonds, which takes enormous force.
That is why three-dimensional covalent network solids are rigid and hard: their covalent bond angles are fixed, so atoms cannot shift position without breaking bonds.
Explain that three-dimensional covalent network solids are rigid and hard because their covalent bond angles are fixed, so atoms cannot shift position without breaking bonds.
Check — multiple choice
A diamond-tipped drill bit keeps its shape while grinding through rock. Why?
Feedback if chosen: This option is wrong — you gave the atoms mobility they lack. Fixed bond angles are precisely what stop the atoms from rearranging.
Feedback if chosen: This option is wrong — you gave diamond discrete molecules. Diamond is one continuous covalent network; there are no separate molecules to attract each other.
Feedback if chosen: This option is wrong — you treated diamond as metallic. Its electrons sit in fixed covalent bonds, not in a delocalized sea.
Feedback if chosen: Correct. The covalent bond angles are fixed, so deforming the tip would mean breaking bonds throughout the network — and grinding rock never supplies that.
Check — multiple choice
Pressing hard on a quartz crystal leaves no dent. What would forming a dent require at the particle level?
Feedback if chosen: This option is wrong — you gave quartz discrete molecules. Its covalent bonding is continuous, so there are no intermolecular gaps to squeeze.
Feedback if chosen: Correct. A dent means atoms shifting position — and in a three-dimensional network with fixed bond angles, that requires breaking covalent bonds, which hand pressure cannot do.
Feedback if chosen: This option is wrong — you treated quartz as ionic. Its Si and O atoms are neutral and covalently bonded; there are no ions to push past one another.
Feedback if chosen: This option is wrong — you treated quartz as metallic. There is no delocalized electron sea; the valence electrons sit in fixed covalent bonds.
Outcome: State that graphite is soft even though it is a covalent network solid.
Drag a pencil tip across paper and it leaves a dark line: graphite flakes off onto the page.
The pressure of a hand is enough — graphite is soft.
That should be surprising: graphite is pure carbon, covalently bonded, a covalent network solid — the same class as diamond.
You've already seen that graphite's covalent bonding is arranged as stacked two-dimensional layers rather than one three-dimensional network.
So the claim to carry: graphite is soft, even though it is a covalent network solid.
This is why the hardness claim was stated for three-dimensional networks — the two-dimensional layered form does not share it.
State that graphite is soft even though it is a covalent network solid — graphite flakes off a pencil tip onto paper.
Check — multiple choice
Which pair of claims about graphite is correct?
Feedback if chosen: This option is wrong — you carried the three-dimensional hardness pattern onto graphite. Graphite's class is right here, but graphite is soft.
Feedback if chosen: This option is wrong — you reclassified graphite to explain its softness. Its covalently bonded layers are continuous, so it is a network solid — just a soft one.
Feedback if chosen: Correct. Graphite is a covalent network solid built as stacked layers, and it is soft — soft enough to flake off a pencil tip.
Feedback if chosen: This option is wrong — graphite is not metallic; it is pure carbon with covalent bonding extending through each layer.
Check — multiple choice
A student claims that every covalent network solid is hard. Which substance shows this claim is wrong?
Feedback if chosen: Correct. Graphite is a covalent network solid, yet it is soft — the one counterexample among these four.
Feedback if chosen: This option is wrong — diamond supports the student's claim rather than refuting it; it is the hardest natural substance.
Feedback if chosen: This option is wrong — quartz supports the student's claim rather than refuting it; it is hard enough to scratch steel.
Feedback if chosen: This option is wrong — silicon carbide supports the student's claim rather than refuting it; it is hard enough to grind metal.
Outcome: Explain why graphite is soft even though it is a covalent network solid.
The figure shows one layer of graphite: every C atom covalently bonded to its neighbors within a flat sheet.
Within a layer, graphite is as strongly bonded as any network solid — its covalent bonding lies entirely within the two-dimensional layers.
Between one layer and the next there are no covalent bonds at all.
You've already seen that attractions between separate particles are much weaker than covalent bonding — and that is all that holds neighboring layers together.
So adjacent layers can slide past each other relatively easily: the weak attractions between layers give way while every covalent bond within each layer stays intact.
That is why graphite is soft — pressing on it shears whole layers sideways.
Writing with a pencil does exactly this: the paper shears stacks of graphite layers off the tip, and they stay behind as the mark.
Explain that graphite is soft because its covalent bonding lies within two-dimensional layers, and adjacent layers can slide past each other relatively easily.
Check — multiple choice
When a pencil writes, graphite transfers to the paper. What happens at the particle level?
Feedback if chosen: Correct. The covalent bonding lies within the layers; between layers only weak attractions act, so the paper's rubbing shears whole layers off the tip.
Feedback if chosen: This option is wrong — you broke the strong bonds instead of the weak attractions. Hand pressure cannot break covalent bonds; it only overcomes the weak between-layer attractions.
Feedback if chosen: This option is wrong — you routed the transfer through the vapor. Graphite is simply sheared off as solid flakes; no vapor forms at pencil temperatures.
Feedback if chosen: This option is wrong — you gave graphite ions. It is pure carbon: neutral atoms, covalently bonded within layers.
Check — multiple choice
Diamond and graphite are both pure carbon, yet diamond is hard and graphite is soft. What makes the difference?
Feedback if chosen: This option is wrong — you located the difference in bond strength. The C–C bonds in both are strong; what differs is where the bonding extends.
Feedback if chosen: Correct. In diamond, shifting atoms means breaking bonds in a three-dimensional network; in graphite, whole layers slide on weak between-layer attractions while the in-layer bonds stay intact.
Feedback if chosen: This option is wrong — within each graphite layer the atoms are covalently bonded at close range; only the between-layer attractions are weak.
Feedback if chosen: This option is wrong — both are pure carbon. The difference is structural: three-dimensional network against stacked two-dimensional layers.
Check — multiple choice
Boron nitride can form a solid of stacked, covalently bonded B–N layers, and it is used as a lubricant powder. Why does it lubricate?
Feedback if chosen: This option is wrong — you melted a network solid with friction. Its covalently bonded layers survive far higher temperatures; the sliding happens in the solid state.
Feedback if chosen: This option is wrong — you gave it discrete rolling molecules. Its covalent bonding extends continuously through each layer; there are no separate molecules.
Feedback if chosen: This option is wrong — you made lubrication chemical. The effect is structural: layers shear, no reaction needed.
Feedback if chosen: Correct. Like graphite, the covalent bonding lies within the two-dimensional layers, so adjacent layers slide easily — exactly what a lubricant needs.
Outcome: State that molecular solids are sometimes composed of very large molecules or polymers.
You've already seen molecular solids: discrete molecules — like the I₂ shown — packed in a lattice and held near one another by intermolecular forces.
Nothing limits how big those discrete molecules can be.
Molecular solids are sometimes composed of very large molecules, or of polymers — you've already seen polymers, very long chain molecules.
Paraffin wax is a molecular solid whose molecules are chains of 20 to 40 carbon atoms with their hydrogens.
Polyethylene goes further — chains of thousands of carbon atoms — yet each chain is still one discrete molecule.
However long the chain, the solid stays molecular: covalent bonds run along each chain, and only intermolecular forces act between chains.
That is the line between a molecular solid and a covalent network solid — in wax the covalent bonding ends where each molecule ends; in quartz it never ends.
State that molecular solids are sometimes composed of very large molecules or polymers — paraffin wax is a molecular solid made of long carbon-chain molecules.
Check — multiple choice
Solid polyethylene consists of separate chains, each of thousands of covalently bonded C and H atoms, attracted to neighboring chains by intermolecular forces. What class of solid is it?
Feedback if chosen: This option is wrong — you read thousands of covalent bonds as a network. Each chain's bonding ends where the chain ends, so the chains are discrete molecules, not one continuous network.
Feedback if chosen: This option is wrong — you assigned ions to neutral chains. Polyethylene contains only covalently bonded C and H atoms.
Feedback if chosen: This option is wrong — you assigned a delocalized electron sea. Polyethylene's electrons sit in fixed covalent bonds along each chain.
Feedback if chosen: Correct. Each chain is one discrete molecule, and only intermolecular forces act between chains — a molecular solid, however large its molecules.
Check — multiple choice
What makes paraffin wax a molecular solid, even though each of its molecules contains dozens of atoms?
Feedback if chosen: This option is wrong — that would describe a covalent network solid. Wax's chains are not bonded to each other; each chain is a separate molecule.
Feedback if chosen: This option is wrong — you gave the molecules full charges. Wax molecules are neutral; full charges belong to ionic substances.
Feedback if chosen: Correct. The size of the molecule never matters — what makes a solid molecular is that its covalent bonding ends at each discrete molecule, with only intermolecular forces between them.
Feedback if chosen: This option is wrong — you reached for hydrogen bonding, but wax's C–H chains have no N, O, or F donors; its chains attract only through London dispersion forces.
Outcome: State that metallic solids are good conductors of heat as well as electricity.
Stir hot soup with a metal spoon and the handle soon warms; a wooden spoon stays cool.
Cookware exploits the same behavior — a copper bottom spreads a burner's heat quickly across the whole pan.
You've already seen the electrical half of this pattern: conducting electricity as a solid is evidence of metallic bonding.
Heat completes the pattern: metallic solids are good conductors of heat as well as electricity.
Copper, aluminum, iron — hold one end of a rod in a flame and the other end soon becomes too hot to hold.
Common nonmetal solids behave differently: wood, glass, and wax stay cool far longer under the same flame.
State that metallic solids are good conductors of heat as well as electricity — the copper bottom of a pan spreads heat quickly.
Check — multiple choice
Four rods of equal size are each held over the same flame at one end. Which becomes hot at the far end first?
Feedback if chosen: This option is wrong — glass is a poor conductor of heat; its far end stays cool long after a metal's would burn you.
Feedback if chosen: Correct. Copper is a metallic solid, and metallic solids are good conductors of heat — the far end warms quickly.
Feedback if chosen: This option is wrong — wood conducts heat poorly, which is exactly why spoon handles and tool grips are made of it.
Feedback if chosen: This option is wrong — plastics conduct heat poorly; the far end of the plastic rod stays comfortable long after the metal's does not.
Check — multiple choice
A solid conducts heat well and also conducts electricity in the solid state. Which class of solid fits both observations?
Feedback if chosen: This option is wrong — ionic solids do not conduct electricity as solids; their ions are locked in the lattice.
Feedback if chosen: This option is wrong — covalent network solids generally do not conduct electricity as solids; their electrons sit in fixed bonds.
Feedback if chosen: This option is wrong — molecular solids conduct neither heat nor electricity well; nothing in them is free to carry charge.
Feedback if chosen: Correct. Good conduction of both heat and electricity in the solid state is the metallic signature.
Outcome: Explain why metals conduct heat well.
You've already seen why metals conduct electricity: their delocalized valence electrons are free to move throughout the lattice.
Those same free electrons explain heat conduction.
Hold one end of a copper rod in a flame and the particles at that end gain energy — they move and jostle faster.
The free electrons at the hot end do not stay put: they move throughout the solid, and they carry their extra energy with them.
Wherever they travel and collide, they hand energy on — from the rod's hotter regions to its cooler regions.
So the far end warms quickly: metals conduct heat well because their free (delocalized) valence electrons move throughout the solid, carrying energy from its hotter regions to its cooler regions.
In wood or glass there are no such free carriers — energy passes slowly from one fixed particle to the next.
Explain that metals conduct heat well because their free (delocalized) valence electrons move throughout the solid, carrying energy from its hotter regions to its cooler regions.
Check — multiple choice
The bottom of an aluminum pan sits on a burner, and the rim soon becomes hot. What carries the energy to the rim?
Feedback if chosen: Correct. The metal's free valence electrons move throughout the solid, gaining energy where it is hot and handing it on where it is cooler.
Feedback if chosen: This option is wrong — you moved the ions instead of the electrons. The Al ions stay in the lattice; it is the delocalized electrons that travel.
Feedback if chosen: This option is wrong — you gave a metal covalent bonding. Aluminum's valence electrons are delocalized, not fixed in bonds — that freedom is what makes the conduction fast.
Feedback if chosen: This option is wrong — you routed the energy outside the pan. Air conducts heat poorly; the rim heats quickly because energy travels through the metal itself.
Check — multiple choice
Copper conducts heat far better than glass. What does copper have that glass lacks?
Feedback if chosen: This option is wrong — you reasoned from atomic mass. Mass does not create a transport mechanism; mobile carriers do.
Feedback if chosen: Correct. Copper's delocalized valence electrons move throughout the solid and carry energy from hotter regions to cooler regions; glass has no free carriers.
Feedback if chosen: This option is wrong — you gave copper covalent bonds. Copper's bonding is metallic, and it is the electron mobility, not bond strength, that moves the energy.
Feedback if chosen: This option is wrong — you freed the ions. Copper's ions stay in place in the lattice; the mobile particles are the delocalized electrons.
Outcome: State that an interstitial alloy tends to be less malleable and less ductile than the pure host metal.
Pure iron is a fairly soft metal — it bends, dents, and hammers into shape with ease.
Steel, made by mixing a small amount of carbon into iron, deforms much less readily than the pure metal.
You've already seen this kind of mixture: carbon in steel is an interstitial alloy — smaller atoms occupying the spaces between the metal atoms.
You've already seen that a malleable solid can be hammered or rolled into sheets without shattering.
You've already seen that a ductile solid can be drawn into a wire without shattering.
The pattern: an interstitial alloy tends to be less malleable and less ductile than the pure host metal.
Steel resists the hammer and the wire-drawer more than pure iron does — which is exactly why builders reach for steel when stiffness is the point.
State that an interstitial alloy tends to be less malleable and less ductile than the pure host metal — steel deforms less readily than pure iron.
Check — multiple choice
To harden iron machine parts, nitrogen atoms are diffused into the metal, where they occupy spaces between the Fe atoms. Compared with pure iron, what should the treated metal be?
Feedback if chosen: This option is wrong — you reversed the pattern. Small atoms in the spaces between metal atoms make an interstitial alloy, and interstitial alloys deform less readily, not more.
Feedback if chosen: This option is wrong — you assumed the guest atoms change nothing. N atoms between the Fe atoms make an interstitial alloy, which tends to be less malleable than the pure metal.
Feedback if chosen: This option is wrong — you traded conductivity for stiffness, but malleability and conduction are separate properties; the interstitial pattern concerns how readily the metal deforms.
Feedback if chosen: Correct. N atoms in the spaces between Fe atoms form an interstitial alloy, and an interstitial alloy tends to be less malleable than the pure host metal.
Check — multiple choice
Which comparison between steel and pure iron is correct?
Feedback if chosen: Correct. Steel is an interstitial alloy of iron, and an interstitial alloy tends to be less malleable and less ductile than its pure host metal.
Feedback if chosen: This option is wrong — you reversed the comparison. It is the interstitial alloy, steel, that resists deformation more.
Feedback if chosen: This option is wrong — the interstitial carbon changes the mechanical behavior; the two do not deform equally.
Feedback if chosen: This option is wrong — pure iron is a malleable metal; it flattens rather than shatters. The difference is that steel deforms less readily.
Outcome: Explain why interstitial atoms make an alloy's lattice more rigid, decreasing malleability and ductility.
You've already seen why metals are malleable and ductile: the positive metal ions can shift position within the delocalized electron sea without breaking the metallic bonding.
Every hammer blow and every drawn wire is rows of metal ions sliding to new positions like this.
Now put the interstitial atoms in the picture: in steel, small C atoms sit in the spaces between the Fe atoms.
A row that tries to slide must now push past the C atoms lodged between the rows.
The interstitial atoms restrict how the metal cores can rearrange, so the lattice is more rigid.
With rearrangement restricted, the metal resists hammering and wire-drawing more — malleability and ductility both decrease.
That is why steel deforms less readily than pure iron: the C atoms keep the Fe layers from sliding as freely.
Explain that interstitial atoms make an alloy's lattice more rigid because they sit between the metal atoms and restrict how the metal cores can rearrange, decreasing malleability and ductility.
Check — multiple choice
Why does steel deform less readily than pure iron?
Feedback if chosen: This option is wrong — you drained the electron sea. The delocalized electrons remain; what changes is how freely the metal cores can rearrange around the lodged C atoms.
Feedback if chosen: This option is wrong — you rebuilt steel as a covalent network. Steel keeps its metallic bonding; the C atoms simply obstruct the sliding of the Fe layers.
Feedback if chosen: Correct. Deforming a metal means rows of metal ions sliding — and the interstitial C atoms lodged between the rows restrict that rearrangement, stiffening the lattice.
Feedback if chosen: This option is wrong — you reasoned from mass. The tiny amount of carbon barely changes the mass; it is the obstruction of sliding, not weight, that stiffens steel.
Check — multiple choice
Hydrogen atoms absorbed into palladium metal occupy the spaces between the Pd atoms. Why would this make the metal resist being drawn into wire?
Feedback if chosen: Correct. Drawing a wire means the metal cores rearranging — and interstitial atoms between them restrict that rearrangement, making the alloy less ductile than pure Pd.
Feedback if chosen: This option is wrong — you expelled the electron sea. The metallic bonding survives; the change is mechanical, in how freely the cores can slide.
Feedback if chosen: This option is wrong — you made the alloy ionic. No full charges form; the lattice stiffens because the guest atoms obstruct rearrangement.
Feedback if chosen: This option is wrong — you rebuilt the metal as a covalent network. The Pd atoms keep their metallic bonding; they are simply blocked from sliding freely.
Outcome: State that alloys typically remain good electrical conductors.
Brass doorknobs, bronze statues, sterling silver spoons — the metals of daily life are usually alloys.
You've already seen that an alloy is a metallic material containing a mixture of a metal with one or more other elements.
You've already seen that conducting electricity in the solid state is evidence of metallic bonding.
Mixing in a second element does not spoil that conduction: alloys typically remain good electrical conductors.
Brass — Cu mixed with Zn — conducts electricity well, and so do bronze (Cu with Sn) and steel.
'Typically' is the honest word: conduction in an alloy is often somewhat weaker than in the pure metal, but it remains good.
State that alloys typically remain good electrical conductors — brass conducts electricity well.
Check — multiple choice
A plumbing fitting is made of brass rather than pure copper. What should be expected of its electrical behavior?
Feedback if chosen: This option is wrong — you applied the ionic pattern. Conducting only when molten is how ionic substances behave; brass is a metallic material and conducts as a solid.
Feedback if chosen: Correct. Alloys typically remain good electrical conductors — brass carries current well, which is why it is used in electrical fittings.
Feedback if chosen: This option is wrong — you turned the Zn atoms into insulation. Mixing Zn into Cu leaves the material metallic, and it keeps conducting.
Feedback if chosen: This option is wrong — you split the two conductions apart. A metallic material that conducts heat well conducts electricity too.
Check — multiple choice
Sterling silver is Ag mixed with a small amount of Cu. Which claim about it fits the pattern for alloys?
Feedback if chosen: This option is wrong — mixing metals does not produce an insulator; alloys typically remain good conductors.
Feedback if chosen: This option is wrong — you confined the current to one element's atoms. The mobile electrons flow throughout the metallic material, not along chains of a single element.
Feedback if chosen: This option is wrong — you added a heating requirement. Metallic materials conduct in the solid state at room temperature.
Feedback if chosen: Correct. Sterling silver is an alloy, and alloys typically remain good electrical conductors.
Outcome: Explain why alloys typically remain conducting.
You've already seen why metals conduct electricity: their delocalized valence electrons are free to move throughout the lattice.
Look at what mixing changed in the figure — and what it did not.
In brass, Zn atoms stand in some of the Cu positions, but both metals still pool their valence electrons into the shared sea.
The sea of mobile valence electrons survives the mixing, so charge still flows freely through the solid.
That is why alloys typically remain conducting: they retain a sea of mobile valence electrons.
Brass conducts even though it mixes Cu and Zn — the mobile electrons never belonged to any one atom in the first place.
Explain that alloys typically remain conducting because they retain a sea of mobile valence electrons — brass conducts even though it mixes Cu and Zn.
Check — multiple choice
Bronze mixes Cu with Sn, yet it conducts electricity well. Why?
Feedback if chosen: This option is wrong — you confined the current to one element. The delocalized electrons move throughout the whole material, past Cu and Sn atoms alike.
Feedback if chosen: This option is wrong — you made the atoms the charge carriers. The atoms stay in the lattice; it is the mobile valence electrons that carry the current.
Feedback if chosen: This option is wrong — you made the alloy ionic. Bronze has no full charges; it is a metallic material with delocalized electrons.
Feedback if chosen: Correct. Cu and Sn both pool their valence electrons into the shared sea, so the alloy retains mobile electrons — and keeps conducting.
Check — multiple choice
When Zn atoms are mixed into copper to make brass, what happens to the delocalized electrons?
Feedback if chosen: This option is wrong — you converted the metallic bonding into covalent bonds. The mixed lattice keeps its delocalized sea; no fixed bonds form.
Feedback if chosen: This option is wrong — you made Zn an electron acceptor forming ions. Zn is a metal: it contributes valence electrons to the sea rather than trapping them.
Feedback if chosen: Correct. The sea of mobile valence electrons survives the mixing — Zn atoms pool their valence electrons just as the Cu atoms do, so brass remains a conductor.
Feedback if chosen: This option is wrong — you localized the electrons onto single atoms. Delocalized electrons belong to the whole lattice, in the alloy just as in the pure metal.
Outcome: Classify a solid as ionic, covalent network, molecular, or metallic from measured property evidence, citing the particle-level interaction that explains the deciding evidence.
You've already seen how to weigh melting-point, conductivity, and deformation evidence together rather than relying on one cue.
This unit adds two solid classes to that toolkit — covalent network and molecular — and sharper property claims for all four.
The table gathers the patterns in one place.
| Class | Melting point | Conducts as solid? | Conducts as liquid? | Deformation |
|---|---|---|---|---|
| Ionic | high | no | yes | hard but brittle |
| Covalent network | very high | generally no | no | very hard (3-D forms) |
| Molecular | low | no | no | soft |
| Metallic | varies widely | yes | yes | malleable and ductile |
Conduction in both the solid and the liquid state points metallic.
Conduction only in the liquid state points ionic.
A very high melting point with no conduction in either state points covalent network.
A low melting point with no conduction points molecular.
A classification is complete when it names the particle-level interaction the deciding evidence reveals.
Metallic conduction in both states reveals a sea of mobile delocalized electrons.
Ionic conduction only as a liquid reveals ions — locked in the lattice as a solid, freed to move by melting.
A network's extreme melting point and hardness reveal covalent bonds extending throughout the structure.
A molecular solid's low melting point reveals discrete molecules held only by weak intermolecular forces.
Try substance Q.
| Property | Observation |
|---|---|
| Melting point | above 1600 °C |
| Conducts as solid? | no |
| Conducts as liquid? | no |
| Hardness | scratches glass |
The deciding evidence is the very high melting point with no conduction in either state — the ionic and metallic classes both conduct as liquids, and molecular solids melt low.
Substance Q is a covalent network solid, and the deciding interaction is covalent bonds extending throughout the structure — melting it means breaking them.
Classify a solid as ionic, covalent network, molecular, or metallic from measured property evidence, citing the particle-level interaction that explains the deciding evidence.
Check — multiple choice
| Property | Observation |
|---|---|
| Melting point | 770 °C |
| Conducts as solid? | no |
| Conducts as liquid? | yes |
| Deformation | shatters when hammered |
Use the measured properties of solid W shown in the table. What class of solid is W?
Feedback if chosen: This option is wrong — you matched the liquid conduction but missed the solid's silence. A metallic solid conducts in both states; W conducts only as a liquid.
Feedback if chosen: Correct. Conduction only in the liquid state is the ionic signature: the ions are locked in the lattice as a solid and freed to move by melting — and the high melting point and brittleness agree.
Feedback if chosen: This option is wrong — you weighed the high melting point alone. A covalent network solid would not conduct even as a liquid, but W does.
Feedback if chosen: This option is wrong — a molecular solid would melt low and never conduct; W melts at 770 °C and conducts as a liquid.
Check — multiple choice
| Property | Observation |
|---|---|
| Melting point | 44 °C |
| Conducts as solid? | no |
| Conducts as liquid? | no |
| Hardness | soft, easily cut |
Use the measured properties of solid X shown in the table. What class of solid is X?
Feedback if chosen: Correct. A low melting point with no conduction in either state is the molecular signature — the deciding interaction is weak intermolecular forces between discrete molecules.
Feedback if chosen: This option is wrong — an ionic solid would conduct once molten and would melt hundreds of degrees higher; X does neither.
Feedback if chosen: This option is wrong — a metallic solid conducts in the solid state; X never conducts.
Feedback if chosen: This option is wrong — a covalent network solid melts very high and resists cutting; X melts at 44 °C and cuts easily.
Check — multiple choice
Solid Y melts at 1085 °C, conducts electricity in both the solid and the liquid state, and can be hammered into sheets. Which particle-level interaction explains the deciding evidence?
Feedback if chosen: This option is wrong — freed ions explain conduction that begins at melting, but Y conducts already as a solid.
Feedback if chosen: This option is wrong — continuous covalent bonds explain extreme hardness and no conduction, but Y conducts in both states and hammers into sheets.
Feedback if chosen: This option is wrong — weak forces between molecules explain low melting and no conduction; Y conducts in both states.
Feedback if chosen: Correct. Conduction in both states is the deciding evidence, and mobile delocalized electrons explain it — they carry charge in the solid and the liquid alike, and they let the metal deform without shattering.
Check — multiple choice
| Property | Observation |
|---|---|
| Melting point | above 2000 °C |
| Conducts as solid? | no |
| Conducts as liquid? | no |
| Hardness | scratches steel |
Use the measured properties of solid Z shown in the table. What class of solid is Z?
Feedback if chosen: This option is wrong — an ionic solid would conduct once molten; Z stays silent even as a liquid.
Feedback if chosen: This option is wrong — a metallic solid conducts in the solid state and dents rather than scratches; Z conducts in neither state.
Feedback if chosen: Correct. An extreme melting point, no conduction in either state, and hardness that scratches steel — the covalent network signature, decided by covalent bonds extending throughout the structure.
Feedback if chosen: This option is wrong — a molecular solid melts low and is soft; Z melts above 2000 °C and scratches steel.
Outcome: Rank substances by melting or boiling point by first comparing solid class and then comparing intermolecular force strength within the molecular substances.
You've already seen how to rank molecular substances by intermolecular force strength, using hydrogen bonding, polarity, and size or polarizability evidence.
You've already seen that stronger intermolecular forces mean higher boiling points.
Across solid classes, the class itself generally decides first: for the substances you will meet in this course, covalent network substances sit highest, ionic substances next, molecular substances lowest.
Metallic melting and boiling points vary too widely for a simple rule, so rankings here compare the other three classes.
The class-first order is a strong tendency, not a law — real melting and boiling ranges overlap between classes, so where a ranking matters, the reasoning goes through the particle-level interactions of the actual substances.
Within the molecular substances, the intermolecular-force ranking takes over.
Rank SiO₂, NaCl, H₂O, and CH₄ by boiling point.
Classify first: SiO₂ is a covalent network substance, NaCl is ionic, and H₂O and CH₄ are molecular.
Class settles the top: SiO₂ boils highest, NaCl next.
Forces settle the rest: H₂O hydrogen-bonds while CH₄ has only London dispersion forces, so H₂O boils above CH₄.
Predicted order: SiO₂ above NaCl above H₂O above CH₄.
The measured values agree.
| Substance | Class | Boiling point (°C) |
|---|---|---|
| SiO₂ | covalent network | above 2200 |
| NaCl | ionic | about 1465 |
| H₂O | molecular (hydrogen bonding) | 100 |
| CH₄ | molecular (dispersion only) | −162 |
Rank substances by melting or boiling point by first comparing solid class and then comparing intermolecular force strength within the molecular substances.
Check — multiple choice
Rank SiC, KBr, CH₃OH, and C₂H₆ by melting point, highest first.
Feedback if chosen: Correct. Class first: covalent network SiC above ionic KBr above the molecular pair; then forces: hydrogen-bonding CH₃OH above dispersion-only C₂H₆.
Feedback if chosen: This option is wrong — you placed the ionic substance above the covalent network. Melting SiC means breaking covalent bonds throughout, which outranks overcoming ionic attractions.
Feedback if chosen: This option is wrong — you ranked the molecular pair by size or mass alone. CH₃OH hydrogen-bonds and C₂H₆ cannot, so CH₃OH melts higher despite the similar mass.
Feedback if chosen: This option is wrong — you put the molecular substances on top. Class decides first here, and the molecular pair sits below ionic KBr and covalent network SiC.
Check — multiple choice
Which of these substances boils at the highest temperature?
Feedback if chosen: This option is wrong — NH₃ hydrogen-bonds, but it is still molecular; escape from ionic KF costs far more than escape from any of these molecular liquids.
Feedback if chosen: This option is wrong — HF is a strong hydrogen-bonder, but hydrogen bonding between neutral molecules cannot outrank the full-charge attractions in ionic KF.
Feedback if chosen: Correct. Class decides first: KF is ionic, and its full-charge Coulombic attractions put its boiling point far above every molecular substance listed.
Feedback if chosen: This option is wrong — CH₄ has only London dispersion forces; it boils lowest of the four, not highest.
Check — multiple choice
Which of these substances boils at the lowest temperature?
Feedback if chosen: This option is wrong — a covalent network element sits at the top of the ranking, not the bottom; boiling it means breaking covalent bonds.
Feedback if chosen: This option is wrong — an ionic substance boils high; overcoming full-charge attractions costs far more than overcoming intermolecular forces.
Feedback if chosen: This option is wrong — you stopped at the molecular class. Within the molecular pair, polar CH₃F has dipole-dipole attractions on top of its dispersion forces, so it boils above CH₄.
Feedback if chosen: Correct. Class puts the two molecular substances lowest, and forces finish the job: nonpolar CH₄ has only London dispersion forces, so it boils below polar CH₃F.
Check — multiple choice
KBr boils far above CH₃OH, even though CH₃OH hydrogen-bonds. Why?
Feedback if chosen: This option is wrong — you applied molecular reasoning to an ionic substance. KBr's particles are ions with full charges; dispersion forces are not what holds its liquid together.
Feedback if chosen: Correct. The ranking starts with class: KBr's full ionic charges must be completely overcome for escape, and that dwarfs even the strongest intermolecular force between neutral molecules.
Feedback if chosen: This option is wrong — hydrogen bonding acts in the liquid too; CH₃OH boils at 65 °C precisely because those liquid-state attractions must be overcome.
Feedback if chosen: This option is wrong — KBr has no molecules; it is an ionic lattice of K⁺ and Br⁻ ions, and its full charges are what set the high boiling point.
Outcome: Write justifications that earn the point: claim, named principle, this system's particles.
On the free-response section of the exam, a justification point is all-or-nothing: a correct answer with a vague or missing justification earns zero.
Every point-earning justification makes three moves: state the claim, name the governing principle or force, and apply it to this system's actual particles.
Naming the principle alone fails — graders reject responses that say 'stronger forces' without naming which force and which particles it acts between.
Repeating what the prompt already told you also fails — the point pays for the new causal link you supply, not for restated givens.
Name the species every time you refer to one — a response built on 'it' or 'the substance' leaves the grader unsure what you mean.
When a part says justify or explain, write three short lines — the claim, the named principle or force, and this system's particles — and check all three are present before moving on.
Free-response question
Source note (not shown as part of the question): 2021 FRQ Q2(b); rubric from the 2021 scoring guidelines
Two compounds that contain Si are SiO₂ and SiH₄.
(b) At 161 K, SiH₄ boils but SiO₂ remains as a solid. Using principles of interparticle forces, explain the difference in boiling points. [1 point]
✓ Model response (the key)
(b)
SiH₄ is composed of molecules, and the only intermolecular forces between the molecules are London dispersion forces.
SiO₂ is a network covalent solid, with covalent bonds between silicon and oxygen atoms continuing throughout the structure.
London dispersion forces are much weaker than covalent bonds, so SiH₄ boils at a much lower temperature than SiO₂.
Scoring rubric — 1 point available
Check — multiple choice
Four students wrote the following responses to part (b) of the question you just answered. Which response earns the point?
Feedback if chosen: This option is wrong — you accepted a response that never names the force or the particles: 'stronger intermolecular forces' is the vague wording graders reject, and it also mislabels SiO₂, whose atoms are held by covalent bonds rather than by intermolecular forces.
Feedback if chosen: Correct. This response makes all three moves — the claim (SiH₄ boils far lower), the named forces (London dispersion forces versus covalent bonds), and this system's particles (separate SiH₄ molecules versus the bonded SiO₂ network).
Feedback if chosen: This option is wrong — you accepted a response that only repeats what the prompt already said; restating that SiH₄ boils lower supplies no principle and no particles, so it earns nothing.
Feedback if chosen: This option is wrong — you accepted a response that names the principle without applying it; the grader needs the specific forces in each substance, not the phrase 'interparticle forces' echoed back.
Check — multiple choice
Sulfur dioxide, SO₂, dissolves in liquid acetone, CH₃COCH₃. Both molecules are polar, and acetone's H atoms are all bonded to C. Which forces act between an SO₂ molecule and an acetone molecule in this mixture?
Feedback if chosen: Correct. Both molecules are polar, so their permanent dipoles attract — and LDFs act between every pair of particles, whatever else is present.
Feedback if chosen: This option is wrong — you dropped the LDFs. Temporary, fluctuating dipoles form in every electron cloud, so LDFs act between every pair of particles.
Feedback if chosen: This option is wrong — you added hydrogen bonding, but neither molecule has an H bonded to N, O, or F: SO₂ has no H at all, and acetone's H atoms sit on C.
Feedback if chosen: This option is wrong — you treated one of the molecules as nonpolar. A dipole-induced dipole interaction needs a nonpolar partner, and both SO₂ and acetone have permanent dipoles.
Check — multiple choice
| Liquid | Vapor pressure (atm) |
|---|---|
| diethyl ether, C₂H₅OC₂H₅ | 0.71 |
| 1-propanol, CH₃CH₂CH₂OH | 0.03 |
The table gives the vapor pressure of two liquids at 25 °C. Which liquid has the stronger intermolecular forces between its molecules?
Feedback if chosen: This option is wrong — you gave the stronger forces to the liquid whose molecules escape more easily. A high vapor pressure means many molecules overcome their attractions, the mark of weaker forces.
Feedback if chosen: This option is wrong — you treated temperature as what sets the vapor pressure. At the same temperature, the difference between the two vapor pressures reflects only the strength of each liquid's attractions.
Feedback if chosen: This option is wrong — vapor pressures at one temperature are exactly the evidence needed: fewer escaping molecules means stronger attractions holding them back.
Feedback if chosen: Correct. Stronger attractions mean fewer molecules escape to the vapor, so the liquid with the far lower vapor pressure — 1-propanol — has the stronger intermolecular forces.
Check — multiple choice
The figure shows a student's drawing of the hydrogen bond between two NH₃ molecules. Which feature of the drawing, if any, is wrong?
Feedback if chosen: This option is wrong — you rejected a correct feature. A hydrogen bond starts at an H bonded to N, O, or F, and this dashed line does start at an N–H hydrogen.
Feedback if chosen: Correct. The dashed line ends on an H atom of the right molecule. A hydrogen bond must end on the acceptor's N, O, or F atom — here, the right molecule's N.
Feedback if chosen: This option is wrong — you rejected a correct feature. Covalent bonds within each molecule are drawn as solid lines; only the hydrogen bond between the molecules is dashed.
Feedback if chosen: This option is wrong — you accepted a hydrogen bond drawn from one molecule's H to another molecule's H. The dashed line must end on the neighboring molecule's N, which carries the δ⁻ charge.
Check — multiple choice
A sealed flask holds solid MgO at room temperature. No measurable vapor forms above the solid. Why?
Feedback if chosen: This option is wrong — you blamed the arrangement rather than the attractions. What keeps the ions in place is the strength of the Coulombic attractions between full charges, not a lack of paths to the surface.
Feedback if chosen: This option is wrong — MgO is ionic, not covalent. The attractions holding its particles are Coulombic attractions between Mg²⁺ and O²⁻ ions, not covalent bonds.
Feedback if chosen: Correct. Leaving the solid for the vapor requires completely overcoming the Coulombic attractions between full ionic charges — so essentially nothing leaves the solid, and the vapor stays too small to measure.
Feedback if chosen: This option is wrong — you blamed the flask. A sealed flask holds vapor readily — a liquid sealed the same way shows a measurable vapor pressure; what stops MgO is the strength of the Coulombic attractions between its full ionic charges.
Check — multiple choice
Powdered graphite is puffed into a sticky door lock, and the key then turns smoothly. Why does graphite work as a lubricant?
Feedback if chosen: Correct. Graphite's covalent bonding lies within two-dimensional layers, and only weak attractions act between adjacent layers — so whole layers shear past one another and let the parts slide.
Feedback if chosen: This option is wrong — you had the grains roll as rigid balls. Graphite lubricates by shearing: only weak attractions act between its covalently bonded layers, so whole layers slide past one another.
Feedback if chosen: This option is wrong — you gave the delocalized electrons a repelling job. They move within each layer; the easy sliding comes from the weak attractions between the layers.
Feedback if chosen: This option is wrong — the covalent bonds within each layer hold. What gives way are the weak attractions between the layers, letting whole layers slide.
Check — multiple choice
Boron atoms diffused into nickel metal sit in the spaces between the Ni atoms. The treated nickel is much harder to bend than pure nickel. Why?
Feedback if chosen: This option is wrong — the sea of delocalized electrons remains. The added atoms change how easily the metal cores rearrange, not whether metallic bonding exists.
Feedback if chosen: This option is wrong — you rebuilt the alloy as a covalent network. The nickel keeps its metallic bonding; the lodged B atoms simply obstruct the Ni cores when layers try to slide.
Feedback if chosen: This option is wrong — no ions form. The lattice stays metallic; the added atoms simply sit between the metal cores and get in the way when layers try to slide.
Feedback if chosen: Correct. Atoms lodged between the metal atoms restrict how the cores can rearrange, so layers slide less freely and the metal resists bending.
Check — multiple choice
| Property | Observation |
|---|---|
| Melting point | 962 °C |
| Conducts as solid? | yes |
| Conducts as liquid? | yes |
| Deformation | flattens into sheets when hammered |
Use the measured properties of solid Q shown in the table. What class of solid is Q?
Feedback if chosen: This option is wrong — you stopped at the liquid's conductivity. Solid Q conducts before melting too, and an ionic solid cannot: its ions stay locked in place until it melts.
Feedback if chosen: Correct. Conducting in the solid state is the deciding evidence — only mobile delocalized electrons carry current through an intact solid — and layers of metal cores shifting past one another explain the flattening.
Feedback if chosen: This option is wrong — covalent bonds extending throughout a structure hold electrons in those bonds rather than carrying current, and such solids are very hard, not solids that flatten into sheets.
Feedback if chosen: This option is wrong — you matched only the deformation. A solid held together by weak attractions between molecules melts low and conducts in neither state.
Check — multiple choice
Rank silicon (Si, a covalent network solid), CaBr₂, C₂H₅OH, and H₂S by boiling point, highest first.
Feedback if chosen: This option is wrong — you put the ionic substance above the covalent network. Boiling Si means breaking covalent bonds that run throughout the structure, which costs even more than overcoming CaBr₂'s full-charge attractions.
Feedback if chosen: This option is wrong — you missed C₂H₅OH's hydrogen bonding. Its O–H group hydrogen bonds to neighboring molecules, which outranks the dipole-dipole attractions between H₂S molecules.
Feedback if chosen: Correct. Solid class ranks first — the covalent network (Si) above the ionic substance (CaBr₂), both far above the molecular pair — and within the molecular pair, hydrogen-bonding C₂H₅OH beats dipole-dipole H₂S.
Feedback if chosen: This option is wrong — you listed the substances lowest-boiling first. The comparison is inverted: the question asks for the highest boiling point first.
8 lessons
Outcome: Classify a solid as crystalline or amorphous from the arrangement of its particles.
Quartz and window glass are both hard, transparent solids, and both are built mainly from SiO₂.
Yet quartz grows as crystals with flat faces and sharp edges, while glass never does.
Since the building blocks match, the difference must lie in how the blocks are arranged.
The figure shows a particle-level view of each solid.
In quartz, one arrangement repeats in all three dimensions, throughout the entire solid.
You've already seen this kind of order: a crystal lattice is a systematic, repeating three-dimensional array of particles.
In glass, the particles are packed just as tightly, but no arrangement repeats across the solid.
Neighboring particles can still sit in similar local arrangements — what glass lacks is a pattern that repeats over long distances.
A solid whose particles repeat in a regular three-dimensional arrangement is called 'crystalline'.
A solid whose particles have no long-range repeating pattern is called 'amorphous'.
The classification reads the arrangement, not the look or feel — glass is hard and clear, and still amorphous.
Quartz, diamond, and table salt are crystalline; glass, rubber, and obsidian are amorphous.
Classify a solid as crystalline when its particles repeat in a regular three-dimensional arrangement, or amorphous when its particles have no long-range repeating pattern.
Check — multiple choice
The figure shows particle-level views of two solids. Which classification is correct?
Feedback if chosen: This option is wrong — you swapped the two labels. Solid Y is the one whose particles repeat in a regular arrangement, so solid Y is the crystalline solid.
Feedback if chosen: This option is wrong — you read tight packing as order. Both boxes show touching particles, but only solid Y's repeat in a regular arrangement.
Feedback if chosen: Correct. Solid X's particles touch with no long-range repeating pattern — amorphous — while solid Y's repeat in a regular arrangement — crystalline.
Feedback if chosen: This option is wrong — you set the drawings aside. The arrangement is exactly what a particle-level view shows, and the arrangement is all the classification needs.
Check — multiple choice
In solid NaCl, the Na⁺ and Cl⁻ ions repeat in the same arrangement throughout the entire solid. How is solid NaCl classified?
Feedback if chosen: Correct. The ions repeat in a regular three-dimensional arrangement throughout the solid, which is exactly what crystalline means.
Feedback if chosen: This option is wrong — you classified a repeating arrangement as disordered. Amorphous is reserved for solids whose particles have no long-range repeating pattern.
Feedback if chosen: This option is wrong — you split one solid into two classes. The stem says the same arrangement repeats throughout the entire solid, so the whole sample is crystalline.
Feedback if chosen: This option is wrong — you reached for a property that is not the test. The classification reads the particle arrangement, and the arrangement is given: it repeats.
Check — multiple choice
When lava cools too quickly for any repeating arrangement to form, the particles are trapped in a jumbled arrangement. How is the resulting solid, obsidian, classified?
Feedback if chosen: This option is wrong — you used hardness as the test. Hard solids can be either class; the classification reads the arrangement, and obsidian's is jumbled.
Feedback if chosen: Correct. Obsidian's particles are trapped with no long-range repeating pattern, so obsidian is amorphous.
Feedback if chosen: This option is wrong — you tied the class to how the solid formed. Glass also forms by cooling, and it is the textbook amorphous solid; what matters is the arrangement that results.
Feedback if chosen: This option is wrong — you made order a requirement for being solid. Glass, rubber, and obsidian are all genuinely solid, and all amorphous.
Outcome: Describe the particles of any solid — crystalline or amorphous — as vibrating about fixed positions, with no movement from place to place.
A granite block can hold its shape for centuries.
It is tempting to conclude that its particles sit perfectly still.
They do not — the particles of every solid are in constant motion.
You've already seen the two solid arrangements — crystalline and amorphous — shown again here.
In both arrangements, each particle vibrates — it jiggles back and forth about its own fixed position.
What no particle does is travel: each stays at its own position instead of moving from place to place.
In solid NaCl, for example, every ion vibrates about its own fixed position in the lattice.
The attractions between the particles pull each one toward its neighbors, holding it near its position.
How the particles arrange themselves is influenced by those attractions and by how well the particles pack together.
Describe the particles of any solid — crystalline or amorphous — as vibrating about fixed positions, with no overall movement from place to place.
Check — multiple choice
Which best describes the motion of the ions in solid KBr?
Feedback if chosen: This option is wrong — you let the ions travel. In a solid each ion stays at its own position; the only motion is vibration about that position.
Feedback if chosen: This option is wrong — you stopped the motion entirely. The ions jiggle constantly about their positions, even while the solid holds its shape.
Feedback if chosen: This option is wrong — you let the ions trade places. In a solid each ion keeps its own position; none slides past a neighbor.
Feedback if chosen: Correct. In a solid the particles vibrate about fixed positions without moving from place to place.
Check — multiple choice
Glass is an amorphous solid. Which describes the motion of its particles?
Feedback if chosen: Correct. Vibration about fixed positions is the motion in every solid, crystalline or amorphous alike.
Feedback if chosen: This option is wrong — you made the repeating pattern the anchor. It is the attractions between particles that hold each one near its position, and those act in both arrangements.
Feedback if chosen: This option is wrong — you read rigidity as stillness. The particles vibrate constantly; the solid stays rigid because no particle travels anywhere new.
Feedback if chosen: This option is wrong — you added travel to the vibration. In a solid, no particle trades positions; each vibrates about its own spot.
Check — multiple choice
A student argues: 'If the particles of a solid never stop moving, the solid's shape ought to drift over time.' What actually keeps the shape fixed?
Feedback if chosen: This option is wrong — you invented a cancellation. The vibrations are not coordinated; the shape holds because every particle stays at its own position.
Feedback if chosen: This option is wrong — you stopped the motion. The vibration never stops; it simply never carries a particle away from its position.
Feedback if chosen: Correct. Vibration is motion about a fixed position — the particle returns again and again, so the arrangement, and the shape, never drift.
Feedback if chosen: This option is wrong — you turned vibration into slow travel. The particles of a solid do not travel at all; each one only jiggles about its own position.
Outcome: Describe the particles of a liquid as staying in close contact while continually moving, colliding, and sliding past one another.
Water poured into a glass rearranges itself to match the glass within a moment.
You've already seen that a solid's particles only vibrate about fixed positions.
Vibration alone cannot rearrange anything — flowing needs particles that actually go somewhere new.
The figure shows a particle-level snapshot of a liquid.
The particles are in close contact — each one touches its neighbors.
But none of them is fixed: the particles are continually moving, colliding, and sliding past one another.
In liquid water, the molecules slide past one another while staying in contact — that is what lets the sample rearrange itself.
A liquid's snapshot looks much like an amorphous solid's; the difference is the motion.
In glass each particle keeps vibrating about the same spot; in a liquid the particles keep changing neighbors.
The stronger the attractions between the particles, the less freely the particles slide past one another.
And as the temperature rises, the particles move faster.
Describe the particles of a liquid as staying in close contact while continually moving, colliding, and sliding past one another — with the motion shaped by the strength of the attractions between particles and by the temperature.
Check — multiple choice
Which best describes the motion of the molecules in liquid ethanol?
Feedback if chosen: This option is wrong — you kept the solid's motion. A liquid's molecules are not fixed; they continually change neighbors.
Feedback if chosen: Correct. Liquid particles remain in close contact while continually moving, colliding, and sliding past one another.
Feedback if chosen: This option is wrong — you pulled the molecules apart. A liquid's molecules stay in close contact, touching their neighbors as they move.
Feedback if chosen: This option is wrong — you separated the flow from the particles. A sample can only rearrange itself because its individual molecules move.
Check — multiple choice
One of these particle-level snapshots shows a liquid; the other shows an amorphous solid. Can the snapshots alone settle which is which?
Feedback if chosen: This option is wrong — you invented a difference between the boxes. The two jumbled arrangements are equally disordered; nothing in a still image separates them.
Feedback if chosen: Correct. A liquid and an amorphous solid share the same disordered, touching arrangement; only the motion — sliding past one another versus vibrating in place — tells them apart.
Feedback if chosen: This option is wrong — you invented a difference between the boxes. The two jumbled arrangements are equally disordered; nothing in a still image separates them.
Feedback if chosen: This option is wrong — you pulled the liquid's particles apart. Liquid particles do stay in close contact; both boxes show a perfectly good liquid arrangement.
Check — multiple choice
Liquid A's particles attract one another more strongly than liquid B's particles do, at the same temperature. In which liquid do the particles slide past one another less freely?
Feedback if chosen: Correct. The stronger the attractions between the particles, the less freely the particles slide past one another — and liquid A's attractions are the stronger ones.
Feedback if chosen: This option is wrong — you reversed the relationship. Weaker attractions let particles slide more freely, so liquid B is the freer one.
Feedback if chosen: This option is wrong — you made temperature the only influence. The strength of the attractions between particles also shapes the motion.
Feedback if chosen: This option is wrong — you brought in the sample size. How much liquid there is does not change how freely each particle moves among its neighbors.
Outcome: Explain why a substance's solid and liquid phases typically have similar molar volume: the particles remain in close contact in both.
Melt an ice cube, and the water takes up about as much room as the cube did.
Boil that water instead, and it demands a container over a thousand times larger.
Why does one phase change barely alter the room a sample needs, while the other multiplies it enormously?
You've already seen that a mole is a fixed count of particles — 6.022 × 10²³ of them.
A fair comparison across phases is the volume that one mole of the substance occupies.
That quantity — the volume occupied per mole — is called the 'molar volume'.
The table gives water's molar volume in each phase.
| Phase | Molar volume |
|---|---|
| ice (0 °C) | 19.7 cm³ |
| liquid water (25 °C) | 18.1 cm³ |
| water vapor (100 °C, 1 atm) | 30,100 cm³ |
The solid and liquid values differ by under 10%.
The vapor value is more than a thousand times larger than either.
You've already seen the reason the first two match: in both a solid and a liquid, the particles are in close contact.
Melting changes the arrangement and the motion of the particles — not the distance between neighbors.
Because the particles remain in close contact in both phases, a substance's solid and liquid phases typically have similar molar volume.
In the vapor, the molecules are no longer in contact, so the same mole of molecules occupies a far larger volume.
Water is slightly unusual in one detail — its solid's molar volume is a little larger than its liquid's, where most substances run the other way — but either way, the two stay close.
Explain that a substance's solid and liquid phases typically have similar molar volume — the volume occupied per mole — because the particles remain in close contact in both phases.
Check — multiple choice
A mole of solid naphthalene and a mole of liquid naphthalene occupy similar volumes. Why?
Feedback if chosen: This option is wrong — you carried the solid's order into the liquid. The liquid's molecules touch with no repeating pattern; it is the contact, not the order, that sets the volume.
Feedback if chosen: This option is wrong — you changed the count. A mole is a fixed count of particles, the same 6.022 × 10²³ molecules in either phase.
Feedback if chosen: This option is wrong — you removed the attractions. Melting loosens the arrangement while the molecules stay in contact; nothing about that keeps the volumes similar except the contact itself.
Feedback if chosen: Correct. With the molecules in close contact in both phases, the distance between neighbors barely changes on melting — so the volume per mole barely changes.
Check — multiple choice
| Phase of benzene | Molar volume |
|---|---|
| liquid | 89 cm³ |
| solid | ? |
The table gives the molar volume of liquid benzene. Which is the best estimate of the molar volume of solid benzene?
Feedback if chosen: Correct. The particles remain in close contact in both phases, so the solid's molar volume sits close to the liquid's — a little smaller, as it is for most substances.
Feedback if chosen: This option is wrong — you shrank the volume tenfold. Freezing keeps the molecules in contact; it does not pack them ten times closer.
Feedback if chosen: This option is wrong — you grew the volume tenfold. A solid packs its molecules about as closely as the liquid does, not ten times more loosely.
Feedback if chosen: This option is wrong — you gave the solid a gas-sized volume. A thousand-fold jump belongs to the phase whose particles leave contact entirely.
Check — multiple choice
| Phase | Molar volume |
|---|---|
| phase A | 34 cm³ |
| phase B | 6,300 cm³ |
| phase C | 31 cm³ |
The table gives the molar volumes of substance X's three phases, in scrambled order. Which phase is the gas?
Feedback if chosen: This option is wrong — you picked one of the paired values. At 34 cm³, phase A sits within 10% of phase C: those two are the solid and the liquid, whose particles stay in close contact.
Feedback if chosen: Correct. Phase B's molar volume dwarfs the other two — its particles are out of contact, and only the gas spreads a mole over so much volume.
Feedback if chosen: This option is wrong — you picked one of the paired values. At 31 cm³, phase C sits within 10% of phase A: those two are the solid and the liquid, whose particles stay in close contact.
Feedback if chosen: This option is wrong — the gas is exactly the phase these numbers can identify. The solid and the liquid cannot be told apart this way, but the thousand-fold value singles out the gas.
Outcome: Describe gas particles as far apart and in constant motion, with their average spacing and how often they collide depending on temperature, pressure, and volume.
You've already seen that a mole of water vapor occupies more than a thousand times the volume of a mole of liquid water.
All that extra room means the particles of a gas are far apart — most of a gas is empty space.
The figure shows the particle-level picture: sparse particles, each moving in its own direction.
Gas particles are in constant motion — nothing holds them in contact with a neighbor, and they never settle.
As they crisscross the container, the particles collide with one another and with the walls.
You've already seen what the wall collisions produce: gas pressure is the force per unit area of gas particles colliding with the walls of their container.
How far apart the particles sit on average, and how often they collide, depend on the sample's temperature, pressure, and volume.
Squeeze the same sample into a smaller volume, and the particles sit closer on average and collide more often.
As the temperature rises, the particles move faster, so the collisions come more often.
Under higher pressure, the particles are crowded closer together.
Describe gas particles as far apart and in constant motion, with their average spacing and how often they collide depending on temperature, pressure, and volume.
Check — multiple choice
Which best describes the motion of the molecules in a sealed flask of air?
Feedback if chosen: This option is wrong — you gave the gas a solid's motion. Nothing fixes a gas particle's position; each one travels through the flask.
Feedback if chosen: This option is wrong — you gave the gas a liquid's contact. Gas molecules are far apart; most of the flask is empty space between them.
Feedback if chosen: Correct. Gas particles are far apart and in constant motion, colliding with one another and with the container's walls.
Feedback if chosen: This option is wrong — you removed the motion. Gas particles never settle; their constant collisions with the walls are what produce the flask's pressure.
Check — multiple choice
A sealed sample of N₂ is squeezed into half its volume at a steady temperature. What happens to the molecules' average spacing and to how often they collide?
Feedback if chosen: This option is wrong — you held everything to the particle count. The count is fixed, but the room shrank: the same molecules now share less space.
Feedback if chosen: This option is wrong — you uncoupled the two changes. Closer neighbors are met more often, not less: shrinking the spacing makes collisions more frequent.
Feedback if chosen: This option is wrong — you grew the spacing while the room shrank. Less space for the same molecules pushes them closer together on average.
Feedback if chosen: Correct. Halving the volume leaves the same molecules less space, so their average spacing shrinks and they collide more often.
Check — multiple choice
A sealed, rigid container of argon is warmed. What changes for the atoms inside?
Feedback if chosen: This option is wrong — you let the sample expand. The container is rigid and sealed: same atoms, same space, so the average spacing cannot grow.
Feedback if chosen: Correct. As the temperature rises, the atoms move faster and so collide more often — while the fixed volume and fixed count keep the average spacing about the same.
Feedback if chosen: This option is wrong — you swelled the atoms. Warming changes how fast the atoms move, not how big they are.
Feedback if chosen: This option is wrong — you left the atoms untouched. Temperature is about the particles themselves: as it rises, they move faster.
Outcome: Explain why a gas has neither a definite volume nor a definite shape: its particles move constantly, and the attractions between them have minimal effect at gas spacings.
Open the valve of a helium cylinder in a sealed room, and helium is soon everywhere in the room.
A solid keeps its own shape; a liquid keeps its own volume; a gas keeps neither.
Whatever container a gas is given, it takes on that container's shape and fills its entire volume.
What lets the particles of a gas do this?
You've already seen the gas picture: far-apart particles in constant motion.
You've already seen that as the distance between particles grows, the attraction between them weakens.
At gas spacings, the particles sit so far apart that the attractions between them have minimal effect on the motion.
So each particle keeps traveling until a wall — not a neighbor — turns it around.
Constant motion with nothing reining it in carries the particles into every corner: a gas expands to fill any container it is placed in.
That is why a gas has no definite volume and no definite shape — both are set by the container, not by the gas.
A liquid's particles, by contrast, stay close enough for the attractions to hold the sample together at one volume.
Explain that a gas has neither a definite volume nor a definite shape because its particles move constantly and the attractions between them have minimal effect at gas spacings.
Check — multiple choice
Why does a sample of neon expand to fill any container it is placed in?
Feedback if chosen: This option is wrong — you built the spreading out of repulsions. It is each atom's own constant motion that carries it through the space; the atoms are too far apart to push on one another.
Feedback if chosen: Correct. The constant motion carries the atoms everywhere, and the far-apart spacing leaves the attractions between them too weak to rein that motion in.
Feedback if chosen: This option is wrong — you let the atoms change size. The atoms stay tiny; it is their spreading motion that reaches every part of the container.
Feedback if chosen: This option is wrong — you gave the walls a pull. The walls only turn the atoms around; nothing draws the atoms outward.
Check — multiple choice
Liquid water keeps a definite volume; water vapor does not. What accounts for the difference?
Feedback if chosen: This option is wrong — you changed the molecule. An H₂O molecule has the same mass in either phase; what differs is the spacing between molecules.
Feedback if chosen: This option is wrong — you gave the liquid a solid's order. Liquid molecules touch with no repeating pattern; it is the contact, and the attractions acting across it, that hold the volume.
Feedback if chosen: This option is wrong — you destroyed the attractions. They still exist; the wide spacing just leaves them with minimal effect — bring the molecules back into contact and the attractions take hold again.
Feedback if chosen: Correct. In the liquid, the close contact lets the attractions hold the sample at one volume; at gas spacings, the attractions have minimal effect, so the motion spreads the molecules through any space.
Check — multiple choice
A 1 L cylinder of N₂ is opened inside an otherwise empty, sealed 200 L tank. What volume does the N₂ end up occupying?
Feedback if chosen: This option is wrong — you gave the gas a volume of its own. A gas has no definite volume; its constant motion carries the molecules through the whole tank.
Feedback if chosen: This option is wrong — you let the attractions call a halt. At gas spacings the attractions have minimal effect, so nothing stops the spreading partway.
Feedback if chosen: Correct. The molecules move constantly, and with the attractions between them having minimal effect at gas spacings, they spread through the tank's entire volume.
Feedback if chosen: This option is wrong — you gave the gas a liquid's behavior. Nothing holds gas molecules in contact, so they cannot pool; the motion spreads them through all the space.
Outcome: Draw particulate models that distinguish the solid, liquid, and gas phases of one substance by particle spacing, ordering, and contact.
You've already seen that a particulate model shows the identities and relative positions of particles, not the substance at visible scale.
The figure gathers the three phase pictures you've already seen — solid, liquid, and gas — for a single substance.
One substance means one particle: draw every particle as the same-sized circle in all three boxes.
Solid box: draw the solid as a crystalline solid — the particles touch, and the arrangement repeats in a regular pattern.
Crystalline or amorphous, the particles of any solid stay closely packed and vibrate about fixed positions — the repeating pattern is what marks the drawing as crystalline.
Liquid box: the particles still touch, but the arrangement has no pattern.
Gas box: a few particles, widely separated and scattered through the whole box.
Short arrows on the gas particles are a good touch — they record the constant motion.
Three features carry the whole comparison: spacing, ordering, and contact.
Draw particulate models that distinguish the solid, liquid, and gas phases of one substance by particle spacing, ordering, and contact.
Check — drawing task (pen and paper, self-marked)
Take a pen and paper and draw particulate models of one substance, bromine, in three boxes labeled solid, liquid, and gas, drawing each Br₂ molecule as a small circle and drawing the solid as a crystalline solid. When you are done, select Check to compare your drawing with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
Check — multiple choice
A classmate has drawn a correct solid box and wants to turn a copy of it into the liquid box. Which one change is right?
Feedback if chosen: Correct. Melting removes the order but not the contact: the liquid box keeps touching particles with no repeating pattern.
Feedback if chosen: This option is wrong — you changed the wrong feature. A liquid keeps the contact and loses the order; this change loses the contact and keeps the order.
Feedback if chosen: This option is wrong — you followed the gas recipe. Widely separated, scattered particles belong in the gas box, not the liquid box.
Feedback if chosen: This option is wrong — you changed the particle size. One substance is drawn as one particle size in every phase; size never changes between boxes.
Check — multiple choice
Which set of features belongs in the gas box?
Feedback if chosen: This option is wrong — you described the solid box: touching plus a repeating pattern is the crystalline-solid recipe these boxes use.
Feedback if chosen: This option is wrong — you described the liquid box: touching with no pattern is the liquid recipe.
Feedback if chosen: This option is wrong — you spaced the particles but kept the order. Gas particles are scattered with no arrangement at all; a regular grid belongs to the solid.
Feedback if chosen: Correct. Wide separation, no pattern, and coverage of the whole box are the gas box's features.
Outcome: Evaluate whether a particulate drawing matches its labeled phase, and identify the specific flaw when it does not.
You've already drawn the three phase pictures; now judge drawings someone else made.
Here are the three correct pictures again, for one substance.
Read the drawing's label first, then test the drawing against the three features: spacing, ordering, and contact.
A drawing matches its label only when all three features fit the labeled phase.
When a drawing fails, name the specific feature that is off — the exact flaw.
Evaluate whether a supplied particulate drawing matches its labeled phase, identifying the specific flaw if it does not.
Check — multiple choice
The figure shows a student's drawing labeled gas. One feature of the drawing does not fit a gas — which one?
Feedback if chosen: Correct. A gas's particles are widely separated and scattered through the whole box; a touching clump is the flaw.
Feedback if chosen: This option is wrong — you flagged a feature that fits: a gas box properly holds only a few particles, since so few fit in any small region of a real gas.
Feedback if chosen: This option is wrong — you flagged a feature that fits: motion arrows properly record the constant motion of gas particles.
Feedback if chosen: This option is wrong — you flagged a feature that fits: one substance is drawn as one particle size in every phase.
Check — multiple choice
This drawing is labeled liquid. One feature of the drawing does not fit the arrangement of a liquid's particles — which one?
Feedback if chosen: This option is wrong — you flagged a feature that fits: a liquid's particles do stay in close contact with their neighbors.
Feedback if chosen: This option is wrong — you flagged a feature that fits: a liquid's touching particles properly fill most of their box.
Feedback if chosen: Correct. A liquid's particles touch with no repeating pattern; the regular repeating grid drawn here belongs in a solid box.
Feedback if chosen: This option is wrong — you flagged the missing arrows, but arrows would describe motion, not arrangement — and no phase's drawing requires them.
Check — multiple choice
This drawing is labeled solid. One feature of the drawing does not fit a solid — which one?
Feedback if chosen: This option is wrong — you flagged a feature that fits: one substance is drawn as one particle size in every phase.
Feedback if chosen: This option is wrong — you flagged a feature that fits: arrows are an optional touch for gas particles, and their absence is never the flaw in a solid box.
Feedback if chosen: This option is wrong — you flagged a feature that fits: every one of these particulate models is drawn inside a box.
Feedback if chosen: Correct. A solid's particles touch, repeating in a regular pattern; wide separation is the flaw in this drawing.
Check — multiple choice
One more drawing, also labeled liquid, is shown in the figure. Which evaluation is correct?
Feedback if chosen: This option is wrong — you flagged the contact. Touching particles are exactly right for a liquid.
Feedback if chosen: This option is wrong — you flagged the disorder. No repeating pattern is exactly right for a liquid.
Feedback if chosen: Correct. Touching particles with no repeating pattern is the liquid picture — all three features fit the label.
Feedback if chosen: This option is wrong — you demanded arrows. They are an optional touch for gas particles; a liquid drawing does not need them.
10 lessons
Outcome: Describe the kinetic molecular theory model of an ideal gas — tiny, far-apart particles in constant random motion with negligible forces between them — and connect the particles' motion to the gas's measurable properties.
A bicycle tire holds up a rider, yet the air inside it is nothing but tiny particles separated by empty space.
How can mostly empty space push back hard enough to carry a person?
You've already seen that gas particles are far apart and constantly moving, colliding as they travel.
You've already seen that gas pressure is the force per unit area of particles colliding with the walls of their container.
The figure shows that picture inside a cylinder of gas: widely spaced particles striking the walls from every direction.
Chemists sharpen that picture into a model built on three deliberate simplifications.
First: each particle is treated as a moving point — its own volume is negligible compared with the empty space around it.
Second: the particles are in constant random motion — always moving, in no preferred direction.
Third: the forces between the particles are negligible — each particle travels in a straight line until it collides with something.
A gas that fits all three simplifications is called an 'ideal gas'.
This model of an ideal gas is called the 'kinetic molecular theory', or KMT.
The collisions in this model waste nothing: particles bounce off one another and off the walls with none of their energy of motion lost — such collisions are called 'elastic'.
The point of the model is the connection it makes: what the particles do fixes what you can measure.
The tire pushes back because gas pressure is produced by particles colliding with the container walls, so more frequent or more forceful collisions mean higher pressure.
Described with the model, a sample of N₂ is an enormous crowd of tiny particles in constant random motion — and every measurable property of the sample traces back to that motion.
Describe the kinetic molecular theory (KMT) model of an ideal gas — point-like particles in constant random motion with negligible forces between them — and connect the particle motion to the gas's measurable properties.
Check — multiple choice
Which statement matches the kinetic molecular theory picture of an ideal gas?
Feedback if chosen: This option is wrong — you let the particles fill the container. The model treats each particle as a moving point whose own volume is negligible next to the empty space around it.
Feedback if chosen: Correct. Constant random motion is one of the model's three simplifications — the particles never stop and favor no direction.
Feedback if chosen: This option is wrong — you gave the particles strong attractions. The model takes the forces between particles as negligible, so each travels in a straight line until it collides with something.
Feedback if chosen: This option is wrong — you let the motion stop and start. In the model the motion is continuous: particles rebound from collisions without pausing.
Check — multiple choice
The figure shows a student's particulate drawing of a sample of N₂ gas. Which one change would bring the drawing in line with the kinetic molecular theory picture?
Feedback if chosen: This option is wrong — you drew the ordered arrangement of a solid. Gas particles move constantly and at random, so they show no ordered pattern.
Feedback if chosen: This option is wrong — you let the particles settle. Constant random motion keeps gas particles spread through the entire container, not resting at the bottom.
Feedback if chosen: This option is wrong — you read 'a gas fills its container' as the particles swelling. The particles stay tiny; it is their spreading motion that reaches every part of the container.
Feedback if chosen: Correct. In the KMT picture the particles are separated by lots of empty space and found throughout the whole container — not gathered in a touching clump.
Check — multiple choice
According to the kinetic molecular theory, what produces the pressure a gas exerts on its container?
Feedback if chosen: Correct. Gas pressure is produced by particles colliding with the container walls — more frequent or more forceful collisions mean higher pressure.
Feedback if chosen: This option is wrong — you built the push out of particle-particle forces. The model takes those forces as negligible; only collisions with the walls deliver the push.
Feedback if chosen: This option is wrong — you let the particles rest. They are in constant random motion, striking the walls and rebounding, never leaning on them.
Feedback if chosen: This option is wrong — you let the particles change size. They stay tiny points; the pressure comes from their collisions with the walls.
Outcome: Predict that a particle's kinetic energy increases as its speed increases, when its mass is unchanged.
Catch a gently tossed ball and it taps your hand; catch the same ball thrown fast and it stings.
Same ball, same mass — the only difference is speed.
The faster ball carries more energy of motion.
That energy of motion is called 'kinetic energy'.
The figure shows one N₂ molecule at two moments: moving slowly, then moving fast.
As the molecule's speed increases, its kinetic energy increases.
Nothing else about the molecule changes — same molecule, same mass, more kinetic energy.
As a particle slows down, the reverse happens: its kinetic energy falls.
Predict how a particle's kinetic energy changes when its speed changes and its mass stays the same.
Check — multiple choice
One O₂ molecule is moving faster now than it was a moment ago. How does its kinetic energy now compare with its kinetic energy before?
Feedback if chosen: Correct. As a particle's speed increases, its kinetic energy increases — and the mass here is the same at both moments.
Feedback if chosen: This option is wrong — you reversed the relationship. Kinetic energy grows with speed; the faster moment is the more energetic one.
Feedback if chosen: This option is wrong — you let the constant mass freeze the energy. Mass is only half the story: with the mass fixed, a higher speed alone means more kinetic energy.
Feedback if chosen: This option is wrong — you reached for a calculation. The mass is unchanged between the two moments, so the increase in speed settles the comparison by itself.
Check — multiple choice
Which change increases a gas particle's kinetic energy?
Feedback if chosen: This option is wrong — you reversed the relationship. As a particle slows down, its kinetic energy falls.
Feedback if chosen: This option is wrong — you counted direction as part of the energy. Kinetic energy depends on how fast the particle moves, not on which way it is heading.
Feedback if chosen: Correct. As a particle's speed increases, its kinetic energy increases — with its mass unchanged, speed is the deciding feature.
Feedback if chosen: This option is wrong — with nothing about the motion changing, the kinetic energy stays exactly what it was.
Check — multiple choice
Three identical N₂ molecules are moving at 300 m/s, 500 m/s, and 800 m/s. Which lists them from most kinetic energy to least?
Feedback if chosen: This option is wrong — you ranked the slowest molecule as the most energetic. Kinetic energy grows with speed, so the fastest molecule leads.
Feedback if chosen: Correct. The masses are identical, so speed alone decides: the 800 m/s molecule has the most kinetic energy and the 300 m/s molecule the least.
Feedback if chosen: This option is wrong — you let equal masses force equal energies. Equal masses are exactly why the different speeds decide the ranking.
Feedback if chosen: This option is wrong — you reached for a number you don't need. Identical molecules share one mass, whatever its value, so the speeds alone give the ranking.
Outcome: Predict that at the same speed, a more massive particle has more kinetic energy than a less massive one.
You've already seen that a particle's kinetic energy increases as its speed increases, when its mass is unchanged.
Now hold the speed fixed instead, and let the mass differ.
A slowly rolling bowling ball is far harder to stop than a table-tennis ball rolling at the same speed.
At the same speed, the more massive object carries more kinetic energy.
The figure compares the masses of two gas particles: a He atom and an SF₆ molecule.
Picture the two moving at exactly the same speed.
At 146 amu against 4 amu, the SF₆ molecule has far more kinetic energy than the He atom.
At equal speeds, the more massive particle always has the more kinetic energy.
Predict which of two particles moving at the same speed has more kinetic energy: the more massive one.
Check — multiple choice
| Particle | Mass (amu) |
|---|---|
| H₂ | 2.0 |
| O₂ | 32.0 |
The table gives the masses of an H₂ molecule and an O₂ molecule. The two molecules are moving at the same speed. Which has more kinetic energy?
Feedback if chosen: This option is wrong — you gave the lighter molecule the energy advantage. At equal speeds, kinetic energy grows with mass, and O₂ is the more massive molecule.
Feedback if chosen: This option is wrong — you let equal speeds force equal energies. Speed is only half the story: at the same speed, the more massive particle carries more kinetic energy.
Feedback if chosen: This option is wrong — you reached for a number you don't need. Whatever the shared speed is, it is the same for both molecules, so the masses alone settle the comparison.
Feedback if chosen: Correct. At the same speed, the more massive particle has more kinetic energy — and O₂, at 32.0 amu, is sixteen times the mass of H₂.
Check — multiple choice
| Particle | Mass (amu) |
|---|---|
| CH₄ | 16.0 |
| CO₂ | 44.0 |
| Xe | 131.3 |
The table gives the masses of three particles, all moving at the same speed. Which particle has the most kinetic energy?
Feedback if chosen: This option is wrong — you gave the lightest particle the energy advantage. At equal speeds the ranking follows mass, and CH₄ has the least mass in the table.
Feedback if chosen: This option is wrong — you may have ranked by the size or complexity of the molecule. The mass column decides, and 131.3 amu is well above 44.0 amu.
Feedback if chosen: Correct. At the same speed, kinetic energy grows with mass, and Xe is the most massive particle in the table at 131.3 amu.
Feedback if chosen: This option is wrong — you let equal speeds force equal energies. At one shared speed, the differing masses make the kinetic energies differ.
Outcome: State the kinetic-energy equation KE = ½mv², and predict the factor by which a particle's kinetic energy changes when its speed or mass changes.
You've already seen that a particle's kinetic energy increases as its speed increases, and increases as its mass increases.
One equation captures both effects at once.
Mass appears to the first power: doubling the mass doubles the kinetic energy.
Speed appears squared: doubling the speed quadruples the kinetic energy.
The square makes speed the stronger lever — changing the speed by some factor moves the kinetic energy more than changing the mass by the same factor.
The equation is used here for comparisons: apply the change to the equation, then compare with the original.
Double a particle's speed, and its kinetic energy becomes ½m(2v)² = 4 × ½mv² — four times the original.
The mass factor passes straight through: a particle with twice the mass at the same speed has 2 × ½mv², twice the kinetic energy.
State the kinetic-energy equation KE = ½mv², and find the factor by which the kinetic energy changes when the speed or the mass changes by a given factor.
Check — numeric entry
A molecule's speed triples while its mass stays the same. By what factor does its kinetic energy increase?
Feedback when correct: Correct. Speed is squared in KE = ½mv², so tripling the speed multiplies the kinetic energy by 3² = 9.
Anticipated wrong answers (feedback shown):
| 3 | You let the kinetic energy grow in step with the speed. Speed is squared in KE = ½mv²: (3v)² = 9v², so the kinetic energy becomes nine times the original. |
| 6 | You doubled the factor instead of squaring it. The ² in KE = ½mv² means the factor itself is squared: 3² = 9. |
Worked solution revealed after a second miss
Write down the equation
KE = ½mv²
Apply the change to the speed
new KE = ½m(3v)²
new KE = 9 × ½mv²
Compare with the original
the kinetic energy is 9 times the original
Check — numeric entry
Molecule B has three times the mass of molecule A, and the two move at the same speed. Molecule B's kinetic energy is how many times molecule A's?
Feedback when correct: Correct. Mass appears to the first power in KE = ½mv², so three times the mass at the same speed means three times the kinetic energy.
Anticipated wrong answers (feedback shown):
| 9 | You squared the mass factor. Only the speed is squared in KE = ½mv²; mass appears to the first power, so three times the mass means three times the kinetic energy. |
Worked solution revealed after a second miss
Write down the equation
KE = ½mv²
Apply the change to the mass
KE of B = ½(3m)v²
KE of B = 3 × ½mv²
Compare with the original
molecule B has 3 times the kinetic energy of molecule A
Check — multiple choice
Which change increases a particle's kinetic energy by the larger factor?
Feedback if chosen: This option is wrong — you gave the two factors equal weight. Mass enters KE = ½mv² to the first power, so doubling it only doubles the energy; the squared speed does more.
Feedback if chosen: Correct. Speed is squared in KE = ½mv², so doubling the speed quadruples the kinetic energy, while doubling the mass merely doubles it.
Feedback if chosen: This option is wrong — you overlooked the square. Doubling the mass gives a factor of 2; doubling the speed gives a factor of 2² = 4.
Feedback if chosen: This option is wrong — both mass and speed sit inside KE = ½mv², so changing either one changes the kinetic energy.
Outcome: Calculate a Kelvin temperature from a Celsius temperature using K = °C + 273.
Zero on the Celsius scale is just the temperature at which water freezes — the thermometer's zero, not nature's.
Cool a gas below 0 °C and its particles keep moving; there is colder to go.
There is a true bottom: the coldest possible temperature, below which nothing can be cooled.
The Kelvin scale puts its zero at that coldest possible temperature and counts up in steps the same size as Celsius degrees.
A temperature on this scale is called a 'Kelvin temperature', with the unit K.
The figure lines the two scales up at three familiar temperatures.
| Celsius | Kelvin | |
|---|---|---|
| water freezes | 0 °C | 273 K |
| room temperature | 25 °C | 298 K |
| water boils | 100 °C | 373 K |
Each Kelvin value in the table is 273 more than its Celsius partner.
The exact offset between the scales is 273.15, and this course's calculations round it to 273.
One addition converts any Celsius temperature.
Q: What is 25 °C in kelvin?
Write down the values in the question
Celsius temperature = 25 °C
Write down the equation
K = °C + 273
Substitute in the values, and calculate
K = 25 + 273
K = 298 K
Q: A freezer holds a sample at −18 °C. What is that temperature in kelvin?
Write down the values in the question
Celsius temperature = −18 °C
Write down the equation
K = °C + 273
Substitute in the values, and calculate
K = −18 + 273
K = 255 K
Calculate a Kelvin temperature from a Celsius temperature using K = °C + 273.
Check — numeric entry
A summer afternoon reaches 31 °C. What is that temperature in kelvin?
Feedback when correct: Correct. K = 31 + 273 = 304 K.
Anticipated wrong answers (feedback shown):
| 31 | You gave the Celsius number unchanged. Kelvin and Celsius values differ: add 273 to convert, giving 304 K. |
| 242 | You subtracted 273 instead of adding it: 273 − 31 = 242. The conversion adds 273 to the Celsius value, so K = 31 + 273 = 304 K. |
Worked solution revealed after a second miss
Write down the values in the question
Celsius temperature = 31 °C
Write down the equation
K = °C + 273
Substitute in the values, and calculate
K = 31 + 273
K = 304 K
Check — numeric entry
A pizza oven operates at 260 °C. What is that temperature in kelvin?
Also accepted: 533.15
Feedback when correct: Correct. K = 260 + 273 = 533 K.
Anticipated wrong answers (feedback shown):
| 260 | You gave the Celsius number unchanged. Add 273 to convert: K = 260 + 273 = 533 K. |
| 13 | You subtracted 260 from 273. The conversion adds 273 to the Celsius value: K = 260 + 273 = 533 K. |
Worked solution revealed after a second miss
Write down the values in the question
Celsius temperature = 260 °C
Write down the equation
K = °C + 273
Substitute in the values, and calculate
K = 260 + 273
K = 533 K
Check — numeric entry
Dry ice — solid CO₂ — sits at −78 °C. What is that temperature in kelvin?
Feedback when correct: Correct. K = −78 + 273 = 195 K.
Anticipated wrong answers (feedback shown):
| 351 | You dropped the minus sign and added 78: 273 + 78 = 351. The Celsius value here is negative: K = −78 + 273 = 195 K. |
| -78 | You gave the Celsius number unchanged. Add 273 to convert: K = −78 + 273 = 195 K. |
Worked solution revealed after a second miss
Write down the values in the question
Celsius temperature = −78 °C
Write down the equation
K = °C + 273
Substitute in the values, and calculate
K = −78 + 273
K = 195 K
Outcome: State that the Kelvin temperature of a sample is proportional to the average kinetic energy of its particles, and predict how the average kinetic energy changes when the Kelvin temperature changes.
A thermometer in a flask of gas reads one number for a sample of countless invisible, moving particles.
What about the particles is that one number reporting?
You've already seen that each particle carries kinetic energy set by its mass and speed, KE = ½mv².
Across a whole sample, those kinetic energies have an average.
The thermometer is reporting that average.
The figure shows one sealed sample at 300 K and again at 600 K.
The relationship is exact on the Kelvin scale: doubling the Kelvin temperature doubles the average kinetic energy, because Kelvin temperature is proportional to the average kinetic energy of the particles.
The Celsius scale cannot play this role — its zero sits at water's freezing point, not at the true bottom of the scale.
So a comparison of average kinetic energies always starts by putting both temperatures in kelvin.
One more consequence: any two samples at the same Kelvin temperature have the same average kinetic energy, whatever the gases are.
State that the Kelvin temperature of a sample is proportional to the average kinetic energy of its particles, and use the proportionality to compare samples.
Check — numeric entry
A sealed sample of argon is heated from 200 K to 800 K. By what factor does the average kinetic energy of its particles increase?
Feedback when correct: Correct. Average kinetic energy is proportional to Kelvin temperature, and 800 / 200 = 4.
Anticipated wrong answers (feedback shown):
| 600 | You found the temperature increase, 800 − 200 = 600 K, instead of the factor. Average kinetic energy scales with the Kelvin temperature itself: 800 / 200 = 4. |
| 3 | You compared the increase with the starting temperature: 600 / 200 = 3. The factor comes from the ratio of the two Kelvin temperatures: 800 / 200 = 4. |
Worked solution revealed after a second miss
Write down the relationship
average kinetic energy ∝ Kelvin temperature
Compare the two Kelvin temperatures
800 / 200 = 4
State the change
the average kinetic energy increases by a factor of 4
Check — multiple choice
A gas sample warms from 25 °C to 50 °C. What happens to the average kinetic energy of its particles?
Feedback if chosen: This option is wrong — you doubled the Celsius reading. The proportionality holds on the Kelvin scale: 25 °C is 298 K and 50 °C is 323 K, nowhere near double.
Feedback if chosen: This option is wrong — you let a small change count as none. Any rise in Kelvin temperature raises the average kinetic energy in proportion.
Feedback if chosen: Correct. In kelvin the change is 298 K to 323 K — an increase of under 10%, so the average kinetic energy rises by well under double.
Feedback if chosen: This option is wrong — you turned warming into cooling. A higher temperature always means a higher average kinetic energy.
Check — multiple choice
One container of He and one container of SF₆ are both held at 400 K. How do the average kinetic energies of their particles compare?
Feedback if chosen: Correct. Kelvin temperature is proportional to the average kinetic energy of the particles, so two samples at 400 K match in average kinetic energy — whatever the gases are.
Feedback if chosen: This option is wrong — you gave the lighter gas an energy advantage. Temperature alone fixes the average kinetic energy, and both samples sit at 400 K.
Feedback if chosen: This option is wrong — you gave the heavier gas an energy advantage. Temperature alone fixes the average kinetic energy, and both samples sit at 400 K.
Feedback if chosen: This option is wrong — you reached for the masses. The particle masses never enter this comparison: the shared Kelvin temperature settles it.
Outcome: Predict which of two gases at the same temperature has the faster-moving particles on average, by combining equal average kinetic energies with KE = ½mv².
A balloon of He and a balloon of N₂ sit in the same room, both at 300 K.
You've already seen that samples at the same Kelvin temperature have the same average kinetic energy.
Does the same average kinetic energy mean the same average speed?
You've already seen the kinetic-energy equation, KE = ½mv² — the energy depends on mass as well as speed.
The figure compares the two particle masses.
You've already seen where such numbers come from: a particle's mass in amu has the same numerical value as the substance's molar mass in grams per mole.
Hold the two averages equal, and let the equation speak.
A He atom has a seventh of the mass of an N₂ molecule, so matching the N₂ molecule's kinetic energy takes a larger v².
Smaller mass with the same ½mv² — the speed must be higher.
So at 300 K the He atoms move faster than the N₂ molecules, on average.
The rule generalizes: at the same temperature, the gas with the lower molar mass has the faster-moving particles on average.
Predict which of two gases at the same temperature has the faster particles on average — the gas with the lower molar mass — by combining equal average kinetic energies with KE = ½mv².
Check — multiple choice
| Gas | Particle mass (amu) |
|---|---|
| Ne | 20.2 |
| Ar | 39.9 |
The table gives the particle masses of two gases, both held at 350 K. Which gas's particles move faster on average?
Feedback if chosen: This option is wrong — you gave the heavier atom the speed advantage. With the average kinetic energies matched, the more massive atom must move more slowly.
Feedback if chosen: This option is wrong — you matched the speeds instead of the energies. Equal ½mv² with different masses forces different speeds.
Feedback if chosen: This option is wrong — the masses, together with the shared temperature, are exactly what decide it: same average kinetic energy, so the lighter atom is faster.
Feedback if chosen: Correct. Ne atoms have about half the mass of Ar atoms, so matching the average kinetic energy at 350 K requires the Ne atoms to move faster on average.
Check — multiple choice
| Gas | Particle mass (amu) |
|---|---|
| H₂ | 2.0 |
| O₂ | 32.0 |
Both gases in the table are held at 300 K. Which statement compares them correctly?
Feedback if chosen: Correct. The shared temperature fixes one average kinetic energy for both, and at a sixteenth of the mass, the H₂ molecules need higher speeds to match it.
Feedback if chosen: This option is wrong — you matched the speeds as well as the energies. Equal ½mv² with unequal masses forces unequal speeds: the lighter H₂ must be faster.
Feedback if chosen: This option is wrong — you gave the lighter gas an energy advantage. Temperature alone sets the average kinetic energy, and both gases sit at 300 K.
Feedback if chosen: This option is wrong — you gave the heavier gas both advantages. The average kinetic energies match at one temperature, and the heavier molecules are the slower ones.
Check — multiple choice
| Gas | Particle mass (amu) |
|---|---|
| CH₄ | 16.0 |
| SF₆ | 146.1 |
A container of CH₄ and a container of SF₆ are both at 298 K. Which gas's particles are slower on average?
Feedback if chosen: This option is wrong — you gave the lighter molecule the slower speed. At matched average kinetic energy, the lighter molecule is the faster one.
Feedback if chosen: This option is wrong — you matched the speeds instead of the energies. One temperature fixes the average kinetic energy, and unequal masses then demand unequal speeds.
Feedback if chosen: Correct. SF₆ molecules are about nine times the mass of CH₄ molecules, so matching the average kinetic energy at 298 K leaves them moving more slowly on average.
Feedback if chosen: This option is wrong — you brought in the sample size. The number of particles sets how much gas there is, not how fast each particle moves.
Outcome: Read a Maxwell-Boltzmann distribution: say what each axis shows, find the most probable speed at the peak, and state what the high-speed tail and the area under the curve represent.
A gas sample holds an enormous number of particles, all in constant random motion.
You've already seen that Kelvin temperature is proportional to the average kinetic energy of the particles — but an average says nothing about any single particle.
The figure shows one sample of N₂ at 300 K.
The horizontal axis is particle speed.
The height of the curve at each speed is the number of particles moving at speeds close to that speed.
Almost no particles are nearly still, so the curve starts at zero.
It climbs to a peak, then falls away slowly toward high speeds.
The peak sits over the speed that more particles share than any other.
That speed is called the 'most probable speed'.
On the curve shown, the peak sits over 400 m/s.
The curve is not symmetric — it stretches much further to the right of the peak than to the left.
That long right-hand stretch is called the 'high-speed tail': a small number of particles moving much faster than most of the sample.
The whole curve describes a sample at a single temperature.
One temperature still allows many speeds, because Kelvin temperature is proportional to the average kinetic energy of the particles — it fixes the average, not each particle's speed.
One more feature carries meaning: the area under the entire curve is the total number of particles in the sample.
A graph of this kind — particle count against speed for a sample at one temperature — is called a 'Maxwell-Boltzmann distribution'.
Read a Maxwell-Boltzmann distribution: say what each axis shows, find the most probable speed at the peak, and state what the high-speed tail and the area under the curve represent.
Check — numeric entry
The figure shows the Maxwell-Boltzmann distribution for a sample of neon gas. What is the most probable particle speed, in m/s?
Feedback when correct: Correct. The dashed guide from the peak meets the speed axis at 500 m/s — more particles move at speeds close to 500 m/s than close to any other speed.
Anticipated wrong answers (feedback shown):
| 1250 | You read the far end of the speed axis, where the tail flattens out, instead of the peak. The most probable speed sits directly below the curve's highest point. |
| 300 | You answered the temperature from the figure caption instead of reading a speed from the axis. 300 is a temperature in kelvin, not a speed. |
Check — numeric entry
The figure shows the Maxwell-Boltzmann distribution for a sample of argon gas. What is the most probable particle speed, in m/s?
Feedback when correct: Correct. The dashed guide from the peak meets the speed axis at 350 m/s — more particles move at speeds close to 350 m/s than close to any other speed.
Anticipated wrong answers (feedback shown):
| 1000 | You read the far end of the speed axis, where the tail flattens out, instead of the peak. The most probable speed sits directly below the curve's highest point. |
| 300 | You answered the temperature from the figure caption instead of reading a speed from the axis. 300 is a temperature in kelvin, not a speed. |
Check — multiple choice
On the Maxwell-Boltzmann distribution shown, point P marks the curve at one speed. What does the height of the curve at P show?
Feedback if chosen: This option is wrong — you read one height as a running total. A count over a range of speeds is an area under the curve, not a single height.
Feedback if chosen: Correct. The height of the curve at any speed is the number of particles moving at speeds close to that speed.
Feedback if chosen: This option is wrong — you read one height as the size of the high-speed tail. That count is the area under the tail, not a single height.
Feedback if chosen: This option is wrong — you read one height as the whole sample. The total is the area under the entire curve.
Outcome: Sketch Maxwell-Boltzmann curves for one gas at two temperatures: the higher-temperature curve peaks at a higher speed, sits lower, and encloses the same total area.
You've already seen how to read this graph — the Maxwell-Boltzmann distribution for a sample of N₂ at 300 K.
Warm the same sample to 600 K — what happens to the curve?
You've already seen the starting point of the answer: heating the sample raises the average kinetic energy, because Kelvin temperature is proportional to the average kinetic energy of the particles.
The figure adds the 600 K curve for the same N₂ sample.
As the temperature rises, the average kinetic energy rises, so a larger share of the particles move at high speeds.
The peak moves toward higher speed: the most probable speed is higher at 600 K.
The number of particles has not changed, and the area under the curve is the total number of particles — so the two areas must be equal.
With the same area spread across a wider range of speeds, the 600 K peak must sit lower.
Higher temperature: peak further right, peak lower, same area — the curve flattens and spreads.
Every sketch keeps the shape you've already seen: each curve starts at zero, rises to a single peak, and falls away slowly along its high-speed tail.
Sketch Maxwell-Boltzmann curves for one gas at two temperatures, with the higher-temperature curve peaked at a higher speed, sitting lower, and equal in total area.
Check — drawing task (pen and paper, self-marked)
Take a pen and paper and draw the Maxwell-Boltzmann distribution of one sample of O₂ at 200 K and at 400 K — two curves on one set of axes, each labeled with its temperature. When you are done, select Check to compare your drawing with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
Check — multiple choice
The figure shows a student's sketch of one N₂ sample at 300 K and at 600 K. Which statement identifies an error in the sketch?
Feedback if chosen: This option is wrong — you treated the height difference as the error. Heating flattens the curve, so the 600 K peak should be lower; the heights are drawn correctly for the labels.
Feedback if chosen: Correct. Heating shifts the peak toward higher speed, so the 600 K peak belongs to the right of the 300 K peak — this sketch puts it on the left.
Feedback if chosen: This option is wrong — starting at zero is required: almost no particles are nearly still, at either temperature.
Feedback if chosen: This option is wrong — the gradual fall along the high-speed tail is required: a few particles always move much faster than most.
Check — multiple choice
One gas sample is sketched at two temperatures on one set of axes. What must be the same about the two curves?
Feedback if chosen: This option is wrong — the peak speed is exactly what shifts: as the temperature rises, the most probable speed rises.
Feedback if chosen: This option is wrong — the peak height is exactly what drops: as the curve spreads toward higher speeds, its peak must sit lower to keep the area fixed.
Feedback if chosen: This option is wrong — as the temperature rises, fewer particles remain at slow speeds, so that count changes.
Feedback if chosen: Correct. The area under each curve is the total number of particles, and heating does not change how many particles the sample holds.
Outcome: Sketch Maxwell-Boltzmann speed curves for two gases of different molar mass at one temperature, with the lighter gas's curve shifted toward higher speeds.
You've already sketched one gas at two temperatures; now hold the temperature fixed and change the gas.
You've already seen that at the same temperature, the gas with the lower molar mass has the faster-moving particles on average.
The figure shows the Maxwell-Boltzmann curves of He and N₂, both at 300 K.
Because the He atoms are lighter, they move faster on average, so the whole He curve sits shifted toward higher speeds.
Equal-sized samples hold the same number of particles, so the two curves enclose equal areas.
Spread across a wider range of speeds, the He curve's peak must sit lower.
The recipe for any pair: same axes, equal areas, lighter gas peaked further right.
Sketch Maxwell-Boltzmann speed curves for two gases of different molar mass at one temperature, with the lighter gas's curve shifted toward higher speeds.
Check — drawing task (pen and paper, self-marked)
Take a pen and paper and draw the Maxwell-Boltzmann curves of H₂ (particle mass 2.0 amu) and O₂ (particle mass 32.0 amu), both at 300 K, on one set of axes, labeling each curve. When you are done, select Check to compare your drawing with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
Check — multiple choice
The figure shows a student's sketch of He and Ar samples at the same temperature. Which statement identifies an error in the sketch?
Feedback if chosen: This option is wrong — you treated the height difference as the error. The lighter gas spreads over more speeds, so its peak should be lower; the heights are drawn correctly for the labels.
Feedback if chosen: This option is wrong — a gradual fall toward high speeds is exactly what a speed distribution does; that part of the sketch is correct.
Feedback if chosen: Correct. The lighter gas moves faster on average, so the He peak belongs at the higher speed — this sketch gives the speed advantage to the heavier gas.
Feedback if chosen: This option is wrong — rising from zero is required: almost no particles are nearly still, in either gas.
Check — multiple choice
The two curves show two different gases at the same temperature. Which gas has the higher molar mass?
Feedback if chosen: Correct. Gas X peaks at the lower speed, and at one shared temperature the slower-moving gas is the one with the higher molar mass.
Feedback if chosen: This option is wrong — you gave the faster gas the higher mass. At matched average kinetic energy, heavier particles move more slowly, not faster.
Feedback if chosen: This option is wrong — same temperature matches average kinetic energies, not speeds; gases of different molar mass give different curves, exactly as drawn.
Feedback if chosen: This option is wrong — the peak positions are exactly the evidence needed: at one temperature, the curve peaked at the lower speed belongs to the more massive gas.
Check — multiple choice
Sugar solidifies in two forms. In rock candy, the sugar molecules repeat in a regular three-dimensional arrangement; in cotton candy, they are trapped in a jumbled arrangement with no regular pattern. Which classification is correct?
Feedback if chosen: This option is wrong — you classified by the substance instead of the arrangement. One substance can solidify either way; what decides the class is whether the arrangement repeats.
Feedback if chosen: This option is wrong — you swapped the two labels. Rock candy is the form whose molecules repeat in a regular arrangement, so rock candy is the crystalline one.
Feedback if chosen: Correct. Rock candy's molecules repeat in a regular three-dimensional arrangement — crystalline — while cotton candy's are trapped with no long-range repeating pattern — amorphous.
Feedback if chosen: This option is wrong — you reached for a property that is not the test. The classification reads the particle arrangement, and both arrangements are given.
Check — multiple choice
| Phase of octane | Molar volume |
|---|---|
| liquid | 163 cm³ |
| solid | ? |
The table gives the molar volume of liquid octane. Which is the best estimate of the molar volume of solid octane?
Feedback if chosen: Correct. The particles remain in close contact in both phases, so the solid's molar volume sits close to the liquid's — a little smaller, as it is for most substances.
Feedback if chosen: This option is wrong — you shrank the volume tenfold. Freezing keeps the molecules in contact; it does not pack them ten times closer.
Feedback if chosen: This option is wrong — you grew the volume tenfold. A solid packs its molecules about as closely as the liquid does, not ten times more loosely.
Feedback if chosen: This option is wrong — you gave the solid a gas-sized volume. A thousand-fold jump belongs to the phase whose particles leave contact entirely.
Check — multiple choice
A valve releases a small sample of argon into an empty, sealed 40 L chamber. Why does the argon end up spread through the chamber's entire volume?
Feedback if chosen: This option is wrong — you built the spreading out of repulsions. It is each atom's own constant motion that carries it through the space; the atoms are far too spread out to push on one another.
Feedback if chosen: Correct. The constant motion carries the atoms into every corner, and the wide spacing leaves the attractions between them too weak to rein that motion in — so the gas takes the chamber's whole volume.
Feedback if chosen: This option is wrong — you let the atoms change size. The atoms stay tiny; it is their spreading motion that reaches every part of the chamber.
Feedback if chosen: This option is wrong — you gave the walls a pull. The walls only turn the atoms around; nothing draws the atoms outward.
Check — multiple choice
A student draws particulate models of one substance, CO₂, in three boxes labeled solid, liquid, and gas. The drawing is shown. Which evaluation is correct?
Feedback if chosen: This option is wrong — you flagged the one box that is right by design: a crystalline solid's particles touch and repeat in a regular pattern, exactly as drawn.
Feedback if chosen: This option is wrong — you pulled the liquid's particles apart. A liquid's particles stay in close contact, touching with no repeating pattern — that box matches its label.
Feedback if chosen: This option is wrong — you accepted the gas box, but its particles are drawn touching. Gas particles are widely separated, scattered through the whole box.
Feedback if chosen: Correct. The solid and liquid boxes fit their labels, but the box labeled gas shows touching particles — a gas's particles are widely separated and scattered through the whole box.
Check — multiple choice
A sealed sample of CO₂ is heated from 87 °C to 447 °C. What happens to the average kinetic energy of its particles?
Feedback if chosen: Correct. Converted to kelvin, 87 °C is 360 K and 447 °C is 720 K — the Kelvin temperature doubles, and Kelvin temperature is proportional to the average kinetic energy of the particles.
Feedback if chosen: This option is wrong — you compared the Celsius readings (447 ÷ 87 is about 5). The proportionality holds on the Kelvin scale: 360 K to 720 K is a factor of 2.
Feedback if chosen: This option is wrong — you squared the factor. The proportionality is direct: a doubled Kelvin temperature means a doubled average kinetic energy, not a quadrupled one.
Feedback if chosen: This option is wrong — you assumed converting to kelvin always shrinks the change to a sliver. Here it does not: 87 + 273 = 360 K and 447 + 273 = 720 K, exactly double.
Check — multiple choice
| Gas | Particle mass (amu) |
|---|---|
| N₂ | 28.0 |
| Xe | 131.3 |
The table gives the particle masses of two gases, both held at 320 K. Which gas's particles move faster on average?
Feedback if chosen: This option is wrong — you gave the heavier atom the speed advantage. With the average kinetic energies matched, the more massive particle must move more slowly.
Feedback if chosen: This option is wrong — you matched the speeds instead of the energies. Equal ½mv² with different masses forces different speeds.
Feedback if chosen: This option is wrong — you reached for a number you don't need. Whatever the shared temperature is, it fixes one average kinetic energy for both gases, so the masses alone settle the comparison.
Feedback if chosen: Correct. At 320 K both gases have the same average kinetic energy, and at less than a quarter of the mass, the N₂ molecules must move faster on average to match it.
Check — multiple choice
A sealed sample of gas is held at one temperature. Its Maxwell-Boltzmann distribution is shown. Which statement correctly describes the speeds of the particles in the sample?
Feedback if chosen: This option is wrong — you treated temperature as fixing every particle's speed. It fixes only the average, because Kelvin temperature is proportional to the average kinetic energy of the particles.
Feedback if chosen: Correct. The curve spans a wide stretch of the speed axis, and its long right-hand stretch — the high-speed tail — shows that a small number of particles move much faster than most.
Feedback if chosen: This option is wrong — you read the peak as a maximum. The curve continues well past the peak, and that stretch is the high-speed tail.
Feedback if chosen: This option is wrong — you treated the peak as a halfway point. The curve is not symmetric, so the peak marks the most common speed, not a 50-50 split.
Check — multiple choice
The figure shows a student's sketch of the Maxwell-Boltzmann speed curves of Ne (20.2 amu) and Kr (83.8 amu), both at 300 K. Which statement identifies an error in the sketch?
Feedback if chosen: This option is wrong — you treated the height difference as the error. The lighter gas spreads its particles over more speeds, so the Ne peak should be lower; the heights are drawn correctly for the labels.
Feedback if chosen: This option is wrong — you flagged a required feature: almost no particles are nearly still, in either gas, so both curves start at zero.
Feedback if chosen: Correct. At one temperature the lighter gas moves faster on average, so the Ne peak belongs at the higher speed — this sketch gives the speed advantage to the heavier Kr.
Feedback if chosen: This option is wrong — a gradual fall toward high speeds is exactly what a speed distribution does; that part of the sketch is correct.
10 lessons
Outcome: Predict that decreasing the volume of a fixed amount of gas at constant temperature increases its pressure, and explain why through wall collisions.
Cap the tip of an air-filled syringe and push the plunger in until the trapped air fills half its old space.
The further in you push, the harder the trapped air pushes back.
You've already seen the kinetic molecular theory picture: the trapped air is a huge number of tiny particles in constant, random motion, striking the walls.
Pushing the plunger in does not change the particles or their speeds — the same particles now move in half the space.
In the smaller space, each particle crosses to a wall sooner, so the walls are struck more often each second.
The pressure inside rises, because gas pressure is produced by particles colliding with the container walls, so more frequent or more forceful collisions mean higher pressure.
The temperature did not change, so no collision is any harder than before — there are simply more collisions each second.
The prediction holds for any fixed amount of gas at constant temperature: decreasing the volume increases the pressure.
Predict that decreasing the volume of a fixed amount of gas at constant temperature increases its pressure — the same particles strike the container walls more often in a smaller space.
Check — multiple choice
The sealed cylinder shown is compressed to one-third of its volume at constant temperature, and the pressure inside rises. Why?
Feedback if chosen: This option is wrong — you gave the temperature mechanism. The temperature is constant, so the particles' speeds are unchanged; what changes is how often each particle reaches a wall.
Feedback if chosen: Correct. With the same particles in one-third of the space, each particle reaches a wall sooner, so the walls are struck more often — and more frequent collisions mean higher pressure.
Feedback if chosen: This option is wrong — you counted collisions between particles. Gas pressure is produced by particles colliding with the container walls, not by particles striking one another.
Feedback if chosen: This option is wrong — you reached for attractions between particles. Attractions pull particles toward one another; they do not push on the walls. The pressure here comes from wall collisions.
Check — multiple choice
The plunger of a capped syringe of air is pulled outward at constant temperature until the trapped air fills twice its old space. What happens to the pressure inside?
Feedback if chosen: Correct. In the larger space each particle reaches a wall less often, so the walls are struck less often and the pressure falls.
Feedback if chosen: This option is wrong — you applied the prediction backward. Decreasing the volume raises the pressure; increasing the volume spreads the same particles out, so the walls are struck less often and the pressure falls.
Feedback if chosen: This option is wrong — you kept the pressure fixed while the volume changed. With the same particles at the same temperature, changing the space must change how often the walls are struck.
Feedback if chosen: This option is wrong — you brought the gas's identity into it. The prediction rests only on particles striking walls, so the prediction is the same for any gas.
Outcome: State that pressure and volume are inversely proportional for a fixed amount of gas at constant temperature, and reason with factors.
You've already seen that decreasing the volume of a fixed amount of gas at constant temperature increases its pressure.
The trade-off is exact: halve the volume and the walls are struck twice as often, so the pressure doubles.
Squeeze the volume to a third and the pressure triples; let the volume double and the pressure halves.
Two quantities that trade off by matching factors — one multiplied by whatever factor the other is divided by — are called 'inversely proportional'.
The figure shows this trade-off for one sample: at 4.0 L the pressure is 6.0 atm, and at 8.0 L it is 3.0 atm.
Multiply each pair: 6.0 × 4.0 = 24, and 3.0 × 8.0 = 24 — the product of pressure and volume stays constant.
A constant product is the mark of an inverse proportion: every point on the curve gives the same product.
State that for a fixed amount of gas at constant temperature, pressure and volume are inversely proportional, so the product of pressure and volume stays constant.
Check — numeric entry
A sealed sample of gas is at a pressure of 0.60 atm. At constant temperature, its volume is halved. What is the new pressure, in atm? Give your answer to 2 significant figures.
Feedback when correct: Correct. Pressure and volume are inversely proportional, so halving the volume doubles the pressure: 2 × 0.60 = 1.2 atm.
Anticipated wrong answers (feedback shown):
| 0.30 | You halved the pressure along with the volume — that is the pattern of a direct proportion. Pressure and volume are inversely proportional: halving the volume doubles the pressure, giving 1.2 atm. |
| 0.60 | You left the pressure unchanged. In half the space the same particles strike the walls twice as often, so the pressure doubles to 1.2 atm. |
Check — numeric entry
A gas sample at 4.20 atm expands to three times its starting volume at constant temperature. What is the new pressure, in atm? Give your answer to 3 significant figures.
Feedback when correct: Correct. Tripling the volume cuts the pressure to a third: 4.20 ÷ 3 = 1.40 atm.
Anticipated wrong answers (feedback shown):
| 12.6 | You multiplied the pressure by 3 — that is the pattern of a direct proportion. Pressure and volume trade off: tripling the volume divides the pressure by 3, giving 1.40 atm. |
| 2.10 | You halved the pressure, but the volume tripled rather than doubled. Divide by the same factor the volume was multiplied by: 4.20 ÷ 3 = 1.40 atm. |
Check — multiple choice
A fixed amount of gas is held at constant temperature while a piston changes its volume. Which quantity stays constant throughout?
Feedback if chosen: This option is wrong — you described a direct proportion. A constant ratio belongs to two quantities that grow together; pressure and volume trade off, so it is their product that stays constant.
Feedback if chosen: This option is wrong — you added two quantities that trade off by factors, not by fixed amounts, so their sum changes as the piston moves. Their product is what stays constant.
Feedback if chosen: This option is wrong — you subtracted the two quantities, and their difference changes as the piston moves. Their product is what stays constant.
Feedback if chosen: Correct. Halving the volume doubles the pressure, so pressure multiplied by volume comes out the same at every point on the curve.
Outcome: Predict that heating a fixed amount of gas at constant pressure increases its volume, and explain why the gas must expand.
Carry a balloon from a cold porch into a warm room and it visibly swells — warmth alone made it bigger.
You've already seen that gas pressure is produced by particles colliding with the container walls, so more frequent or more forceful collisions mean higher pressure.
Heating the gas makes its particles move faster on average, because Kelvin temperature is proportional to the average kinetic energy of the particles.
The cylinder below shows the same change with a piston that slides freely, so the pressure inside stays matched to the steady push from outside.
If the gas could not expand, the faster particles would strike the walls harder and more often, and the pressure would rise.
Instead, the stronger push drives the piston outward, and the gas expands until its collisions are spread thinly enough to keep the pressure unchanged.
So the prediction: heating a fixed amount of gas at constant pressure increases its volume — a balloon's stretchy skin plays the part of the sliding piston.
Predict that heating a fixed amount of gas at constant pressure increases its volume, because the faster-moving particles would otherwise strike the walls harder and more often, so the gas must expand to keep the pressure unchanged.
Check — multiple choice
A sealed plastic bag of air, floppy enough to keep the pressure inside unchanged, is laid in the sun and warms up. It puffs outward. Why must its volume increase?
Feedback if chosen: This option is wrong — you resized the particles. Heating does not swell a particle; it makes the particles move faster, and the faster collisions are what drive the expansion.
Feedback if chosen: This option is wrong — you added particles. The bag is sealed, so the particle count is fixed; heating changes the particles' speeds, not their number.
Feedback if chosen: Correct. Faster particles would strike the walls harder and more often, raising the pressure — so the bag gives way and the gas expands until the collisions again match the unchanged pressure.
Feedback if chosen: This option is wrong — you reached for attractions. Gas particles are far apart and the expansion needs no attraction story: harder, more frequent collisions push the walls outward.
Check — multiple choice
An air-filled balloon is placed in a freezer, and the stretch of its skin keeps the pressure inside unchanged as it cools. What happens to the balloon's volume?
Feedback if chosen: This option is wrong — you applied the prediction backward. Cooling slows the particles, so their collisions get softer and less frequent; the gas contracts rather than expands.
Feedback if chosen: Correct. Cooling slows the particles, so they would strike the skin less hard and less often; the balloon shrinks until the collisions again match the unchanged pressure.
Feedback if chosen: This option is wrong — you froze the volume. With slower particles in an unchanged space, the collisions could not keep matching the pressure — the balloon must shrink.
Feedback if chosen: This option is wrong — you required a leak. The particle count stays fixed; slower particles alone are enough to make the gas contract at constant pressure.
Outcome: State that volume is directly proportional to Kelvin temperature for a fixed amount of gas at constant pressure, and reason with Kelvin factors.
You've already seen that heating a fixed amount of gas at constant pressure increases its volume.
Measured on the Kelvin scale, the relationship is exact: double the Kelvin temperature and the volume doubles.
'Doubling the temperature' here always means doubling the Kelvin temperature — 300 K to 600 K — never doubling a Celsius reading.
Two quantities that grow by the same factor together are called 'directly proportional'.
For a fixed amount of gas at constant pressure, volume is directly proportional to the Kelvin temperature, so volume divided by Kelvin temperature stays constant.
The figure shows one sample at constant pressure: 3.0 L at 150 K, and double that — 6.0 L — at 300 K.
Divide each pair: 3.0 ÷ 150 and 6.0 ÷ 300 give the same value — the constant behind the straight line.
On a graph, a direct proportion is a straight line through the origin.
State that for a fixed amount of gas at constant pressure, volume is directly proportional to the Kelvin temperature, so volume divided by Kelvin temperature stays constant.
Check — numeric entry
A gas sample occupies 2.50 L at 200 K. At constant pressure, it is heated to 400 K. What is the new volume, in L? Give your answer to 3 significant figures.
Feedback when correct: Correct. Volume is directly proportional to Kelvin temperature, so doubling 200 K to 400 K doubles the volume: 2 × 2.50 = 5.00 L.
Anticipated wrong answers (feedback shown):
| 1.25 | You halved the volume — that is the inverse-proportion pattern, which belongs to pressure and volume. Volume and Kelvin temperature grow together: doubling the Kelvin temperature doubles the volume to 5.00 L. |
| 2.50 | You left the volume unchanged. Volume divided by Kelvin temperature stays constant, so a doubled Kelvin temperature must bring a doubled volume: 5.00 L. |
Check — numeric entry
A gas sample occupies 9.0 L at 600 K. At constant pressure, it is cooled to 200 K. What is the new volume, in L? Give your answer to 2 significant figures.
Feedback when correct: Correct. The Kelvin temperature falls to a third (600 K to 200 K), so the volume falls to a third: 9.0 ÷ 3 = 3.0 L.
Anticipated wrong answers (feedback shown):
| 27 | You multiplied by 3 while the Kelvin temperature was divided by 3. Volume follows the Kelvin temperature's factor in the same direction: 9.0 ÷ 3 = 3.0 L. |
| 4.5 | You halved the volume, but the Kelvin temperature fell to a third, not to a half. Match the factor exactly: 9.0 ÷ 3 = 3.0 L. |
Check — multiple choice
A balloon of air at 20 °C is warmed to 40 °C at constant pressure. What happens to its volume?
Feedback if chosen: This option is wrong — you doubled a Celsius reading. The proportion uses Kelvin temperatures: 20 °C is 293 K and 40 °C is 313 K, so the Kelvin temperature grows by only a few percent.
Feedback if chosen: This option is wrong — you shrank a warming gas. Heating at constant pressure always grows the volume; the only question is by what Kelvin factor.
Feedback if chosen: Correct. Converted to kelvin, 293 K rises to 313 K — a factor of about 1.07 — so the volume grows by about 7 percent, nowhere near double.
Feedback if chosen: This option is wrong — you ignored the temperature change. Volume divided by Kelvin temperature stays constant, so a higher Kelvin temperature must bring a larger volume.
Outcome: Predict that heating a fixed amount of gas at constant volume increases its pressure, and explain why through collisions.
Aerosol cans carry a warning against strong sun, and the danger is the gas trapped inside the rigid, sealed can.
Heating the trapped gas makes its particles move faster on average, because Kelvin temperature is proportional to the average kinetic energy of the particles.
The cylinder below shows a fixed amount of gas heated with its piston locked in place, so the volume cannot change.
The faster particles strike the walls harder, and they cross the container more quickly, so they also strike more often.
The pressure rises, because gas pressure is produced by particles colliding with the container walls, so more frequent or more forceful collisions mean higher pressure.
Nothing else changed — same particles, same space — so heating alone drove the pressure up.
Predict that heating a fixed amount of gas at constant volume increases its pressure, because the faster-moving particles strike the walls harder and more often.
Check — multiple choice
After a long drive, the air inside a car's tires is much warmer, and each tire's volume has barely changed. Why is the pressure inside higher than before the drive?
Feedback if chosen: Correct. Heating raises the particles' average speed, so each wall collision is harder and the collisions come more often — more frequent or more forceful collisions mean higher pressure.
Feedback if chosen: This option is wrong — you changed the amount of gas. The tire is sealed, so the particle count is fixed; the extra pressure comes from the same particles moving faster.
Feedback if chosen: This option is wrong — you resized the particles. Heating does not swell a particle; it raises the particles' speeds, and the faster collisions raise the pressure.
Feedback if chosen: This option is wrong — you turned attractions into a push. Attractions act between particles, not on the walls; the pressure comes from particles colliding with the walls.
Check — multiple choice
A sealed, rigid flask of gas is moved from a warm room into an ice bath and left to cool. What happens to the pressure inside?
Feedback if chosen: This option is wrong — you applied the prediction backward. Cooling slows the particles, so their wall collisions get softer and less frequent, and the pressure falls.
Feedback if chosen: This option is wrong — you kept the pressure fixed. In a rigid flask nothing can give way, so slower particles must mean softer, less frequent wall collisions — a lower pressure.
Feedback if chosen: This option is wrong — you emptied the gas of motion. At the ice bath's temperature the particles still move quickly; the pressure falls, but it stays well above zero.
Feedback if chosen: Correct. Cooling slows the particles, so they strike the walls less hard and less often, and the pressure falls.
Check — multiple choice
Two identical samples of gas are warmed by the same amount. Sample 1 sits in a sealed, rigid flask; sample 2 sits in a cylinder whose piston slides freely against a steady outside push. Which statement is correct?
Feedback if chosen: This option is wrong — you let the cylinder's pressure rise, but its piston slides outward until the push from inside again matches the steady push from outside; its pressure ends unchanged.
Feedback if chosen: This option is wrong — you let the rigid flask expand. Its walls fix the volume, so the flask's response to the faster particles is a higher pressure, not a larger volume.
Feedback if chosen: Correct. What cannot change decides what does: the rigid flask fixes the volume, so its pressure rises; the free piston keeps the pressure matched, so that gas expands instead.
Feedback if chosen: This option is wrong — you froze both samples. A fixed amount of gas still responds to heating: its particles speed up, and either the pressure or the volume must rise.
Outcome: State that pressure is directly proportional to Kelvin temperature for a fixed amount of gas at constant volume, and reason with Kelvin factors.
You've already seen that heating a fixed amount of gas at constant volume increases its pressure.
On the Kelvin scale this relationship is exact too: doubling the Kelvin temperature doubles the pressure.
As always, 'doubling the temperature' means the Kelvin reading doubles — 250 K to 500 K — not the Celsius reading.
For a fixed amount of gas at constant volume, pressure is directly proportional to the Kelvin temperature, so pressure divided by Kelvin temperature stays constant.
The figure shows one sealed, rigid sample: 0.50 atm at 200 K and 1.00 atm at 400 K — pressure divided by Kelvin temperature is the same at both points.
Because the proportion uses Kelvin temperatures, convert any Celsius reading to kelvin before comparing factors: K = °C + 273.
State that for a fixed amount of gas at constant volume, pressure is directly proportional to the Kelvin temperature, so pressure divided by Kelvin temperature stays constant.
Check — numeric entry
A sealed, rigid tank holds gas at 0.60 atm and 250 K. It is heated to 750 K. What is the new pressure, in atm? Give your answer to 2 significant figures.
Feedback when correct: Correct. The Kelvin temperature triples (250 K to 750 K), so the pressure triples: 3 × 0.60 = 1.8 atm.
Anticipated wrong answers (feedback shown):
| 0.20 | You divided by 3 — that is the inverse-proportion pattern, which belongs to pressure and volume. Pressure and Kelvin temperature grow together: tripling 250 K triples the pressure to 1.8 atm. |
| 0.60 | You left the pressure unchanged. Pressure divided by Kelvin temperature stays constant, so a tripled Kelvin temperature must bring a tripled pressure: 1.8 atm. |
Check — numeric entry
A sealed, rigid flask holds gas at 1.20 atm and 27 °C. It is heated to 327 °C. What is the new pressure, in atm? Give your answer to 3 significant figures.
Feedback when correct: Correct. Convert to kelvin first: 27 + 273 = 300 K and 327 + 273 = 600 K. The Kelvin temperature doubles, so the pressure doubles: 2 × 1.20 = 2.40 atm.
Anticipated wrong answers (feedback shown):
| 14.5 | You compared the Celsius readings (327 ÷ 27 is about 12). The proportion uses Kelvin temperatures: convert first, and 300 K to 600 K is a factor of 2, so the pressure is 2.40 atm. |
| 0.60 | You halved the pressure — that is the inverse-proportion pattern, which belongs to pressure and volume. Here the Kelvin temperature doubles (300 K to 600 K), so the pressure doubles to 2.40 atm. |
Check — multiple choice
The gas in a sealed, rigid can sits at 27 °C. Left near a heater, it warms to 54 °C. What happens to the pressure inside?
Feedback if chosen: This option is wrong — you doubled a Celsius reading. Convert to kelvin first: 27 °C is 300 K and 54 °C is 327 K, so the Kelvin temperature rises by far less than a factor of 2.
Feedback if chosen: Correct. In kelvin, 300 K rises to 327 K — a factor of about 1.09 — so the pressure rises by about 9 percent, nowhere near double.
Feedback if chosen: This option is wrong — you ignored the temperature change. Pressure divided by Kelvin temperature stays constant, so a higher Kelvin temperature must bring a higher pressure.
Feedback if chosen: This option is wrong — you reversed the direction. Warming speeds the particles up, so they strike the walls harder and more often, and the pressure rises.
Outcome: Predict that adding gas particles to a sample at constant temperature and pressure increases its volume, and explain why the gas must expand.
Each breath you blow into a balloon adds more air particles to the sample inside, and the balloon grows.
You've already seen that gas pressure is produced by particles colliding with the container walls, so more frequent or more forceful collisions mean higher pressure.
The cylinder below shows the same change with a sliding piston: gas is added at constant temperature, and the piston rises.
With more particles moving in the same space, the walls would be struck more often, and the pressure would rise.
The free piston — or the stretchy balloon skin — gives way instead: the gas expands until the collisions are spread thinly enough to keep the pressure unchanged.
Adding particles does not change how fast any particle moves — the temperature sets that, and it has not changed.
So the prediction: adding more gas at constant temperature and pressure increases the volume.
Predict that adding more gas particles to a sample at constant temperature and pressure increases its volume, because more particles striking the walls would otherwise raise the pressure, so the gas must expand.
Check — multiple choice
An air mattress is pumped up at constant temperature, its floppy walls keeping the pressure inside steady. Each pump stroke pushes more air in, and the mattress swells. Why does the volume grow?
Feedback if chosen: This option is wrong — you swapped in the temperature mechanism. The temperature is constant here; what changes with each stroke is the number of particles inside.
Feedback if chosen: This option is wrong — you gave the new particles special speeds. At one temperature the particles share one average speed; what changes is how many of them there are.
Feedback if chosen: This option is wrong — you resized the particles. Particles do not swell; there are simply more of them, striking the walls more often.
Feedback if chosen: Correct. More particles means more wall collisions, which would raise the pressure — so the walls give way and the gas expands until the collisions again match the steady pressure.
Check — multiple choice
A balloon with a slow leak loses air overnight at constant temperature, and the stretch of its skin keeps the pressure inside unchanged. What happens to the balloon's volume?
Feedback if chosen: Correct. With fewer particles, the walls would be struck less often, so the skin squeezes the gas into a smaller volume until the collisions again match the unchanged pressure.
Feedback if chosen: This option is wrong — you applied the prediction backward. Removing particles is the reverse of adding them: fewer particles at constant temperature and pressure means a smaller volume.
Feedback if chosen: This option is wrong — you dropped the pressure along with the volume. The stretch of the skin holds the pressure steady while the balloon shrinks; only the volume changes.
Feedback if chosen: This option is wrong — you let the pressure fall, but the balloon's skin keeps the pressure unchanged; the volume is what gives way as particles are lost.
Outcome: State that volume is directly proportional to the amount of gas in moles at constant temperature and pressure, and reason with factors.
You've already seen that adding gas at constant temperature and pressure expands the sample.
The relationship is exact: double the amount of gas — the number of moles — and the volume doubles.
For a gas at constant temperature and pressure, volume is directly proportional to the amount of gas in moles, so volume divided by moles stays constant.
The figure shows one gas at constant temperature and pressure: 2.0 mol occupies 60 L.
Divide: 60 ÷ 2.0 = 30, and every point on the line gives that same value of volume divided by moles.
As with every direct proportion, the graph is a straight line through the origin: no gas, no volume.
State that for a gas at constant temperature and pressure, volume is directly proportional to the amount of gas in moles, so volume divided by moles stays constant.
Check — numeric entry
At constant temperature and pressure, 0.40 mol of gas occupies 10.0 L. More gas is added until the sample holds 1.20 mol. What is the new volume, in L? Give your answer to 3 significant figures.
Feedback when correct: Correct. The amount of gas triples (0.40 mol to 1.20 mol), so the volume triples: 3 × 10.0 = 30.0 L.
Anticipated wrong answers (feedback shown):
| 3.33 | You divided by 3 — that is the inverse-proportion pattern, which belongs to pressure and volume. Volume and amount of gas grow together: tripling the moles triples the volume to 30.0 L. |
| 10.0 | You left the volume unchanged. Volume divided by moles stays constant, so a tripled amount of gas must bring a tripled volume: 30.0 L. |
Check — numeric entry
A balloon holds 0.90 mol of air in a volume of 21.0 L. At constant temperature and pressure, gas is released until 0.30 mol remains. What is the new volume, in L? Give your answer to 3 significant figures.
Feedback when correct: Correct. The amount of gas falls to a third (0.90 mol to 0.30 mol), so the volume falls to a third: 21.0 ÷ 3 = 7.00 L.
Anticipated wrong answers (feedback shown):
| 63.0 | You multiplied by 3 while the amount of gas was divided by 3. Volume follows the amount's factor in the same direction: 21.0 ÷ 3 = 7.00 L. |
| 10.5 | You halved the volume, but the amount of gas fell to a third, not to a half. Match the factor exactly: 21.0 ÷ 3 = 7.00 L. |
Check — multiple choice
More gas is added to a sample at constant temperature and pressure. Which quantity stays the same as the sample grows?
Feedback if chosen: Correct. Volume and amount of gas grow by the same factor together, so their quotient — volume divided by moles — comes out the same at every point.
Feedback if chosen: This option is wrong — you described an inverse proportion. A constant product belongs to two quantities that trade off, the way pressure and volume do; here the two grow together.
Feedback if chosen: This option is wrong — you fixed the volume, but the volume is exactly what grows as particles are added at constant temperature and pressure.
Feedback if chosen: This option is wrong — you fixed the amount of gas, but the amount is what is being changed; the constant is the quotient of volume and moles.
Outcome: Identify which relationship between two gas variables a graph shows, from its axes and its shape.
You've already seen that pressure and volume are inversely proportional for a fixed amount of gas at constant temperature.
You've also seen three direct proportions: volume with Kelvin temperature, pressure with Kelvin temperature, and volume with amount of gas.
A graph of any one of these carries its identity in two places: the axis labels and the shape of the curve.
Read the axes first — they name the two variables being compared, and every other gas variable is held constant.
A curve that falls steeply and then flattens is the picture of an inverse proportion — here, pressure against volume at constant temperature.
A straight line through the origin is the picture of a direct proportion — here, pressure against Kelvin temperature at constant volume.
Shape alone cannot tell the three direct proportions apart — only the axis labels can.
So identify in two steps every time: name the two variables from the axes, then match the shape to an inverse or a direct proportion.
Identify which relationship between two gas variables a supplied graph shows — with the other variables held constant — using its axes and shape.
Check — multiple choice
The graph shows measurements of a fixed amount of gas. Which relationship does it show?
Feedback if chosen: This option is wrong — you matched a rising line to an inverse proportion. An inverse proportion falls as you move right; this line rises.
Feedback if chosen: Correct. The axes name volume and Kelvin temperature, and the straight line through the origin is the shape of a direct proportion.
Feedback if chosen: This option is wrong — you misread the horizontal axis. It is labeled as temperature in kelvin, not pressure — the axes name the variables being compared.
Feedback if chosen: This option is wrong — you gave the line a starting threshold. It passes through the origin, which is exactly what a direct proportion requires.
Check — multiple choice
The graph shows a second set of measurements of a fixed amount of gas. Which relationship does this one show?
Feedback if chosen: This option is wrong — you matched a falling curve to a direct proportion. A direct proportion rises as a straight line through the origin; this curve falls.
Feedback if chosen: Correct. The axes name pressure and volume, and the falling curve that flattens out is the shape of an inverse proportion.
Feedback if chosen: This option is wrong — you misread the horizontal axis. It is labeled as volume in liters, not Kelvin temperature — the axes name the variables being compared.
Feedback if chosen: This option is wrong — you read the fall as a straight line. The drop is steep at small volumes and shallow at large volumes: equal added volumes remove less and less pressure.
Check — multiple choice
The graph shows measurements of a gas sample. Which relationship does it show?
Feedback if chosen: This option is wrong — you matched a rising line to an inverse proportion. An inverse proportion falls as you move right; this line rises with the amount of gas.
Feedback if chosen: This option is wrong — you misread the vertical axis. It is labeled as volume in liters, not pressure — the axes name the variables being compared.
Feedback if chosen: Correct. The axes name volume and amount of gas in moles, and the straight line through the origin is the shape of a direct proportion.
Feedback if chosen: This option is wrong — you flattened the line. A constant volume would plot as a horizontal line; this line climbs steadily as gas is added.
Check — multiple choice
The graph shows the pressure of a gas sample against its Kelvin temperature. For the graph to show this single relationship, which gas variables must be held constant?
Feedback if chosen: This option is wrong — you left the volume free to change. If the volume also varied, many different pressures could pair with one temperature; the volume must be held constant along with the amount of gas.
Feedback if chosen: This option is wrong — you held one of the plotted variables fixed. Kelvin temperature is on an axis, so it is one of the two variables allowed to change.
Feedback if chosen: This option is wrong — you let everything vary. If the volume or the amount of gas also changed, many different pressures could pair with one temperature and no single line would appear.
Feedback if chosen: Correct. The two axis variables are the ones changing; the remaining gas variables — volume and amount of gas — must be held constant for one clean line to appear.
Outcome: Sketch the graph of a specified gas relationship with labeled axes, including the dashed extrapolation of a volume-temperature graph back to 0 K.
You've already seen how to identify a gas relationship from a graph's axes and shape — sketching applies the same two steps in reverse.
Start every sketch by drawing and labeling both axes with the variable names — an unlabeled sketch shows nothing.
For a direct proportion, draw a straight line that passes through the origin.
For volume against Kelvin temperature, real measurements exist only where the sample is still a gas, so draw the line solid there and continue it back to 0 K as a dashed line — the dashes reach the origin.
The dashed stretch says: if the sample stayed a gas all the way down, its volume would shrink to zero at 0 K.
For the inverse proportion, draw a curve that falls steeply at small volumes and flattens at large volumes, approaching each axis without touching it.
A straight falling line is the wrong picture for an inverse proportion — equal added volumes do not remove equal amounts of pressure.
Sketch the graph of a specified relationship between two gas variables, with the other variables held constant and both axes labeled.
Check — drawing task (pen and paper, self-marked)
Take a pen and paper and draw the graph of volume against Kelvin temperature for a fixed amount of gas at constant pressure. Label both axes, and use a dashed line to continue the graph from the measured region back to 0 K. When you are done, select Check to compare your drawing with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
Check — drawing task (pen and paper, self-marked)
Take a pen and paper and draw the graph of pressure against volume for a fixed amount of gas at constant temperature, labeling both axes. When you are done, select Check to compare your drawing with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
Check — multiple choice
A student's sketch of volume against Kelvin temperature for a fixed amount of gas at constant pressure is shown. Which feature of the sketch does not match the relationship?
Feedback if chosen: This option is wrong — you flagged the direction, but a direct proportion does rise: a higher Kelvin temperature brings a larger volume.
Feedback if chosen: This option is wrong — you flagged the straightness, but a direct proportion is a straight line; that part of the sketch is right.
Feedback if chosen: Correct. Volume divided by Kelvin temperature stays constant only for a line through the origin — a sketch that meets the volume axis above zero shows a different relationship.
Feedback if chosen: This option is wrong — you flagged the axis choice, but plotting volume vertically against temperature horizontally is the standard layout.
Check — multiple choice
A student's sketch of pressure against volume for a fixed amount of gas at constant temperature is shown. Which feature of the sketch does not match the relationship?
Feedback if chosen: This option is wrong — you flagged the direction, but an inverse proportion does fall: increasing the volume decreases the pressure.
Feedback if chosen: This option is wrong — you flagged a correct feature: with an inverse proportion, the smallest volume carries the highest pressure.
Feedback if chosen: This option is wrong — you flagged the axis choice, but plotting pressure vertically against volume horizontally is the standard layout.
Feedback if chosen: Correct. Pressure and volume are inversely proportional, so the graph must curve: doubling a small volume removes a lot of pressure, while doubling a large volume removes far less per liter.
3 lessons
Outcome: State the ideal gas law PV = nRT, say what each variable measures, and match the units of pressure, volume, and temperature to the value of the gas constant R.
The sealed cylinder in the figure holds 0.500 mol of gas at 300 K, filling a volume of 12.3 L.
What pressure does the gas exert on the cylinder walls?
You've already seen that for a fixed amount of gas at constant temperature, as the volume decreases, the pressure increases — the two are inversely proportional.
You've already seen that with everything else held constant, raising the Kelvin temperature raises the volume, and raising the Kelvin temperature raises the pressure, each in direct proportion.
You've already seen that adding moles of gas at constant temperature and pressure raises the volume in direct proportion.
None of those relationships answers the question here, because nothing is changing — this is a single state, with three quantities known and the pressure unknown.
Answering it takes one equation that ties all four quantities together at the same time.
That equation is the ideal gas law: PV = nRT.
P is the pressure of the gas, and V is the volume it fills.
n is the amount of gas in moles, and T is its Kelvin temperature.
R is the gas constant — the fixed number that makes the two sides of the equation match.
R has the same value for every gas: helium, nitrogen, and carbon dioxide all follow PV = nRT with the same R.
In this course, R = 0.08206 L·atm·mol⁻¹·K⁻¹.
The relationships you've already seen all live inside this one equation.
Hold n and T constant, and the right-hand side nRT is fixed — so when the volume falls, the pressure must rise to keep the product PV the same.
Hold n and P constant, and the equation forces the volume to rise in step with the Kelvin temperature — the direct proportion you've already seen.
For the cylinder you saw first, three of the four quantities are known — the equation pins the pressure to one definite value.
The units inside R's value dictate the units of every quantity in the equation.
With R = 0.08206 L·atm·mol⁻¹·K⁻¹, pressure must be in atmospheres, volume in liters, and temperature in kelvin.
You've already seen how to convert a Celsius temperature: K = degrees Celsius + 273.
A quantity in any other unit — a volume in milliliters, a pressure in kilopascals — must be converted to match R before it enters the equation.
State the ideal gas law PV = nRT, say what each variable measures, and match the units of pressure, volume, and temperature to the value of the gas constant R.
Check — multiple choice
In the ideal gas law PV = nRT, what does n measure?
Feedback if chosen: This option is wrong — you read n as a raw particle count. n is already the amount in moles; a particle count would have to be divided by Avogadro's number to give n, never multiplied.
Feedback if chosen: Correct. n is the amount of gas, measured in moles.
Feedback if chosen: This option is wrong — you read n as the sample's mass. The equation takes the amount of gas in moles, not a mass in grams.
Feedback if chosen: This option is wrong — you read n as the molar mass. The identity of the gas never enters the equation — only how much of it is present, in moles.
Check — multiple choice
A student is using PV = nRT with R = 0.08206 L·atm·mol⁻¹·K⁻¹. Which set of units must pressure, volume, and temperature carry?
Feedback if chosen: Correct. R carries L·atm·mol⁻¹·K⁻¹, so pressure must be in atmospheres, volume in liters, and temperature in kelvin.
Feedback if chosen: This option is wrong — you matched the pressure and temperature but not the volume. R carries liters, so a volume in milliliters must be converted to liters first.
Feedback if chosen: This option is wrong — you matched the volume and temperature but not the pressure. R carries atmospheres, so a pressure in kilopascals must be converted to atmospheres first.
Feedback if chosen: This option is wrong — you matched the pressure and volume but not the temperature. R carries kelvin, so a Celsius temperature must be converted: K = degrees Celsius + 273.
Check — multiple choice
Which one of these measured values can be substituted into PV = nRT with R = 0.08206 L·atm·mol⁻¹·K⁻¹ without being converted first?
Feedback if chosen: This option is wrong — you passed a Celsius temperature straight into the equation. R carries kelvin, so convert first: 25 + 273 = 298 K.
Feedback if chosen: This option is wrong — you passed a volume in milliliters straight into the equation. R carries liters, so 250 mL must be converted to liters first.
Feedback if chosen: Correct. R carries kelvin, and 298 K is already a Kelvin temperature — it can be substituted as it stands.
Feedback if chosen: This option is wrong — you passed a pressure in kilopascals straight into the equation. R carries atmospheres, so the pressure must be converted to atmospheres first.
Outcome: Calculate any one of pressure, volume, moles, and temperature for a single gas state from the other three, using PV = nRT.
You've already seen the ideal gas law, PV = nRT, and matched its units to the gas constant R = 0.08206 L·atm·mol⁻¹·K⁻¹.
Real problems give you three of the four measurements and ask for the fourth.
One equation covers every one of these calculations: PV = nRT.
There is no separate equation for finding volume, moles, or temperature.
Whatever the question asks for, start from PV = nRT and rearrange it to make that quantity the subject.
Calculate any one of pressure, volume, moles, and temperature for a single gas state from the other three, using PV = nRT.
Match every value to the units of R before substituting.
With R = 0.08206 L·atm·mol⁻¹·K⁻¹, pressure must be in atmospheres, volume in liters, and temperature in kelvin.
If the question gives a temperature in degrees Celsius, convert it to kelvin first: K = degrees Celsius + 273.
The conversion is its own labeled step, done before substituting.
Q: What is the volume of 1.00 mol of an ideal gas at 500 K and 2.00 atm? (R = 0.08206 L·atm·mol⁻¹·K⁻¹)
Write down the values in the question
P = 2.00 atm
n = 1.00 mol
T = 500 K
R = 0.08206 L·atm·mol⁻¹·K⁻¹
Write down the equation
PV = nRT
Make V the subject
V = nRT / P
Substitute in the values, and calculate
V = (1.00 × 0.08206 × 500) / 2.00
V = 41.03 / 2.00
V = 20.5 L
Q: A 8.206 L container holds an ideal gas at 27 °C and 3.00 atm. How many moles of gas are in the container? (R = 0.08206 L·atm·mol⁻¹·K⁻¹)
Write down the values in the question
P = 3.00 atm
V = 8.206 L
T = 27 °C
R = 0.08206 L·atm·mol⁻¹·K⁻¹
Convert the temperature to kelvin
T = 27 + 273
T = 300 K
Write down the equation
PV = nRT
Make n the subject
n = PV / (RT)
Substitute in the values, and calculate
n = (3.00 × 8.206) / (0.08206 × 300)
n = 24.618 / 24.618
n = 1.00 mol
Check — numeric entry
What volume, in liters, does 3.00 mol of an ideal gas occupy at 200 K and 1.00 atm? (R = 0.08206 L·atm·mol⁻¹·K⁻¹.) Give your answer to 3 significant figures.
Tolerance: 0.05
Feedback when correct: Correct. Rearranging PV = nRT to V = nRT / P and substituting the three given values gives 49.2 L.
Anticipated wrong answers (feedback shown):
| 0.00123 | You divided by the temperature instead of multiplying by it. Making V the subject gives V = nRT / P — the temperature belongs in the numerator. |
| 600 | You left the gas constant R out of the calculation. V = nRT / P, and R = 0.08206 L·atm·mol⁻¹·K⁻¹ must be part of the product. |
Worked solution revealed after a second miss
Write down the values in the question
P = 1.00 atm
n = 3.00 mol
T = 200 K
Write down the equation
PV = nRT
Make V the subject
V = nRT / P
Substitute in the values, and calculate
V = (3.00 × 0.08206 × 200) / 1.00
V = 49.2 L
Check — numeric entry
A 10.0 L container holds 1.00 mol of an ideal gas at 227 °C. What is the pressure, in atmospheres, inside the container? (R = 0.08206 L·atm·mol⁻¹·K⁻¹.) Give your answer to 3 significant figures.
Tolerance: 0.005
Feedback when correct: Correct. Converting 227 °C to 500 K first, then rearranging PV = nRT to P = nRT / V, gives 4.10 atm.
Anticipated wrong answers (feedback shown):
| 1.86 | You substituted the Celsius temperature without converting it to kelvin. Convert first: T = 227 + 273 = 500 K, as its own step before substituting. |
| 410 | You multiplied by the volume instead of dividing by it. Making P the subject gives P = nRT / V — the volume belongs in the denominator. |
Worked solution revealed after a second miss
Write down the values in the question
V = 10.0 L
n = 1.00 mol
T = 227 °C
Convert the temperature to kelvin
T = 227 + 273
T = 500 K
Write down the equation
PV = nRT
Make P the subject
P = nRT / V
Substitute in the values, and calculate
P = (1.00 × 0.08206 × 500) / 10.0
P = 4.10 atm
Check — numeric entry
A sample of 0.500 mol of an ideal gas occupies 8.206 L at a pressure of 2.00 atm. What is the temperature of the gas, in kelvin? (R = 0.08206 L·atm·mol⁻¹·K⁻¹.) Give your answer to 3 significant figures.
Tolerance: 0.5
Feedback when correct: Correct. Rearranging PV = nRT to T = PV / (nR) and substituting gives 400 K.
Anticipated wrong answers (feedback shown):
| 0.0025 | You divided nR by PV instead of PV by nR. Making T the subject gives T = PV / (nR) — the pressure and volume belong in the numerator. |
| 200 | You left the moles out of the denominator. Making T the subject gives T = PV / (nR) — both n and R divide the product PV. |
Worked solution revealed after a second miss
Write down the values in the question
P = 2.00 atm
V = 8.206 L
n = 0.500 mol
Write down the equation
PV = nRT
Make T the subject
T = PV / (nR)
Substitute in the values, and calculate
T = (2.00 × 8.206) / (0.500 × 0.08206)
T = 16.412 / 0.04103
T = 400 K
Outcome: Calculate the new pressure, volume, or temperature of a changing gas sample: write PV = nRT for each state, cancel the quantities that do not change, and solve for the one that did.
You've already seen how to calculate any one unknown for a single gas state from the other three, using PV = nRT.
A different kind of question describes one gas sample measured twice: once before a change, and once after.
In the figure, a piston squeezes a sealed cylinder of gas to half of its volume while the temperature is held constant.
If the gas started at 1.20 atm, what exactly is its pressure after the squeeze?
Write PV = nRT twice — once for the state before the change, and once for the state after.
Subscript 1 marks the quantities before the change: P₁V₁ = n₁RT₁.
Subscript 2 marks the quantities after: P₂V₂ = n₂RT₂.
There is no new equation to memorize for a changing sample — both lines are the same ideal gas law.
The wording of the question tells you which quantities do not change.
'Sealed' means no gas enters or leaves, so n is unchanged.
'Rigid' means the container's volume cannot change, so V is unchanged.
R is a constant, so it is always unchanged.
Every unchanged quantity appears identically in both equations — so it cancels, leaving an equation containing only the quantities that change.
Q: A sealed cylinder holds gas at 1.20 atm in a volume of 6.00 L. A piston squeezes the gas to 3.00 L while the temperature is held constant. What is the new pressure?
Write down the values in the question
P₁ = 1.20 atm
V₁ = 6.00 L
V₂ = 3.00 L
n and T are unchanged(sealed; constant temperature)
Write PV = nRT for each state
P₁V₁ = n₁RT₁(state 1)
P₂V₂ = n₂RT₂(state 2)
Cancel the quantities that do not change
n₁ = n₂, T₁ = T₂, and R is constant
nRT is the same in both states
P₁V₁ = P₂V₂(both equal nRT)
Make P₂ the subject
P₂ = P₁V₁ / V₂
Substitute in the values, and calculate
P₂ = (1.20 × 6.00) / 3.00
P₂ = 7.20 / 3.00
P₂ = 2.40 atm
Q: A sealed, rigid 5.00 L container holds gas at 1.10 atm and 300 K. The container is heated to 600 K. What is the new pressure?
Write down the values in the question
P₁ = 1.10 atm
T₁ = 300 K
T₂ = 600 K
n and V are unchanged(sealed; rigid)
Write PV = nRT for each state
P₁V₁ = n₁RT₁(state 1)
P₂V₂ = n₂RT₂(state 2)
Cancel the quantities that do not change
n₁ = n₂, V₁ = V₂, and R is constant
P₁/T₁ = nR/V(state 1, dividing both sides by T₁ and V)
P₂/T₂ = nR/V(state 2, dividing both sides by T₂ and V)
P₁/T₁ = P₂/T₂(both equal nR/V)
Make P₂ the subject
P₂ = P₁T₂ / T₁
Substitute in the values, and calculate
P₂ = (1.10 × 600) / 300
P₂ = 660 / 300
P₂ = 2.20 atm
Sometimes two quantities change at once — a gas can be compressed and heated together.
The method does not change: write PV = nRT for each state, cancel whatever is unchanged, and keep every quantity that changes.
If only n and R cancel, the equation that remains is P₁V₁/T₁ = P₂V₂/T₂ — each side equals nR.
Temperatures in these equations are always kelvin — you've already seen the conversion, K = degrees Celsius + 273, and it is done as its own step before substituting.
Calculate the new pressure, volume, or temperature of a changing gas sample by writing PV = nRT for each state, canceling the unchanged quantities, and solving for the quantity the question asks for.
Check — multiple choice
A sealed, rigid steel tank of nitrogen sits in the sun and warms from 280 K to 310 K. Which quantities in PV = nRT stay the same while it warms?
Feedback if chosen: This option is wrong — you picked the two quantities that change. The tank warms, so T changes, and as the gas warms in a fixed volume, its pressure rises, so P changes too.
Feedback if chosen: This option is wrong — you are right that 'sealed' fixes n, but the temperature does not stay the same: the tank warms from 280 K to 310 K.
Feedback if chosen: This option is wrong — you are right that 'rigid' fixes V, but the pressure does not stay the same: as the gas warms in a fixed volume, its pressure rises. That new pressure is what you would calculate.
Feedback if chosen: Correct. 'Sealed' means no gas enters or leaves, so n is fixed; 'rigid' means the volume cannot change, so V is fixed. P and T are the pair that change.
Check — numeric entry
A sealed, rigid container holds argon at 2.00 atm and 400 K. It is cooled to 300 K. What is the new pressure, in atmospheres? Give your answer to 3 significant figures.
Tolerance: 0.005
Feedback when correct: Correct. With n and V unchanged, PV = nRT for each state reduces to P₁/T₁ = P₂/T₂, so P₂ = (2.00 × 300) / 400 = 1.50 atm.
Anticipated wrong answers (feedback shown):
| 2.67 | You put the temperatures in upside down. Making P₂ the subject of P₁/T₁ = P₂/T₂ gives P₂ = P₁T₂/T₁ — and cooling the gas must lower its pressure, not raise it. |
| 0.667 | You substituted the temperature drop, 100 K, instead of the final temperature. The equation compares whole states: P₂ = P₁T₂/T₁ with T₂ = 300 K. |
Worked solution revealed after a second miss
Write down the values in the question
P₁ = 2.00 atm
T₁ = 400 K
T₂ = 300 K
n and V are unchanged — sealed; rigid
Write PV = nRT for each state
P₁V₁ = n₁RT₁
P₂V₂ = n₂RT₂
Cancel the quantities that do not change
n₁ = n₂, V₁ = V₂, and R is constant
P₁/T₁ = P₂/T₂
Make P₂ the subject
P₂ = P₁T₂ / T₁
Substitute in the values, and calculate
P₂ = (2.00 × 300) / 400
P₂ = 1.50 atm
Check — numeric entry
A sample of helium fills 9.00 L at 1.00 atm. It is squeezed at constant temperature until its pressure is 3.00 atm, and no gas escapes. What is the new volume, in liters? Give your answer to 3 significant figures.
Tolerance: 0.005
Feedback when correct: Correct. With n and T unchanged, the two equations reduce to P₁V₁ = P₂V₂, so V₂ = (1.00 × 9.00) / 3.00 = 3.00 L.
Anticipated wrong answers (feedback shown):
| 27.0 | You put the pressures in upside down. Making V₂ the subject of P₁V₁ = P₂V₂ gives V₂ = P₁V₁/P₂ — and squeezing the gas to a higher pressure must shrink its volume, not triple it. |
| 0.333 | You left the starting volume out of the numerator. Making V₂ the subject gives V₂ = P₁V₁/P₂ = (1.00 × 9.00) / 3.00 = 3.00 L. |
Worked solution revealed after a second miss
Write down the values in the question
P₁ = 1.00 atm
V₁ = 9.00 L
P₂ = 3.00 atm
n and T are unchanged — no gas escapes; constant temperature
Write PV = nRT for each state
P₁V₁ = n₁RT₁
P₂V₂ = n₂RT₂
Cancel the quantities that do not change
n₁ = n₂, T₁ = T₂, and R is constant
P₁V₁ = P₂V₂
Make V₂ the subject
V₂ = P₁V₁ / P₂
Substitute in the values, and calculate
V₂ = (1.00 × 9.00) / 3.00
V₂ = 3.00 L
Check — numeric entry
A weather balloon holds 6.00 L of helium at 27 °C and 1.00 atm. It rises to an altitude where the pressure is 0.250 atm and the temperature is −53 °C, and no gas escapes. What is the new volume, in liters? Give your answer to 3 significant figures.
Tolerance: 0.05
Feedback when correct: Correct. Converting both temperatures to kelvin first, canceling n and R leaves P₁V₁/T₁ = P₂V₂/T₂, so V₂ = (1.00 × 6.00 × 220) / (300 × 0.250) = 17.6 L.
Anticipated wrong answers (feedback shown):
| -47.1 | You substituted the Celsius temperatures without converting them to kelvin — the negative volume is the giveaway. Convert first, each as its own step: T₁ = 27 + 273 = 300 K and T₂ = −53 + 273 = 220 K. |
| 1.10 | You put the pressure ratio in upside down. Making V₂ the subject of P₁V₁/T₁ = P₂V₂/T₂ gives V₂ = P₁V₁T₂ / (T₁P₂) — and a falling outside pressure must let the balloon expand, not shrink. |
| 24.0 | You left the temperature change out. Both the pressure and the temperature change here, so both belong in P₁V₁/T₁ = P₂V₂/T₂; the cooling pulls the volume back down from 24.0 L to 17.6 L. |
| 4.40 | You left the pressure change out. Both the pressure and the temperature change here, so both belong in P₁V₁/T₁ = P₂V₂/T₂; the falling pressure lets the balloon expand well past 4.40 L. |
Worked solution revealed after a second miss
Write down the values in the question
P₁ = 1.00 atm
V₁ = 6.00 L
T₁ = 27 °C
P₂ = 0.250 atm
T₂ = −53 °C
n is unchanged — no gas escapes
Convert the temperatures to kelvin
T₁ = 27 + 273 = 300 K
T₂ = −53 + 273 = 220 K
Write PV = nRT for each state
P₁V₁ = n₁RT₁
P₂V₂ = n₂RT₂
Cancel the quantities that do not change
n₁ = n₂ and R is constant
P₁V₁/T₁ = P₂V₂/T₂
Make V₂ the subject
V₂ = P₁V₁T₂ / (T₁P₂)
Substitute in the values, and calculate
V₂ = (1.00 × 6.00 × 220) / (300 × 0.250)
V₂ = 1320 / 75.0
V₂ = 17.6 L
6 lessons
Outcome: State that each gas in a mixture exerts its own partial pressure — the same pressure it would exert alone in the same container at the same temperature.
A sealed container holds a mixture of two ideal gases, O₂ and N₂, at one temperature.
A gauge on the container reads 1.2 atm, because gas pressure is produced by particles colliding with the container walls, so more frequent or more forceful collisions mean higher pressure.
The gauge feels every collision the same way — it cannot tell an O₂ hit from an N₂ hit.
So one question has no obvious answer yet: how much of that 1.2 atm is the O₂'s doing?
You've already seen that ideal gas particles move independently — they neither attract nor repel one another.
Because the particles do not interact, each O₂ particle hits the walls exactly as it would if the N₂ were not there.
Picture the same container at the same temperature holding only the O₂.
The O₂'s collisions with the walls — and the pressure those collisions produce — are the same in both pictures.
So in the mixture, the O₂ exerts its own definite pressure: the pressure it would exert alone in the same container at the same temperature.
That pressure is called the O₂'s 'partial pressure'.
Every gas in a mixture exerts its own partial pressure, independently of the other gases present.
The O₂'s contribution to the 1.2 atm reading is its partial pressure — a definite quantity, even before you can calculate its value.
State that in a mixture of ideal gases, each gas exerts its own pressure — called its partial pressure — independently of the other gases present.
Check — multiple choice
A sealed container holds a mixture of helium and neon at one temperature. Which statement describes the pressure the neon exerts?
Feedback if chosen: This option is wrong — you pictured the helium blocking some of the neon's collisions with the walls. Ideal gas particles move independently, so the neon's collisions are unchanged by the helium.
Feedback if chosen: This option is wrong — you pictured collisions with the helium driving the neon into the walls more often. Ideal gas particles move independently, so the neon's push is unchanged.
Feedback if chosen: Correct. Because the particles move independently, the neon hits the walls exactly as it would with no helium present — its partial pressure is the pressure it would exert alone.
Feedback if chosen: This option is wrong — you treated pressure as something only the whole mixture can exert. Each gas in a mixture exerts its own definite partial pressure.
Check — multiple choice
A rigid container holds a mixture of N₂ and O₂ at one temperature. The N₂ is pumped out through a valve, leaving the O₂ behind, and the temperature and the volume do not change. What happens to the pressure exerted by the O₂?
Feedback if chosen: Correct. The O₂ particles never depended on the N₂ — they already moved independently, so their collisions with the walls, and the pressure those collisions produce, are unchanged.
Feedback if chosen: This option is wrong — you pictured the O₂ gaining room to move and hitting the walls more often. The O₂ already filled the whole container, so removing the N₂ changes nothing about its motion.
Feedback if chosen: This option is wrong — you counted the loss of the N₂ against the O₂. The container's overall push falls, but the pressure the O₂ itself exerts is unchanged.
Feedback if chosen: This option is wrong — you treated pressure as something only the whole mixture can exert. The O₂ exerts its own partial pressure, with or without the N₂.
Outcome: Calculate the mole fraction of a gas in a mixture by dividing its moles by the total moles of gas present.
A container holds 1.0 mol of N₂ mixed with 3.0 mol of O₂ — one particle in every four is N₂.
'1.0 mol of N₂' describes the N₂ by itself, but it does not tell you how much of the mixture the N₂ makes up.
In 4.0 mol of gas, that 1.0 mol is a quarter of the particles; in 100.0 mol of gas, the same 1.0 mol would be a tiny sliver.
What's missing is a number for one gas's share of the mixture's particles.
To measure the share, divide the moles of that gas by the total moles of gas in the mixture.
Here is that ratio written as an equation, for the N₂ in this mixture.
For this mixture, X(N₂) = 1.0 / 4.0 = 0.25 — a quarter, matching the picture.
This share is called the 'mole fraction' of the N₂.
A mole fraction has no units — it is moles divided by moles, so the units cancel.
A mole fraction is never smaller than 0 and never larger than 1.
Because the gases in a mixture together account for every particle present, their mole fractions add up to exactly 1.
Calculate the mole fraction of a gas in a mixture as the moles of that gas divided by the total moles of gas present.
Q: A container holds 1.0 mol of N₂ mixed with 3.0 mol of O₂. What is the mole fraction of the N₂?
Write down the values from the question
n(N₂) = 1.0 mol
n(O₂) = 3.0 mol
Work out the total moles
n(total) = 1.0 + 3.0
n(total) = 4.0 mol
Write down the equation
X(N₂) = n(N₂) / n(total)
Substitute in the values, and calculate
X(N₂) = 1.0 / 4.0
X(N₂) = 0.25
Q: A mixture holds 0.5 mol of CO₂, 1.5 mol of N₂, and 2.0 mol of O₂. What is the mole fraction of the O₂?
Write down the values from the question
n(CO₂) = 0.5 mol
n(N₂) = 1.5 mol
n(O₂) = 2.0 mol
Work out the total moles
n(total) = 0.5 + 1.5 + 2.0
n(total) = 4.0 mol
Write down the equation
X(O₂) = n(O₂) / n(total)
Substitute in the values, and calculate
X(O₂) = 2.0 / 4.0
X(O₂) = 0.50
Check — numeric entry
A container holds 1.0 mol of helium mixed with 4.0 mol of neon. What is the mole fraction of the helium? Give your answer to 2 decimal places.
Feedback when correct: Correct. The total is 1.0 + 4.0 = 5.0 mol, and X(He) = 1.0 / 5.0 = 0.20 — the helium is a fifth of the particles.
Anticipated wrong answers (feedback shown):
| 0.25 | You divided the helium's moles by the neon's moles instead of by the total. The total is 1.0 + 4.0 = 5.0 mol, so X(He) = 1.0 / 5.0 = 0.20. |
| 5 | You divided the total moles by the helium's moles — that ratio is upside down. A mole fraction is the gas's moles over the total: X(He) = 1.0 / 5.0 = 0.20. |
Worked solution revealed after a second miss
Write down the values from the question
n(He) = 1.0 mol
n(Ne) = 4.0 mol
Work out the total moles
n(total) = 1.0 + 4.0
n(total) = 5.0 mol
Write down the equation
X(He) = n(He) / n(total)
Substitute in the values, and calculate
X(He) = 1.0 / 5.0
X(He) = 0.20
Check — numeric entry
A mixture holds 0.6 mol of CO₂, 0.9 mol of N₂, and 1.5 mol of O₂. What is the mole fraction of the N₂? Give your answer to 2 decimal places.
Feedback when correct: Correct. The total is 0.6 + 0.9 + 1.5 = 3.0 mol, and X(N₂) = 0.9 / 3.0 = 0.30.
Anticipated wrong answers (feedback shown):
| 0.43 | You divided the N₂'s moles by the moles of the other two gases (2.1 mol) instead of by the total. The N₂'s own moles count in the total too: X(N₂) = 0.9 / 3.0 = 0.30. |
| 0.50 | You calculated the O₂'s mole fraction (1.5 / 3.0) instead of the N₂'s. The question asks for the N₂: X(N₂) = 0.9 / 3.0 = 0.30. |
Worked solution revealed after a second miss
Write down the values from the question
n(CO₂) = 0.6 mol
n(N₂) = 0.9 mol
n(O₂) = 1.5 mol
Work out the total moles
n(total) = 0.6 + 0.9 + 1.5
n(total) = 3.0 mol
Write down the equation
X(N₂) = n(N₂) / n(total)
Substitute in the values, and calculate
X(N₂) = 0.9 / 3.0
X(N₂) = 0.30
Check — numeric entry
A mixture contains only two gases, argon and neon. The mole fraction of the argon is 0.375. What is the mole fraction of the neon? Give your answer to 3 decimal places.
Feedback when correct: Correct. The two mole fractions must add up to 1, so X(Ne) = 1 − 0.375 = 0.625.
Anticipated wrong answers (feedback shown):
| 0.5 | You gave each gas an equal share without using the data. The shares must add up to 1 but need not be equal: X(Ne) = 1 − 0.375 = 0.625. |
| 0.725 | You slipped in the subtraction — check the tenths column. 1.000 − 0.375 = 0.625. |
Worked solution revealed after a second miss
Write down the values from the question
X(Ar) = 0.375
only two gases present
Write down the equation
X(Ar) + X(Ne) = 1
Make X(Ne) the subject
X(Ne) = 1 − X(Ar)
Substitute in the values, and calculate
X(Ne) = 1 − 0.375
X(Ne) = 0.625
Outcome: State that the total pressure of a gas mixture is the sum of the partial pressures of its components, and use the sum to find a total or a missing partial pressure.
You've already seen that each gas in a mixture exerts its own partial pressure, independently of the other gases present.
The figure shows a container of N₂ held at a fixed volume, before and after O₂ is pumped in at the same temperature.
Before the O₂ arrives, the N₂ alone exerts 0.5 atm.
When the O₂ is pumped in, the reading rises, because gas pressure is produced by particles colliding with the container walls, so more frequent or more forceful collisions mean higher pressure.
The O₂'s collisions land on top of the N₂'s collisions — and since the particles move independently, neither gas's own push changes.
Suppose the O₂'s partial pressure — the pressure it would exert alone in this container — is 0.3 atm.
The gauge now reads 0.5 + 0.3 = 0.8 atm.
The total pressure of a gas mixture equals the sum of the partial pressures of its components.
For this mixture, P(total) = P(N₂) + P(O₂) — and with more gases present, the sum simply gains a term for each one.
This statement is called 'Dalton's law of partial pressures'.
State that the total pressure of a gas mixture equals the sum of the partial pressures of its components.
Check — numeric entry
A container holds a mixture of N₂ and CO₂. The partial pressure of the N₂ is 1.00 atm, and the partial pressure of the CO₂ is 0.40 atm. What is the total pressure, in atm? Give your answer to 2 decimal places.
Feedback when correct: Correct. Each gas's push lands on the walls in full, so the total is the sum: P(total) = 1.00 + 0.40 = 1.40 atm.
Anticipated wrong answers (feedback shown):
| 0.60 | You subtracted the partial pressures. Both gases push outward on the same walls, so their contributions add: P(total) = 1.00 + 0.40 = 1.40 atm. |
| 0.70 | You averaged the partial pressures. Every collision counts in full, so the contributions add rather than average: P(total) = 1.00 + 0.40 = 1.40 atm. |
Worked solution revealed after a second miss
Write down the values from the question
P(N₂) = 1.00 atm
P(CO₂) = 0.40 atm
Write down the equation
P(total) = P(N₂) + P(CO₂)
Substitute in the values, and calculate
P(total) = 1.00 + 0.40
P(total) = 1.40 atm
Check — numeric entry
A sealed flask holds only helium and argon, at a total pressure of 1.50 atm. The partial pressure of the helium is 0.90 atm. What is the partial pressure of the argon, in atm? Give your answer to 2 decimal places.
Feedback when correct: Correct. The two partial pressures must add up to the total, so P(Ar) = 1.50 − 0.90 = 0.60 atm.
Anticipated wrong answers (feedback shown):
| 2.40 | You added the helium's partial pressure to the total. The total already contains the helium's contribution, so subtract it to leave the argon's: P(Ar) = 1.50 − 0.90 = 0.60 atm. |
Worked solution revealed after a second miss
Write down the values from the question
P(total) = 1.50 atm
P(He) = 0.90 atm
Write down the equation
P(total) = P(He) + P(Ar)
Make P(Ar) the subject
P(Ar) = P(total) − P(He)
Substitute in the values, and calculate
P(Ar) = 1.50 − 0.90
P(Ar) = 0.60 atm
Check — multiple choice
A rigid container holds gas A at one temperature. Gas B is pumped in, and the temperature does not change. What is the total pressure once B has been added?
Feedback if chosen: This option is wrong — you averaged the two contributions. Every collision counts in full, so the pushes add: P(total) = P(A) + P(B).
Feedback if chosen: Correct. B's particles collide with the walls on top of A's collisions, so the total pressure is the sum of the two partial pressures.
Feedback if chosen: This option is wrong — you treated a fixed volume as fixing the pressure. Pressure comes from wall collisions, and B's particles add collisions of their own, so the total rises by B's partial pressure.
Feedback if chosen: This option is wrong — you added an extra contribution from collisions between the two gases. Ideal gas particles move independently, so the total is exactly the sum of the partial pressures.
Outcome: State that a gas's partial pressure equals its mole fraction times the total pressure, and explain why a gas's share of the pressure matches its share of the particles.
You've already seen that the total pressure of a mixture is the sum of the partial pressures of its components.
You've also seen that a gas's mole fraction is its share of the mixture's particles.
Those two ideas connect: a gas's share of the total pressure equals its share of the particles.
In the mixture shown, 3 of the 12 particles — a quarter — are O₂.
Both gases share one temperature, so both have the same average kinetic energy, because Kelvin temperature is proportional to the average kinetic energy of the particles.
With equal average kinetic energies, the lighter particles hit the walls more often and the heavier particles hit harder — and on average, the two effects balance exactly.
So each particle, whichever gas it belongs to, contributes the same amount to the total pressure on average.
A quarter of the particles are O₂, so the O₂ makes a quarter of the collisions and contributes a quarter of the total pressure.
If the total pressure is 2.0 atm, the O₂ contributes 0.25 × 2.0 = 0.5 atm.
Here is the connection written as an equation, for the O₂.
A gas's partial pressure equals its mole fraction times the total pressure of the mixture.
State that a gas's partial pressure equals its mole fraction times the total pressure of the mixture.
Check — multiple choice
A mixture of helium and neon has total pressure P(total), and the mole fraction of the helium is X(He). Which expression gives the partial pressure of the helium?
Feedback if chosen: This option is wrong — you divided by the mole fraction instead of multiplying by it. A mole fraction is a share below 1, and taking a share of the total means multiplying: P(He) = X(He) × P(total).
Feedback if chosen: Correct. The helium's share of the particles is X(He), so its share of the total pressure is X(He) × P(total).
Feedback if chosen: This option is wrong — you divided the share by the pressure instead of taking the share of the pressure. The mole fraction multiplies the total: P(He) = X(He) × P(total).
Feedback if chosen: This option is wrong — you took the helium's share of the neon's partial pressure. The mole fraction takes a share of the total pressure of the whole mixture: P(He) = X(He) × P(total).
Check — multiple choice
A sealed container holds three times as many moles of N₂ as of O₂, and nothing else. How does the partial pressure of the N₂ compare with the partial pressure of the O₂?
Feedback if chosen: This option is wrong — you read 'exerts its own pressure independently' as 'exerts the same pressure'. Independence means the gases don't affect each other; the size of each push still tracks the number of particles.
Feedback if chosen: This option is wrong — you applied the factor of three twice. Partial pressure is proportional to mole fraction, so three times the moles gives three times the partial pressure.
Feedback if chosen: This option is wrong — you inverted the comparison. The N₂ has more particles, so it makes more of the collisions and exerts the larger partial pressure.
Feedback if chosen: Correct. The N₂'s mole fraction is three times the O₂'s, and each multiplies the same total pressure, so the N₂'s partial pressure is three times as large.
Check — multiple choice
A mixture of helium and argon sits at one temperature. On average, why does each particle contribute the same amount to the total pressure, whichever gas it belongs to?
Feedback if chosen: Correct. At one temperature both gases have the same average kinetic energy, because Kelvin temperature is proportional to the average kinetic energy of the particles — the lighter particles hit more often, the heavier ones hit harder, and on average the contributions match.
Feedback if chosen: This option is wrong — you gave the two gases equal molar masses. Helium and argon have different molar masses; it is their average kinetic energy that matches, not their mass.
Feedback if chosen: This option is wrong — you had collisions leveling every particle to one speed. At one temperature the gases share an average kinetic energy, so the lighter particles move faster on average, not at the same speed.
Feedback if chosen: This option is wrong — you reasoned from the space the particles fill. Ideal gas particles take up almost none of the container's volume; the pressure comes from their collisions with the walls.
Outcome: Calculate a gas's partial pressure from its mole fraction and the total pressure of the mixture.
You've already seen that a gas's partial pressure equals its mole fraction times the total pressure.
A typical question supplies a gas's mole fraction and the mixture's total pressure, and asks for that gas's partial pressure.
Multiply the mole fraction by the total pressure — and because a mole fraction has no units, the answer keeps the total pressure's unit, atm.
When a question supplies moles instead of a mole fraction, calculate the mole fraction first, as its own step.
Calculate a gas's partial pressure from its mole fraction and the total pressure of the mixture.
Q: In air at a total pressure of 1.0 atm, the mole fraction of N₂ is 0.78. What is the partial pressure of the N₂?
Write down the values from the question
X(N₂) = 0.78
P(total) = 1.0 atm
Write down the equation
P(N₂) = X(N₂) × P(total)
Substitute in the values, and calculate
P(N₂) = 0.78 × 1.0
P(N₂) = 0.78 atm
Q: A container holds 1.0 mol of helium and 3.0 mol of neon at a total pressure of 2.0 atm. What is the partial pressure of the helium?
Write down the values from the question
n(He) = 1.0 mol
n(Ne) = 3.0 mol
P(total) = 2.0 atm
Work out the mole fraction of the helium
X(He) = n(He) / n(total)
X(He) = 1.0 / 4.0 = 0.25
Write down the equation
P(He) = X(He) × P(total)
Substitute in the values, and calculate
P(He) = 0.25 × 2.0
P(He) = 0.50 atm
Check — numeric entry
A diver's breathing mixture of helium and oxygen has a total pressure of 5.0 atm, and the mole fraction of the O₂ is 0.20. What is the partial pressure of the O₂, in atm? Give your answer to 1 decimal place.
Feedback when correct: Correct. The O₂'s share of the particles is 0.20, so its share of the pressure is P(O₂) = 0.20 × 5.0 = 1.0 atm.
Anticipated wrong answers (feedback shown):
| 25 | You divided the total pressure by the mole fraction instead of multiplying. Taking the O₂'s share of the total means multiplying: P(O₂) = 0.20 × 5.0 = 1.0 atm. |
| 0.04 | You divided the mole fraction by the total pressure. The mole fraction multiplies the total: P(O₂) = 0.20 × 5.0 = 1.0 atm. |
| 4.0 | You used the helium's mole fraction (0.80) and calculated the helium's partial pressure. The question asks for the O₂: P(O₂) = 0.20 × 5.0 = 1.0 atm. |
Worked solution revealed after a second miss
Write down the values from the question
X(O₂) = 0.20
P(total) = 5.0 atm
Write down the equation
P(O₂) = X(O₂) × P(total)
Substitute in the values, and calculate
P(O₂) = 0.20 × 5.0
P(O₂) = 1.0 atm
Check — numeric entry
In a gas mixture with a total pressure of 2.4 atm, the mole fraction of the argon is 0.375. What is the partial pressure of the argon, in atm? Give your answer to 2 decimal places.
Feedback when correct: Correct. P(Ar) = 0.375 × 2.4 = 0.90 atm — a bit over a third of the total, matching the mole fraction.
Anticipated wrong answers (feedback shown):
| 6.4 | You divided the total pressure by the mole fraction instead of multiplying. Taking the argon's share means multiplying: P(Ar) = 0.375 × 2.4 = 0.90 atm. |
| 1.5 | You used the rest of the mixture's share (1 − 0.375 = 0.625) and calculated everything except the argon. The question asks for the argon: P(Ar) = 0.375 × 2.4 = 0.90 atm. |
Worked solution revealed after a second miss
Write down the values from the question
X(Ar) = 0.375
P(total) = 2.4 atm
Write down the equation
P(Ar) = X(Ar) × P(total)
Substitute in the values, and calculate
P(Ar) = 0.375 × 2.4
P(Ar) = 0.90 atm
Check — numeric entry
A container holds 0.5 mol of CO₂ and 1.5 mol of N₂ at a total pressure of 3.0 atm. What is the partial pressure of the CO₂, in atm? Give your answer to 2 decimal places.
Feedback when correct: Correct. The mole fraction comes first: X(CO₂) = 0.5 / 2.0 = 0.25, and then P(CO₂) = 0.25 × 3.0 = 0.75 atm.
Anticipated wrong answers (feedback shown):
| 0.25 | You stopped at the mole fraction. X(CO₂) = 0.25 is the CO₂'s share of the particles; its partial pressure is that share of the total: P(CO₂) = 0.25 × 3.0 = 0.75 atm. |
| 1.0 | You divided the CO₂'s moles by the N₂'s moles instead of by the total. X(CO₂) = 0.5 / 2.0 = 0.25, so P(CO₂) = 0.25 × 3.0 = 0.75 atm. |
| 2.25 | You calculated the N₂'s partial pressure (0.75 × 3.0). The question asks for the CO₂: P(CO₂) = 0.25 × 3.0 = 0.75 atm. |
Worked solution revealed after a second miss
Write down the values from the question
n(CO₂) = 0.5 mol
n(N₂) = 1.5 mol
P(total) = 3.0 atm
Work out the mole fraction of the CO₂
X(CO₂) = n(CO₂) / n(total)
X(CO₂) = 0.5 / 2.0 = 0.25
Write down the equation
P(CO₂) = X(CO₂) × P(total)
Substitute in the values, and calculate
P(CO₂) = 0.25 × 3.0
P(CO₂) = 0.75 atm
Outcome: Calculate a gas's mole fraction, or the total pressure of a mixture, from supplied partial pressures.
You've already seen that a gas's partial pressure equals its mole fraction times the total pressure.
The same equation answers the reversed question: given partial pressures, find a mole fraction or the total pressure.
To find a mole fraction, make it the subject: divide the gas's partial pressure by the total pressure.
You've already seen that the total pressure is the sum of the partial pressures — so when a question supplies every partial pressure, add them up to get the total.
Calculate a gas's mole fraction, or the total pressure of a mixture, from supplied partial pressures.
Q: In a gas mixture with a total pressure of 1.6 atm, the O₂ contributes a partial pressure of 0.4 atm. What is the mole fraction of the O₂?
Write down the values from the question
P(O₂) = 0.4 atm
P(total) = 1.6 atm
Write down the equation
P(O₂) = X(O₂) × P(total)
Make X(O₂) the subject
X(O₂) = P(O₂) / P(total)
Substitute in the values, and calculate
X(O₂) = 0.4 / 1.6
X(O₂) = 0.25
Q: A mixture of helium and argon has partial pressures P(He) = 0.9 atm and P(Ar) = 0.3 atm. What is the mole fraction of the helium?
Write down the values from the question
P(He) = 0.9 atm
P(Ar) = 0.3 atm
Work out the total pressure
P(total) = P(He) + P(Ar)
P(total) = 0.9 + 0.3 = 1.2 atm
Write down the equation
P(He) = X(He) × P(total)
Make X(He) the subject
X(He) = P(He) / P(total)
Substitute in the values, and calculate
X(He) = 0.9 / 1.2
X(He) = 0.75
Check — numeric entry
In a mixture with a total pressure of 1.60 atm, the argon contributes a partial pressure of 0.60 atm. What is the mole fraction of the argon? Give your answer to 3 decimal places.
Feedback when correct: Correct. X(Ar) = P(Ar) / P(total) = 0.60 / 1.60 = 0.375 — the argon supplies that share of the pressure because it is that share of the particles.
Anticipated wrong answers (feedback shown):
| 2.67 | You divided the total pressure by the argon's partial pressure — that ratio is upside down, and a mole fraction can never exceed 1. X(Ar) = 0.60 / 1.60 = 0.375. |
| 1.0 | You subtracted the argon's partial pressure from the total, which gives the rest of the mixture's partial pressure in atm, not a mole fraction. X(Ar) = 0.60 / 1.60 = 0.375. |
Worked solution revealed after a second miss
Write down the values from the question
P(Ar) = 0.60 atm
P(total) = 1.60 atm
Write down the equation
P(Ar) = X(Ar) × P(total)
Make X(Ar) the subject
X(Ar) = P(Ar) / P(total)
Substitute in the values, and calculate
X(Ar) = 0.60 / 1.60
X(Ar) = 0.375
Check — numeric entry
A mixture holds three gases, with partial pressures P(O₂) = 0.30 atm, P(N₂) = 1.20 atm, and P(He) = 0.50 atm. What is the mole fraction of the N₂? Give your answer to 2 decimal places.
Feedback when correct: Correct. The total is 0.30 + 1.20 + 0.50 = 2.00 atm, and X(N₂) = 1.20 / 2.00 = 0.60.
Anticipated wrong answers (feedback shown):
| 1.5 | You divided the N₂'s partial pressure by the other two gases' combined pressure (0.80 atm) instead of by the total. The N₂'s own contribution counts in the total too: X(N₂) = 1.20 / 2.00 = 0.60. |
| 0.80 | You left the helium out of the total. All three partial pressures add into the total: P(total) = 0.30 + 1.20 + 0.50 = 2.00 atm, so X(N₂) = 1.20 / 2.00 = 0.60. |
Worked solution revealed after a second miss
Write down the values from the question
P(O₂) = 0.30 atm
P(N₂) = 1.20 atm
P(He) = 0.50 atm
Work out the total pressure
P(total) = 0.30 + 1.20 + 0.50
P(total) = 2.00 atm
Write down the equation
P(N₂) = X(N₂) × P(total)
Make X(N₂) the subject
X(N₂) = P(N₂) / P(total)
Substitute in the values, and calculate
X(N₂) = 1.20 / 2.00
X(N₂) = 0.60
Check — numeric entry
Four gases make up a mixture, with partial pressures P(N₂) = 0.60 atm, P(O₂) = 0.15 atm, P(Ar) = 0.35 atm, and P(CO₂) = 0.40 atm. What is the total pressure, in atm? Give your answer to 2 decimal places.
Feedback when correct: Correct. Every gas's push lands on the walls in full, so the total is the sum: P(total) = 0.60 + 0.15 + 0.35 + 0.40 = 1.50 atm.
Anticipated wrong answers (feedback shown):
| 0.375 | You averaged the four partial pressures. Every collision counts in full, so the contributions add rather than average: P(total) = 0.60 + 0.15 + 0.35 + 0.40 = 1.50 atm. |
| 1.10 | You left the CO₂'s 0.40 atm out of the sum. Every gas present contributes: P(total) = 0.60 + 0.15 + 0.35 + 0.40 = 1.50 atm. |
Worked solution revealed after a second miss
Write down the values from the question
P(N₂) = 0.60 atm
P(O₂) = 0.15 atm
P(Ar) = 0.35 atm
P(CO₂) = 0.40 atm
Write down the equation
P(total) = P(N₂) + P(O₂) + P(Ar) + P(CO₂)
Substitute in the values, and calculate
P(total) = 0.60 + 0.15 + 0.35 + 0.40
P(total) = 1.50 atm
4 lessons
Outcome: Explain why attractions between particles make a real gas's measured pressure lower than PV = nRT predicts, and identify conditions close to condensation as where the shortfall is largest.
You've already seen the ideal gas law: PV = nRT predicts the pressure of a gas sample from its amount, volume, and temperature.
The figure puts that prediction to the test: 1.00 mol of ammonia, NH₃, held in a 4.92 L tank at 300 K.
| Pressure | Value (atm) |
|---|---|
| Predicted by PV = nRT | 5.00 |
| Measured by the gauge | about 4.7 |
PV = nRT predicts a pressure of 5.00 atm.
The gauge reads about 4.7 atm — the ammonia pushes on the walls more weakly than the prediction says.
Ammonia in a tank is a real gas, and real gases do not follow PV = nRT exactly.
Any gap between a real gas's measured values and the PV = nRT prediction is called a 'deviation' from ideal behavior.
You've already seen that the ideal gas model treats its particles as if they do not attract one another.
Real NH₃ particles do attract one another — you've already seen how to rank substances by overall intermolecular force strength, and ammonia's hydrogen bonding places it high in that ranking.
You've already seen where pressure comes from: gas pressure is produced by particles colliding with the container walls, so more frequent or more forceful collisions mean higher pressure.
Now picture one particle heading for the wall.
As it moves toward the wall, attractions to the neighboring particles behind it pull it backward.
Because of that backward pull, the particle strikes the wall more softly than it would without attractions.
Softer wall collisions across the whole sample mean a lower measured pressure.
So attractions between particles pull them together, and the measured pressure comes out lower than PV = nRT predicts.
For a gas that follows PV = nRT exactly, the combination PV/nRT works out to exactly 1 at every pressure.
For a real gas, chemists compute PV/nRT from the measured values and plot it against pressure.
The figure shows this plot for NH₃: the dashed line at 1 is the ideal gas, and the solid curve is the measured NH₃.
Where attractions lower the measured pressure, measured PV/nRT falls below 1 — that is the dip in the curve.
The deeper the dip, the larger the pressure shortfall.
At the far right the curve climbs back above 1 — that climb has a different cause.
As a gas is cooled or squeezed toward the point where it turns into a liquid, its particles move more slowly and sit closer together.
Slow, crowded particles are the ones attractions pull backward most effectively.
The change from gas to liquid is called 'condensation', and close to condensation the pressure shortfall is at its largest.
That is why NH₃ near its boiling point exerts noticeably less pressure than PV = nRT predicts.
Explain why a real gas's measured pressure is lower than PV = nRT predicts — attractions between particles pull them together — and state that the shortfall is largest close to condensation, where the gas is about to become a liquid.
Check — multiple choice
At 300 K, the measured pressure of a CO₂ sample is lower than PV = nRT predicts. What causes the shortfall?
Feedback if chosen: This option is wrong — you assumed collisions between particles drain kinetic energy from the sample. At a fixed temperature the average kinetic energy stays fixed; collisions pass energy between particles without losing it.
Feedback if chosen: This option is wrong — you turned the interparticle forces into pushes. The forces between CO₂ particles are attractions, and attractions pull an outbound particle backward, away from the wall.
Feedback if chosen: Correct. A particle heading for the wall is pulled backward by attractions to its neighbors, so it strikes the wall more softly — and softer collisions mean a lower measured pressure.
Feedback if chosen: This option is wrong — you tied reaching the walls to temperature. Particles at 300 K move at hundreds of meters per second and strike the walls constantly; the shortfall comes from attractions weakening those strikes.
Check — multiple choice
The figure shows measured PV/nRT for a CH₄ sample. In the region where the curve sits below the dashed line, what is true of the sample?
Feedback if chosen: Correct. PV/nRT below 1 means the measured pressure falls short of the prediction — attractions between the CH₄ particles weaken their wall collisions.
Feedback if chosen: This option is wrong — you read the dip backwards. A measured pressure higher than the prediction would push PV/nRT above 1, not below it.
Feedback if chosen: This option is wrong — you blamed the measurement instead of the gas. T in the ratio is the measured temperature, so a change in temperature is already accounted for; the dip is the gas deviating from the model.
Feedback if chosen: This option is wrong — you blamed a leak instead of the gas. n in the ratio is the measured amount, so a loss of gas is already accounted for; the dip is the gas deviating from the model.
Check — multiple choice
Four samples of propane, C₃H₈, are prepared. In which sample is the measured pressure furthest below the PV = nRT prediction?
Feedback if chosen: This option is wrong — you picked fast-moving, spread-out particles. Far above the boiling point, attractions barely deflect the particles, so the shortfall is small.
Feedback if chosen: Correct. Just short of condensation the particles are slow and close together, so attractions pull them together most effectively — the pressure shortfall is at its largest.
Feedback if chosen: This option is wrong — you treated expansion as increasing the deviation. Doubling the volume spreads the particles apart, so attractions act less, not more.
Feedback if chosen: This option is wrong — you treated any temperature change as increasing the deviation. Warming the sample carries it further from condensation, so the shortfall shrinks.
Outcome: Explain why the particles' own volume makes a real gas's measured volume larger than PV = nRT predicts, especially at extremely high pressures.
You've already seen one way a real gas deviates from PV = nRT: attractions between particles make the measured pressure lower than predicted.
The figure shows a different experiment: 1.00 mol of hydrogen, H₂, held at 300 K under a pressure of 100 atm.
| Volume | Value (L) |
|---|---|
| Predicted by PV = nRT | 0.246 |
| Measured | about 0.26 |
PV = nRT predicts a volume of 0.246 L.
The measured volume is about 0.26 L — larger than the prediction, not smaller.
Attractions cannot explain this deviation: pulling particles together makes measured values fall short, and this measurement overshoots.
A deviation in the opposite direction needs a different cause.
You've already seen that the ideal gas model treats its particles as points that take up no space of their own.
Real particles are not points — each H₂ particle fills a small but real volume.
Packed side by side as a liquid, the particles of one mole of H₂ fill about 0.03 L.
At ordinary pressures that 0.03 L is negligible: at 1.00 atm and 300 K, the same mole spreads through 24.6 L, almost all of it empty space.
Squeeze the sample to 1000 atm, and PV = nRT predicts just 0.0246 L — less space than the particles themselves fill.
The measured volume can never shrink below the space the particles themselves fill.
So at extremely high pressures the measured volume is larger than PV = nRT predicts: the prediction ignores the particles' own volume, and the measurement cannot.
You've already seen the deviation graph — measured PV/nRT against pressure — and the figure shows it for H₂ at 300 K.
When the measured volume is larger than the prediction, measured PV/nRT rises above 1.
The higher the pressure, the larger the particles' own volume looms as a share of the whole, so the curve climbs further above the line.
H₂'s attractions are so weak that its curve never dips below 1 — the particle-volume effect is the only one that shows.
Explain why a real gas's measured volume is larger than PV = nRT predicts — the particles themselves occupy volume — and state that the effect matters most at extremely high pressures.
Check — multiple choice
At 300 K and 500 atm, the measured volume of a helium sample is larger than PV = nRT predicts. Why?
Feedback if chosen: This option is wrong — you made attractions push the sample outward. Attractions pull particles together, and their effect is a pressure shortfall, not extra volume.
Feedback if chosen: This option is wrong — you tied particle speed to pressure. Particle speed follows temperature, and the temperature here is fixed at 300 K.
Feedback if chosen: This option is wrong — you let the particles change size. Each particle's size is fixed; what changes at high pressure is how large that fixed size is as a share of the whole sample.
Feedback if chosen: Correct. PV = nRT treats the particles as points, but the measured volume must include the space the particles themselves fill — so the measurement comes out larger.
Check — multiple choice
The figure shows measured PV/nRT for a neon sample. In the region where the curve sits above the dashed line, what is true of the sample?
Feedback if chosen: Correct. PV/nRT above 1 means the measured values overshoot the prediction — at high pressure the neon particles' own volume keeps the sample from shrinking as far as PV = nRT says it should.
Feedback if chosen: This option is wrong — you read the climb backwards. A measured volume smaller than the prediction would drag PV/nRT below 1, not above it.
Feedback if chosen: This option is wrong — you blamed the measurement instead of the gas. T in the ratio is the measured temperature, so a change in temperature is already accounted for; the climb is the gas deviating from the model.
Feedback if chosen: This option is wrong — you blamed added gas instead of the gas itself. n in the ratio is the measured amount, so extra gas is already accounted for; the climb is the gas deviating from the model.
Check — multiple choice
A gas sample is held at four different pressures, all at the same temperature. At which pressure do the particles' own volumes matter most, compared with the space the sample fills?
Feedback if chosen: This option is wrong — you picked the most spread-out sample. At 1 atm the sample is almost entirely empty space, so the particles' own volumes are a negligible share of it.
Feedback if chosen: This option is wrong — you stopped the squeeze too early. At 10 atm the empty space has shrunk tenfold but still dwarfs the particles' own volumes.
Feedback if chosen: This option is wrong — you picked a high pressure, but not the highest. The particles' fixed volumes take up an even larger share of the sample at 1000 atm than at 100 atm.
Feedback if chosen: Correct. The particles' own volumes are fixed, so the harder the sample is squeezed, the larger the share of the remaining space they take up — the effect is greatest at the highest pressure.
Outcome: Predict that a real gas deviates most from ideal behavior at low temperature and/or very high pressure.
You've already seen that attractions between particles make a real gas's measured pressure lower than PV = nRT predicts.
You've already seen that the particles' own volume makes a real gas's measured volume larger than PV = nRT predicts.
Whether these deviations are tiny or serious depends on the gas's conditions: its temperature and its pressure.
You've already seen that Kelvin temperature is proportional to the average kinetic energy of the particles.
At low temperature, the particles move slowly.
As a particle moves more slowly, the same attractions bend its path more — a slow particle is easier to pull backward from the wall.
As a particle moves faster, it flies past its neighbors before their attractions can deflect it much.
So cooling a gas makes the attraction deviation grow, and heating it makes that deviation shrink.
The figure shows measured PV/nRT for N₂ at 200 K: the cold gas dips well below the ideal line.
The figure shows the same gas at 600 K: the hot gas never dips below the line.
Same gas, same attractions — but at 600 K the particles move too fast for those attractions to cut the pressure noticeably.
At very high pressure, the sample is squeezed into a small space, so the particles sit close together.
As the particles get closer, the attractions between them act more strongly.
Squeezing also removes empty space, so the particles' own volume becomes a bigger share of the whole.
So raising the pressure strengthens both causes of deviation at once.
At low pressure the particles are far apart: attractions barely reach, the particles' own volume is negligible, and the gas behaves nearly ideally.
A real gas deviates most from ideal behavior at low temperature and/or very high pressure.
CO₂ behaves least ideally when it is cold and highly compressed — both causes at full strength.
Turned around: a real gas follows PV = nRT most closely when it is hot and at low pressure.
Predict the conditions where a real gas deviates most from ideal behavior: low temperature and/or very high pressure.
Check — multiple choice
Under which set of conditions does a sample of CH₄ deviate most from PV = nRT?
Feedback if chosen: Correct. Cold particles are slow, so attractions deflect them strongly; high pressure crowds the particles, strengthening attractions further and making their own volume a large share of the sample.
Feedback if chosen: This option is wrong — you picked the conditions that weaken both causes. Hot particles outrun the attractions, and at 1 atm the particles are far apart.
Feedback if chosen: This option is wrong — you kept the high pressure but lost the low temperature. At 600 K the fast particles resist the attraction effect, so the colder crowded sample deviates more.
Feedback if chosen: This option is wrong — you kept the low temperature but lost the high pressure. At 1 atm the particles are far apart, so both causes stay weak compared with the crowded sample.
Check — multiple choice
The figure shows measured PV/nRT for O₂ at 250 K. The same sample is measured again at 500 K. How does the new curve compare with the one shown?
Feedback if chosen: This option is wrong — you made heating strengthen the attraction effect. Faster particles fly past their neighbors before attractions can deflect them, so heating weakens the dip.
Feedback if chosen: Correct. At 500 K the particles move much faster, so the attractions cut the measured pressure far less — the dip shrinks toward the ideal line.
Feedback if chosen: This option is wrong — you made the deviations independent of temperature. Particle speed rises with temperature, and speed decides how much the attractions can deflect each particle.
Feedback if chosen: This option is wrong — you made the hot gas perfectly ideal. Heating weakens the attraction effect, but the particles still occupy volume, so the curve still climbs above 1 at high pressure.
Check — multiple choice
Four argon samples are prepared. Which one follows PV = nRT most closely?
Feedback if chosen: This option is wrong — you picked the conditions where both causes are strongest: slow particles that attractions deflect easily, crowded into almost no empty space.
Feedback if chosen: This option is wrong — you fixed the pressure but kept the low temperature. The spread-out particles are still slow, so attractions still deflect them noticeably.
Feedback if chosen: Correct. Hot particles outrun the attractions, and at 0.5 atm the particles are so far apart that their own volume is a negligible share of the sample — both causes are at their weakest.
Feedback if chosen: This option is wrong — you fixed the temperature but kept the very high pressure. The fast particles are still crowded together, so their own volume still swells the measured values.
Outcome: Predict which of two gases deviates more from ideal behavior under the same conditions, using their intermolecular force strengths and particle sizes.
You've already seen the conditions that make any one gas deviate most: low temperature and/or very high pressure.
Hold the conditions fixed and compare two different gases, and the deviations still differ — the identity of the gas matters.
You've already seen how to rank substances by overall intermolecular force strength, using hydrogen bonding, polarity, and size or polarizability evidence from their structures.
The first property is intermolecular force strength: as the attractions between its particles get stronger, a gas's measured pressure falls further below the PV = nRT prediction.
The second property is particle size: as its particles get larger, they fill more of the sample's space themselves, and the measured volume rises further above the prediction.
You've already seen that larger electron clouds are more polarizable and give stronger London dispersion forces — so larger particles feed both causes at once.
So the gas with stronger intermolecular forces and larger particles deviates more, and the gas with weaker forces and smaller particles stays closer to ideal behavior.
Compare NH₃ and He under the same conditions.
NH₃ particles hydrogen-bond to one another; He particles share only the weakest London dispersion forces.
NH₃ particles are also much larger than He particles.
Both comparisons point the same way, so NH₃ deviates far more from ideal behavior than He.
The figure shows measured PV/nRT for NH₃: strong attractions drag the curve deep below the ideal line before crowding pushes it back up.
The figure shows He under the same conditions: the curve stays close to the ideal line, drifting only slightly above it at high pressure.
If two gases are close in intermolecular force strength, particle size decides: Ar deviates more than He, because its larger, more polarizable particles bring stronger London dispersion forces and fill more space.
If two gases are close in particle size, intermolecular force strength decides: H₂O vapor deviates far more than N₂, because hydrogen bonding between H₂O particles is far stronger than N₂'s London dispersion forces.
Predict which of two gases deviates more from ideal behavior under the same conditions, using their intermolecular force strengths and particle sizes.
Check — multiple choice
Four gases are held under identical conditions. Which one deviates most from PV = nRT?
Feedback if chosen: This option is wrong — you picked the gas at the bottom of both comparisons. He has the smallest particles and the weakest London dispersion forces, so it stays closest to ideal behavior.
Feedback if chosen: This option is wrong — you stopped at a small nonpolar particle. N₂'s London dispersion forces are weak, and two of the listed gases have larger particles and stronger attractions.
Feedback if chosen: This option is wrong — you picked a fairly large particle but missed the polarity. CO₂ is nonpolar, while SO₂'s bent shape makes it polar — and SO₂'s particles are larger as well.
Feedback if chosen: Correct. SO₂ has the largest, most polarizable particles here, and its bent, polar shape adds dipole-dipole attractions — the strongest intermolecular forces of the four, so the largest deviation.
Check — multiple choice
The figure shows measured PV/nRT for an unknown gas, X, under conditions where its curve dips far below the ideal line. Which gas could X be?
Feedback if chosen: This option is wrong — you matched a deep dip to the gas with the weakest attractions. He's tiny particles and feeble London dispersion forces barely lower its measured pressure.
Feedback if chosen: Correct. A deep dip means attractions cut the measured pressure sharply, and H₂O's hydrogen bonding is by far the strongest attraction among these gases.
Feedback if chosen: This option is wrong — you matched a deep dip to very weak attractions. H₂'s London dispersion forces are so weak that its curve barely dips at all.
Feedback if chosen: This option is wrong — you matched a deep dip to weak attractions. Ne's small, hard-to-polarize particles attract one another feebly, so its measured pressure falls only slightly short.
Check — multiple choice
CH₄, C₂H₆, C₃H₈, and C₄H₁₀ are all nonpolar gases. Under the same conditions, which deviates most from PV = nRT?
Feedback if chosen: This option is wrong — you stopped partway up the size ranking. C₂H₆ is larger than CH₄, but two of the listed particles are larger still.
Feedback if chosen: Correct. With polarity equal, size decides: C₄H₁₀ has the largest, most polarizable particles, so the strongest London dispersion forces and the most particle volume — the largest deviation.
Feedback if chosen: This option is wrong — you ranked the smallest particle as the largest deviator. CH₄'s small, weakly polarizable particles give it the weakest attractions of the four, so it stays closest to ideal behavior.
Feedback if chosen: This option is wrong — you stopped one particle short of the top. C₃H₈ is large, but C₄H₁₀ is larger still, with stronger London dispersion forces and more particle volume.
Check — multiple choice
Four gases are held under identical conditions. Which one follows PV = nRT most closely?
Feedback if chosen: This option is wrong — you picked the gas with the strongest attractions. NH₃'s hydrogen bonding pulls its measured pressure far below the prediction.
Feedback if chosen: This option is wrong — you picked a polar gas. HCl's dipole-dipole attractions deflect its particles more than the London dispersion forces of the two smallest gases here.
Feedback if chosen: Correct. H₂ has the smallest particles and the weakest London dispersion forces of the four, so both causes of deviation are at their weakest and it stays closest to the prediction.
Feedback if chosen: This option is wrong — you picked a nonpolar gas but not the smallest one. CH₄'s particles are larger and more polarizable than H₂'s, so its attractions and particle volume both deviate more.
Check — multiple choice
A sealed sample of gas has a pressure of 3.00 atm. At constant temperature, its volume is reduced to one quarter of its starting value. What is the new pressure?
Feedback if chosen: This option is wrong — you divided the pressure by 4 along with the volume — that is the pattern of a direct proportion. Pressure and volume trade off: a quarter of the volume means four times the pressure.
Feedback if chosen: Correct. Pressure and volume are inversely proportional, so cutting the volume to a quarter multiplies the pressure by 4: 4 × 3.00 = 12.0 atm.
Feedback if chosen: This option is wrong — you doubled the pressure, but the volume fell to a quarter, not to a half. Multiply by the same factor the volume was divided by: 4 × 3.00 = 12.0 atm.
Feedback if chosen: This option is wrong — you left the pressure unchanged. In a quarter of the space the same particles strike the walls four times as often, so the pressure quadruples.
Check — multiple choice
A gas sample occupies 1.50 L at 120 K. At constant pressure, it is heated to 360 K. What is the new volume?
Feedback if chosen: Correct. Volume is directly proportional to Kelvin temperature, and 120 K to 360 K is a factor of 3, so the volume triples: 3 × 1.50 = 4.50 L.
Feedback if chosen: This option is wrong — you divided by 3 — that is the inverse-proportion pattern, which belongs to pressure and volume. Volume and Kelvin temperature grow together: 3 × 1.50 = 4.50 L.
Feedback if chosen: This option is wrong — you doubled the volume, but the Kelvin temperature tripled (120 K to 360 K), not doubled. Match the factor exactly: 3 × 1.50 = 4.50 L.
Feedback if chosen: This option is wrong — you left the volume unchanged. Volume divided by Kelvin temperature stays constant, so a tripled Kelvin temperature must bring a tripled volume.
Check — multiple choice
The graph shows measurements of a fixed amount of gas. Which relationship does it show?
Feedback if chosen: This option is wrong — you matched a rising line to an inverse proportion. An inverse proportion falls as you move right; this line rises.
Feedback if chosen: This option is wrong — you misread the vertical axis. It is labeled as pressure in atmospheres, not volume — the axes name the variables being compared.
Feedback if chosen: Correct. The axes name pressure and Kelvin temperature, and the straight line through the origin is the shape of a direct proportion.
Feedback if chosen: This option is wrong — you gave the line a starting threshold. It passes through the origin, which is exactly what a direct proportion requires.
Check — multiple choice
What is the volume of 0.250 mol of an ideal gas at 127 °C and 1.00 atm? (R = 0.08206 L·atm·mol⁻¹·K⁻¹.)
Feedback if chosen: This option is wrong — you substituted the Celsius temperature without converting it to kelvin. Convert first, as its own step: T = 127 + 273 = 400 K.
Feedback if chosen: This option is wrong — you used R = 8.314 J·mol⁻¹·K⁻¹ with the pressure in atmospheres, instead of the R given in L·atm units.
Feedback if chosen: This option is wrong — you inverted the rearrangement, dividing P by nRT instead of dividing nRT by P. Making V the subject gives V = nRT / P.
Feedback if chosen: Correct. Convert the temperature first — 127 + 273 = 400 K — then V = nRT / P = (0.250 × 0.08206 × 400) / 1.00 = 8.21 L.
Check — multiple choice
A cylinder fitted with a piston holds 6.00 L of N₂ at 27 °C and 1.00 atm. The gas is heated to 177 °C while the piston compresses it to a pressure of 3.00 atm, and no gas enters or leaves. What is the new volume?
Feedback if chosen: Correct. Convert both temperatures to kelvin (300 K and 450 K), write PV = nRT for each state, and cancel n and R: P₁V₁/T₁ = P₂V₂/T₂, so V₂ = (1.00 × 6.00 × 450) / (300 × 3.00) = 3.00 L.
Feedback if chosen: This option is wrong — you left the pressure change out. Both the pressure and the temperature change here, so both belong in P₁V₁/T₁ = P₂V₂/T₂; the tripled pressure squeezes the volume back down from 9.00 L.
Feedback if chosen: This option is wrong — you substituted the Celsius temperatures without converting them to kelvin. Convert first, each as its own step: T₁ = 27 + 273 = 300 K and T₂ = 177 + 273 = 450 K.
Feedback if chosen: This option is wrong — you left the temperature change out. Both the pressure and the temperature change here, so both belong in P₁V₁/T₁ = P₂V₂/T₂; the heating swells the volume beyond 2.00 L.
Check — multiple choice
A sealed tank holds a mixture of gases at a total pressure of 4.00 atm. The mole fraction of the Ne in the mixture is 0.15. What partial pressure does the Ne contribute?
Feedback if chosen: This option is wrong — you read the mole fraction itself as a pressure. A mole fraction is a unitless share; the Ne's partial pressure is that share of the total: 0.15 × 4.00 = 0.60 atm.
Feedback if chosen: This option is wrong — you calculated the partial pressure of everything except the Ne (0.85 × 4.00). The question asks for the Ne's own contribution: 0.15 × 4.00 = 0.60 atm.
Feedback if chosen: Correct. A gas's partial pressure equals its mole fraction times the total pressure: P(Ne) = 0.15 × 4.00 = 0.60 atm.
Feedback if chosen: This option is wrong — you divided the total pressure by the mole fraction instead of multiplying. Taking the Ne's share of the total means multiplying: 0.15 × 4.00 = 0.60 atm.
Check — multiple choice
A sealed flask holds only CH₄ and He. The partial pressure of the CH₄ is 0.80 atm, and the partial pressure of the He is 2.40 atm. What is the mole fraction of the CH₄?
Feedback if chosen: This option is wrong — you divided the CH₄'s partial pressure by the He's partial pressure instead of by the total. The CH₄'s own contribution counts in the total too: X(CH₄) = 0.80 / 3.20 = 0.25.
Feedback if chosen: Correct. The total pressure is the sum of the partial pressures, 0.80 + 2.40 = 3.20 atm, and X(CH₄) = P(CH₄) / P(total) = 0.80 / 3.20 = 0.25.
Feedback if chosen: This option is wrong — you calculated the He's mole fraction (2.40 / 3.20). The question asks for the CH₄: X(CH₄) = 0.80 / 3.20 = 0.25.
Feedback if chosen: This option is wrong — you divided the total pressure by the CH₄'s partial pressure — that ratio is upside down, and a mole fraction can never exceed 1. X(CH₄) = 0.80 / 3.20 = 0.25.
Check — multiple choice
A sample of CO deviates noticeably from PV = nRT. Which change in its conditions brings its behavior closest to ideal?
Feedback if chosen: This option is wrong — you fixed the pressure but cooled the gas. Slow particles are the ones attractions deflect most, so cooling strengthens the attraction deviation.
Feedback if chosen: This option is wrong — you fixed the temperature but raised the pressure. Crowding the particles strengthens the attractions and makes the particles' own volume a larger share of the sample.
Feedback if chosen: This option is wrong — you picked the conditions where both causes of deviation grow: slow particles that attractions deflect easily, crowded into little empty space.
Feedback if chosen: Correct. Fast particles outrun the attractions, and at low pressure the particles sit so far apart that the attractions barely reach and their own volume is a negligible share — both causes of deviation are at their weakest.
Check — multiple choice
F₂ and Cl₂ are held under identical conditions. A student claims the Cl₂ deviates more from PV = nRT than the F₂ does. Is the claim correct?
Feedback if chosen: Correct. Both are nonpolar, so size decides: Cl₂'s larger electron cloud is more polarizable, giving stronger London dispersion forces — a bigger pressure shortfall — and its larger particles fill more of the sample's space.
Feedback if chosen: This option is wrong — the verdict is right, but you made Cl₂ polar. Both F₂ and Cl₂ are nonpolar; Cl₂ deviates more because its larger, more polarizable particles bring stronger London dispersion forces and more particle volume.
Feedback if chosen: This option is wrong — you treated one kind of force as one strength. London dispersion forces grow with polarizability, and Cl₂'s larger electron cloud gives it noticeably stronger attractions than F₂'s.
Feedback if chosen: This option is wrong — you gave the advantage to the lighter gas. Smaller, lighter particles have weaker London dispersion forces and less volume of their own, so F₂ stays closer to ideal behavior.
6 lessons
Outcome: Classify a mixture as a solution or as heterogeneous, from whether its properties are the same everywhere in the sample.
You've already seen how to classify a sample as a pure substance or a mixture: a pure substance contains one type of particle, a mixture more than one.
Two jars stand on a bench: one holds muddy water, the other holds salt water.
Both are mixtures — but they behave very differently.
Take a 10 mL spoonful from the top of each jar and another from the bottom, dry each spoonful out, and weigh the solid left behind — the figure shows the results.
| Spoonful drawn from | Muddy water | Salt water |
|---|---|---|
| Top of the jar | 0.2 g of solid | 0.35 g of solid |
| Bottom of the jar | 1.6 g of solid | 0.35 g of solid |
In the muddy water, what you get depends on where you dip: the bottom spoonful carries eight times as much solid as the top one.
In the salt water, every spoonful is the same: 0.35 g of solid, whether it came from the top or the bottom.
A mixture whose properties are the same everywhere in the sample is called a 'solution', or a 'homogeneous mixture'.
Salt water is a solution: the salt has spread through the water so completely that every part of the sample matches every other part.
A mixture whose properties depend on where in the sample you look is called 'heterogeneous'.
Muddy water is heterogeneous: a spoonful from the bottom is nothing like a spoonful from the top.
Air is a solution of gases: nitrogen, oxygen, and small amounts of other gases, mixed so evenly that every breath has the same composition.
Solids can be solutions too.
You've already seen brass, a substitutional alloy: zinc atoms replace copper atoms at random positions throughout the metal.
Because the zinc atoms are spread evenly through the copper, every piece of a brass bar has the same composition and the same properties — brass is a solution.
A mixture's state does not decide its classification: solid, liquid, or gas, the question is always whether the properties are the same everywhere.
Classify a mixture as a solution (a homogeneous mixture, whose properties are the same everywhere in the sample) or as heterogeneous (whose properties depend on location in the mixture).
Check — multiple choice
| Spoonful drawn from | Solid left behind |
|---|---|
| Top of the jar | 0.1 g |
| Bottom of the jar | 1.8 g |
A jar of pond water is sampled with a 10 mL spoonful at the top and another at the bottom; the figure shows the solid left after drying each spoonful. Which classification fits the pond water?
Feedback if chosen: This option is wrong — you classified without using the data. A spoonful from the bottom leaves eighteen times as much solid as one from the top, so the properties depend on location in the sample.
Feedback if chosen: Correct. The solid left behind depends on where the spoonful was drawn, so the properties depend on location — the pond water is a heterogeneous mixture.
Feedback if chosen: This option is wrong — you treated the sample as one substance. Drying leaves solid behind, and the amount differs by location, so the jar holds more than one type of particle, mixed unevenly.
Feedback if chosen: This option is wrong — you classified each region separately. A classification applies to the whole sample: because its properties depend on location, the whole jar is one heterogeneous mixture.
Check — multiple choice
Which of these samples is a solution?
Feedback if chosen: This option is wrong — you overlooked how the mud is distributed. Mud hangs and settles unevenly, so a sample from the bottom differs from one at the top: the mixture is heterogeneous.
Feedback if chosen: This option is wrong — you let the solid state decide. The visible grains mean the composition changes from point to point, so granite is heterogeneous even though it is solid.
Feedback if chosen: Correct. The zinc is spread evenly through the copper, so every piece of brass has the same composition and properties — a solid mixture that is the same everywhere is a solution.
Feedback if chosen: This option is wrong — you counted mixed liquids as automatically uniform. The oil layer and the vinegar layer have different properties, so the dressing's properties depend on location.
Check — multiple choice
A diver's tank is filled with helium and oxygen, mixed so evenly that a sample drawn from any part of the tank has the same composition. Which classification fits the gas in the tank?
Feedback if chosen: This option is wrong — you reversed the definition. Heterogeneous means the properties depend on location, and here every sample from the tank matches every other one.
Feedback if chosen: This option is wrong — you treated even mixing as purity. The tank holds two types of particle, helium and oxygen, so it is a mixture however evenly they are mixed.
Feedback if chosen: This option is wrong — you assumed the gases settle into layers by identity. The stem says the mixing is even: a sample from any part of the tank has the same composition.
Feedback if chosen: Correct. A gas mixture that is the same everywhere is a solution — solutions can be solid, liquid, or gas.
Outcome: Name the solute and the solvent in a solution, and identify a solution as aqueous when its solvent is water.
You've already seen that salt water is a solution: the salt spreads through the water until every part of the sample is the same.
The two substances play different roles: the salt is taken apart and spread out, while the water does the spreading.
The dissolved substance is called the 'solute'.
The substance doing the dissolving is called the 'solvent'.
In salt water, NaCl is the solute and water is the solvent.
The figure names the parts of three solutions.
| Solution | Solute (dissolved) | Solvent (does the dissolving) |
|---|---|---|
| Salt water | NaCl | Water |
| Glucose in water | Glucose | Water |
| Brass | Zinc | Copper |
In brass, the copper does the dissolving: the zinc atoms are taken into the copper and spread through it, so zinc is the solute and copper is the solvent.
In each of these solutions, the substance present in the larger amount is the one doing the dissolving.
Water is by far the most common solvent in the laboratory.
A solution whose solvent is water is called an 'aqueous' solution.
Salt water and glucose in water are aqueous; brass is a solution, but it is not aqueous, because its solvent is copper, not water.
Name the parts of a solution: the dissolved substance is the solute, the substance doing the dissolving is the solvent, and a solution whose solvent is water is aqueous.
Check — multiple choice
KNO₃ is stirred into a beaker of water until the solid disappears and the mixture is the same throughout. Which substance is the solute?
Feedback if chosen: This option is wrong — you named the substance doing the dissolving. The water takes the KNO₃ apart and spreads it out, so the water is the solvent.
Feedback if chosen: This option is wrong — you named the solution itself. The solution is solute plus solvent together; the solute is the dissolved part alone.
Feedback if chosen: This option is wrong — you have the roles reversed. The substance present in the larger amount is doing the dissolving — that is the solvent. The solute is the substance that gets dissolved.
Feedback if chosen: Correct. The KNO₃ is taken apart and spread through the water — it is the dissolved substance, the solute.
Check — multiple choice
Which of these solutions is aqueous?
Feedback if chosen: This option is wrong — you looked only at the solute. 'Aqueous' names the solvent, and here the solvent is ethanol, not water.
Feedback if chosen: This option is wrong — you counted every solution as aqueous. Brass is a solution, but its solvent is copper; aqueous means the solvent is water.
Feedback if chosen: Correct. The solvent — the substance doing the dissolving — is water, so this solution is aqueous.
Feedback if chosen: This option is wrong — you classified a gas solution as aqueous. Air contains no water solvent at all; the gases dissolve in one another.
Check — multiple choice
Sterling silver is a solid solution: a small amount of copper is spread evenly through a much larger amount of silver. Which substance is the solvent?
Feedback if chosen: Correct. The silver is present in the much larger amount and takes the copper in — the silver does the dissolving, so it is the solvent.
Feedback if chosen: This option is wrong — you named the dissolved substance. The copper is spread out through the silver, so the copper is the solute.
Feedback if chosen: This option is wrong — you named the solution itself. The solution is solute plus solvent together; the solvent is the component doing the dissolving.
Feedback if chosen: This option is wrong — you assumed every solution's solvent is water. Only aqueous solutions have water as the solvent, and sterling silver contains no water.
Outcome: State that molarity is the moles of solute per liter of solution, and read a solution's molarity as the moles of solute dissolved in each liter.
You've already seen that in an aqueous solution the solute spreads evenly through the whole volume.
The figure shows two bottles of salt water made in a lab.
| Bottle | Moles of NaCl dissolved | Volume of solution |
|---|---|---|
| A | 0.50 mol | 2.0 L |
| B | 0.30 mol | 1.0 L |
Bottle A holds more NaCl than bottle B — 0.50 mol against 0.30 mol.
But bottle A's NaCl is spread through twice the volume.
Which solution is saltier, sip for sip?
The mole counts alone cannot say, because the two amounts are spread through different volumes.
To compare fairly, count the moles of solute in each liter of solution.
Bottle A: 0.50 mol spread through 2.0 L puts 0.25 mol in each liter.
Bottle B: 0.30 mol spread through 1.0 L puts 0.30 mol in each liter.
Liter for liter, bottle B holds more dissolved NaCl — bottle B is the saltier solution.
The moles of solute per liter of solution is called the 'molarity' — the most common laboratory measure of a solution's composition.
As an equation: the molarity is the moles of solute divided by the volume of the solution in liters.
Molarity is written with the unit symbol M: bottle B is a 0.30 M solution, spoken as '0.30 molar'.
A full liter of bottle B's solution and a single drop of it have the same molarity — the solute is spread evenly, so the ratio does not depend on the sample size.
The volume in the ratio is the volume of the finished solution, not the volume of water added.
Dissolving solid NaCl in exactly 1.00 L of water gives slightly more than 1.00 L of solution, because the dissolved solid takes up some room.
So chemists dissolve the solute in less water first, then add water up to the final volume: 0.10 mol of NaCl made up to exactly 1.00 L of solution gives a 0.10 M solution.
State that molarity, the most common laboratory measure of solution composition, is the moles of solute per liter of solution, written with the unit symbol M.
Check — numeric entry
A flask holds 1.00 L of solution with 0.25 mol of NaCl dissolved in it. What is the molarity of the solution? Give your answer to 2 significant figures.
Tolerance: 0.005
Feedback when correct: Correct. Molarity is the moles of solute per liter of solution: 0.25 mol in 1.00 L is 0.25 mol in each liter, a 0.25 M solution.
Anticipated wrong answers (feedback shown):
| 4.0 | You divided the volume by the moles, which is the ratio upside down. Molarity is moles of solute per liter of solution: M = n/V. |
Worked solution revealed after a second miss
Write down the values from the question
n = 0.25 mol
V = 1.00 L
Write down the equation
M = n/V
Substitute in the values, and calculate
M = 0.25 / 1.00
M = 0.25 M
Check — multiple choice
A label reads '0.30 M glucose solution'. What does the label say about the solution?
Feedback if chosen: Correct. Molarity is the moles of solute per liter of solution, so every liter of this solution carries 0.30 mol of dissolved glucose.
Feedback if chosen: This option is wrong — you measured per liter of water added. Molarity counts per liter of finished solution, whose volume includes the room the dissolved solute takes up.
Feedback if chosen: This option is wrong — you read a ratio as a total. Molarity is moles per liter; the total moles in the container depends on how many liters it holds.
Feedback if chosen: This option is wrong — you swapped grams in for moles. The unit M counts moles of solute per liter of solution, not grams.
Check — multiple choice
A student dissolves 0.20 mol of KNO₃ in 1.00 L of water; the finished solution's volume is 1.01 L. When the molarity is calculated, which volume divides the moles of solute?
Feedback if chosen: This option is wrong — you used the solvent volume. Molarity is per liter of solution, and the finished solution's volume is 1.01 L, because the dissolved solid takes up some room.
Feedback if chosen: This option is wrong — you used the change in volume. The denominator is the whole volume the solute is spread through: the 1.01 L of finished solution.
Feedback if chosen: This option is wrong — you split the difference between the two volumes. The definition names one volume only: the volume of the finished solution.
Feedback if chosen: Correct. Molarity is the moles of solute per liter of solution, so the volume of the finished solution — 1.01 L — divides the moles.
Outcome: Calculate a solution's molarity from the moles of solute — or from a mass converted to moles — and the solution volume.
You've already seen molarity: the moles of solute per liter of solution, M = n/V.
A question gives the amount of solute and the volume of solution, and asks for the molarity.
Write the amount in moles and the volume in liters, then substitute into M = n/V.
Q: 0.25 mol of KNO₃ is dissolved in enough water to make 0.50 L of solution. What is the molarity of the solution?
Write down the values from the question
n = 0.25 mol
V = 0.50 L
Write down the equation
M = n/V
Substitute in the values, and calculate
M = 0.25 / 0.50
M = 0.50 M
Questions often give the solute as a mass in grams.
You've already seen how to convert a mass to moles: divide the mass by the molar mass.
The conversion is its own labeled step, done before the molarity equation.
Q: 29.22 g of NaCl (molar mass 58.44 g/mol) is dissolved in enough water to make 2.00 L of solution. What is the molarity of the solution?
Write down the values from the question
mass of NaCl = 29.22 g
molar mass of NaCl = 58.44 g/mol
V = 2.00 L
Convert the mass to moles
n = 29.22 / 58.44(mass divided by molar mass)
n = 0.500 mol
Write down the equation
M = n/V
Substitute in the values, and calculate
M = 0.500 / 2.00
M = 0.250 M
Calculate the molarity of a solution from the amount of solute — given in moles, or as a mass converted to moles first — and the volume of the solution.
Check — numeric entry
0.80 mol of glucose is dissolved in enough water to make 2.00 L of solution. What is the molarity of the solution? Give your answer to 3 significant figures.
Tolerance: 0.0005
Feedback when correct: Correct. M = n/V = 0.80 / 2.00 = 0.400 M — each liter of the solution carries 0.400 mol of glucose.
Anticipated wrong answers (feedback shown):
| 2.50 | You divided the volume by the moles, which is the ratio upside down. Molarity is moles of solute per liter of solution: M = n/V, with the moles on top. |
| 1.60 | You multiplied the moles by the volume instead of dividing by it. Molarity is a per-liter ratio: M = n/V. |
Worked solution revealed after a second miss
Write down the values from the question
n = 0.80 mol
V = 2.00 L
Write down the equation
M = n/V
Substitute in the values, and calculate
M = 0.80 / 2.00
M = 0.400 M
Check — numeric entry
20.0 g of NaOH (molar mass 40.00 g/mol) is dissolved in enough water to make 2.50 L of solution. What is the molarity of the solution? Give your answer to 3 significant figures.
Tolerance: 0.0005
Feedback when correct: Correct. Converting the mass first, 20.0 / 40.00 = 0.500 mol, then M = n/V = 0.500 / 2.50 = 0.200 M.
Anticipated wrong answers (feedback shown):
| 8.00 | You divided the mass by the volume without converting to moles. Convert first — n = 20.0 / 40.00 = 0.500 mol — then divide by the volume. |
| 1.25 | You multiplied the moles by the volume instead of dividing by it. Molarity is a per-liter ratio: M = n/V. |
| 5.00 | You divided the volume by the moles, which is the ratio upside down. Molarity is moles of solute per liter of solution: M = n/V, with the moles on top. |
Worked solution revealed after a second miss
Write down the values from the question
mass of NaOH = 20.0 g
molar mass of NaOH = 40.00 g/mol
V = 2.50 L
Convert the mass to moles
n = 20.0 / 40.00
n = 0.500 mol
Write down the equation
M = n/V
Substitute in the values, and calculate
M = 0.500 / 2.50
M = 0.200 M
Check — multiple choice
A student dissolves 20.22 g of KNO₃ (molar mass 101.11 g/mol) in enough water to make 0.400 L of solution. Which expression gives the molarity, in M?
Feedback if chosen: Correct. The mass is converted to moles first — 20.22 / 101.11 — and the moles are then divided by the solution volume: M = n/V.
Feedback if chosen: This option is wrong — you skipped the mass-to-moles conversion. Dividing grams by liters gives grams per liter, not molarity.
Feedback if chosen: This option is wrong — you multiplied the moles by the volume instead of dividing by it. Molarity is a per-liter ratio: M = n/V.
Feedback if chosen: This option is wrong — you divided the molar mass by the mass, which is the conversion upside down. Moles are the mass divided by the molar mass.
Outcome: Calculate the moles of solute in a given volume of solution from the molarity, and convert those moles to a number of solute particles.
You've already seen molarity: M = n/V, the moles of solute per liter of solution.
This lesson uses the ratio the other way around: the molarity and the volume are given, and the amount of solute is the unknown.
Make n the subject: n = M × V — the molarity times the volume in liters.
You've already seen how to count particles from moles: multiply the moles by Avogadro's number, 6.022 × 10²³ particles per mole.
So the path goes from the volume of solution to the moles of solute to the number of particles: two multiplications, each its own labeled step.
Q: How many moles of NaCl, and how many formula units of NaCl, are in 0.200 L of a 0.10 M NaCl solution? (Avogadro's number = 6.022 × 10²³ mol⁻¹.)
Write down the values from the question
M = 0.10 M
V = 0.200 L
Write down the equation
M = n/V
Make n the subject
n = M × V
Substitute in the values, and calculate
n = 0.10 × 0.200
n = 0.020 mol
Multiply by Avogadro's number to count the formula units
formula units = 0.020 × (6.022 × 10²³)
= 0.1204 × 10²³(0.020 × 6.022 = 0.1204)
= 1.2 × 10²² formula units
The particle name follows the solute: formula units for an ionic solid like NaCl, molecules for a molecular substance like glucose.
Calculate the moles of solute in a given volume of a solution of known molarity, and from the moles the number of solute particles.
Check — numeric entry
How many moles of KNO₃ are in 0.250 L of a 0.400 M KNO₃ solution? Give your answer to 3 significant figures.
Tolerance: 0.0005
Feedback when correct: Correct. Making n the subject of M = n/V gives n = M × V = 0.400 × 0.250 = 0.100 mol.
Anticipated wrong answers (feedback shown):
| 1.60 | You divided the molarity by the volume instead of multiplying them. Making n the subject of M = n/V gives n = M × V. |
| 0.625 | You divided the volume by the molarity instead of multiplying them. Making n the subject of M = n/V gives n = M × V. |
Worked solution revealed after a second miss
Write down the values from the question
M = 0.400 M
V = 0.250 L
Write down the equation
M = n/V
Make n the subject
n = M × V
Substitute in the values, and calculate
n = 0.400 × 0.250
n = 0.100 mol
Check — numeric entry
How many molecules of glucose are in 0.500 L of a 0.200 M glucose solution? (Avogadro's number = 6.022 × 10²³ mol⁻¹.) Give your answer in units of 10²² molecules, to 3 significant figures.
Tolerance: 0.005
Feedback when correct: Correct. n = M × V = 0.200 × 0.500 = 0.100 mol, and 0.100 × 6.022 × 10²³ = 6.02 × 10²² molecules.
Anticipated wrong answers (feedback shown):
| 0.602 | Your arithmetic is right, but the answer is not matched to the 10²² supplied by the answer box: 0.602 × 10²³ molecules is 6.02 × 10²² molecules. |
| 12.0 | You multiplied the molarity by Avogadro's number without finding the moles first. Find n = M × V = 0.100 mol, then multiply by Avogadro's number. |
| 0.100 | You stopped after the moles step. 0.100 is the moles of glucose; multiply by Avogadro's number to count the molecules. |
Worked solution revealed after a second miss
Write down the values from the question
M = 0.200 M
V = 0.500 L
Write down the equation
M = n/V
Make n the subject
n = M × V
Substitute in the values, and calculate
n = 0.200 × 0.500
n = 0.100 mol
Multiply by Avogadro's number to count the molecules
molecules = 0.100 × (6.022 × 10²³)
= 6.02 × 10²²
Match the answer to the units of the answer box
the box counts units of 10²² — enter 6.02
Check — multiple choice
Which expression gives the number of formula units of NaCl in 0.300 L of a 0.20 M NaCl solution?
Feedback if chosen: This option is wrong — you divided the molarity by the volume instead of multiplying them. The moles are n = M × V, and the moles are then multiplied by Avogadro's number.
Feedback if chosen: Correct. The moles come first — n = M × V = 0.20 × 0.300 — and multiplying the moles by Avogadro's number counts the formula units.
Feedback if chosen: This option is wrong — you divided by Avogadro's number instead of multiplying by it. Each mole holds 6.022 × 10²³ formula units, so the moles are multiplied by it.
Feedback if chosen: This option is wrong — you divided the volume by the molarity instead of multiplying them. The moles are n = M × V, and the moles are then multiplied by Avogadro's number.
Outcome: Calculate the volume of a solution of known molarity that contains a required amount of solute.
A procedure in the lab calls for 0.10 mol of NaOH.
The shelf bottle is labeled 0.50 M NaOH — every liter of it carries 0.50 mol of NaOH.
The question is how much of the solution to measure out so that exactly 0.10 mol of NaOH comes with it.
You've already seen molarity as M = n/V; this time the volume is the unknown.
Make V the subject: V = n/M — the moles needed divided by the molarity.
Q: What volume of a 0.50 M NaOH solution contains 0.10 mol of NaOH?
Write down the values from the question
n = 0.10 mol
M = 0.50 M
Write down the equation
M = n/V
Make V the subject
V = n/M
Substitute in the values, and calculate
V = 0.10 / 0.50
V = 0.20 L
A sense check: each liter of the bottle carries 0.50 mol, and 0.10 mol is a fifth of that, so a fifth of a liter is needed.
The more concentrated the solution, the less of it delivers the same amount of solute.
From a 1.0 M NaOH bottle, 0.10 mol takes only 0.10 L; from the 0.50 M bottle, it takes 0.20 L.
Calculate the volume of a solution of known molarity needed to deliver a required amount of solute, using V = n/M.
Check — numeric entry
What volume of a 0.250 M KNO₃ solution contains 0.0500 mol of KNO₃? Give your answer to 3 significant figures.
Tolerance: 0.0005
Feedback when correct: Correct. Making V the subject of M = n/V gives V = n/M = 0.0500 / 0.250 = 0.200 L.
Anticipated wrong answers (feedback shown):
| 0.0125 | You multiplied the moles by the molarity instead of dividing by it. Making V the subject of M = n/V gives V = n/M. |
| 5.00 | You divided the molarity by the moles, which is the rearrangement upside down. V = n/M — the moles needed go on top. |
Worked solution revealed after a second miss
Write down the values from the question
n = 0.0500 mol
M = 0.250 M
Write down the equation
M = n/V
Make V the subject
V = n/M
Substitute in the values, and calculate
V = 0.0500 / 0.250
V = 0.200 L
Check — numeric entry
A procedure calls for 1.50 mol of NaCl, supplied from a 0.750 M NaCl solution. What volume of the solution must be measured out? Give your answer to 3 significant figures.
Tolerance: 0.005
Feedback when correct: Correct. V = n/M = 1.50 / 0.750 = 2.00 L — each liter carries 0.750 mol, so two liters carry the required 1.50 mol.
Anticipated wrong answers (feedback shown):
| 1.13 | You multiplied the moles by the molarity instead of dividing by it. Making V the subject of M = n/V gives V = n/M. |
| 0.500 | You divided the molarity by the moles, which is the rearrangement upside down. V = n/M — the moles needed go on top. |
Worked solution revealed after a second miss
Write down the values from the question
n = 1.50 mol
M = 0.750 M
Write down the equation
M = n/V
Make V the subject
V = n/M
Substitute in the values, and calculate
V = 1.50 / 0.750
V = 2.00 L
Check — multiple choice
Two bottles sit on a shelf: bottle 1 holds a 0.10 M NaCl solution, and bottle 2 holds a 1.0 M NaCl solution. A student must measure out 0.050 mol of NaCl from one bottle. Which bottle delivers it in the smaller poured volume?
Feedback if chosen: This option is wrong — you picked the less concentrated bottle. Each liter of bottle 1 carries less NaCl, so more of it must be poured: V = n/M = 0.50 L against 0.050 L.
Feedback if chosen: This option is wrong — you treated the volume as independent of the molarity. V = n/M, so the same moles take ten times the volume from the bottle that is ten times less concentrated.
Feedback if chosen: Correct. V = n/M: the more concentrated solution packs the same moles into less volume — 0.050 L from bottle 2 against 0.50 L from bottle 1.
Feedback if chosen: This option is wrong — you treated the labels as insufficient. A molarity ties amount to volume exactly: V = n/M gives the volume for any required moles.
2 lessons
Outcome: Draw a particulate model of a dissolved ionic solid: separated ions, each surrounded by water molecules with the correct ends pointing in, in counts that match the molarity.
You've already seen that salt water is an aqueous solution: NaCl is the solute and water is the solvent.
What happens to the Na⁺ and Cl⁻ ions when the solid disappears into the water?
The figure follows the ions through dissolving: the solid on the left, the dissolved ions on the right.
In the solid, the ions sit locked in an alternating lattice — the drawing you've already seen for an ionic solid.
In the solution, that lattice is gone: each Na⁺ and each Cl⁻ has left it and moved into the water on its own.
Separation is the first feature of the drawing: no intact NaCl pair floats in the water.
You've already seen the ion-dipole interaction: an ion attracts the oppositely charged partial end of a polar molecule.
Each dissolved ion holds on to the water molecules around it through exactly that attraction.
Because water's O end carries the partial negative charge, the waters around Na⁺ point their O ends at it.
Because water's H ends carry the partial positive charge, the waters around Cl⁻ point their H ends at it.
The finished drawing: each ion separated, ringed by water molecules with the attracting ends facing in — here four around each ion.
You've already calculated molarity: the moles of solute dissolved in each liter of solution.
A particulate drawing carries that information as a count — the number of dissolved ions drawn in the box.
The box stands for a fixed volume, so a solution with twice the molarity is drawn with twice as many ions in the same box.
NaCl supplies one Na⁺ for every Cl⁻, so any NaCl drawing holds equal numbers of the two ions.
Draw a particulate model of a solution: ions separated, each surrounded by water molecules with the correct ends pointing in, and ion counts that match the relative molarities.
Check — drawing task (pen and paper, self-marked)
Take a pen and paper and draw a particulate model of dissolved NaCl: one Na⁺ and one Cl⁻, separated, each surrounded by four water molecules with the correct ends pointing at the ion. When you are done, select Check to compare your drawing with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
Check — drawing task (pen and paper, self-marked)
Solution A is 0.10 M KBr and solution B is 0.20 M KBr. Take a pen and paper and draw two same-size boxes, one for each solution, showing the dissolved ions as labeled circles — 2 K⁺ and 2 Br⁻ in box A, and the counts that match box B's molarity — with the ions spread through each box; take the water as filling the rest of each box. When you are done, select Check to compare your drawing with the model answer.
Model answer shown to the student
Self-mark checklist — the drawing should show:
Check — multiple choice
A drawing of 0.10 M NaCl shows 2 Na⁺ and 2 Cl⁻ in its box. How many ions belong in a same-size box drawn for 0.30 M NaCl?
Feedback if chosen: This option is wrong — you kept the counts fixed, and a same-size box stands for the same volume, so three times the molarity means three times the dissolved ions.
Feedback if chosen: This option is wrong — you added one ion of each instead of scaling: 0.30 M packs three times the ions of 0.10 M into the same volume, so multiply each count by three.
Feedback if chosen: Correct. 0.30 M packs three times the moles of solute into the same volume as 0.10 M, so the box shows 3 × 2 = 6 of each ion.
Feedback if chosen: This option is wrong — you scaled only the Na⁺ count, and NaCl supplies one Na⁺ for every Cl⁻, so the two counts stay equal.
Outcome: Evaluate a particulate drawing of a solution, and name the one specific feature that is wrong when the drawing has a flaw.
You've already seen the model drawing of dissolved NaCl: separated ions, each ringed by water molecules with the attracting ends facing in.
Evaluating someone else's drawing means applying that model as a checklist and pointing at the one feature that breaks it.
Check one — separation: dissolving pulls every ion out of the lattice, so no intact solute pairs float in the water.
This drawing fails check one: its Na⁺ and Cl⁻ still travel as touching pairs, as if the solid had merely broken into fragments.
Check two — orientation: because opposite partial charges attract, the waters around a cation point their partially negative O ends in, and the waters around an anion point their partially positive H ends in.
A drawing with those ends flipped shows like partial charges facing each other — pairs that push apart rather than attract.
Check three — spread: a solution is the same everywhere, so the dissolved ions are spread through the whole box rather than gathered in one clump.
Check four — counts: same-size boxes hold ion counts in the same ratio as the molarities, and the two ions keep the ratio the formula gives them.
A drawing that passes all four checks is correct — reject a drawing only when you can name the specific check it fails.
Evaluate whether a particulate drawing of a solution shows the separation, the water orientation, the spread, and the relative counts correctly, and identify the one specific flaw when it does not.
Check — multiple choice
The figure shows a student's drawing of a KCl solution. One feature of the drawing is wrong — which one?
Feedback if chosen: This option is wrong — equal numbers are correct here: KCl supplies one K⁺ for every Cl⁻.
Feedback if chosen: Correct. Dissolving pulls every ion out of the lattice, so each K⁺ and each Cl⁻ moves through the water on its own — touching K⁺–Cl⁻ units show solid fragments, not a solution.
Feedback if chosen: This option is wrong — spreading through the whole box is required: a solution is the same everywhere in the sample.
Feedback if chosen: This option is wrong — the size difference is correct: a Cl⁻ ion is larger than a K⁺ ion.
Check — multiple choice
The figure shows a student's drawing of dissolved LiCl. One feature of the drawing is wrong — which one?
Feedback if chosen: This option is wrong — separated ions are required: dissolving pulls each Li⁺ and each Cl⁻ into the water on its own.
Feedback if chosen: This option is wrong — the surrounding waters are required: each dissolved ion attracts the water molecules around it.
Feedback if chosen: This option is wrong — one Li⁺ for every Cl⁻ is correct: that is the ratio the formula LiCl gives.
Feedback if chosen: Correct. Opposite partial charges attract, so the O ends belong toward Li⁺ and the H ends toward Cl⁻ — every water in this drawing points the wrong end at its ion.
Check — multiple choice
The figure shows a student's drawing of a NaBr solution. One feature of the drawing is wrong — which one?
Feedback if chosen: Correct. A solution is the same everywhere, so the dissolved ions belong spread through the water — a single pile of touching ions with the waters far away shows solid sitting in water, not a solution.
Feedback if chosen: This option is wrong — equal numbers are correct: NaBr supplies one Na⁺ for every Br⁻.
Feedback if chosen: This option is wrong — the size difference is correct: a Br⁻ ion is larger than a Na⁺ ion.
Feedback if chosen: This option is wrong — showing the water is required: the solvent is part of the solution, so its molecules belong in the drawing.
Check — multiple choice
One student draws 0.10 M NaCl as 3 Na⁺ and 3 Cl⁻, separated and spread through a box, each ion ringed by waters with the O ends toward Na⁺ and the H ends toward Cl⁻. A second student draws 0.30 M NaCl the same way — 3 Na⁺ and 3 Cl⁻ — in a same-size box. Which check does the second drawing fail?
Feedback if chosen: This option is wrong — the orientation described is correct: partially negative O ends toward Na⁺ and partially positive H ends toward Cl⁻.
Feedback if chosen: This option is wrong — drawn-apart ions are required: dissolved NaCl travels as separate Na⁺ and Cl⁻ ions, never as pairs.
Feedback if chosen: Correct. Same-size boxes stand for the same volume, so the 0.30 M drawing needs three times the ions of the 0.10 M drawing — 9 Na⁺ and 9 Cl⁻.
Feedback if chosen: This option is wrong — one check fails: with three times the molarity in the same volume, the ion counts must also be three times as large.
Outcome: Draw particulate answers that pass the three checks graders score: conservation, proportion, orientation.
Graders score a particulate drawing on what the particles show — counts, charges, and directions — never on artistic quality.
The first check is conservation: every atom and every unit of charge in the formula must appear in your drawing, so the total charge of the drawn ions comes out to zero.
The second check is proportion: the count of each particle must follow the formula's ratio — one Ca²⁺ for every two Cl⁻ in dissolved CaCl₂, for example — because a wrong ratio and an unbalanced charge each lose the same point.
The third check is orientation: every water molecule you draw must point its partially negative O end at a cation and one of its partially positive H atoms at an anion.
Before you call a drawing finished, check it three ways — conservation, proportion, orientation — because on the exam each can be a separate point.
Free-response question
Source note (not shown as part of the question): 2023 FRQ Q5(b), adapted for pen and paper (the booklet's answer box and water-molecule image legend are replaced with a written description); rubric from the 2023 scoring guidelines
HCl is a molecular gas as a pure substance but acts as an acid in aqueous solution.
(b) When HCl ionizes in aqueous solution, Cl⁻(aq) ions are formed. Draw three water molecules with proper orientation around the Cl⁻ ion. Show each water molecule as its one O atom and its two H atoms. [1 point]
✓ Model response (the key)
(b)
The drawing shows the Cl⁻ ion with three separate water molecules around it.
Each water molecule has one of its H atoms — its partially positive end — pointed toward the Cl⁻ ion.
Scoring rubric — 1 point available
The model drawing looks like this — three water molecules, each with an H atom pointed at the Cl⁻ ion.
7 lessons
Outcome: Explain why filtration separates an undissolved solid from a liquid but cannot separate the components of a solution.
A beaker holds sand stirred into salt water: solid grains suspended in a solution.
You've already seen that salt water is a solution — the dissolved salt is spread so evenly that every part of the sample matches every other part.
Pour the beaker's contents through a paper cone folded into a funnel: the paper is called 'filter paper', and the method is called 'filtration'.
The sand stays on the paper — but the liquid that drips through is still salt water: dry a drop of it, and salt is left behind.
Filtration caught one component and completely missed the other — why?
Filter paper is a mat of fibers, and between the fibers are tiny gaps.
A sand grain is a cluster of an enormous number of particles — vastly larger than those gaps — so the paper holds it back.
You've already seen the particulate picture of the dissolved salt: individual Na⁺ and Cl⁻ ions, each surrounded by water molecules.
A single ion or molecule is far smaller than any gap in the paper, so the dissolved salt rides through with the water.
Reaching for a finer paper does not help: dissolved particles are so small that they pass through the pores of any ordinary filter paper.
In a solution, every component is molecule- or ion-sized, so filter paper — which separates by particle size — cannot tell the components apart.
What the components still differ in is their intermolecular attractions — how strongly each one holds to its neighbors and to other substances.
Separation methods work by exploiting those differences in attraction.
Explain why filtration cannot separate the components of a solution: dissolved particles are molecule- or ion-sized and pass through the pores of any ordinary filter paper, so solution components are separated instead by exploiting differences in their intermolecular attractions.
Check — multiple choice
Sugar is stirred into water until it dissolves completely, and the mixture is poured through fine filter paper. What collects in the flask below?
Feedback if chosen: This option is wrong — you treated dissolved sugar like suspended grains. Dissolved sugar is single molecules, far smaller than the gaps in the paper, so it passes through with the water.
Feedback if chosen: This option is wrong — you sorted the dissolved sugar by size. Dissolved sugar is already single molecules, all the same size, and every one is far smaller than the gaps in the paper.
Feedback if chosen: Correct. Every dissolved sugar molecule is far smaller than the gaps in the paper, so the whole solution passes through unchanged.
Feedback if chosen: This option is wrong — you had molecule-sized particles clog the paper. Particles small enough to slip through the gaps cannot block them, so the solution drips through freely.
Check — multiple choice
A student pours an aqueous KNO₃ solution through the finest filter paper available, hoping to catch the dissolved KNO₃. Why does the KNO₃ pass straight through?
Feedback if chosen: Correct. Dissolving takes the KNO₃ apart into individual ions, and single ions are far smaller than the pores of any ordinary filter paper — they pass through with the water.
Feedback if chosen: This option is wrong — you made the dissolved substance vanish. The K⁺ and NO₃⁻ ions are still there, spread among the water molecules; they pass through because they are far smaller than the paper's gaps.
Feedback if chosen: This option is wrong — you blamed the flow speed. However slowly the solution drips, the ions still pass: each one is far smaller than any gap in the paper.
Feedback if chosen: This option is wrong — you made mass the deciding factor. The paper stops particles by size, not weight, and single ions are far smaller than its gaps.
Check — multiple choice
Which of these mixtures can filtration separate?
Feedback if chosen: This option is wrong — you picked a solution. The dissolved Na⁺ and Cl⁻ ions are far smaller than the paper's gaps, so they pass through with the water.
Feedback if chosen: Correct. Suspended chalk grains are clusters of enormous numbers of particles — far larger than the paper's gaps — so the paper holds them back while the water passes.
Feedback if chosen: This option is wrong — you picked a solution. Dissolved sugar is single molecules, far smaller than the paper's gaps, so it passes through with the water.
Feedback if chosen: This option is wrong — you picked a solution. Both dyes are dissolved and molecule-sized, so both pass through the paper with the water.
Outcome: Describe how chromatography works: a mobile phase carries a dissolved mixture across a stationary phase.
Is the black ink in a marker a single black dye, or a mixture of colored dyes?
Every dye in the ink is dissolved, and you've already seen that filtration cannot separate dissolved components — molecule-sized particles pass through the paper's pores.
Separating dissolved components takes a method built on their attractions — and the simplest one needs only a strip of paper and a solvent.
Put a small spot of the ink near the bottom of a paper strip, and stand the strip's bottom edge in a shallow pool of solvent, below the spot.
The solvent climbs the paper, and as it climbs past the spot, it carries the ink's dissolved components up the strip with it.
The figure shows the strip after the solvent has climbed most of the way up.
The single black spot has become two separate spots — the black ink is a mixture of a blue dye and a yellow dye.
The moving liquid (or gas) that carries the mixture is called the 'mobile phase'.
The fixed surface the mobile phase moves across — here, the paper — is called the 'stationary phase'.
Any method that passes a dissolved mixture in a mobile phase across a stationary phase is called 'chromatography'.
On a paper strip, the paper is the stationary phase — paper chromatography.
On a thin solid coating spread over a glass plate, the coating is the stationary phase — thin-layer chromatography.
In a column, the mobile phase flows through a tube packed with a solid powder, and the packing is the stationary phase — column chromatography.
Describe the chromatography setup: a mobile phase (a moving liquid or gas) carries a dissolved mixture across a stationary phase (a fixed surface such as paper, a thin coating, or a packed column).
Check — multiple choice
A drop of leaf extract is spotted onto a plate coated with a thin solid layer, and the plate's edge is stood in hexane; the hexane climbs the coating and carries the extract's components with it. What is the mobile phase?
Feedback if chosen: This option is wrong — you named the fixed surface. The coating stays put while the hexane moves across it, so the coating is the stationary phase.
Feedback if chosen: This option is wrong — you named the sample. The extract is the mixture being separated; it is carried by the phases, but it is not a phase itself.
Feedback if chosen: This option is wrong — you named the support. The glass only holds the coating in place; it neither moves nor interacts with the components.
Feedback if chosen: Correct. The hexane is the moving liquid that carries the dissolved components across the fixed coating — the mobile phase.
Check — multiple choice
In a column chromatography experiment, a liquid flows through a tube packed with solid powder, carrying a dissolved mixture through it. What is the stationary phase?
Feedback if chosen: This option is wrong — you named the moving carrier. The flowing liquid is the mobile phase; the stationary phase is the surface that stays fixed.
Feedback if chosen: Correct. The packed powder is the fixed surface the mobile phase flows across — the stationary phase.
Feedback if chosen: This option is wrong — you named the sample. The mixture is what gets separated; it is neither the moving carrier nor the fixed surface.
Feedback if chosen: This option is wrong — you named the container. The tube only holds the packing; the fixed surface the components actually cling to is the powder itself.
Check — multiple choice
During a paper chromatography run, what carries the dissolved components of the spot up the strip?
Feedback if chosen: Correct. The climbing solvent is the mobile phase: as it moves past the spot, it carries the dissolved components up the strip with it.
Feedback if chosen: This option is wrong — you made the stationary phase do the carrying. The paper stays fixed; it is the moving solvent that carries the components.
Feedback if chosen: This option is wrong — you brought in heat. Nothing is heated in this setup; the components move only because the climbing solvent carries them.
Feedback if chosen: This option is wrong — you credited the components' own motion. On their own they would only blur outward around the spot; the directed climb comes from the moving solvent.
Outcome: Explain why the components of a mixture separate during chromatography: each component's attractions to the stationary and mobile phases differ, so the components travel at different speeds.
A black ink separates on a paper strip into a blue spot high on the paper and a yellow spot much lower.
Both dyes started in the same spot, and the same solvent climbed past both — so why did they stop at different heights?
You've already seen how to rank the strength of the interactions between different pairs of species from their structures.
On the strip, every dye particle feels two pulls at once: attraction to the stationary phase beneath it, and attraction to the mobile phase moving past it.
A component attracted more strongly to the stationary phase spends more of its time held on the surface, so it travels more slowly.
A component attracted more strongly to the mobile phase spends more of its time dissolved in the moving liquid, so it travels more quickly.
The two dyes have different structures, so their attractions to the paper and to the solvent differ in strength.
The yellow dye stopped lower: of the two, it is the one held to the paper more strongly, relative to its attraction to the solvent.
Different speeds pull the spots apart — by the time the solvent nears the top, the faster dye is far ahead of the slower one.
Chromatography separates a mixture because each component's attractions to the stationary phase and to the mobile phase differ in strength.
Explain why chromatography separates: each component's attractions to the stationary and mobile phases differ in strength, so components attracted more strongly to the stationary phase travel more slowly.
Check — multiple choice
Two dyes from one ink are carried up a paper strip by the same solvent; dye R stops much lower than dye G. What explains dye R's lower stopping point?
Feedback if chosen: This option is wrong — you made weight the deciding factor. The solvent does not lift particles against their weight; position is set by how strongly each dye is held by the paper compared with the solvent.
Feedback if chosen: This option is wrong — you invented a head start. Both dyes sit in the same spot and meet the climbing solvent at the same moment; only their attractions to the two phases differ.
Feedback if chosen: Correct. A component attracted more strongly to the stationary phase spends more of its time held on the paper, so it travels more slowly and stops lower.
Feedback if chosen: This option is wrong — you turned a weaker attraction into repulsion. Dye R still dissolves in and moves with the solvent — just less of the time, because the paper holds it more strongly.
Check — multiple choice
A component of a mixture is attracted far more strongly to the mobile phase than to the stationary phase. How does it behave during a chromatography run?
Feedback if chosen: This option is wrong — you reversed the attractions. Staying near the start line is what a component strongly held by the stationary phase does; this one is mostly held by the moving phase.
Feedback if chosen: This option is wrong — you left the attractions out. The solvent carries each component only while that component is dissolved in it, and how much time that is depends on each component's attractions.
Feedback if chosen: This option is wrong — you made the mobile phase an anchor. The mobile phase moves, so time spent attracted to it is time spent traveling, not time spent held back.
Feedback if chosen: Correct. Strong attraction to the mobile phase keeps the component dissolved in the moving liquid most of the time, so it travels quickly along the strip.
Check — multiple choice
Two components of a mixture travel partway up a thin-layer plate together and never separate. Which statement explains this?
Feedback if chosen: This option is wrong — you read shared speed as a chemical bond. Separate substances travel together whenever the two phases pull on them in the same proportions.
Feedback if chosen: Correct. Speed along the plate is set by the balance of each component's attractions to the two phases; if two components balance the same way, they travel at the same speed and never separate.
Feedback if chosen: This option is wrong — you blamed the run conditions. The components did interact with the coating — they stopped partway up; they stayed together because their attractions balance the same way.
Feedback if chosen: This option is wrong — you gave the components opposite attractions. Opposite pulls would produce very different speeds and a clear separation, not two components moving together.
Outcome: Rank the components of a mixture by relative polarity from their positions on a chromatogram, given the polarities of both phases.
The developed strip or plate — the paper after the run, showing where each component stopped — is called a 'chromatogram'.
A chromatogram records one fact per component: how far it traveled before the run ended.
You've already seen that a component attracted more strongly to the stationary phase travels more slowly.
To turn positions into polarities, you need one more piece of information: the polarity of each phase.
Take the most common case — the paper is polar, and the solvent is nonpolar.
A polar component is strongly attracted to the polar paper and only weakly carried by the nonpolar solvent, so it travels least.
A nonpolar component is only weakly held by the polar paper and rides easily in the nonpolar solvent, so it travels farthest.
On polar paper developed with a nonpolar solvent, the strip reads as a polarity ranking: the lowest spot is the most polar component, and the highest spot is the least polar.
On this chromatogram, C traveled least, so C is the most polar component; A traveled farthest, so A is the least polar; B sits between them.
If the phases were the other way around — a nonpolar stationary phase developed with a polar solvent — the same positions would read in reverse.
So a chromatogram ranks polarity only when both phase polarities are given — check them before reading a single spot.
Rank the components of a mixture by relative polarity from their positions on a chromatogram, using the given polarities of the two phases.
Check — multiple choice
Three food dyes were carried up polar paper by a nonpolar solvent; the chromatogram is shown. Which dye is the most polar?
Feedback if chosen: Correct. The polar paper grips the most polar dye hardest while the nonpolar solvent carries it least, so the most polar dye travels least — the lowest spot, Z.
Feedback if chosen: This option is wrong — you picked the middle spot. The strongest grip shows at the bottom of the strip: on polar paper with a nonpolar solvent, the most polar dye is the one that traveled least.
Feedback if chosen: This option is wrong — you ranked the farthest-traveling spot as most polar. A dye that rides the nonpolar solvent to the top is the one held least by the polar paper — the least polar dye.
Feedback if chosen: This option is wrong — you set the positions aside. Both phase polarities are given here, so the positions do rank the dyes: the lowest spot is the most polar.
Check — multiple choice
Two pen inks, S and T, were run on the same polar thin-layer coating with a nonpolar solvent; the chromatogram is shown. Which ink is more polar?
Feedback if chosen: This option is wrong — you ranked the higher spot as more polar. S traveled far because the polar coating held it back least — S is the less polar ink.
Feedback if chosen: This option is wrong — you ignored the height difference. Equal polarity would mean the two phases pull on both inks in the same proportions, so the spots would stop at the same height.
Feedback if chosen: Correct. T traveled least, so the polar coating grips it hardest relative to the nonpolar solvent — T is the more polar ink.
Feedback if chosen: This option is wrong — you let the solvent's polarity stand in for the inks'. The solvent's polarity is a property of the mobile phase; the inks' polarities are read from how far each spot traveled.
Check — multiple choice
The chromatogram shows three plant pigments, D, E, and F, run on polar paper with a nonpolar solvent. Which list orders them from most polar to least polar?
Feedback if chosen: This option is wrong — you ranked the farthest-traveling pigment as most polar. On polar paper with a nonpolar solvent, the most polar pigment is held back hardest and travels least — the ranking starts at the bottom of the strip.
Feedback if chosen: This option is wrong — you listed the pigments in label order instead of reading their positions. The ranking comes from the spots: lowest is most polar, highest is least.
Feedback if chosen: This option is wrong — you found the most polar pigment correctly, then read the remaining spots from the top down. Keep reading upward from the bottom: F, then D, then E.
Feedback if chosen: Correct. Reading upward from the bottom of the strip, F traveled least (most polar), D is next, and E traveled farthest (least polar).
Outcome: Describe how distillation works: heat a liquid mixture so that vapor forms, then collect and condense that vapor in a separate container.
A ship's crew has tanks of seawater — salt water — and needs drinking water.
You've already seen that filtration cannot help: the dissolved ions pass through the filter paper with the water.
The water has to be pulled out of the solution itself — and heating gives a way to do it.
The figure shows the apparatus: a flask of salt water sits over a burner, and its only exit is a side arm near the top of the neck.
Heating the flask makes the liquid boil, and vapor forms in the space above it.
The vapor has nowhere to go but the side arm, which slopes away from the flask through a tube wrapped in a jacket of cold water.
The cold tube cools the vapor until it turns back into liquid.
That change — vapor turning back into liquid — is called 'condensation', and the cooled tube is called the 'condenser'.
The condensed liquid drips out of the condenser into a receiving flask, drop by drop.
Run it long enough and the receiving flask fills with nearly pure water, while the salt stays behind in the boiling flask.
The whole method — heating a liquid mixture so that vapor forms, then collecting and condensing that vapor — is called 'distillation'.
Describe distillation: heat a liquid mixture so that vapor forms, then collect that vapor and condense it (return it to liquid) in a separate container.
Check — multiple choice
In the distillation apparatus shown, what happens inside the highlighted tube?
Feedback if chosen: This option is wrong — you put the burner's job in the condenser. The highlighted tube is jacketed in cold water; it cools what passes through, it never heats it.
Feedback if chosen: Correct. The tube is the condenser: its cold-water jacket cools the vapor until it condenses, and the liquid drips on into the receiving flask.
Feedback if chosen: This option is wrong — you had the vapor stay a gas. The cold jacket takes energy out of the vapor, so it condenses to liquid before it reaches the receiving flask.
Feedback if chosen: This option is wrong — you turned the condenser into a filter. Nothing is caught on the walls; the tube's job is to cool the vapor back into liquid.
Check — multiple choice
Which sequence puts the steps of a distillation in order?
Feedback if chosen: Correct. Distillation is heat, then transport, then condense: vapor forms over the boiling liquid, travels to the condenser, and returns to liquid in a separate container.
Feedback if chosen: This option is wrong — you put the steps in reverse order. Distillation begins with heating the liquid mixture; condensation comes at the end, when the vapor is cooled.
Feedback if chosen: This option is wrong — you described a cooling separation. Distillation never freezes anything: it makes vapor by heating, then condenses that vapor.
Feedback if chosen: This option is wrong — you skipped the condensation step. A distillation is only complete when the vapor is cooled back into liquid and collected.
Check — multiple choice
Sugar water is distilled until half of the liquid has boiled away. What has collected in the receiving flask?
Feedback if chosen: This option is wrong — you sent the whole solution across. Only the vapor travels through the side arm, and the dissolved sugar stays behind in the boiling flask.
Feedback if chosen: This option is wrong — you sent the sugar across instead of the water. The dissolved sugar stays in the boiling flask; what reaches the condenser is the vapor from the water.
Feedback if chosen: Correct. Only the vapor travels to the condenser, and the dissolved sugar stays behind in the boiling flask — so the condensed liquid is nearly pure water.
Feedback if chosen: This option is wrong — you let the sugar begin to travel once the solution concentrated. However concentrated the solution becomes, the dissolved sugar cannot enter the vapor; only water reaches the condenser.
Outcome: Explain why the vapor above a heated liquid mixture is enriched in the component with the weaker intermolecular attractions, and why distillation therefore separates.
Distilling salt water collects nearly pure water — but why does the salt stay behind?
Everything in the receiving flask arrived as vapor, so the real question is which particles escape the boiling liquid.
You've already seen that vapor pressure decreases as intermolecular force strength increases, because vaporizing requires completely overcoming intermolecular attractions, so stronger attractions mean fewer particles escape and more energy is needed.
In salt water, each Na⁺ and Cl⁻ ion is gripped by strong ion-dipole attractions to the water molecules around it.
The water molecules hold each other through hydrogen bonds — strong for molecules, but far weaker than the grip on the ions.
So when the solution is heated, the particles that escape into the vapor are almost entirely water molecules; the ions essentially never break free.
The vapor above heated salt water is nearly pure water — and condensing that vapor collects nearly pure water.
The same reasoning covers any liquid mixture: the component with the weaker intermolecular attractions escapes more easily, so the vapor is enriched in that component.
You've already seen that acetone's vapor pressure at 25 °C is about ten times water's, because acetone's attractions are weaker.
Heat an acetone-water mixture, and the vapor above it is much richer in acetone than the liquid below it is.
Condense that vapor, and the collected liquid is enriched in acetone — distillation separates by exploiting the difference in attractions.
Explain why distillation separates: components with weaker intermolecular attractions have higher vapor pressures, so the vapor above a heated mixture is enriched in the weaker-attraction component.
Check — multiple choice
When salt water boils, why is the vapor above it nearly pure water?
Feedback if chosen: This option is wrong — you placed the selection at the return step. The ions almost never escape in the first place: leaving the liquid means completely overcoming their strong attractions to the water molecules, and almost none can.
Feedback if chosen: This option is wrong — you made size the deciding factor. What keeps an ion in the liquid is the strength of its attractions, not its fit: escaping requires completely overcoming those attractions.
Feedback if chosen: This option is wrong — you settled the salt out. The dissolved ions are spread evenly through the solution, surface included; they stay because their attractions are too strong to break.
Feedback if chosen: Correct. Each ion is gripped by strong ion-dipole attractions, far stronger than the hydrogen bonds between water molecules — so the particles that escape into the vapor are almost all water molecules.
Check — multiple choice
| Component | Intermolecular attractions |
|---|---|
| A | hydrogen bonds plus London dispersion forces |
| B | London dispersion forces only |
Liquids A and B are mixed and heated; the figure compares their intermolecular attractions. Which statement describes the vapor above the mixture?
Feedback if chosen: This option is wrong — you enriched the vapor in the more strongly held component. A's hydrogen bonds hold its particles back; the component that escapes more easily is B.
Feedback if chosen: Correct. B's particles are held only by London dispersion forces, weaker than A's hydrogen-bonded network, so more B particles escape — the vapor is enriched in B.
Feedback if chosen: This option is wrong — you gave both components the same escape rate. Their attractions differ in strength, so their escape rates differ, and the vapor's composition shifts toward B.
Feedback if chosen: This option is wrong — you stretched 'enriched' to 'pure'. Some A particles still escape; B's weaker attractions make its share of the vapor larger, not total.
Check — multiple choice
A mixture of acetone and water is distilled slowly. As the run goes on, what happens to the liquid left in the boiling flask?
Feedback if chosen: Correct. Acetone's weaker attractions mean acetone particles escape into the vapor faster than water molecules do, so the liquid left behind shifts toward water.
Feedback if chosen: This option is wrong — you enriched the flask in the component that is leaving fastest. Acetone escapes preferentially, so the flask holds proportionally less of it as the run goes on.
Feedback if chosen: This option is wrong — you gave both components the same escape rate. Acetone's attractions are weaker, so it leaves the liquid faster than water does, and the flask's composition shifts.
Feedback if chosen: This option is wrong — you turned 'escapes more easily' into 'escapes instantly'. Water molecules escape too, so the acetone leaves gradually over the whole run, not all at once.
Outcome: Explain the outcome of a separation experiment — a spot pattern, an elution order, or a distillation order — from the intermolecular interactions of the components.
You've already seen why chromatography separates: a component attracted more strongly to the stationary phase travels more slowly.
You've already seen why distillation separates: the vapor above a heated mixture is enriched in the component with the weaker intermolecular attractions.
Every separation result — where the spots stopped, what left the column first, what boiled over first — is explained the same way: compare the strengths of the attractions involved.
In column chromatography, the components eventually wash out of the bottom of the column, one after another.
A component is said to 'elute' when it leaves the column, and the order in which the components leave is the 'elution order'.
The reasoning is the paper strip's reasoning turned on its side: a component held weakly by the packing and carried readily by the solvent elutes first.
On a polar column packing run with a nonpolar solvent, the least polar component elutes first and the most polar elutes last.
Given a result, work backward: name the competing attractions, decide which component wins each, and state how that produces the observed order.
For a spot pattern or an elution order, the competition is stationary phase against mobile phase; for a distillation order, it is each component's own attractions against escape into the vapor.
Explain the outcome of a separation experiment — a spot pattern, an elution order, or a distillation order — from the intermolecular interactions of the components.
Check — multiple choice
Two marker dyes, M and K, were run on polar paper with a nonpolar solvent; the chromatogram is shown. Why did dye M stop above dye K?
Feedback if chosen: This option is wrong — you made weight the deciding factor. Position on the strip is set by the balance of attractions to the two phases, not by how heavy the molecules are.
Feedback if chosen: This option is wrong — you turned a strong attraction into a push. A polar dye is pulled toward the polar paper and held back by it; nothing on the strip pushes a component upward.
Feedback if chosen: Correct. The polar paper grips the more polar dye (K) hardest, while the less polar dye (M) rides the nonpolar solvent farther before stopping.
Feedback if chosen: This option is wrong — you invented two different destinations. Both dyes travel the same route — held by the paper, carried by the solvent — and only the balance of those attractions differs.
Check — multiple choice
A mixture of two dyes is passed through a polar column packing with a nonpolar solvent; dye V elutes first and dye W elutes long after. What explains the order?
Feedback if chosen: This option is wrong — you reversed the polarity reading. A more polar dye is gripped by the polar packing and elutes late; the early dye is the one the packing holds least.
Feedback if chosen: This option is wrong — you made size the deciding factor. Elution order is set by the balance of each dye's attractions to the packing and the solvent, not by particle size.
Feedback if chosen: This option is wrong — you explained the order with a head start. The dyes enter together in one sample, and a long gap between elutions can only come from different attractions to the two phases.
Feedback if chosen: Correct. The polar packing grips the more polar dye (W) hardest, so W spends most of its time held in place while the less polar V rides the nonpolar solvent out first.
Check — multiple choice
| Liquid | Boiling point |
|---|---|
| ethanol | 78 °C |
| water | 100 °C |
A mixture of ethanol and water is distilled; the figure gives both boiling points. The first liquid collected is far richer in ethanol than the mixture was. Why?
Feedback if chosen: Correct. The boiling points show ethanol's attractions are the weaker ones, so ethanol particles escape more readily — the vapor, and the first liquid condensed from it, is enriched in ethanol.
Feedback if chosen: This option is wrong — you made size the deciding factor. An ethanol molecule, C₂H₅OH, is larger than a water molecule; ethanol escapes more readily because its attractions are weaker.
Feedback if chosen: This option is wrong — you had the components leave one at a time. Both escape throughout the run; ethanol's weaker attractions just give it the larger share of the vapor.
Feedback if chosen: This option is wrong — you made the condenser selective. The cold tube condenses whatever vapor reaches it; the enrichment happens earlier, when ethanol particles escape the liquid more readily.
Check — multiple choice
| Liquid | Intermolecular attractions |
|---|---|
| hexane | London dispersion forces only |
| water | hydrogen bonds plus London dispersion forces |
A flask holding hexane and water is warmed gently, and the vapor above the liquid is sampled; it is far richer in hexane. The figure gives each liquid's attractions. Why is the vapor richer in hexane?
Feedback if chosen: This option is wrong — you aimed water's attractions at the hexane. Hydrogen bonds hold water molecules to each other; hexane escapes readily because its own attractions are weak.
Feedback if chosen: Correct. The vapor is enriched in hexane because vaporizing requires completely overcoming intermolecular attractions, so stronger attractions mean fewer particles escape and more energy is needed — and hexane's London-dispersion-only attractions are far weaker than water's hydrogen-bonded network.
Feedback if chosen: This option is wrong — you made mass the deciding factor. A hexane molecule, C₆H₁₄, is heavier than a water molecule; hexane escapes more readily because its attractions are weaker.
Feedback if chosen: This option is wrong — you switched vapor off below the boiling point. Liquids form vapor at any temperature — water has a vapor pressure at 25 °C — so the enrichment comes from hexane's weaker attractions, not from water contributing nothing.
4 lessons
Outcome: State why substances with similar types and strengths of intermolecular interactions tend to dissolve in one another, and use the terms miscible and soluble.
Shake olive oil with water in one jar, and ethanol with water in another: the figure shows what each pair looks like a minute later.
| Liquid pair | One minute after shaking |
|---|---|
| Olive oil + water | Two separate layers |
| Ethanol + water | One uniform liquid — still mixed |
However hard you shake the first jar, the oil gathers back into its own layer.
Ethanol never separates: it mixes with water completely, in any proportions.
Both are liquids poured into the same water — so what decides which pair mixes?
You've already seen how to work out which intermolecular interactions a pair of species can form, and how to compare their strengths.
Before any mixing, each liquid holds itself together with its own interactions: water hydrogen-bonds with water, and ethanol's O–H groups hydrogen-bond with ethanol.
Mixing forces a trade: ethanol-ethanol and water-water contacts are given up, and ethanol-water contacts form in their place.
Ethanol offers water a fair trade, because its O–H group hydrogen-bonds with water about as strongly as the hydrogen bonds each liquid gave up.
The mixture holds together as well as the separate liquids did, so the two liquids mix completely.
Olive oil offers water a bad trade: its nonpolar molecules bring only London dispersion forces, far weaker than the hydrogen bonds water would have to give up.
Water molecules hold on to each other instead, and the oil is pushed out into its own layer.
Substances with similar types and strengths of intermolecular interactions tend to dissolve in one another, because dissolving replaces solute-solute and solvent-solvent interactions with solute-solvent interactions, which is only favorable when the interactions are of similar type and strength.
Chemists sum this rule up as 'like dissolves like'.
Two liquids that mix completely in any proportions, like ethanol and water, are called 'miscible'.
A substance that dissolves in a given solvent is 'soluble' in that solvent.
State why substances with similar types and strengths of intermolecular interactions tend to be miscible or soluble in one another: dissolving replaces solute-solute and solvent-solvent interactions with solute-solvent interactions of similar type and strength.
Check — multiple choice
Methanol hydrogen-bonds with its own molecules, just as water does. When methanol dissolves in water, what happens to the intermolecular interactions?
Feedback if chosen: This option is wrong — you left the mixture with no attractions at all. The old contacts are given up, but new methanol-water contacts form in their place.
Feedback if chosen: Correct. Dissolving is a trade: solute-solute and solvent-solvent interactions are replaced by solute-solvent interactions, and here the trade is fair because both pairs hydrogen-bond.
Feedback if chosen: This option is wrong — you turned dissolving into a chemical change. The methanol molecules stay intact; only the attractions between molecules change.
Feedback if chosen: This option is wrong — you invented a strengthening that doesn't happen. The hydrogen bonds in the mixture are of similar strength to the ones given up, not stronger.
Check — multiple choice
Acetone and water mix completely, in any proportions. Which statement describes this pair of liquids?
Feedback if chosen: This option is wrong — you picked the term for a mixture whose properties depend on location. A pair that mixes completely forms one uniform liquid throughout.
Feedback if chosen: This option is wrong — immiscible describes the opposite behavior: liquids that settle into separate layers instead of mixing.
Feedback if chosen: This option is wrong — insoluble means a substance does not dissolve, but the acetone plainly does mix into the water.
Feedback if chosen: Correct. Liquids that mix completely in any proportions are miscible.
Check — multiple choice
Mineral oil is a nonpolar liquid. Shaken with water, it settles back into a separate layer. Why do the two liquids stay separate?
Feedback if chosen: Correct. Water would have to give up strong hydrogen bonds and gain only much weaker attractions to the oil — a bad trade, so the liquids stay separate.
Feedback if chosen: This option is wrong — you blamed molecular size. Ethanol molecules are larger than water molecules and still mix completely; the interaction match, not size, decides.
Feedback if chosen: This option is wrong — you turned the separation into a chemical change. No reaction happens; each liquid simply attracts its own molecules more strongly than the other's.
Feedback if chosen: This option is wrong — you have the strength backwards. Water's hydrogen bonds are strong, and giving them up for weaker oil-water attractions is exactly what makes mixing unfavorable.
Outcome: Predict whether a molecular substance dissolves in a given solvent by matching the interaction types of the solute and the solvent.
You've already seen that substances with similar interactions tend to dissolve in one another, because dissolving replaces solute-solute and solvent-solvent interactions with solute-solvent interactions, which is only favorable when the interactions are of similar type and strength.
That rule turns solubility prediction into a matching problem: list the interactions the solute can form, list the solvent's, and see whether they match.
A nonpolar molecular substance attracts its neighbors only through London dispersion forces.
A polar molecular substance adds dipole-dipole attractions — and hydrogen bonding, if it has hydrogen bonded directly to N, O, or F.
Vegetable oil is nonpolar, hexane is nonpolar, and water hydrogen-bonds — the figure shows what the oil does in each solvent.
| Solvent | Solvent's interactions | Does the oil dissolve? |
|---|---|---|
| Hexane | London dispersion forces only | Yes |
| Water | Hydrogen bonds | No |
In hexane the match is good: oil-oil and hexane-hexane London dispersion forces are traded for oil-hexane attractions of similar strength, so the oil dissolves.
In water the match fails: water would have to give up hydrogen bonds and gain only weak oil-water attractions, so the oil does not dissolve.
The same matching works for any molecular solute: nonpolar solutes dissolve in nonpolar solvents, and polar or hydrogen-bonding solutes dissolve in polar or hydrogen-bonding solvents.
Predict whether a molecular substance will dissolve in a given solvent by matching the intermolecular interaction types of the solute and the solvent.
Check — multiple choice
Naphthalene, the strong-smelling solid in old-fashioned mothballs, is made of nonpolar molecules. In which solvent will naphthalene dissolve?
Feedback if chosen: This option is wrong — you matched a nonpolar solute to a hydrogen-bonding solvent. Water would have to give up hydrogen bonds for weak naphthalene-water attractions, so it refuses the trade.
Feedback if chosen: This option is wrong — you ignored the mismatch with water. Only hexane offers naphthalene the same kind of London dispersion forces it already forms.
Feedback if chosen: Correct. Naphthalene and hexane both attract through London dispersion forces of similar strength, so the trade is fair and the solid dissolves.
Feedback if chosen: This option is wrong — you treated a solid as undissolvable. A solid dissolves whenever the solvent's interactions match its own, and hexane's do.
Check — multiple choice
Citric acid molecules carry several O–H groups that form hydrogen bonds. In which solvent will citric acid dissolve?
Feedback if chosen: Correct. Citric acid's O–H groups hydrogen-bond with water, replacing like with like; hexane offers nothing that could replace those hydrogen bonds.
Feedback if chosen: This option is wrong — you matched a hydrogen-bonding solute to a solvent with only London dispersion forces. Hexane cannot replace citric acid's hydrogen bonds.
Feedback if chosen: This option is wrong — you overlooked the match with water. Citric acid and water both hydrogen-bond, so solute-solvent contacts replace like with like.
Feedback if chosen: This option is wrong — you gave hexane a match it cannot make. Its molecules form no hydrogen bonds, so citric acid gains nothing by mixing into hexane.
Check — multiple choice
Candle wax is made of large nonpolar molecules, and it does not dissolve in water. Why not?
Feedback if chosen: This option is wrong — you blamed the mass of the molecules. Heavy molecules dissolve readily when the interactions match; the mismatch with water is what stops the wax.
Feedback if chosen: This option is wrong — you made the wax's own forces the obstacle. A nonpolar solvent with similar London dispersion forces does dissolve wax; the failure here is the mismatch with water.
Feedback if chosen: This option is wrong — you invented a chemical change. No reaction occurs; the wax and the water simply attract their own molecules more strongly than each other's.
Feedback if chosen: Correct. The trade is uneven: water would give up strong hydrogen bonds and gain only weak attractions to the wax, so the wax stays undissolved.
Outcome: Predict whether an ionic compound dissolves in a solvent from whether the solvent can form ion-dipole attractions with the ions.
An ionic solid is held together by strong attractions between its cations and anions.
For the solid to dissolve, the solvent must offer the ions new attractions that can replace those ion-ion attractions.
You've already seen ion-dipole interactions: attractions between an ion and the oppositely charged partial end of a polar molecule.
Only a polar solvent can make that offer, because only a polar molecule has oppositely charged partial ends for the ions to attract.
The figure shows NaCl dissolving in water: each Na⁺ attracts the partially negative O ends of the water molecules around it, and each Cl⁻ attracts the partially positive H ends.
Many ion-dipole attractions form around each ion, and together they replace the ion-ion attractions of the lattice — NaCl dissolves in water.
Hexane's nonpolar molecules have no charged partial ends, so hexane cannot form ion-dipole attractions with the ions.
Nothing can replace the strong ion-ion attractions, so the ions stay in the lattice — NaCl does not dissolve in hexane.
Ion-dipole attractions are what a solvent must offer for an ionic solid to dissolve at all: without them, as in a nonpolar solvent, the solid stays undissolved.
With them, dissolving becomes possible but not guaranteed — how much solid actually dissolves also depends on how strongly the lattice holds its ions, a comparison left for later study.
So to predict whether an ionic compound can dissolve in a solvent, ask whether the solvent can form ion-dipole attractions with the ions.
Predict whether an ionic compound can dissolve in a given solvent from whether the solvent can form the ion-dipole attractions that dissolving requires.
Check — multiple choice
Potassium bromide, KBr, is an ionic solid. In which liquid will it dissolve?
Feedback if chosen: This option is wrong — you have the match reversed. Hexane's nonpolar molecules cannot form ion-dipole attractions, so they offer the K⁺ and Br⁻ ions nothing.
Feedback if chosen: This option is wrong — you let any liquid dissolve an ionic solid. Only a polar solvent can form the ion-dipole attractions the ions need.
Feedback if chosen: This option is wrong — you treated the ion-ion attractions as unbeatable. Water's many ion-dipole attractions can replace them, so water does dissolve KBr.
Feedback if chosen: Correct. Polar water forms ion-dipole attractions with K⁺ and Br⁻ that replace the ion-ion attractions; nonpolar hexane cannot form any.
Check — multiple choice
Calcium chloride, CaCl₂, is an ionic solid, and it does not dissolve in hexane. Why not?
Feedback if chosen: This option is wrong — you blamed the space available. Ions dissolve or stay put according to the attractions on offer, not the room between molecules; hexane has room but nothing to attract the ions with.
Feedback if chosen: Correct. With no polar ends, hexane offers the ions no ion-dipole attractions, so nothing replaces the strong ion-ion attractions of the lattice.
Feedback if chosen: This option is wrong — you restricted London dispersion forces to nonpolar partners. LDFs act between all particles, ions included; they are simply far too weak to replace the ion-ion attractions of the lattice.
Feedback if chosen: This option is wrong — you added a melting requirement that doesn't exist. NaCl dissolves in room-temperature water without melting; the solvent's interactions, not the solid's state, decide.
Check — multiple choice
Which property of a liquid decides whether an ionic compound can dissolve in it?
Feedback if chosen: This option is wrong — you made molecular size the test. A solvent of any molecular size fails if it cannot attract the ions.
Feedback if chosen: This option is wrong — you ranked the wrong interaction. London dispersion forces act between any particles, but they are far too weak to replace ion-ion attractions.
Feedback if chosen: Correct. A polar solvent's oppositely charged partial ends form ion-dipole attractions with the ions — the replacement that lets the lattice come apart.
Feedback if chosen: This option is wrong — you reasoned from density. Density does not measure attractions; a solvent dissolves an ionic solid only if its molecules are polar.
Outcome: Explain how well a substance dissolves in water compared with a nonpolar solvent from the intermolecular interactions involved.
You've already seen how to predict whether a molecular substance dissolves: match the interaction types of the solute and the solvent.
You've already seen that an ionic compound dissolves only where the solvent can form ion-dipole attractions with its ions.
The same logic compares two solvents: a substance dissolves better in whichever solvent offers interactions closer in type and strength to the ones the solute already makes.
Iodine, I₂, is nonpolar: its molecules attract each other only through London dispersion forces.
The figure shows what iodine crystals do in each solvent.
| Solvent | Observation |
|---|---|
| Hexane | Dissolves readily — the solution turns purple |
| Water | Barely dissolves — the crystals mostly remain |
Hexane offers iodine London dispersion forces of similar strength to the ones it gives up — a fair trade, so iodine dissolves readily.
Water offers a bad trade: it would have to give up hydrogen bonds and gain only weak iodine-water attractions, so iodine barely dissolves.
So iodine is far more soluble in hexane than in water.
Sucrose — table sugar — picks the other solvent: its many O–H groups hydrogen-bond with water, so it dissolves readily in water and not at all in hexane.
Ethanol picks both: its O–H group hydrogen-bonds with water, while its carbon chain offers hexane London dispersion forces of similar strength — ethanol is miscible with water and with hexane.
To explain any relative solubility, compare the two trades and ask which solvent replaces the solute's interactions with the closest match.
Explain how well a substance dissolves in water compared with a nonpolar solvent, using the intermolecular interactions between the particles involved.
Check — multiple choice
| Solvent | Observation |
|---|---|
| Hexane | Mixes in readily |
| Water | Only a little dissolves; the rest stays as a separate layer |
Bromine, Br₂, is a nonpolar liquid. The figure shows how it behaves in two solvents. Why does bromine dissolve so much better in hexane than in water?
Feedback if chosen: Correct. Nonpolar bromine offers only London dispersion forces; hexane trades like for like, while water would swap strong hydrogen bonds for weak bromine-water attractions.
Feedback if chosen: This option is wrong — you reasoned from density. Density decides where an undissolved layer sits, not whether a substance dissolves; the interaction match does that.
Feedback if chosen: This option is wrong — you made Br₂ polar. Its two identical atoms share electrons evenly, so the molecule is nonpolar and attracts only through London dispersion forces.
Feedback if chosen: This option is wrong — you reasoned from empty space. Solvents don't hold ready-made gaps; the trade of attractions, not room, decides solubility.
Check — multiple choice
| Solvent | Observation |
|---|---|
| Water | Mixes in completely |
| Hexane | Does not mix — two layers form |
Glycerol's molecules carry three O–H groups. The figure shows its behavior in two solvents. Why does glycerol mix completely with water but not with hexane?
Feedback if chosen: This option is wrong — you made fit the test. Glycerol molecules are larger than water molecules and mix anyway; the match of interactions, not size, decides.
Feedback if chosen: This option is wrong — you turned London dispersion forces into a barrier. They are weak attractions, not a defense; the problem is that they cannot replace glycerol's hydrogen bonds.
Feedback if chosen: Correct. In water the trade swaps hydrogen bonds for hydrogen bonds; in hexane glycerol would trade its hydrogen bonds for much weaker attractions, so it refuses.
Feedback if chosen: This option is wrong — you reasoned from density. Density plays no part; the match of interaction types decides where a solute dissolves.
Check — multiple choice
1-Propanol has an O–H group at one end of a three-carbon chain. It mixes completely with water and also with hexane. Why can it mix with both?
Feedback if chosen: This option is wrong — you let all liquids mix. Olive oil and water are both liquids and refuse to mix; being liquid guarantees nothing.
Feedback if chosen: Correct. The molecule carries a matching offer for each solvent: hydrogen bonding for water, and London dispersion forces of similar strength for hexane.
Feedback if chosen: This option is wrong — you turned dissolving into a chemical change. The molecules stay intact; no new compounds form when 1-propanol mixes into either solvent.
Feedback if chosen: This option is wrong — you made fit the test. Room between molecules is not what decides; 1-propanol mixes because it can make a fair trade of attractions with either solvent.
Check — multiple choice
Lithium chloride, LiCl, is an ionic solid. It dissolves readily in water but not at all in hexane. Which comparison explains the difference?
Feedback if chosen: This option is wrong — you have the strengths reversed. Hexane's London dispersion forces are weak; the real difference is that hexane has no polar ends to attract the ions at all.
Feedback if chosen: This option is wrong — you broke covalent bonds that stay intact. Water molecules survive dissolving unchanged; their oppositely charged partial ends, not freed atoms, attract the ions.
Feedback if chosen: This option is wrong — you made fit the test. Hexane has room between its molecules but no attraction to offer the ions; the attractions, not the space, decide.
Feedback if chosen: Correct. Only polar water can replace the lattice's ion-ion attractions with ion-dipole attractions, so LiCl dissolves in water and stays solid in hexane.
Check — multiple choice
Which of these samples is a heterogeneous mixture?
Feedback if chosen: This option is wrong — you classified by the sample being a mixture at all. Its composition is the same everywhere in the sample, so it is a solution — a homogeneous mixture.
Feedback if chosen: This option is wrong — you ruled a solid out of being a solution. The gold and copper are mixed evenly, so every part of the ring has the same properties — a solid solution.
Feedback if chosen: Correct. The juice's properties depend on location — more pulp near the bottom than near the top — which is exactly what makes a mixture heterogeneous.
Feedback if chosen: This option is wrong — you kept the dissolved sugar as a separate region. Fully dissolved, it is spread evenly, so the tea's macroscopic properties are the same everywhere — a solution.
Check — multiple choice
A bottle holds 0.500 L of a solution labeled '0.220 M fructose'. What does the 0.220 M on the label state?
Feedback if chosen: This option is wrong — you read the molarity as a total amount. It is a ratio — moles of solute per liter of solution — so this 0.500 L bottle holds only 0.110 mol.
Feedback if chosen: Correct. Molarity is the moles of solute per liter of solution, so 0.220 M means 0.220 mol of fructose in each liter of the solution.
Feedback if chosen: This option is wrong — you divided by the water that went in. Molarity divides by the volume of the finished solution, which can differ from the volume of water added.
Feedback if chosen: This option is wrong — you swapped grams in for moles. The unit M counts moles of solute per liter of solution, not grams.
Check — multiple choice
How many formula units of NaNO₃ are in 0.250 L of a 0.080 M NaNO₃ solution? (Avogadro's number = 6.022 × 10²³ mol⁻¹.)
Feedback if chosen: Correct. Moles first: 0.080 M × 0.250 L = 0.020 mol; then 0.020 × 6.022 × 10²³ = 1.20 × 10²² formula units.
Feedback if chosen: This option is wrong — you treated the molarity as the moles present and skipped the volume. First find the moles: 0.080 × 0.250 = 0.020 mol.
Feedback if chosen: This option is wrong — you divided the molarity by the volume. Moles of solute = molarity × volume: 0.080 × 0.250 = 0.020 mol.
Feedback if chosen: This option is wrong — you divided the volume by the molarity, inverting the ratio. Moles of solute = molarity × volume: 0.080 × 0.250 = 0.020 mol.
Check — multiple choice
A student is asked to draw particulate models of two solutions in same-size boxes: box A for 0.20 M KI and box B for 0.60 M KI. The drawing is shown. Which evaluation is correct?
Feedback if chosen: This option is wrong — you read the comparison qualitatively. The counts must scale with the molarity: three times the molarity means three times the ions in the same-size box.
Feedback if chosen: This option is wrong — you re-paired the ions. Dissolved KI exists as separate K⁺ and I⁻ ions spread through the solution, exactly as both boxes show.
Feedback if chosen: This option is wrong — you let concentration change the size of a particle. Each K⁺ and I⁻ stays its own size; a higher concentration means more ions in the same volume, not bigger ones.
Feedback if chosen: Correct. Relative concentration is carried by the counts: 0.60 M is three times 0.20 M, so a same-size box needs three times box A's ions — 6 K⁺ and 6 I⁻ — but this drawing shows only 4 of each.
Check — multiple choice
A student has four mixtures. From which one can filtration remove a component?
Feedback if chosen: This option is wrong — you set the paper to catch dissolved ions. The dissolved ions are far smaller than the gaps between the paper's fibers, so they pass through with the water.
Feedback if chosen: Correct. Sand grains are enormous compared with a filter's gaps, so the paper holds them back while the seawater passes through — filtration separates particles, not dissolved substances.
Feedback if chosen: This option is wrong — you set the paper to catch a dissolved substance. The coloring's particles are molecule-sized, far smaller than the paper's gaps; separating them from the water means exploiting differences in intermolecular attractions instead.
Feedback if chosen: This option is wrong — you set the paper to catch a dissolved liquid. The ethanol molecules pass through the paper with the water; separating them means exploiting differences in intermolecular attractions instead.
Check — multiple choice
Three candy-shell colorings, P, Q, and R, were carried up a polar coating by a nonpolar solvent; the developed chromatogram is shown. Which list orders the colorings from most polar to least polar?
Feedback if chosen: This option is wrong — you gave the most polar coloring the longest ride. On a polar coating with a nonpolar solvent, the most polar coloring grips the coating hardest and stops lowest — that is R, not P.
Feedback if chosen: This option is wrong — you read one carrier as one polarity. The three spots stopped at different heights precisely because the colorings grip the polar coating with different strengths.
Feedback if chosen: Correct. The polar coating holds the most polar coloring back hardest, so the lowest spot, R, is the most polar and the highest spot, P, is the least polar.
Feedback if chosen: This option is wrong — you asked for more than the question needs. With both phase polarities given, the stopping heights alone rank the colorings' polarities.
Check — multiple choice
Heptane (C₇H₁₆) and decane (C₁₀H₂₂) are nonpolar liquids that mix completely. When the mixture is warmed, the vapor above it is much richer in heptane. Why?
Feedback if chosen: Correct. Molecules of both liquids reach the surface, but heptane's smaller electron cloud gives it weaker London dispersion forces, so at the same temperature more heptane molecules break free into the vapor.
Feedback if chosen: This option is wrong — you made mass the price of escape. What a molecule must overcome to leave the liquid is the attractions to its neighbors, and it is heptane's weaker London dispersion forces, not its lighter mass, that let more of its molecules break free.
Feedback if chosen: This option is wrong — you broke covalent bonds to make vapor. Vaporizing overcomes only the attractions between molecules; every C–H bond stays intact.
Feedback if chosen: This option is wrong — you separated the liquids by density. The two liquids mix completely, so molecules of both are at the surface; decane simply escapes less readily.
Check — multiple choice
Two dyes dissolved in water are passed through a column packed with nonpolar beads, with water flowing through the column as the solvent. Dye 1, whose molecules are polar, comes out almost immediately; dye 2, whose molecules are nonpolar, comes out much later. What explains the order?
Feedback if chosen: This option is wrong — you made size the separator. The column separates by attraction: whichever dye is held to the beads more strongly, relative to the water, comes out later.
Feedback if chosen: This option is wrong — you turned a weak attraction into a repulsion. The beads do attract dye 1 — every particle attracts every other through London dispersion forces — but dye 1's stronger pull toward the polar water wins.
Feedback if chosen: This option is wrong — you gave the polar water the wrong partner. Polar water interacts more strongly with the polar dye; the nonpolar dye is the one it holds least.
Feedback if chosen: Correct. Dye 2 spends the run gripped by the nonpolar beads through London dispersion forces of matched type, while polar dye 1 sticks with the moving water — so dye 1 exits first.
Check — multiple choice
Bicycle grease is made of large nonpolar molecules. A mechanic can soak a greasy part in kerosene, whose hydrocarbon molecules are nonpolar, or in water. In which liquid will the grease dissolve?
Feedback if chosen: Correct. Grease and kerosene trade London dispersion forces of similar type and strength, so mixing is favorable; water would have to give up strong hydrogen bonds for weak grease-water attractions.
Feedback if chosen: This option is wrong — you matched the nonpolar grease to the polar solvent. Dissolving in water would replace water's hydrogen bonds with much weaker grease-water attractions — an unfavorable trade.
Feedback if chosen: This option is wrong — you let both solvents work. Kerosene works, but water would have to give up its strong hydrogen bonds for far weaker grease-water attractions, so the grease stays out of water.
Feedback if chosen: This option is wrong — you shut both solvents out. Kerosene's nonpolar molecules swap London dispersion forces of similar strength with the grease, so that trade is favorable and the grease dissolves.
Check — multiple choice
Magnesium bromide, MgBr₂, is an ionic solid. Two liquids are available: water, whose molecules are polar, and cyclohexane, whose molecules are nonpolar. In which liquid will MgBr₂ dissolve?
Feedback if chosen: This option is wrong — you sent the ions into the nonpolar liquid. Cyclohexane's molecules have no oppositely charged partial ends, so no ion-dipole attractions can form to replace the ion-ion attractions in the solid.
Feedback if chosen: This option is wrong — you let cyclohexane dissolve an ionic solid. Only the polar water offers charged partial ends; cyclohexane cannot form ion-dipole attractions with the ions.
Feedback if chosen: This option is wrong — you shut water out. Water's polar molecules form many ion-dipole attractions with Mg²⁺ and Br⁻, replacing the ion-ion attractions that dissolving breaks.
Feedback if chosen: Correct. Water's polar molecules point oppositely charged partial ends at the ions and form ion-dipole attractions that replace the ion-ion attractions; nonpolar cyclohexane cannot form them at all.
Check — multiple choice
Solid iodine, I₂, barely dissolves in water yet dissolves readily in hexane. Which comparison of the intermolecular interactions explains the difference?
Feedback if chosen: This option is wrong — you dissolved I₂ as if it were ionic. I₂ enters solution as whole molecules — and if ions did form, water, not hexane, would be the better partner, through ion-dipole attractions.
Feedback if chosen: Correct. Dissolving swaps interactions: I₂-hexane London dispersion forces match the ones both substances give up, but water would trade away strong hydrogen bonds for weak I₂-water attractions — an unfavorable exchange.
Feedback if chosen: This option is wrong — you made solvent size the test. Many small water molecules can easily ring a large solute; what stops the dissolving is the unfavorable trade of water's hydrogen bonds for weak I₂-water attractions.
Feedback if chosen: This option is wrong — you turned dissolving into a reaction. The I₂ molecules stay I₂ in hexane; they simply mingle with hexane molecules through London dispersion forces.
8 lessons
Outcome: Describe electromagnetic radiation as a wave, giving its wavelength as the crest-to-crest distance and its frequency as the number of waves passing a point each second.
A microwave oven, a heat lamp, and a flashlight all send energy across space as electromagnetic radiation.
One warms soup, one warms skin, and one lights a room — so the radiation itself must differ from source to source.
The figure draws a stretch of radiation from two different sources, to the same scale.
Electromagnetic radiation travels as a wave: a repeating pattern of crests and troughs moving through space.
The two drawings clearly differ — and the difference can be measured.
Along each wave, the distance from one crest to the next crest stays the same.
That crest-to-crest distance is called the wave's 'wavelength', written with the symbol λ.
On each wave in the figure, the bracket labeled λ spans exactly one wavelength.
The upper wave carries the longer wavelength — that is how the two drawings differ.
Red light, for example, has a wavelength near 700 nm.
A drawing freezes the wave, but a real wave moves.
Stand at one fixed point and count the complete waves that pass you in one second.
That count is called the wave's 'frequency'.
Wavelength is a distance you could mark on a still picture.
Frequency is a count per second — it only shows up while the wave is moving.
Describe electromagnetic radiation as a wave, giving its wavelength (the crest-to-crest distance) and its frequency (the number of waves passing a point each second).
Check — multiple choice
The figure shows two electromagnetic waves, A and B, drawn to the same scale. Which statement about their wavelengths is correct?
Feedback if chosen: Correct. The crest-to-crest distance of wave A is about twice that of wave B, so wave A has the longer wavelength.
Feedback if chosen: This option is wrong — you counted wave B's larger number of crests as more wavelength. Fitting more crests into the same width means the crest-to-crest distance is shorter, not longer.
Feedback if chosen: This option is wrong — you compared the heights of the crests. Both waves rise equally high, but wavelength is measured along the wave, from one crest to the next.
Feedback if chosen: This option is wrong — wavelength is a distance, and the two waves are drawn to the same scale, so their crest-to-crest distances can be compared directly.
Check — multiple choice
Which of these is the frequency of an electromagnetic wave?
Feedback if chosen: This option is wrong — the crest-to-crest distance is the wavelength. Frequency is a count per second, not a distance.
Feedback if chosen: Correct. Frequency counts how many complete waves pass a fixed point in one second.
Feedback if chosen: This option is wrong — the height of a crest describes how tall the wave is, not how often the wave repeats at a point.
Feedback if chosen: This option is wrong — a distance covered per second is a speed. Frequency counts passing waves; it does not measure how far the wave moves.
Check — multiple choice
Wave X and wave Y are electromagnetic waves. Each second, more complete waves of X than of Y pass a fixed detector. Which quantity is larger for wave X than for wave Y?
Feedback if chosen: This option is wrong — the detector counted passing waves; it measured no distance along the wave. A count per second is a frequency, not a wavelength.
Feedback if chosen: This option is wrong — how many waves pass each second does not set how tall the crests are.
Feedback if chosen: This option is wrong — you turned a count into a distance covered per second. Counting complete waves passing a fixed point each second is the definition of frequency.
Feedback if chosen: Correct. The number of complete waves passing a fixed point each second is the wave's frequency, so wave X has the higher frequency.
Outcome: State that light behaves as a stream of photons, individual packets each carrying a definite amount of light energy.
Shine a red beam on a wall, then turn the source down until the beam is barely visible.
The wall now receives far less light energy each second.
There are two ways that could happen: the light could arrive weakened, or less of the same light could arrive.
Sensitive detectors settle the question.
However dim the beam gets, its energy arrives in separate bursts of a fixed size — never in ever-smaller amounts.
Light behaves as a stream of individual packets, each carrying a definite amount of light energy.
Each packet is called a 'photon'.
A photon delivers its energy all at once, as one whole packet.
Dimming the beam does not weaken each photon.
A dim beam delivers fewer photons each second than a bright beam of the same color.
You've already seen that light travels as a wave.
The photon picture does not replace the wave picture — light behaves both ways, and chemistry uses both.
State that light also behaves as a stream of photons — individual packets each carrying a definite amount of light energy.
Check — multiple choice
A dim beam and a bright beam of the same color shine on the same surface. How do the two beams differ?
Feedback if chosen: This option is wrong — dimming does not weaken each packet. The beams have the same color, so their photons are alike; the dim beam simply delivers fewer of them each second.
Feedback if chosen: Correct. Light comes in whole packets, so a dimmer beam of the same color means fewer photons arriving each second.
Feedback if chosen: This option is wrong — light never arrives as a smooth, continuous flow. However dim the beam, its energy still arrives photon by photon.
Feedback if chosen: This option is wrong — you tied dimness to wavelength. The two beams have the same color; dimness changes how many photons arrive each second, not the wave that describes them.
Check — multiple choice
Which of these describes a photon?
Feedback if chosen: Correct. A photon is one individual packet of light, carrying a definite amount of light energy.
Feedback if chosen: This option is wrong — the crest-to-crest distance is the wavelength, a property of the wave picture, not a packet of energy.
Feedback if chosen: This option is wrong — a photon is part of the light itself, not a piece of the object the light strikes.
Feedback if chosen: This option is wrong — the count of waves passing a point each second is the frequency, not an energy packet.
Check — multiple choice
A detector receives an extremely dim beam of one color and records its energy arriving as separate, equal-size bursts. Which statement explains the record?
Feedback if chosen: This option is wrong — you kept the smooth-flow picture. The bursts are real: light delivers its energy in whole packets, and a sensitive detector shows exactly that.
Feedback if chosen: This option is wrong — you blamed a mixture of colors. The beam is a single color, and equal-size bursts show the beam arrives as identical packets.
Feedback if chosen: Correct. Light is a stream of photons, so even the dimmest beam delivers its energy one whole packet at a time.
Feedback if chosen: This option is wrong — photons do not split. A photon delivers its energy all at once or not at all, which is why every burst is the same size.
Outcome: Rank the spectral regions microwave, infrared, visible, and ultraviolet by frequency.
You've already seen that electromagnetic radiation travels as a wave described by its wavelength and its frequency.
Radiation from different sources can differ enormously in frequency.
The figure lines up electromagnetic radiation from the lowest frequencies on the left to the highest on the right.
Each named stretch of the bar is called a 'spectral region'.
The whole ordered bar is called the 'electromagnetic spectrum'.
Neighboring regions blend into each other — there is no sharp edge where one region ends and the next begins.
Four regions matter most in this course: microwave, infrared, visible, and ultraviolet.
The figure outlines two of the four: microwave and ultraviolet.
Of the four, microwave sits lowest in frequency.
Ultraviolet sits highest.
Infrared and visible sit between them, in that order.
So, in order of increasing frequency, the four regions run: microwave, infrared, visible, ultraviolet.
The names carry a memory aid.
'Infrared' means below red — red is the lowest-frequency light the eye can see, and infrared sits just below it.
'Ultraviolet' means beyond violet — violet is the highest-frequency light the eye can see, and ultraviolet sits just above it.
Rank the spectral regions microwave, infrared, visible, and ultraviolet by frequency: microwave lowest, then infrared, then visible, then ultraviolet.
Check — multiple choice
On the spectrum shown, the infrared and visible regions are outlined. Which statement about them is correct?
Feedback if chosen: This option is wrong — you read the bar against the arrow. Visible sits further along the direction of increasing frequency than infrared.
Feedback if chosen: This option is wrong — the two regions occupy different stretches of the bar, so their frequencies differ.
Feedback if chosen: Correct. The arrow marks the direction of increasing frequency, and visible sits further along it than infrared.
Feedback if chosen: This option is wrong — you tied frequency to brightness. Brightness sets how much radiation arrives, not which region it belongs to; each region keeps its own frequency range whatever the source.
Check — multiple choice
Ranking the four regions microwave, infrared, visible, and ultraviolet by frequency, which region has the second-lowest frequency?
Feedback if chosen: This option is wrong — microwave has the lowest frequency of the four, not the second-lowest.
Feedback if chosen: Correct. In order of increasing frequency the four run microwave, infrared, visible, ultraviolet — infrared is second-lowest.
Feedback if chosen: This option is wrong — visible sits above both microwave and infrared, making it second-highest, not second-lowest.
Feedback if chosen: This option is wrong — ultraviolet has the highest frequency of the four, at the opposite end of the ranking.
Check — multiple choice
Of the four regions microwave, infrared, visible, and ultraviolet, which statement about the two ends of the frequency ranking is correct?
Feedback if chosen: This option is wrong — you wrote the ranking in reverse. Frequency increases toward ultraviolet, so microwave sits at the bottom and ultraviolet at the top.
Feedback if chosen: This option is wrong — you stopped the ranking at visible. Ultraviolet sits above visible in frequency, so ultraviolet, not visible, holds the top end.
Feedback if chosen: This option is wrong — you swapped microwave and infrared at the bottom. Microwave sits below infrared, so microwave holds the low end.
Feedback if chosen: Correct. Of the four regions, microwave has the lowest frequency and ultraviolet the highest.
Outcome: State that atoms and molecules can hold only specific allowed energies, with allowed levels for rotation and vibration as well as for their electrons.
Pluck a guitar string gently, then slightly harder: the string can vibrate with any energy in between.
A vibrating molecule seems like it should behave the same way.
It does not.
The figure shows the vibration energies actually available to one molecule, drawn as horizontal lines on an energy scale.
Each line marks an energy the molecule is allowed to have.
The gaps between the lines are energies the molecule can never hold.
So the molecule's vibration energy changes in steps rather than continuously.
An energy restricted to specific allowed values like this is called 'quantized'.
The allowed values themselves are called 'energy levels'.
You've already seen this picture for atoms: an atom's electrons sit in shells, occupying only specific allowed energy levels like the ones shown.
Molecules keep that restriction on their electrons and add two more.
A molecule's rotation — how fast it turns end over end — is quantized.
A molecule's vibration — its atoms moving back and forth along their bonds — is quantized as well.
Rotation, vibration, and the electrons' energies all come in allowed levels.
State that atoms and molecules can hold only specific allowed energies — quantized levels for rotation and vibration as well as for their electrons.
Check — multiple choice
Which statement describes the vibration energy of a CO₂ molecule?
Feedback if chosen: This option is wrong — you gave the guitar-string picture. A molecule's vibration energy is quantized: values between the allowed levels are never available.
Feedback if chosen: This option is wrong — changing slowly does not unlock the in-between values. However the energy changes, it can only land on allowed levels.
Feedback if chosen: This option is wrong — this describes which partners the molecule trades energy with, not which energies it is allowed to hold.
Feedback if chosen: Correct. Only specific vibration energies are allowed, so the energy moves between levels in steps.
Check — multiple choice
The figure shows the allowed vibration energies of a molecule. The molecule currently holds energy E₂. Which energy could it hold at some later moment?
Feedback if chosen: Correct. E₃ is one of the allowed levels, so the molecule can hold it; every value between the lines is off limits.
Feedback if chosen: This option is wrong — the halfway point falls in a gap. The gaps between the lines are energies the molecule can never hold.
Feedback if chosen: This option is wrong — slightly above E₂ still lands in a gap. The next available value going up is E₃ itself.
Feedback if chosen: This option is wrong — staying below E₄ is not enough. Only the drawn levels are allowed, not the values between them.
Check — multiple choice
Which energies of a molecule are quantized?
Feedback if chosen: This option is wrong — you carried the atomic picture over unchanged. Molecules add allowed levels for rotation and vibration on top of the electronic ones.
Feedback if chosen: This option is wrong — vibration energy is quantized, but so are the energies of rotation and of the electrons.
Feedback if chosen: Correct. A molecule's rotation, vibration, and electron energies all come in specific allowed levels.
Feedback if chosen: This option is wrong — smooth change is exactly what quantization rules out. Each kind of energy moves between allowed levels in steps.
Outcome: State that microwave radiation is associated with transitions between molecular rotational energy levels.
You've already seen that a molecule's rotation is quantized — it turns end over end only with specific allowed energies.
You've also seen that light delivers its energy as photons.
The figure shows a molecule's allowed rotational energy levels, with a photon arriving.
When the molecule absorbs the photon, it jumps from its rotational energy level up to a higher one.
At the higher level, the molecule rotates faster.
A jump between two allowed energy levels is called a 'transition'.
The photons that drive rotational transitions come from one spectral region: microwave.
Microwave radiation is associated with transitions between molecular rotational energy levels.
A microwave oven uses these photons, but the water in food is liquid, where each molecule constantly pushes on its neighbors — so the absorbed energy spreads through the food as heat rather than leaving each molecule simply spinning faster.
The clean picture — one photon, one molecule jumping to a faster rotation — belongs to water molecules in the gas phase, far from any neighbors.
The association covers both directions: a molecule dropping from a higher rotational energy level to a lower one releases microwave radiation.
State that microwave radiation is associated with transitions between molecular rotational energy levels.
Check — multiple choice
A water molecule in a sample of water vapor absorbs microwave radiation. What changes in the molecule?
Feedback if chosen: This option is wrong — back-and-forth motion along the bonds is vibration, and that is not the motion microwave radiation is associated with. Microwave absorption changes rotation.
Feedback if chosen: This option is wrong — you reached for the electrons. Microwave photons drive transitions between rotational energy levels, not electronic ones.
Feedback if chosen: Correct. Absorbing microwave radiation moves the molecule to a higher rotational energy level, so it rotates faster.
Feedback if chosen: This option is wrong — absorbing microwave radiation changes the molecule's rotation, not its makeup.
Check — multiple choice
Which spectral region is associated with transitions between a molecule's rotational energy levels?
Feedback if chosen: This option is wrong — you paired rotation with the highest-frequency region of the four. Rotational transitions pair with microwave radiation.
Feedback if chosen: This option is wrong — visible light is not the rotation region. Rotational transitions pair with microwave radiation.
Feedback if chosen: This option is wrong — infrared sits next to microwave on the spectrum, but the rotation pairing belongs to microwave itself.
Feedback if chosen: Correct. Microwave radiation is associated with transitions between rotational energy levels.
Outcome: State that infrared radiation is associated with transitions between molecular vibrational energy levels.
You've already seen that a molecule's vibration is quantized — its atoms move back and forth along the bonds only with specific allowed energies.
You've also seen that microwave radiation pairs with rotational transitions.
The figure shows a molecule's allowed vibrational energy levels, with a photon arriving.
When the molecule absorbs the photon, it jumps from its vibrational energy level up to a higher one.
At the higher level, the atoms swing back and forth more energetically.
The photons that drive vibrational transitions come from the infrared region.
Infrared radiation is associated with transitions between molecular vibrational energy levels.
Absorbing infrared radiation makes a CO₂ molecule vibrate more energetically.
The association covers both directions: a molecule dropping from a higher vibrational energy level to a lower one releases infrared radiation.
State that infrared radiation is associated with transitions between molecular vibrational energy levels.
Check — multiple choice
A CO₂ molecule in the air absorbs infrared radiation. What changes in the molecule?
Feedback if chosen: This option is wrong — you reached for the electrons. Infrared photons drive transitions between vibrational energy levels, not electronic ones.
Feedback if chosen: Correct. Absorbing infrared radiation moves the molecule to a higher vibrational energy level, so its atoms vibrate more energetically.
Feedback if chosen: This option is wrong — turning end over end is rotation, and you've already seen that rotational transitions pair with microwave radiation. Infrared absorption changes vibration.
Feedback if chosen: This option is wrong — absorbing infrared radiation changes the molecule's vibration, not its makeup.
Check — multiple choice
Absorbing radiation moves a molecule from one vibrational energy level up to the next. Which spectral region is that radiation from?
Feedback if chosen: Correct. Infrared radiation is associated with transitions between vibrational energy levels.
Feedback if chosen: This option is wrong — you've already seen that microwave radiation pairs with rotational transitions. Vibrational transitions pair with infrared.
Feedback if chosen: This option is wrong — visible light is not the vibration region. Vibrational transitions pair with infrared radiation.
Feedback if chosen: This option is wrong — you paired vibration with the highest-frequency region of the four. Vibrational transitions pair with infrared radiation.
Outcome: State that ultraviolet and visible radiation are associated with transitions between electronic energy levels.
A drop of dye in water absorbs some of the light passing through it.
Which change inside a dye molecule takes in that light energy?
You've already seen that a molecule's electrons, like an atom's, occupy specific allowed energy levels.
The figure shows electronic energy levels, with a photon arriving.
When the molecule absorbs the photon, one electron moves from its energy level up to a higher one.
Moving an electron up to a higher energy level is called 'promoting' the electron.
The photons that drive electronic transitions come from two neighboring spectral regions: visible and ultraviolet.
Ultraviolet and visible radiation are associated with transitions between electronic energy levels.
A dye absorbs visible light when an electron is promoted to a higher energy level.
The association covers both directions: an electron dropping from a higher electronic energy level to a lower one releases ultraviolet or visible radiation.
State that ultraviolet and visible radiation are associated with transitions between electronic energy levels.
Check — multiple choice
A dye molecule absorbs visible light. What changes in the molecule?
Feedback if chosen: This option is wrong — faster turning is a rotational transition, and you've already seen that pairing belongs to microwave radiation. Visible light promotes an electron.
Feedback if chosen: This option is wrong — more energetic back-and-forth motion is a vibrational transition, which pairs with infrared radiation. Visible light promotes an electron.
Feedback if chosen: This option is wrong — the molecule keeps its own electrons. Absorption promotes one of them to a higher energy level; it does not add one.
Feedback if chosen: Correct. Visible light drives electronic transitions: absorbing it promotes one electron to a higher energy level.
Check — multiple choice
Which pair of spectral regions is associated with transitions between electronic energy levels?
Feedback if chosen: Correct. Ultraviolet and visible radiation are both associated with electronic transitions.
Feedback if chosen: This option is wrong — these are the two lowest-frequency regions of the four, and you've already seen they pair with rotational and vibrational transitions, not electronic ones.
Feedback if chosen: This option is wrong — visible belongs in the pair, but infrared pairs with vibrational transitions. Visible's partner is ultraviolet.
Feedback if chosen: This option is wrong — ultraviolet belongs in the pair, but microwave pairs with rotational transitions. Ultraviolet's partner is visible.
Check — multiple choice
A molecule absorbs ultraviolet radiation. Which description fits what happens?
Feedback if chosen: This option is wrong — electron energies are quantized, so no smooth, shared gain is available. One electron jumps between allowed levels.
Feedback if chosen: This option is wrong — faster rotation is a rotational transition, which pairs with microwave radiation. Ultraviolet drives electronic transitions.
Feedback if chosen: Correct. Ultraviolet radiation is associated with electronic transitions: one electron is promoted between allowed levels.
Feedback if chosen: This option is wrong — more energetic vibration is a vibrational transition, which pairs with infrared radiation. Ultraviolet drives electronic transitions.
Outcome: Identify the transition type associated with a given spectral region, and the spectral region associated with a given transition type.
You've already seen each spectral pairing on its own.
The table collects all three.
| Spectral region | Transition type |
|---|---|
| microwave | rotational |
| infrared | vibrational |
| ultraviolet and visible | electronic |
Read the table in either direction.
Given a spectral region, name the transition it drives: infrared absorption, for example, means a vibrational transition.
Given a transition, name the spectral region: a molecule starting to rotate faster has absorbed microwave radiation.
Whichever direction a transition goes, the region stays the same.
A drop between two levels releases radiation from the same region whose absorption drives the upward jump.
Identify the transition type associated with a given spectral region, and the spectral region associated with a given transition type.
Check — multiple choice
The figure shows a molecule dropping between two of its rotational energy levels and releasing a photon. From which spectral region is the released radiation?
Feedback if chosen: This option is wrong — infrared pairs with vibrational transitions. A drop between rotational energy levels releases microwave radiation.
Feedback if chosen: Correct. Rotational transitions pair with microwave radiation in both directions — absorption on the way up, emission on the way down.
Feedback if chosen: This option is wrong — visible pairs with electronic transitions. A drop between rotational energy levels releases microwave radiation.
Feedback if chosen: This option is wrong — ultraviolet pairs with electronic transitions. A drop between rotational energy levels releases microwave radiation.
Check — multiple choice
Sunscreen is tested with ultraviolet radiation, and a molecule in it absorbs some. Which change in the molecule is associated with that absorption?
Feedback if chosen: This option is wrong — faster turning is a rotational transition, which pairs with microwave radiation, not ultraviolet.
Feedback if chosen: This option is wrong — more energetic back-and-forth motion is a vibrational transition, which pairs with infrared radiation, not ultraviolet.
Feedback if chosen: Correct. Ultraviolet radiation pairs with electronic transitions, so absorbing it promotes an electron to a higher energy level.
Feedback if chosen: This option is wrong — you turned absorption into breakage. The ultraviolet pairing is an electronic transition: an electron promoted between allowed levels, with the molecule intact.
Check — multiple choice
In a lab instrument, radiation makes CO molecules jump to higher vibrational energy levels. Which spectral region is the instrument using?
Feedback if chosen: This option is wrong — microwave radiation pairs with rotational transitions. Vibrational jumps pair with infrared radiation.
Feedback if chosen: This option is wrong — visible light pairs with electronic transitions. Vibrational jumps pair with infrared radiation.
Feedback if chosen: This option is wrong — ultraviolet pairs with electronic transitions. Vibrational jumps pair with infrared radiation.
Feedback if chosen: Correct. Vibrational transitions pair with infrared radiation in both directions.
Check — multiple choice
A glowing sign releases visible light as its molecules lose energy. Which transition inside the molecules releases that light?
Feedback if chosen: Correct. Visible light pairs with electronic transitions: an electron dropping between energy levels releases the light.
Feedback if chosen: This option is wrong — a drop between rotational energy levels releases microwave radiation, not visible light.
Feedback if chosen: This option is wrong — a drop between vibrational energy levels releases infrared radiation, not visible light.
Feedback if chosen: This option is wrong — molecular energies are quantized, so energy leaves in whole photons as transitions happen, not as a smooth, continuous flow.
9 lessons
Outcome: Predict which of two electromagnetic radiations has the shorter wavelength from their frequencies, using the fact that all electromagnetic radiation travels at the same speed.
You've already seen that electromagnetic radiation travels as a wave, with a wavelength (the crest-to-crest distance) and a frequency (the number of waves passing a point each second).
In a vacuum, every electromagnetic wave — radio, visible, ultraviolet, all of them — travels at exactly the same speed.
The figure shows two electromagnetic waves drawn to the same scale, one with twice the wavelength of the other.
Imagine both waves streaming past a fixed point at the same speed.
Because the shorter wave packs less length into each cycle, its crests pass the point more often.
More crests passing each second is a higher frequency.
So, because all electromagnetic radiation travels at the same speed, radiation of higher frequency always has a shorter wavelength.
You've already seen the frequency ranking microwave < infrared < visible < ultraviolet — the wavelength order is the exact reverse.
Ultraviolet light, with its higher frequency, has a shorter wavelength than infrared light.
Predict which of two electromagnetic radiations has the shorter wavelength, from their frequencies and the fact that both travel at the same speed.
Check — multiple choice
The figure shows two electromagnetic waves, A and B, drawn to the same scale and traveling at the same speed. Which wave has the higher frequency?
Feedback if chosen: This option is wrong — you matched the longer wavelength with the higher frequency. At a shared speed, longer waves pass a point less often, so wave A has the lower frequency.
Feedback if chosen: Correct. Both waves travel at the same speed, so the wave with the shorter wavelength delivers more crests each second — a higher frequency.
Feedback if chosen: This option is wrong — you treated frequency as independent of wavelength. At a shared speed, equal frequencies would need equal wavelengths, and these differ.
Feedback if chosen: This option is wrong — you assumed the speeds might differ. All electromagnetic radiation in a vacuum travels at the same speed, so the drawn wavelengths decide it.
Check — multiple choice
Radio waves have a lower frequency than microwaves. Which statement about their wavelengths is correct?
Feedback if chosen: Correct. Both travel at the same speed, so the lower-frequency radio waves must be the longer ones — fewer, longer waves pass a point each second.
Feedback if chosen: This option is wrong — you matched the lower frequency with the shorter wavelength. At a shared speed, fewer waves passing each second means each wave must be longer.
Feedback if chosen: This option is wrong — you treated wavelength as independent of frequency. At a shared speed, different frequencies force different wavelengths.
Feedback if chosen: This option is wrong — you assumed the speeds might differ. All electromagnetic radiation in a vacuum travels at the same speed, so the given frequencies decide it.
Check — multiple choice
Two electromagnetic waves travel through a vacuum. Wave X delivers 4 × 10⁶ crests to a detector each second, and wave Y delivers 8 × 10⁶ crests each second. Which statement is correct?
Feedback if chosen: This option is wrong — you matched more crests per second with a longer wavelength. A higher frequency at the same speed forces a shorter wavelength.
Feedback if chosen: This option is wrong — you treated the crest counts as irrelevant. Equal wavelengths at the same speed would deliver equal crest counts, and these differ.
Feedback if chosen: Correct. Wave X has half the frequency of wave Y, so at the shared speed each of its waves must be twice as long.
Feedback if chosen: This option is wrong — you turned the frequency difference into a speed difference. All electromagnetic radiation in a vacuum travels at the same speed.
Outcome: State that wavelength and frequency of electromagnetic radiation are related to the speed of light by c = λν, with c = 3.00 × 10⁸ m/s.
You've already seen that, because all electromagnetic radiation travels at the same speed, higher-frequency radiation has a shorter wavelength.
That trade is exact: for any electromagnetic radiation in a vacuum, the wavelength multiplied by the frequency always gives the same number.
That number is the shared speed itself, called the speed of light: c = 3.00 × 10⁸ m/s.
Written as an equation, this is the 'wave equation': c = λν.
The symbol λ (lambda) stands for the wavelength, measured in meters.
The symbol ν (nu) stands for the frequency, measured in s⁻¹ — waves per second.
The product must be the speed: each second, ν full waves pass a point, and each of those waves is λ meters long.
The wave therefore moves forward λ × ν meters every second — and that distance covered per second is its speed, c.
Because c is fixed, doubling the wavelength must halve the frequency, and halving the wavelength must double the frequency.
State that wavelength and frequency are related to the speed of light by c = λν, where c = 3.00 × 10⁸ m/s.
Check — multiple choice
In a vacuum, radiation P has twice the wavelength of radiation Q. What does c = λν say about the frequency of P?
Feedback if chosen: Correct. The product λν must equal the fixed value c for both radiations, so doubling the wavelength halves the frequency.
Feedback if chosen: This option is wrong — you doubled the frequency along with the wavelength. The product λν is fixed at c, so one doubling forces the other to halve.
Feedback if chosen: This option is wrong — you let the product λν change. It cannot: for both radiations, λν = c.
Feedback if chosen: This option is wrong — you halved the frequency twice over. Doubling λ requires halving ν exactly once to keep λν = c.
Check — numeric entry
For any electromagnetic radiation traveling through a vacuum, the product λ × ν always comes out to the same fixed value. What is that value, in units of 10⁸ m/s? Give your answer to 3 significant figures.
Tolerance: 0.005
Feedback when correct: Correct. Wavelength × frequency = the speed of light, c = 3.00 × 10⁸ m/s, for every electromagnetic radiation in a vacuum.
Anticipated wrong answers (feedback shown):
| 186000 | You gave the speed of light in miles per second. The wave equation uses meters per second: c = 3.00 × 10⁸ m/s, so the box takes 3.00. |
| 300 | You folded the wrong power of ten into the box. The speed is 3.00 × 10⁸ m/s, and the box already supplies the 10⁸ — enter 3.00. |
| 340 | That is close to the speed of sound in air, in m/s. Light travels almost a million times faster: c = 3.00 × 10⁸ m/s. |
Check — multiple choice
Orange light has a wavelength of 600 nm; blue light has a wavelength of 450 nm. In a vacuum, how do their λ × ν products compare?
Feedback if chosen: This option is wrong — you let the longer wavelength win the product. Its frequency is proportionally lower, so the product comes out the same.
Feedback if chosen: This option is wrong — you let the higher frequency win the product. Its wavelength is proportionally shorter, so the product comes out the same.
Feedback if chosen: This option is wrong — you treated the product as an unknown. For any electromagnetic radiation in a vacuum, λν equals c — no further measurement needed.
Feedback if chosen: Correct. Both products equal the speed of light, 3.00 × 10⁸ m/s — that is exactly what c = λν says.
Outcome: Calculate a wavelength from a frequency, or a frequency from a wavelength, by rearranging c = λν.
You've already seen the wave equation: for electromagnetic radiation in a vacuum, c = λν, with c = 3.00 × 10⁸ m/s.
A question gives you one of the two wave measurements and asks for the other.
Whatever it asks for, start from c = λν and rearrange it to make that quantity the subject.
Wavelengths of light are usually quoted in nanometers, and 1 nm = 10⁻⁹ m.
The equation needs the wavelength in meters, because c is a distance in meters covered each second.
If the question gives the wavelength in nanometers, convert it to meters first, as its own labeled step.
The powers of ten get their own line: divide or multiply the leading numbers first, then combine the powers of ten.
Q: What is the frequency of light with a wavelength of 500 nm? (c = 3.00 × 10⁸ m/s.)
Write down the values from the question
λ = 500 nm
c = 3.00 × 10⁸ m/s
Convert the wavelength to meters
λ = 500 × 10⁻⁹ m
λ = 5.00 × 10⁻⁷ m
Write down the equation
c = λν
Make ν the subject
ν = c / λ
Substitute in the values, and calculate
ν = (3.00 × 10⁸) / (5.00 × 10⁻⁷)
ν = (3.00 / 5.00) × 10⁸⁺⁷(dividing by 10⁻⁷ is multiplying by 10⁷)
ν = 0.600 × 10¹⁵
ν = 6.00 × 10¹⁴ s⁻¹
Q: An FM radio station broadcasts at a frequency of 1.00 × 10⁸ s⁻¹. What is the wavelength of its radio waves? (c = 3.00 × 10⁸ m/s.)
Write down the values from the question
ν = 1.00 × 10⁸ s⁻¹
c = 3.00 × 10⁸ m/s
Write down the equation
c = λν
Make λ the subject
λ = c / ν
Substitute in the values, and calculate
λ = (3.00 × 10⁸) / (1.00 × 10⁸)
λ = (3.00 / 1.00) × 10⁸⁻⁸(the powers of ten cancel)
λ = 3.00 m
Calculate a wavelength from a frequency, or a frequency from a wavelength, by rearranging c = λν.
Check — numeric entry
Light of wavelength 600 nm shines from a lamp. What is the frequency of this light, in units of 10¹⁴ s⁻¹? (c = 3.00 × 10⁸ m/s.) Give your answer to 3 significant figures.
Tolerance: 0.005
Feedback when correct: Correct. Converting 600 nm to 6.00 × 10⁻⁷ m and dividing c by the wavelength gives ν = 0.500 × 10¹⁵ = 5.00 × 10¹⁴ s⁻¹.
Anticipated wrong answers (feedback shown):
| 1.80 | You multiplied c by the wavelength instead of dividing by it. Making ν the subject gives ν = c / λ. |
| 2.00 | You divided the wavelength by c, which is the rearrangement upside down. Making ν the subject gives ν = c / λ — the speed goes on top. |
| 0.500 | Your arithmetic is right, but the answer is not matched to the 10¹⁴ supplied by the answer box: 0.500 × 10¹⁵ s⁻¹ is 5.00 × 10¹⁴ s⁻¹. |
Worked solution revealed after a second miss
Write down the values from the question
λ = 600 nm
c = 3.00 × 10⁸ m/s
Convert the wavelength to meters
λ = 600 × 10⁻⁹ m
λ = 6.00 × 10⁻⁷ m
Write down the equation
c = λν
Make ν the subject
ν = c / λ
Substitute in the values, and calculate
ν = (3.00 × 10⁸) / (6.00 × 10⁻⁷)
ν = (3.00 / 6.00) × 10⁸⁺⁷
ν = 0.500 × 10¹⁵
ν = 5.00 × 10¹⁴ s⁻¹
Check — numeric entry
An electromagnetic wave has a frequency of 4.00 × 10¹⁴ s⁻¹. What is its wavelength, in units of 10⁻⁷ m? (c = 3.00 × 10⁸ m/s.) Give your answer to 3 significant figures.
Tolerance: 0.005
Feedback when correct: Correct. λ = c / ν = (3.00 × 10⁸) / (4.00 × 10¹⁴) = 0.750 × 10⁻⁶ = 7.50 × 10⁻⁷ m.
Anticipated wrong answers (feedback shown):
| 1.33 | You divided the frequency by c, which is the rearrangement upside down. Making λ the subject gives λ = c / ν — the speed goes on top. |
| 1.20 | You multiplied c by the frequency instead of dividing by it. Making λ the subject gives λ = c / ν. |
| 0.750 | Your arithmetic is right, but the answer is not matched to the 10⁻⁷ supplied by the answer box: 0.750 × 10⁻⁶ m is 7.50 × 10⁻⁷ m. |
Worked solution revealed after a second miss
Write down the values from the question
ν = 4.00 × 10¹⁴ s⁻¹
c = 3.00 × 10⁸ m/s
Write down the equation
c = λν
Make λ the subject
λ = c / ν
Substitute in the values, and calculate
λ = (3.00 × 10⁸) / (4.00 × 10¹⁴)
λ = (3.00 / 4.00) × 10⁸⁻¹⁴
λ = 0.750 × 10⁻⁶
λ = 7.50 × 10⁻⁷ m
Check — multiple choice
A question gives a wavelength of 350 nm, already converted to 3.50 × 10⁻⁷ m, and asks for the frequency. Which expression gives the frequency, in s⁻¹? (c = 3.00 × 10⁸ m/s.)
Feedback if chosen: This option is wrong — you multiplied by the wavelength instead of dividing by it. Making ν the subject of c = λν gives ν = c / λ.
Feedback if chosen: This option is wrong — you divided the wavelength by the speed, which is the rearrangement upside down: ν = c / λ, with the speed on top.
Feedback if chosen: Correct. Making ν the subject of c = λν gives ν = c / λ, the speed divided by the wavelength in meters.
Feedback if chosen: This option is wrong — you treated c = λν as a sum. The equation is a product, so the rearrangement divides: ν = c / λ.
Outcome: Predict which of two photons carries more energy from the frequencies of their radiations.
The figure marks two regions of the electromagnetic spectrum: infrared, and higher-frequency ultraviolet.
An infrared heat lamp can warm your skin all afternoon without harming it, yet a few minutes of ultraviolet sunlight is enough to start a burn.
Brightness cannot be the reason — the heat lamp delivers far more light energy in total than the faint ultraviolet edge of sunlight does.
You've already seen that light arrives as photons, and that a dim beam simply delivers fewer photons per second than a bright one.
So the difference must sit inside the individual photons themselves.
Photons of higher-frequency radiation each carry more energy.
As a radiation's frequency increases, the energy of each of its photons increases.
An ultraviolet photon therefore carries more energy than an infrared photon — enough, one photon at a time, to damage the molecules in skin.
Brightness sets only how many photons arrive each second, while frequency sets how much energy each one carries.
You've already seen the frequency ranking microwave < infrared < visible < ultraviolet — that is also the photon-energy ranking, lowest to highest.
Predict that a photon's energy increases as the radiation's frequency increases.
Check — multiple choice
One photon of visible light and one photon of microwave radiation arrive at a detector. Which photon carries more energy?
Feedback if chosen: Correct. Visible light sits above microwave radiation in the frequency ranking, and higher frequency means each photon carries more energy.
Feedback if chosen: This option is wrong — you ranked microwave radiation above visible light in frequency. The ranking is microwave < infrared < visible < ultraviolet.
Feedback if chosen: This option is wrong — you treated all photons as identical packets. A photon's energy depends on its radiation's frequency.
Feedback if chosen: This option is wrong — you tied photon energy to brightness. Brightness sets how many photons arrive each second, not the energy of each one.
Check — multiple choice
A dim beam of ultraviolet light and a bright beam of infrared light shine on the same surface. Which statement is correct?
Feedback if chosen: This option is wrong — you let the beam's brightness set its photons' energy. Frequency alone sets photon energy, and ultraviolet is the higher-frequency radiation.
Feedback if chosen: This option is wrong — you treated photons as identical packets. Photon energy rises with the radiation's frequency.
Feedback if chosen: This option is wrong — you flipped the brightness. The dim beam delivers less light energy each second, so it delivers far fewer photons.
Feedback if chosen: Correct. Photon energy depends only on the radiation's frequency, and ultraviolet sits above infrared — brightness cannot change what each photon carries.
Check — multiple choice
The figure shows two electromagnetic waves, R and S, drawn to the same scale and traveling at the same speed. Photons of which wave carry more energy?
Feedback if chosen: This option is wrong — you matched the longer wavelength with the higher photon energy. At the same speed, the longer wave has the lower frequency, so its photons carry less energy.
Feedback if chosen: Correct. The shorter wavelength at the same speed means a higher frequency, and higher frequency means higher photon energy.
Feedback if chosen: This option is wrong — you treated the two radiations as alike. Equal photon energies would need equal frequencies, and these wavelengths differ.
Feedback if chosen: This option is wrong — you set the drawing aside. The shared speed lets wavelength stand in for frequency, so the drawing decides it.
Outcome: State Planck's equation E = hν, which ties a photon's energy to the radiation's frequency through Planck's constant.
You've already seen that a photon's energy increases as the radiation's frequency increases.
That relationship is exact, and it is a straight proportion: a photon's energy equals the frequency multiplied by one fixed number.
Written as an equation, this is 'Planck's equation': E = hν.
The fixed number h is called 'Planck's constant': h = 6.626 × 10⁻³⁴ J·s.
The same h works for every kind of electromagnetic radiation, from radio waves to gamma rays.
E is the energy of a single photon, in joules — not the energy of the whole beam.
Because h is so small, one photon carries a very small number of joules, even for high-frequency radiation.
Doubling the frequency doubles each photon's energy — the proportion is straight through the origin.
State Planck's equation E = hν, which relates a photon's energy to the radiation's frequency through Planck's constant h.
Check — multiple choice
In Planck's equation E = hν, what does E stand for?
Feedback if chosen: This option is wrong — you read E as the whole beam's energy. The beam's total depends on brightness; E in Planck's equation is the energy of one photon.
Feedback if chosen: Correct. Planck's equation gives the energy carried by one photon of radiation with frequency ν.
Feedback if chosen: This option is wrong — you tied E to the source. Planck's equation describes the radiation itself: each photon of frequency ν carries E = hν.
Feedback if chosen: This option is wrong — you tied E to the absorber. Planck's equation sets each photon's energy before anything absorbs it.
Check — multiple choice
The frequency of one radiation is three times the frequency of another. According to Planck's equation, how do their photon energies compare?
Feedback if chosen: This option is wrong — you inverted the proportion. E = hν climbs with frequency, so the higher-frequency photons carry more energy, not less.
Feedback if chosen: This option is wrong — you squared the factor of three. E = hν is a straight proportion in ν, so tripling ν triples E.
Feedback if chosen: Correct. E = hν is a straight proportion: tripling the frequency triples each photon's energy.
Feedback if chosen: This option is wrong — you read frequency as a delivery rate of identical packets. Frequency sets each photon's energy, not how often photons arrive.
Check — multiple choice
Which quantity in E = hν has the same value for every photon of every electromagnetic radiation?
Feedback if chosen: This option is wrong — you froze the quantity the equation is built to vary. E changes with frequency; that is the whole content of the equation.
Feedback if chosen: Correct. h = 6.626 × 10⁻³⁴ J·s is a fixed constant of nature — the same for radio waves and gamma rays alike.
Feedback if chosen: This option is wrong — you froze the frequency, but ν is exactly what distinguishes one radiation from another.
Feedback if chosen: This option is wrong — you let the constant drift. E and ν vary together, but h never changes.
Outcome: Calculate a photon's energy from the frequency, or the frequency from the photon energy, by rearranging E = hν.
You've already seen Planck's equation: E = hν, with Planck's constant h = 6.626 × 10⁻³⁴ J·s.
A question gives you the frequency or the photon energy and asks for the other.
Whatever it asks for, start from E = hν and rearrange it to make that quantity the subject.
The powers of ten get their own line: multiply or divide the leading numbers first, then combine the powers of ten.
Q: What is the energy of a photon of radiation whose frequency is 6.00 × 10¹⁴ s⁻¹? (h = 6.626 × 10⁻³⁴ J·s.)
Write down the values from the question
ν = 6.00 × 10¹⁴ s⁻¹
h = 6.626 × 10⁻³⁴ J·s
Write down the equation
E = hν(E is already the subject)
Substitute in the values, and calculate
E = (6.626 × 10⁻³⁴) × (6.00 × 10¹⁴)
E = (6.626 × 6.00) × 10⁻³⁴⁺¹⁴(add the powers of ten when multiplying)
E = 39.756 × 10⁻²⁰
E = 3.9756 × 10⁻¹⁹
E = 3.98 × 10⁻¹⁹ J(to 3 significant figures)
Q: A photon has an energy of 3.313 × 10⁻¹⁹ J. What is the frequency of the radiation? (h = 6.626 × 10⁻³⁴ J·s.)
Write down the values from the question
E = 3.313 × 10⁻¹⁹ J
h = 6.626 × 10⁻³⁴ J·s
Write down the equation
E = hν
Make ν the subject
ν = E / h
Substitute in the values, and calculate
ν = (3.313 × 10⁻¹⁹) / (6.626 × 10⁻³⁴)
ν = (3.313 / 6.626) × 10⁻¹⁹⁺³⁴(dividing by 10⁻³⁴ is multiplying by 10³⁴)
ν = 0.500 × 10¹⁵
ν = 5.00 × 10¹⁴ s⁻¹
Calculate a photon's energy from the frequency, or the frequency from the photon energy, by rearranging E = hν.
Check — numeric entry
What is the energy of one photon of radiation with a frequency of 4.50 × 10¹⁴ s⁻¹, in units of 10⁻¹⁹ J? (h = 6.626 × 10⁻³⁴ J·s.) Give your answer to 3 significant figures.
Tolerance: 0.005
Feedback when correct: Correct. E = hν = (6.626 × 4.50) × 10⁻³⁴⁺¹⁴ = 29.8 × 10⁻²⁰ = 2.98 × 10⁻¹⁹ J.
Anticipated wrong answers (feedback shown):
| 29.8 | Your multiplication is right, but the answer is not matched to the 10⁻¹⁹ supplied by the answer box: 29.8 × 10⁻²⁰ J is 2.98 × 10⁻¹⁹ J. |
| 1.47 | You divided h by the frequency instead of multiplying. E is already the subject of E = hν — a product. |
Worked solution revealed after a second miss
Write down the values from the question
ν = 4.50 × 10¹⁴ s⁻¹
h = 6.626 × 10⁻³⁴ J·s
Write down the equation
E = hν
Substitute in the values, and calculate
E = (6.626 × 10⁻³⁴) × (4.50 × 10¹⁴)
E = (6.626 × 4.50) × 10⁻³⁴⁺¹⁴
E = 29.817 × 10⁻²⁰
E = 2.98 × 10⁻¹⁹ J
Check — numeric entry
A photon carries 2.6504 × 10⁻¹⁹ J of energy. What is the frequency of the radiation, in units of 10¹⁴ s⁻¹? (h = 6.626 × 10⁻³⁴ J·s.) Give your answer to 3 significant figures.
Tolerance: 0.005
Feedback when correct: Correct. ν = E / h = (2.6504 / 6.626) × 10⁻¹⁹⁺³⁴ = 0.400 × 10¹⁵ = 4.00 × 10¹⁴ s⁻¹.
Anticipated wrong answers (feedback shown):
| 0.400 | Your division is right, but the answer is not matched to the 10¹⁴ supplied by the answer box: 0.400 × 10¹⁵ s⁻¹ is 4.00 × 10¹⁴ s⁻¹. |
| 1.76 | You multiplied the energy by h instead of dividing by it. Making ν the subject gives ν = E / h. |
Worked solution revealed after a second miss
Write down the values from the question
E = 2.6504 × 10⁻¹⁹ J
h = 6.626 × 10⁻³⁴ J·s
Write down the equation
E = hν
Make ν the subject
ν = E / h
Substitute in the values, and calculate
ν = (2.6504 × 10⁻¹⁹) / (6.626 × 10⁻³⁴)
ν = (2.6504 / 6.626) × 10⁻¹⁹⁺³⁴
ν = 0.400 × 10¹⁵
ν = 4.00 × 10¹⁴ s⁻¹
Check — multiple choice
A question gives a photon's energy E and asks for the radiation's frequency. Which expression gives the frequency?
Feedback if chosen: This option is wrong — you multiplied when the frequency is found by dividing. Making ν the subject of E = hν gives ν = E / h.
Feedback if chosen: Correct. Dividing both sides of E = hν by h gives ν = E / h.
Feedback if chosen: This option is wrong — you divided the constant by the energy, which is the rearrangement upside down: ν = E / h.
Feedback if chosen: This option is wrong — you treated E = hν as a sum. The equation is a product, so the rearrangement divides: ν = E / h.
Outcome: State that absorbing or emitting one photon changes an atom's or molecule's energy by exactly that photon's energy.
You've already seen that atoms and molecules can hold only specific allowed energies — their rotational, vibrational, and electronic levels are quantized.
You've also seen that light arrives as photons, each carrying a definite amount of energy, E = hν.
When an atom or molecule absorbs a photon, the photon disappears and the absorber's energy increases by exactly the photon's energy.
Nothing of the photon is left over — the whole photon energy, hν, becomes the absorber's energy.
On the ladder, absorption lifts the molecule from one allowed level to a higher one, and the photon's energy equals the size of that jump.
A molecule absorbing one infrared photon, for example, gains exactly that photon's energy as vibrational energy.
Emission is the exact reverse: when an atom or molecule emits a photon, its energy decreases by exactly the energy of the photon it releases.
A photon is all-or-nothing: an atom or molecule cannot absorb half a photon or keep a fraction of its energy.
State that when an atom or molecule absorbs or emits a photon, its energy increases or decreases by exactly that photon's energy.
Check — multiple choice
A molecule absorbs one photon carrying 4.0 × 10⁻²⁰ J. What happens to the molecule's energy?
Feedback if chosen: Correct. Absorption is all-or-nothing: the photon disappears and the molecule's energy rises by exactly the photon's energy, 4.0 × 10⁻²⁰ J.
Feedback if chosen: This option is wrong — you let part of the photon's energy go missing. A photon is absorbed whole, so the molecule gains exactly 4.0 × 10⁻²⁰ J.
Feedback if chosen: This option is wrong — you swapped absorption and emission. Absorbing a photon raises the absorber's energy.
Feedback if chosen: This option is wrong — you tied the energy change to brightness. Brightness sets how many photons arrive; each absorbed photon delivers exactly its own energy.
Check — multiple choice
An excited molecule emits one photon of energy 3.0 × 10⁻¹⁹ J. What happens to the molecule's energy?
Feedback if chosen: This option is wrong — you swapped emission and absorption. An emitted photon carries energy away from the molecule.
Feedback if chosen: This option is wrong — you created the photon's energy from nothing. The photon's energy comes out of the molecule's own energy.
Feedback if chosen: Correct. The emitted photon carries away exactly 3.0 × 10⁻¹⁹ J, so the molecule's energy drops by exactly that amount.
Feedback if chosen: This option is wrong — you let the molecule hold back a fraction. A photon carries its full energy, hν, nothing less.
Check — numeric entry
A molecule sits in an energy level 2.0 × 10⁻²⁰ J above its lowest level. It absorbs a photon carrying 5.0 × 10⁻²⁰ J, lifting it to a higher level. How far above the lowest level, in units of 10⁻²⁰ J, is the molecule now? Give your answer to 2 significant figures.
Tolerance: 0.05
Feedback when correct: Correct. Absorption adds exactly the photon's energy to what the molecule already had: 2.0 + 5.0 = 7.0, so it now sits 7.0 × 10⁻²⁰ J above its lowest level.
Anticipated wrong answers (feedback shown):
| 5.0 | You replaced the molecule's energy with the photon's energy. The photon's energy is added to what the molecule already had: 2.0 + 5.0 = 7.0. |
| 3.0 | You subtracted the energies, which describes emission. Absorbing a photon increases the molecule's energy: 2.0 + 5.0 = 7.0. |
Outcome: Explain why ultraviolet/visible, infrared, and microwave photons pair with electronic, vibrational, and rotational transitions, using level spacing and photon-energy matching.
You've already seen the pairings: microwave radiation with rotational transitions, infrared with vibrational transitions, and ultraviolet or visible light with electronic transitions.
The pairing is not arbitrary — it follows from how far apart the allowed levels sit.
Electronic levels are the most widely spaced: promoting an electron to a higher level takes a large jump in energy.
Vibrational levels sit much closer together: making a molecule vibrate more energetically takes a smaller step.
Rotational levels are the most closely spaced of all — the smallest energy steps a molecule can take.
So the spacing between allowed levels decreases in the order electronic, then vibrational, then rotational.
You've already seen that an absorbed photon delivers its energy all-or-nothing — exactly hν.
An atom or molecule can only land on an allowed level, so a photon is absorbed only when its energy matches a gap between levels.
For the transition in question, the photon is absorbed only when its energy matches that gap — a photon with too little or too much energy for the gap cannot drive that transition.
You've already seen that photon energy rises with frequency: microwave photons carry the least energy, then infrared, then visible and ultraviolet.
Low-energy microwave photons match only the tiny rotational gaps, so absorbing microwave radiation changes how fast molecules rotate.
Infrared photons carry energy the size of vibrational gaps, so absorbing infrared radiation makes molecules vibrate more energetically.
Ultraviolet and visible photons carry the most energy, matching the wide electronic gaps, so they promote electrons to higher levels.
A vibrational transition absorbs infrared rather than ultraviolet light because an ultraviolet photon carries far more energy than the vibrational gap — it does not match.
Explain the region-transition pairings: level spacing decreases from electronic to vibrational to rotational, and a photon is absorbed only when its energy matches the gap.
Check — multiple choice
A dye molecule's electronic gap matches the energy of a visible photon. Why is microwave radiation unable to drive this electronic transition?
Feedback if chosen: This option is wrong — you gave the two radiations different speeds. All electromagnetic radiation travels at the same speed; what differs is each photon's energy.
Feedback if chosen: This option is wrong — you turned the long wavelength into a physical size. Absorption is about energy matching, not about fitting a wave inside a molecule.
Feedback if chosen: This option is wrong — you made brightness the barrier. However many microwave photons arrive, each one carries far too little energy to match the electronic gap.
Feedback if chosen: Correct. A photon is absorbed only when its energy matches the gap, and a microwave photon's energy falls far short of any electronic gap.
Check — multiple choice
Which list ranks the level spacings from widest to narrowest?
Feedback if chosen: Correct. Electronic gaps are the widest and rotational gaps the narrowest, which is why their matching photons run ultraviolet/visible, then infrared, then microwave.
Feedback if chosen: This option is wrong — you reversed the order. Promoting an electron takes the biggest energy step; changing a molecule's rotation takes the smallest.
Feedback if chosen: This option is wrong — you put vibrational gaps above electronic ones. Electronic transitions absorb the highest-energy photons, so their gaps are the widest.
Feedback if chosen: This option is wrong — you swapped vibrational and rotational. Rotational gaps are the narrowest — the low-energy microwave photons match them.
Check — multiple choice
An infrared photon arrives at a molecule whose smallest electronic gap is far larger than the photon's energy. Considering the electronic levels only, what happens?
Feedback if chosen: This option is wrong — you let the molecule land between rungs. The allowed energies are quantized — there is no level partway up.
Feedback if chosen: Correct. A photon is absorbed only when its energy matches a gap; this photon matches no electronic gap, so the electronic ladder ignores it.
Feedback if chosen: This option is wrong — you let the molecule bank a fraction of a jump. Absorption is one photon matching one gap — there is no saving up between levels.
Feedback if chosen: This option is wrong — you patched the mismatch with leftover energy. The photon alone must match the gap for absorption to occur.
Outcome: Calculate any one of photon energy, wavelength, and frequency by combining E = hν with c = λν, and connect the result to the transition involved.
You've already seen the two equations separately: c = λν gives the frequency from the wavelength, and E = hν gives the photon energy from the frequency.
Frequency appears in both, so it is the bridge between wavelength and photon energy.
To get from a wavelength to a photon energy, find the frequency from c = λν first, then feed it into E = hν.
To get from a photon energy to a wavelength, run the same chain in reverse: find the frequency from E = hν first, then feed it into c = λν.
Q: Green light in the visible region has a wavelength of 500 nm. What is the energy of one of its photons? (c = 3.00 × 10⁸ m/s; h = 6.626 × 10⁻³⁴ J·s.)
Write down the values from the question
λ = 500 nm
c = 3.00 × 10⁸ m/s
h = 6.626 × 10⁻³⁴ J·s
Convert the wavelength to meters
λ = 500 × 10⁻⁹ m
λ = 5.00 × 10⁻⁷ m
Write down the wave equation, and make ν the subject
c = λν
ν = c / λ
Substitute in the values, and calculate ν
ν = (3.00 × 10⁸) / (5.00 × 10⁻⁷)
ν = (3.00 / 5.00) × 10⁸⁺⁷(dividing by 10⁻⁷ is multiplying by 10⁷)
ν = 0.600 × 10¹⁵
ν = 6.00 × 10¹⁴ s⁻¹
Write down Planck's equation
E = hν(E is already the subject)
Substitute in the values, and calculate E
E = (6.626 × 10⁻³⁴) × (6.00 × 10¹⁴)
E = (6.626 × 6.00) × 10⁻³⁴⁺¹⁴(add the powers of ten when multiplying)
E = 39.756 × 10⁻²⁰
E = 3.98 × 10⁻¹⁹ J(to 3 significant figures)
Because this photon is visible light, absorbing it drives an electronic transition — 3.98 × 10⁻¹⁹ J is the size of an electronic gap.
Q: A photon carries 1.9878 × 10⁻¹⁹ J of energy. What is the wavelength of the radiation? (c = 3.00 × 10⁸ m/s; h = 6.626 × 10⁻³⁴ J·s.)
Write down the values from the question
E = 1.9878 × 10⁻¹⁹ J
c = 3.00 × 10⁸ m/s
h = 6.626 × 10⁻³⁴ J·s
Write down Planck's equation, and make ν the subject
E = hν
ν = E / h
Substitute in the values, and calculate ν
ν = (1.9878 × 10⁻¹⁹) / (6.626 × 10⁻³⁴)
ν = (1.9878 / 6.626) × 10⁻¹⁹⁺³⁴(dividing by 10⁻³⁴ is multiplying by 10³⁴)
ν = 0.300 × 10¹⁵
ν = 3.00 × 10¹⁴ s⁻¹
Write down the wave equation, and make λ the subject
c = λν
λ = c / ν
Substitute in the values, and calculate λ
λ = (3.00 × 10⁸) / (3.00 × 10¹⁴)
λ = (3.00 / 3.00) × 10⁸⁻¹⁴(subtract the powers of ten when dividing)
λ = 1.00 × 10⁻⁶ m
A wavelength of 1.00 × 10⁻⁶ m lies in the infrared region, so a photon like this one is absorbed as vibrational energy.
Once the numbers are in hand, the radiation's region names the transition: ultraviolet or visible photons drive electronic transitions, infrared photons vibrational ones, and microwave photons rotational ones.
Calculate any one of photon energy, wavelength, and frequency by combining E = hν with c = λν, and connect the result to the transition involved.
Check — numeric entry
A dye absorbs visible light of wavelength 600 nm. What is the energy of one absorbed photon, in units of 10⁻¹⁹ J? (c = 3.00 × 10⁸ m/s; h = 6.626 × 10⁻³⁴ J·s.) Give your answer to 3 significant figures.
Tolerance: 0.005
Feedback when correct: Correct. ν = c / λ = 5.00 × 10¹⁴ s⁻¹, then E = hν = (6.626 × 5.00) × 10⁻³⁴⁺¹⁴ = 3.31 × 10⁻¹⁹ J — a visible photon, absorbed by an electronic transition.
Anticipated wrong answers (feedback shown):
| 1.33 | You divided the wavelength by c when finding the frequency, which is upside down: ν = c / λ. That inverted frequency then carried through E = hν. |
| 33.1 | Your chain is right, but the answer is not matched to the 10⁻¹⁹ supplied by the answer box: 33.1 × 10⁻²⁰ J is 3.31 × 10⁻¹⁹ J. |
Worked solution revealed after a second miss
Write down the values from the question
λ = 600 nm
c = 3.00 × 10⁸ m/s
h = 6.626 × 10⁻³⁴ J·s
Convert the wavelength to meters
λ = 600 × 10⁻⁹ m
λ = 6.00 × 10⁻⁷ m
Write down the wave equation, and make ν the subject
c = λν
ν = c / λ
Substitute in the values, and calculate ν
ν = (3.00 × 10⁸) / (6.00 × 10⁻⁷)
ν = 0.500 × 10¹⁵
ν = 5.00 × 10¹⁴ s⁻¹
Write down Planck's equation
E = hν
Substitute in the values, and calculate E
E = (6.626 × 10⁻³⁴) × (5.00 × 10¹⁴)
E = 33.13 × 10⁻²⁰
E = 3.31 × 10⁻¹⁹ J
Check — numeric entry
A molecule absorbs an infrared photon carrying 1.3252 × 10⁻¹⁹ J. What is the wavelength of the radiation, in units of 10⁻⁶ m? (c = 3.00 × 10⁸ m/s; h = 6.626 × 10⁻³⁴ J·s.) Give your answer to 3 significant figures.
Tolerance: 0.005
Feedback when correct: Correct. ν = E / h = 2.00 × 10¹⁴ s⁻¹, then λ = c / ν = 1.50 × 10⁻⁶ m — an infrared wavelength, absorbed as vibrational energy.
Anticipated wrong answers (feedback shown):
| 2.00 | That is the frequency, in 10¹⁴ s⁻¹ — the middle of the chain. Feed it into the wave equation to finish: λ = c / ν = 1.50 × 10⁻⁶ m. |
| 6.67 | You divided the frequency by c when finding the wavelength, which is upside down: λ = c / ν — the speed goes on top. |
Worked solution revealed after a second miss
Write down the values from the question
E = 1.3252 × 10⁻¹⁹ J
c = 3.00 × 10⁸ m/s
h = 6.626 × 10⁻³⁴ J·s
Write down Planck's equation, and make ν the subject
E = hν
ν = E / h
Substitute in the values, and calculate ν
ν = (1.3252 × 10⁻¹⁹) / (6.626 × 10⁻³⁴)
ν = 0.200 × 10¹⁵
ν = 2.00 × 10¹⁴ s⁻¹
Write down the wave equation, and make λ the subject
c = λν
λ = c / ν
Substitute in the values, and calculate λ
λ = (3.00 × 10⁸) / (2.00 × 10¹⁴)
λ = 1.50 × 10⁻⁶ m
Check — multiple choice
A hydrogen chloride molecule, HCl, in the gas phase absorbs one photon of microwave radiation. Which change does the absorption produce?
Feedback if chosen: This option is wrong — you paired microwave photons with vibrational gaps. Vibrational transitions take the higher-energy infrared photons.
Feedback if chosen: This option is wrong — you paired microwave photons with electronic gaps. Electronic gaps are the widest and take ultraviolet or visible photons.
Feedback if chosen: This option is wrong — you flipped absorption into emission. Absorbing a photon raises the molecule's energy by exactly the photon's energy.
Feedback if chosen: Correct. A microwave photon's energy matches the narrow rotational gaps, and absorption adds exactly that photon's energy — the molecule rotates more energetically.
9 lessons
Outcome: State what absorbance is: a spectrophotometer's measure of how much light of a chosen wavelength a solution absorbs.
Here are two samples of the same blue dye, one pale and one deep blue, each with the same beam of light shining through it.
Your eye can tell that the deep blue sample absorbs more of the light passing through it.
But an eye cannot say how much more — comparing shades gives no number.
You've already seen why solutions absorb light at all: a molecule that absorbs a photon gains exactly that photon's energy.
The instrument in the figure shines a beam of light of one chosen wavelength through the sample and measures how strong the beam is on the far side.
The sample sits in a small transparent container called a 'cuvette'.
The beam entering the cuvette is labeled I₀, and the beam leaving it is labeled I.
The more light the sample absorbs, the weaker the beam that reaches the detector.
From the strength of the entering beam and the strength of the leaving beam, the instrument works out one number, called the 'absorbance'.
A sample that absorbs more of the incoming light gives a higher absorbance reading.
Absorbance is a bare number — it carries no unit.
An instrument that measures absorbance this way is called a 'spectrophotometer'.
The two dye samples now compare cleanly: the deep blue sample gives the higher absorbance reading.
State what absorbance is: a spectrophotometer's measure of how much light of a chosen wavelength a solution absorbs.
Check — multiple choice
Light of one chosen wavelength passes through solutions X and Y, as shown. Which solution gives the higher absorbance reading?
Feedback if chosen: Correct. X's leaving beam is far weaker than its entering beam, so X absorbed more of the light — and a sample that absorbs more gives a higher absorbance reading.
Feedback if chosen: This option is wrong — you picked the solution with the stronger leaving beam. A strong leaving beam means little of the light was absorbed, so Y reads lower.
Feedback if chosen: This option is wrong — you compared the entering beams, which are the same. Absorbance compares each leaving beam with its entering beam, and X's leaving beam is much weaker.
Feedback if chosen: This option is wrong — the figure shows everything needed: X's leaving beam is weaker, so X absorbed more light and reads the higher absorbance.
Check — multiple choice
At one wavelength, solution P reads an absorbance of 0.90 and solution Q reads an absorbance of 0.30. Which statement follows?
Feedback if chosen: This option is wrong — you read the lower absorbance as more absorption. A higher absorbance means more light absorbed, and P's 0.90 is the higher reading.
Feedback if chosen: This option is wrong — you treated two different absorbance readings as equal absorption. The readings differ, so the absorbed amounts differ: P's 0.90 means P absorbed more.
Feedback if chosen: Correct. A higher absorbance means more of the incoming light was absorbed, and 0.90 is higher than 0.30.
Feedback if chosen: This option is wrong — you inverted the reading. A high absorbance means more light was absorbed, so less got through: P let less light through than Q, not more.
Check — multiple choice
A spectrophotometer's absorbance reading reports which of these?
Feedback if chosen: This option is wrong — you described the entering beam alone. Absorbance compares the entering beam with the leaving beam to report what the sample absorbed.
Feedback if chosen: Correct. The instrument compares the beam entering the cuvette with the beam leaving it and reports how much of that light the sample absorbed, as one unitless number.
Feedback if chosen: This option is wrong — you inverted the reading. A higher absorbance means more of the light was absorbed, so less passed through, not more.
Feedback if chosen: This option is wrong — the wavelength is chosen by the operator before measuring; the reading reports what the sample absorbed at that wavelength.
Outcome: Predict that at fixed path length and wavelength, absorbance increases as the concentration of the absorbing species increases.
You've already seen that a solution's absorbance reports how much light of the chosen wavelength it absorbs.
You've already seen concentration measured as molarity: moles of dissolved substance per liter of solution, written M.
The figure shows one dye at two concentrations, 0.05 M and 0.20 M, in identical cuvettes at the same wavelength.
The beam crosses the same distance of solution in each cuvette — but in the 0.20 M sample, four times as many dye particles sit in its way.
As the concentration increases, more absorbing particles sit in the light's path, so more of the light is absorbed.
More light absorbed means a higher absorbance: the 0.20 M sample reads higher than the 0.05 M sample.
The distance the light travels through the solution is called the 'path length' — using identical cuvettes keeps it fixed.
At a fixed path length and a fixed wavelength, raising the concentration raises the absorbance.
Predict that at fixed path length and wavelength, the absorbance increases as the concentration of the absorbing species increases, because more absorbing particles sit in the light path.
Check — multiple choice
One dye is prepared at 0.10 M and at 0.30 M and measured in identical cuvettes at one wavelength, as shown. Which sample gives the higher absorbance?
Feedback if chosen: This option is wrong — you picked the more dilute sample. It puts fewer dye particles in the beam's path, so it absorbs less light and reads lower.
Feedback if chosen: This option is wrong — identical cuvettes fix the path length, but the concentrations still differ, and the more concentrated sample absorbs more.
Feedback if chosen: This option is wrong — a shared wavelength makes the comparison fair; the readings still differ because the concentrations differ.
Feedback if chosen: Correct. At the same path length and wavelength, the 0.30 M sample puts three times as many absorbing particles in the beam's path, so it absorbs more light and reads the higher absorbance.
Check — multiple choice
Which change increases a dye solution's absorbance, at fixed path length and wavelength?
Feedback if chosen: This option is wrong — lowering the concentration removes absorbing particles from the light's path, so the absorbance falls.
Feedback if chosen: Correct. Raising the concentration puts more absorbing particles in the light's path, so more light is absorbed and the absorbance rises.
Feedback if chosen: This option is wrong — adding water spreads the same amount of dye through a larger volume, so the concentration falls and the absorbance falls with it.
Feedback if chosen: This option is wrong — stirring changes neither the concentration nor the path length, so the reading stays where it was.
Check — multiple choice
At fixed path length and wavelength, why does a more concentrated solution give a higher absorbance?
Feedback if chosen: This option is wrong — you changed each particle's absorbing strength. Each particle absorbs as before; there are simply more of them in the beam's path.
Feedback if chosen: This option is wrong — you changed the path length. The cuvette is the same, so the distance the light travels is fixed; only the number of particles in the path grew.
Feedback if chosen: Correct. Raising the concentration packs more absorbing particles into the same light path, so more of the light is absorbed.
Feedback if chosen: This option is wrong — you swapped absorption for reflection. Absorbance reports light absorbed by particles inside the solution, not light bouncing off it.
Outcome: Predict that at fixed concentration and wavelength, absorbance increases as the path length increases.
You've already seen the path length: the distance the light travels through the solution.
The figure shows one solution, at one concentration, in a 1 cm cuvette and in a 2 cm cuvette.
In the 2 cm cuvette, the beam crosses twice as much solution before it gets out.
As the path length increases, the light passes more absorbing particles, so more of the light is absorbed.
More light absorbed means a higher absorbance: the 2 cm cuvette reads higher than the 1 cm cuvette.
The comparison is fair because the concentration and the wavelength were held fixed — only the path length changed.
Path length is written b and is measured in centimeters.
Predict that at fixed concentration and wavelength, the absorbance increases as the path length increases, because the light passes more absorbing particles.
Check — multiple choice
One dye solution fills both cuvettes shown, measured at the same wavelength. Which cuvette gives the higher absorbance?
Feedback if chosen: This option is wrong — you picked the shorter path. The beam crosses less solution there, passes fewer absorbing particles, and reads lower.
Feedback if chosen: Correct. In the 4 cm cuvette the beam crosses four times as much solution, passes four times as many absorbing particles, and so absorbs more light — a higher absorbance.
Feedback if chosen: This option is wrong — a shared concentration is not enough; the path lengths differ, and the longer path puts more absorbing particles in the beam's way.
Feedback if chosen: This option is wrong — a unitless quantity can still take different values; the readings still differ when the path lengths differ.
Check — multiple choice
At fixed concentration and wavelength, why does a longer path length give a higher absorbance?
Feedback if chosen: Correct. A longer crossing puts more absorbing particles in the beam's way, so more of the light is absorbed before it gets out.
Feedback if chosen: This option is wrong — you changed the concentration. It is the same solution; only the distance the light travels through it grew.
Feedback if chosen: This option is wrong — you credited the container. The absorbing is done by the dissolved particles in the beam's path, not by the cuvette's walls.
Feedback if chosen: This option is wrong — you changed the light. The wavelength is set by the instrument and stays fixed; only the crossing distance grew.
Check — multiple choice
A sample is so dilute that its absorbance is barely above zero. Without changing the solution itself, which change raises the reading?
Feedback if chosen: This option is wrong — a shorter path puts fewer absorbing particles in the beam's way, so the reading falls further.
Feedback if chosen: This option is wrong — a brighter lamp strengthens the entering beam and the leaving beam together; absorbance compares the two, so the reading does not rise.
Feedback if chosen: Correct. A longer path length sends the beam past more of the absorbing particles, so more light is absorbed and the reading rises — no change to the solution needed.
Feedback if chosen: This option is wrong — you changed the volume, not the path length. What matters is the distance the beam crosses through solution, not how much solution there is.
Outcome: State that molar absorptivity (ε) measures how intensely a chemical species absorbs light at a specific wavelength.
You've already seen two ways to raise a solution's absorbance: raise the concentration, or increase the path length.
The figure shows two different dyes, P and Q, at the same concentration, in identical cuvettes, measured at the same wavelength.
Concentration, path length, and wavelength are all matched — yet P absorbs far more of the light than Q.
The difference sits in the species itself: some chemical species absorb light of a given wavelength intensely, and others barely absorb it at all.
How intensely a species absorbs light at a specific wavelength is measured by its 'molar absorptivity', written ε.
A strongly colored dye has a large molar absorptivity at the wavelength it absorbs; a species that barely absorbs there has a small one.
Molar absorptivity belongs to the species and the wavelength — it does not change when the solution is diluted or the cuvette is swapped.
Real values span an enormous range: at 525 nm, permanganate ions (MnO₄⁻) have ε = 2400 L·mol⁻¹·cm⁻¹, while at 810 nm, dissolved Cu²⁺ ions manage only ε = 12 L·mol⁻¹·cm⁻¹.
State that molar absorptivity (ε) measures how intensely a chemical species absorbs light at a specific wavelength.
Check — multiple choice
| Species | Wavelength (nm) | ε (L·mol⁻¹·cm⁻¹) |
|---|---|---|
| MnO₄⁻ | 525 | 2400 |
| Cu²⁺ | 810 | 12 |
Equal concentrations of the two species in the table are measured in identical cuvettes, each at the wavelength shown. Which solution absorbs more of its light?
Feedback if chosen: This option is wrong — you picked the species with the smaller molar absorptivity. An ε of 12 marks a weak absorber; 2400 marks an intense one.
Feedback if chosen: Correct. With concentration and path length matched, the species with the larger molar absorptivity absorbs more of its light — and 2400 L·mol⁻¹·cm⁻¹ dwarfs 12 L·mol⁻¹·cm⁻¹.
Feedback if chosen: This option is wrong — equal concentrations only make the comparison fair; the two species' molar absorptivities still differ by a factor of two hundred.
Feedback if chosen: This option is wrong — absorbance compares the leaving beam with the entering beam, so the lamp's brightness cancels out of the reading.
Check — multiple choice
At 640 nm, a species has a very large molar absorptivity. What does that tell you?
Feedback if chosen: Correct. Molar absorptivity is the species' own absorbing strength at that wavelength — a large ε means intense absorption there, whatever the concentration.
Feedback if chosen: This option is wrong — you folded concentration into ε. Molar absorptivity belongs to the species; it does not change with how much of it happens to be dissolved.
Feedback if chosen: This option is wrong — you folded path length into ε. A strong absorber can be measured in any cuvette; ε describes the species, not the container.
Feedback if chosen: This option is wrong — molar absorptivity is tied to a specific wavelength; the same species can absorb intensely at 640 nm and barely at all elsewhere.
Outcome: State the Beer-Lambert law, A = εbc: absorbance is proportional to the molar absorptivity, the path length, and the concentration.
You've already seen that raising the concentration raises the absorbance, at fixed path length and wavelength.
You've already seen that increasing the path length raises the absorbance, at fixed concentration and wavelength.
You've already seen that the species' own molar absorptivity, ε, sets how intensely it absorbs at the chosen wavelength.
One equation combines all three influences into the absorbance reading — it is called the 'Beer-Lambert law'.
Absorbance is proportional to each of the three factors on the right.
If the concentration doubles — same species, same cuvette, same wavelength — the absorbance doubles.
If the path length doubles — same solution moved to a cuvette twice as wide — the absorbance doubles.
If a species with twice the molar absorptivity is measured — same concentration, same path length — the absorbance doubles.
The units multiply away: L·mol⁻¹·cm⁻¹ × cm × mol/L cancels completely, which is why absorbance carries no unit.
The law comes with a working range: it describes readings taken with fixed instrument settings, across the span of concentrations the measurement has been checked on — far beyond that span, real readings drift away from the proportionality.
State the Beer-Lambert law, A = εbc: absorbance is proportional to the molar absorptivity, the path length, and the concentration of the absorbing species.
Check — multiple choice
Which change doubles a solution's absorbance?
Feedback if chosen: Correct. A = εbc is proportional to each factor, so doubling the concentration alone doubles the absorbance.
Feedback if chosen: This option is wrong — you doubled two factors at once. Absorbance is proportional to each, so doubling both multiplies the reading by four.
Feedback if chosen: This option is wrong — you moved the path length the wrong way. Halving b halves the absorbance rather than doubling it.
Feedback if chosen: This option is wrong — pouring in more of the same solution changes neither the concentration nor the beam's path length, so the absorbance stays the same.
Check — multiple choice
A solution reads A = 0.30 in a 1 cm cuvette. The same solution is measured at the same wavelength in a 3 cm cuvette. What does it read?
Feedback if chosen: This option is wrong — you divided by three. A longer path length raises the absorbance: A is proportional to b.
Feedback if chosen: This option is wrong — you treated absorbance as a property of the solution alone. It also tracks the path length, which tripled.
Feedback if chosen: This option is wrong — you doubled instead of tripling. The path length went from 1 cm to 3 cm, a factor of three.
Feedback if chosen: Correct. A = εbc is proportional to the path length, so tripling b triples the reading: 3 × 0.30 = 0.90.
Check — multiple choice
Two dyes are prepared at the same concentration and measured in the same cuvette, each at a wavelength where its molar absorptivity is known: dye R has ε = 500 L·mol⁻¹·cm⁻¹ and dye S has ε = 1000 L·mol⁻¹·cm⁻¹. How do the absorbances compare?
Feedback if chosen: This option is wrong — you inverted the comparison. The larger molar absorptivity gives the larger absorbance, and S's ε is the larger one.
Feedback if chosen: This option is wrong — equal concentrations fix only one factor. The molar absorptivities differ, and absorbance is proportional to ε.
Feedback if chosen: Correct. At fixed b and c, absorbance is proportional to ε: 1000 is twice 500, so S reads twice R.
Feedback if chosen: This option is wrong — you squared the factor. Doubling ε doubles the absorbance; it does not quadruple it.
Outcome: Calculate any one of absorbance, molar absorptivity, path length, or concentration from the other three, using A = εbc.
You've already seen the Beer-Lambert law, A = εbc, with each symbol's units fixed.
A problem gives three of the four quantities and asks for the fourth.
There is no separate equation for concentration, path length, or molar absorptivity.
Whatever the question asks for, start from A = εbc and rearrange it to make that quantity the subject.
Keep the units matched: with ε in L·mol⁻¹·cm⁻¹, the path length goes in as centimeters, the concentration as molarity, and the absorbance as a bare number.
Calculate any one of absorbance, molar absorptivity, path length, or concentration from the other three, using A = εbc.
Q: A 0.0020 M solution of a dye (ε = 250 L·mol⁻¹·cm⁻¹ at the measuring wavelength) is measured in a 1.0 cm cuvette. What is the absorbance?
Write down the values in the question
ε = 250 L·mol⁻¹·cm⁻¹
b = 1.0 cm
c = 0.0020 M
Write down the equation
A = εbc
Substitute in the values, and calculate
A = 250 × 1.0 × 0.0020
A = 0.50
Q: A dye solution in a 1.0 cm cuvette reads A = 0.45; the dye's molar absorptivity at this wavelength is ε = 150 L·mol⁻¹·cm⁻¹. What is the concentration?
Write down the values in the question
A = 0.45
ε = 150 L·mol⁻¹·cm⁻¹
b = 1.0 cm
Write down the equation
A = εbc
Make c the subject
c = A / (εb)
Substitute in the values, and calculate
c = 0.45 / (150 × 1.0)
c = 0.0030 M
Check — numeric entry
A dye solution in a 2.0 cm cuvette reads A = 0.60. At this wavelength the dye's molar absorptivity is ε = 150 L·mol⁻¹·cm⁻¹. What is the concentration of the dye, in M? Give your answer to 2 significant figures.
Tolerance: 5e-05
Feedback when correct: Correct. Making c the subject gives c = A / (εb) = 0.60 / (150 × 2.0) = 0.0020 M.
Anticipated wrong answers (feedback shown):
| 0.0040 | You left the path length out of the denominator. c = A / (εb), and b = 2.0 cm: c = 0.60 / (150 × 2.0) = 0.0020 M. |
| 0.0080 | You multiplied by the path length instead of dividing by it. Making c the subject gives c = A / (εb) — the path length belongs in the denominator. |
Worked solution revealed after a second miss
Write down the values in the question
A = 0.60
ε = 150 L·mol⁻¹·cm⁻¹
b = 2.0 cm
Write down the equation
A = εbc
Make c the subject
c = A / (εb)
Substitute in the values, and calculate
c = 0.60 / (150 × 2.0)
c = 0.60 / 300
c = 0.0020 M
Check — numeric entry
A 0.0010 M solution in a 2.0 cm cuvette reads A = 0.80. What is the molar absorptivity of the dissolved species at this wavelength, in L·mol⁻¹·cm⁻¹? Give your answer to 3 significant figures.
Tolerance: 0.5
Feedback when correct: Correct. Making ε the subject gives ε = A / (bc) = 0.80 / (2.0 × 0.0010) = 400 L·mol⁻¹·cm⁻¹.
Anticipated wrong answers (feedback shown):
| 800 | You left the path length out of the denominator. ε = A / (bc), and b = 2.0 cm: ε = 0.80 / (2.0 × 0.0010) = 400 L·mol⁻¹·cm⁻¹. |
| 0.0016 | You multiplied the three given values together instead of rearranging. Making ε the subject gives ε = A / (bc) = 0.80 / (2.0 × 0.0010) = 400 L·mol⁻¹·cm⁻¹. |
Worked solution revealed after a second miss
Write down the values in the question
A = 0.80
b = 2.0 cm
c = 0.0010 M
Write down the equation
A = εbc
Make ε the subject
ε = A / (bc)
Substitute in the values, and calculate
ε = 0.80 / (2.0 × 0.0010)
ε = 0.80 / 0.0020
ε = 400 L·mol⁻¹·cm⁻¹
Check — numeric entry
A 0.0015 M dye solution reads A = 0.36; the dye's molar absorptivity at this wavelength is ε = 120 L·mol⁻¹·cm⁻¹. What is the path length of the cuvette, in cm? Give your answer to 2 significant figures.
Tolerance: 0.005
Feedback when correct: Correct. Making b the subject gives b = A / (εc) = 0.36 / (120 × 0.0015) = 2.0 cm.
Anticipated wrong answers (feedback shown):
| 0.5 | You divided εc by A instead of A by εc. Making b the subject gives b = A / (εc) = 0.36 / (120 × 0.0015) = 2.0 cm. |
| 0.0030 | You left the concentration out of the denominator. b = A / (εc), and c = 0.0015 M: b = 0.36 / (120 × 0.0015) = 2.0 cm. |
Worked solution revealed after a second miss
Write down the values in the question
A = 0.36
ε = 120 L·mol⁻¹·cm⁻¹
c = 0.0015 M
Write down the equation
A = εbc
Make b the subject
b = A / (εc)
Substitute in the values, and calculate
b = 0.36 / (120 × 0.0015)
b = 0.36 / 0.18
b = 2.0 cm
Outcome: State that at a fixed wavelength and path length, absorbance is proportional to the concentration of the absorbing species alone.
You've already seen the Beer-Lambert law, A = εbc.
In practice, one species is measured at one chosen wavelength, in one cuvette.
One species at one wavelength fixes ε, and one cuvette fixes the path length b.
With ε and b fixed, the law leaves only one moving part: absorbance is proportional to the concentration alone.
As the concentration doubles, the absorbance doubles; as the concentration triples, the absorbance triples.
Plotted, that proportionality is a straight line through the origin: zero concentration reads zero absorbance.
A plot of absorbance against concentration that misses the origin, or bends, signals conditions outside that ideal — the proportionality is not holding there.
A measured straight line is also trusted only across the span of concentrations that produced it — it converts only readings that fall within that tested range.
State that when the wavelength and path length are held constant, absorbance is proportional to the concentration of the absorbing species alone.
Check — numeric entry
In one cuvette at one wavelength, a 0.050 M solution of a dye reads A = 0.20. What absorbance does a 0.150 M solution of the same dye give, in the same cuvette at the same wavelength? Give your answer to 2 significant figures.
Tolerance: 0.005
Feedback when correct: Correct. At fixed wavelength and path length, absorbance is proportional to concentration: 0.150 M is three times 0.050 M, so A = 3 × 0.20 = 0.60.
Anticipated wrong answers (feedback shown):
| 0.40 | You doubled the absorbance, but 0.150 M is three times 0.050 M — the absorbance triples to 0.60. |
| 0.20 | You treated absorbance as a fixed property of the dye. At fixed wavelength and path length it is proportional to concentration, so tripling the concentration triples the reading. |
| 0.067 | You inverted the ratio and scaled the absorbance down. The concentration went up by a factor of three, so the absorbance goes up by the same factor. |
Worked solution revealed after a second miss
Write down the values in the question
A = 0.20 at c = 0.050 M
new c = 0.150 M
Write down the relationship
At fixed wavelength and path length, A is proportional to c
Compare the concentrations
0.150 / 0.050 = 3
Multiply the absorbance by the same factor
A = 3 × 0.20
A = 0.60
Check — multiple choice
A student measures the absorbance of a series of dye solutions at one wavelength in one cuvette and plots the line shown. Which feature of the plot does not match the proportional relationship between absorbance and concentration?
Feedback if chosen: This option is wrong — you flagged the feature the relationship requires: proportionality plots as a straight line.
Feedback if chosen: This option is wrong — rising is exactly right: as the concentration increases, the absorbance increases.
Feedback if chosen: Correct. Proportionality passes through the origin — a solution with zero concentration has nothing in the beam's path to absorb, so it must read zero. A line meeting the absorbance axis above zero is not showing this proportionality; it signals conditions outside the ideal.
Feedback if chosen: This option is wrong — absorbance is a bare number on every correct plot; you flagged normal behavior.
Check — numeric entry
The line shows absorbance against concentration for a dye at one wavelength in one cuvette; one measured point is labeled on both axes. What absorbance does a 0.040 M solution of the same dye give under the same conditions? Give your answer to 2 significant figures.
Tolerance: 0.005
Feedback when correct: Correct. 0.040 M is half of the labeled 0.080 M, and absorbance is proportional to concentration, so A = 0.32 / 2 = 0.16.
Anticipated wrong answers (feedback shown):
| 0.32 | You read the labeled point's absorbance without scaling. That point sits at 0.080 M; the question asks about 0.040 M, half the concentration, so the absorbance halves. |
| 0.64 | You doubled instead of halving. 0.040 M is half of 0.080 M, so the absorbance is half of 0.32. |
Worked solution revealed after a second miss
Write down the values from the graph
A = 0.32 at c = 0.080 M
Write down the relationship
At fixed wavelength and path length, A is proportional to c
Compare the concentrations
0.040 / 0.080 = 1/2
Multiply the absorbance by the same factor
A = 0.32 / 2
A = 0.16
Outcome: Explain why the spectrophotometer is set to the wavelength of maximum absorbance: the reading changes most per change in concentration, so the measurement is most sensitive there.
You've already seen that molar absorptivity is tied to a specific wavelength — the same species absorbs some wavelengths intensely and others barely.
You've already seen that at a fixed wavelength and path length, absorbance is proportional to concentration alone.
A spectrophotometer must be set to one wavelength before it can measure — so which wavelength should it be?
The table shows solutions of one purple species, MnO₄⁻, measured at two different wavelengths in the same 1.0 cm cuvette.
| Concentration (M) | A at 525 nm | A at 635 nm |
|---|---|---|
| 0.00010 | 0.24 | 0.03 |
| 0.00020 | 0.48 | 0.06 |
At 525 nm, doubling the concentration moves the reading by 0.24; at 635 nm, the same doubling moves it by only 0.03.
525 nm is the wavelength this species absorbs most strongly — its wavelength of maximum absorbance.
At the wavelength of maximum absorbance, any change in concentration produces the largest possible change in the reading.
Large changes are easy to measure reliably; concentration differences that would nearly vanish at 635 nm stand out clearly at 525 nm.
That is why the spectrophotometer is typically set to the wavelength of maximum absorbance for the species being analyzed — there the measurement is as sensitive as possible.
Explain why the spectrophotometer is set to the wavelength of maximum absorbance for the species being analyzed: there the reading changes most per change in concentration, so the measurement has maximum sensitivity.
Check — multiple choice
| Wavelength (nm) | Absorbance of the 0.0010 M solution |
|---|---|
| 440 | 0.11 |
| 520 | 0.83 |
| 650 | 0.19 |
The table shows the absorbance of one 0.0010 M dye solution at three wavelengths. To measure the concentrations of other solutions of this dye as sensitively as possible, at which wavelength should the spectrophotometer be set?
Feedback if chosen: This option is wrong — 0.11 is the smallest reading of the three, so a change in concentration would move the reading least there.
Feedback if chosen: Correct. 0.83 is the maximum absorbance of the three, so at 520 nm each change in concentration moves the reading the most — the most sensitive choice.
Feedback if chosen: This option is wrong — 0.19 is well below the 520 nm reading; concentration changes move the reading far less there.
Feedback if chosen: This option is wrong — the three readings differ by up to a factor of nearly eight, so the same concentration change moves them very differently.
Check — multiple choice
Why is the wavelength of maximum absorbance the usual choice for measuring a species' concentration?
Feedback if chosen: Correct. Where the species absorbs most strongly, each change in concentration produces the largest change in the reading, so the measurement is as sensitive as possible.
Feedback if chosen: This option is wrong — the law holds at any fixed wavelength; you gave it a restriction it does not have.
Feedback if chosen: This option is wrong — you inverted the picture: at the wavelength of maximum absorbance the solution absorbs the most and lets the least of that light through.
Feedback if chosen: This option is wrong — the path length always matters; it is held fixed by using one cuvette, not switched off by the wavelength choice.
Outcome: Calculate an unknown concentration from a supplied calibration line, without reading beyond the standards that built it.
A common laboratory job is finding the concentration of a solution — say CuSO₄ — from its absorbance.
You've already seen one route: measure the absorbance and rearrange A = εbc — but that route needs ε.
The laboratory route skips ε: measure solutions whose concentrations are already known, and let them map out the relationship.
Solutions of known concentration prepared for this purpose are called 'standard solutions'.
Four CuSO₄ standard solutions, 0.020 M to 0.080 M, are measured at 810 nm in a 1.00 cm cuvette; each becomes one point on a plot of absorbance against concentration.
The dashed reads mark one of the standards: the 0.040 M solution reads A = 0.48.
You've already seen why the points fall on a straight line through the origin: at a fixed wavelength and path length, absorbance is proportional to concentration alone.
The straight line drawn through the standard solutions' points is called a 'calibration line'.
Measure the unknown solution the same way — same cuvette, same wavelength — and take its absorbance.
The unknown reads A = 0.60: start at 0.60 on the absorbance axis, go across to the calibration line, then drop down to the concentration axis — 0.050 M.
The read is trustworthy because 0.60 lands between the standard solutions' readings — the line is tested on both sides of it.
Beyond the highest standard solution the line was never tested, so a read out there is a guess, not a measurement.
If an unknown's absorbance lands above the highest standard solution's, this line cannot give its concentration.
Calculate an unknown concentration from a supplied calibration line, reading only within the range the standard solutions cover.
Check — numeric entry
The calibration line shown was built from four Co²⁺ standard solutions measured at 510 nm in a 1.00 cm cuvette. An unknown Co²⁺ solution, measured the same way, is marked on the graph. What is its concentration, in M? Give your answer to 2 significant figures.
Tolerance: 0.0005
Feedback when correct: Correct. Following the marked read — across from A = 0.36 to the line, then down — lands at 0.075 M on the concentration axis, and 0.36 sits inside the standards' range, so the read is trustworthy.
Anticipated wrong answers (feedback shown):
| 0.36 | You entered the absorbance instead of the concentration. Follow the arrows from A = 0.36 across to the line and down to the concentration axis: 0.075 M. |
Check — multiple choice
The calibration line shown was built from four MnO₄⁻ standard solutions between 0.00010 M and 0.00040 M, measured at 525 nm in a 1.00 cm cuvette. An unknown MnO₄⁻ solution, measured the same way, gives an absorbance of 1.80. What is its concentration?
Feedback if chosen: This option is wrong — you extended the line far beyond the highest standard and read from the extension. The line is trusted only across the range the standards tested — 0.00010 M to 0.00040 M — and out there it was never checked against a known solution.
Feedback if chosen: This option is wrong — you capped the answer at the highest standard. The unknown's absorbance, 1.80, sits far above the highest standard's reading; it does not equal it.
Feedback if chosen: This option is wrong — you read the absorbance value as a concentration. The two axes carry different quantities.
Feedback if chosen: Correct. A = 1.80 lands far above the highest standard's reading, where the line was never tested — a calibration line only converts readings that fall within its standards' range. Standards covering higher concentrations would be needed.
Check — multiple choice
A calibration line for a dye was built from standard solutions measured in a 1.00 cm cuvette at the dye's wavelength of maximum absorbance. For the line to convert an unknown sample's absorbance into a concentration, how must the unknown be measured?
Feedback if chosen: Correct. The line encodes one molar absorptivity (one species at one wavelength) and one path length; the unknown sits on that same line only if it is measured under exactly those conditions.
Feedback if chosen: This option is wrong — lengthening the path length raises the unknown's reading off the standards' line; the standards were measured at 1.00 cm.
Feedback if chosen: This option is wrong — molar absorptivity changes with wavelength, so a different wavelength puts the unknown on a different line from the one the standards drew.
Feedback if chosen: This option is wrong — the cuvette matches but the wavelength does not; where the dye barely absorbs, the unknown follows a different, much shallower line.
Outcome: Finish a real four-point short question in nine minutes and score it against the real rubric.
The free-response section gives you 7 questions in 105 minutes, so a 4-point short question has a budget of about 9 minutes — a little over 2 minutes per point.
Calculation parts pay only with visible work: write the equation, substitute the values with their units, then report the answer — a bare answer with no setup earns zero.
Justification parts pay fastest when the claim comes first; spend the remaining seconds naming the principle and this system's particles.
Set a 9-minute timer, answer all three parts (four points) of the question below on paper under time, and only then reveal the model response and score yourself.
Free-response question
Source note (not shown as part of the question): 2024 FRQ Q7(a–c), complete; the booklet's materials list and five-row step table are written out in the part (b) prompt, and the results figure is redrawn (spot positions representative); rubric from the 2024 scoring guidelines
A student conducts a chromatography experiment and needs to prepare 100.0 mL of 0.340 M NaCl(aq) to use as the solvent.
(a) Calculate the mass of solid NaCl (molar mass 58.44 g/mol) needed to prepare the 100.0 mL of 0.340 M NaCl(aq). [1 point]
(b) Briefly list the additional steps necessary to prepare the 100.0 mL of 0.340 M NaCl(aq) solution using only materials selected from the choices given. Assume that all appropriate safety measures are already in place. Not all materials in the list may be needed. Materials: solid NaCl; balance; pipet; distilled water; 100.0 mL volumetric flask; 150 mL beakers; weighing paper and scoop; 50.0 mL graduated cylinder; chromatography paper. The steps already written are — Step 1: Use the weighing paper and scoop to measure the correct mass of solid NaCl on the balance. Step 3: Swirl the mixture to dissolve the solid NaCl. Step 5: Stopper and invert the mixture several times to ensure that the mixture is homogeneous. Write step 2 and step 4. [2 points]
(c) The student uses the NaCl(aq) solvent to separate a mixture of compounds X and Y in a chromatography experiment. After 30 minutes, the student removes the chromatography paper from the chamber. The results of the experiment are shown in the figure above. A second student conducts the same chromatography experiment but removes the chromatography paper from the chamber after 15 minutes instead of 30 minutes. Predict the effect, if any, this would have on the separation distance between compounds X and Y in the new experiment. Explain your reasoning. [1 point]
✓ Model response (the key)
(a)
0.1000 L × 0.340 mol/L = 0.0340 mol NaCl
0.0340 mol NaCl × 58.44 g/mol = 1.99 g NaCl
(b)
Step 2: Combine the solid NaCl and some distilled water in the 100.0 mL volumetric flask.
Step 4: Fill the volumetric flask with distilled water to the calibration (100.0 mL) mark.
(c)
The separation distance between X and Y would decrease.
The solvent front will not travel as far in 15 minutes as it does in 30, so X and Y each travel a shorter distance and end up closer together.
Scoring rubric — 4 points available
Check — multiple choice
Propane (CH₃CH₂CH₃), dimethyl ether (CH₃OCH₃), and ethanol (CH₃CH₂OH) all have 26 electrons per molecule, so they are of similar size. Which ranking puts them in order of overall intermolecular force strength, weakest first?
Feedback if chosen: This option is wrong — you put the polar molecule below the nonpolar one. At matched size, dimethyl ether's dipole-dipole attractions add to the London dispersion forces both molecules share, lifting it above propane.
Feedback if chosen: Correct. Similar size makes the London dispersion forces comparable, so the tiebreakers decide: propane is nonpolar (weakest), dimethyl ether is polar but cannot hydrogen-bond, and ethanol's O–H group hydrogen-bonds (strongest).
Feedback if chosen: This option is wrong — you ranked ethanol below dimethyl ether. Dimethyl ether's H atoms are all bonded to C, so it cannot hydrogen-bond with itself; ethanol's O–H can, and hydrogen bonding outweighs the dipole-dipole difference.
Feedback if chosen: This option is wrong — you ranked the list strongest first. The question asks weakest first: nonpolar propane sits lowest and hydrogen-bonding ethanol highest.
Check — multiple choice
| Property of solid J | Observation |
|---|---|
| Melting point | 730 °C |
| Conducts electricity as a solid | No |
| Conducts electricity when melted | Yes |
| Struck sharply | Shatters |
The table gives the measured properties of solid J. Which classification is correct?
Feedback if chosen: Correct. The deciding evidence is conducting only when melted: ions locked in a lattice cannot carry charge, but melting frees them — and a lattice of ions also shatters when a blow shifts its layers and brings like charges together.
Feedback if chosen: This option is wrong — you matched the conduction to a metal, but a sea of mobile electrons carries charge in the solid too. Solid J conducts only after melting, so the mobile charges must be freed ions.
Feedback if chosen: This option is wrong — you let melting turn covalent bonds into charge carriers. Melting merely frees particles to move; it does not create ions from covalent bonds, so a melted network solid does not conduct this way.
Feedback if chosen: This option is wrong — you split molecules into ions at the melting point. Molecules stay intact when a molecular solid melts, so the liquid does not conduct — and attractions between molecules give way far below 730 °C.
Check — multiple choice
On the Maxwell-Boltzmann distribution shown, the entire region under the curve is shaded. What does the shaded area represent?
Feedback if chosen: This option is wrong — you read the area as an average speed. Speeds sit on the horizontal axis, while the area counts particles.
Feedback if chosen: This option is wrong — you read the area as the horizontal extent of the curve. The range of speeds is read along the speed axis, not as an area.
Feedback if chosen: Correct. The height of the curve at each speed is the number of particles moving at speeds close to that speed, so adding those heights across every speed — the area under the entire curve — gives the total number of particles in the sample.
Feedback if chosen: This option is wrong — you read the area as adding up energies. The vertical axis counts particles, so the area counts particles.
Check — multiple choice
A sealed, rigid 8.00 L tank holds gas at 1.20 atm and 47 °C. It is warmed to 207 °C, and no gas enters or leaves. What is the new pressure?
Feedback if chosen: This option is wrong — you left the pressure unchanged. The volume is fixed, so the faster-moving particles strike the walls harder and more often — the pressure must rise with the Kelvin temperature.
Feedback if chosen: This option is wrong — you substituted the Celsius temperatures. Convert each to kelvin first, as its own step: T₁ = 47 + 273 = 320 K and T₂ = 207 + 273 = 480 K.
Feedback if chosen: This option is wrong — you inverted the temperature ratio. With V and n fixed, P grows with Kelvin temperature: P₂ = 1.20 × (480/320) = 1.80 atm.
Feedback if chosen: Correct. Convert both temperatures to kelvin (320 K and 480 K); the rigid, sealed tank fixes V and n, so writing PV = nRT for each state gives P₁/T₁ = P₂/T₂, and P₂ = 1.20 × 480/320 = 1.80 atm.
Check — multiple choice
Ne, O₂, Ar, and HF are held under identical conditions. Which gas deviates most from PV = nRT?
Feedback if chosen: This option is wrong — you ranked by electron count alone. Among the three nonpolar gases Ar's London dispersion forces are indeed the strongest, but HF's hydrogen bonding is far stronger still.
Feedback if chosen: This option is wrong — you picked the gas that behaves most ideally. Ne's few electrons give it the weakest attractions of the four, so it deviates least.
Feedback if chosen: Correct. HF molecules hydrogen-bond — attractions far stronger than the London dispersion forces available to Ne, O₂, and Ar — and stronger attractions mean a larger shortfall from ideal behavior.
Feedback if chosen: This option is wrong — you chose a middling nonpolar gas. O₂'s modest London dispersion forces are no match for HF's hydrogen bonding.
Check — multiple choice
How many formula units of MgSO₄ are in 0.150 L of a 0.240 M MgSO₄ solution? (Avogadro's number = 6.022 × 10²³ mol⁻¹.)
Feedback if chosen: This option is wrong — you treated the molarity as the moles present and skipped the volume. First find the moles: 0.240 × 0.150 = 0.0360 mol.
Feedback if chosen: Correct. Moles first: 0.240 M × 0.150 L = 0.0360 mol; then 0.0360 × 6.022 × 10²³ = 2.17 × 10²² formula units.
Feedback if chosen: This option is wrong — you divided the molarity by the volume. Moles of solute = molarity × volume: 0.240 × 0.150 = 0.0360 mol.
Feedback if chosen: This option is wrong — you divided the volume by the molarity, inverting the ratio. Moles of solute = molarity × volume: 0.240 × 0.150 = 0.0360 mol.
Check — multiple choice
A solution containing caffeine, whose molecules are polar, and limonene, whose molecules are nonpolar, was run on a polar coating with a nonpolar solvent; the chromatogram is shown. Why did the caffeine stop so close to the start line?
Feedback if chosen: Correct. Held to the polar coating by attractions between polar molecules — far stronger than anything the nonpolar solvent offers it — the caffeine spends almost the whole run parked on the coating.
Feedback if chosen: This option is wrong — you weighed the molecules. A chromatogram separates by attraction, not mass: whichever component grips the stationary coating harder, relative to the solvent, stops lower.
Feedback if chosen: This option is wrong — you kept the caffeine out of solution entirely. Its spot sits above the start line, so the solvent did carry it — briefly; the polar coating kept reclaiming it.
Feedback if chosen: This option is wrong — you ended the run early. The limonene spot near the top shows the solvent traveled the whole plate; the caffeine stopped low because the polar coating held it.
Check — multiple choice
| Solvent | Solubility of sucrose |
|---|---|
| water | about 2000 g per liter |
| hexane | essentially zero |
The table gives the measured solubility of table sugar (sucrose), whose molecules carry many O–H groups, in two solvents. Why does sucrose dissolve so well in water and not at all in hexane?
Feedback if chosen: This option is wrong — you dissolved a molecular substance as ions. Sucrose enters solution as whole molecules; no ions form.
Feedback if chosen: This option is wrong — you turned weak attractions into repulsions. Sucrose and hexane do attract — through London dispersion forces — but those attractions are too weak to pay for breaking sucrose's own hydrogen bonds.
Feedback if chosen: This option is wrong — you made solvent size the test. Plenty of hexane molecules could surround a sucrose molecule; what hexane cannot do is replace the hydrogen bonds that dissolving sucrose breaks.
Feedback if chosen: Correct. In water, the hydrogen bonds broken on mixing are replaced by sucrose-water hydrogen bonds of similar strength, so the trade is favorable; hexane offers only weak attractions in exchange, so in hexane the trade never pays.
Check — multiple choice
The figure shows an N₂O molecule dropping from one vibrational energy level to the next level down and releasing a photon. From which spectral region is the released radiation?
Feedback if chosen: Correct. Transitions between vibrational energy levels pair with the infrared region — a molecule stepping down one vibrational level releases an infrared photon.
Feedback if chosen: This option is wrong — you paired the drop with rotation's region. Microwave radiation goes with changes between rotational levels, whose spacing is far smaller than the vibrational spacing shown.
Feedback if chosen: This option is wrong — you paired the drop with an electronic region. Ultraviolet photons carry the large energies of electronic transitions, not a single vibrational step.
Feedback if chosen: This option is wrong — you paired the drop with an electronic region. Visible light goes with electrons moving between energy levels, not with a molecule stepping down one vibrational level.
Check — multiple choice
Radiation from an infrared lamp has a frequency of 2.50 × 10¹⁴ s⁻¹. What is its wavelength? (c = 3.00 × 10⁸ m/s.)
Feedback if chosen: This option is wrong — you slipped a power of ten in the division: 10⁸ divided by 10¹⁴ is 10⁻⁶, and 3.00/2.50 = 1.20, so λ = 1.20 × 10⁻⁶ m.
Feedback if chosen: Correct. Making λ the subject of c = λν gives λ = c/ν = (3.00 × 10⁸)/(2.50 × 10¹⁴) = 1.20 × 10⁻⁶ m.
Feedback if chosen: This option is wrong — you divided the frequency by c, inverting the rearrangement. Making λ the subject of c = λν gives λ = c/ν.
Feedback if chosen: This option is wrong — you multiplied c by the frequency instead of dividing. Making λ the subject of c = λν gives λ = c/ν.
Check — multiple choice
High in the atmosphere, an ozone molecule absorbs one photon of ultraviolet radiation with a wavelength of 250 nm. What is the photon's energy, and what type of transition does the absorption produce? (c = 3.00 × 10⁸ m/s; h = 6.626 × 10⁻³⁴ J·s.)
Feedback if chosen: This option is wrong — you calculated the energy correctly but paired ultraviolet absorption with vibration. Ultraviolet photons promote an electron to a higher electronic level; infrared pairs with vibration.
Feedback if chosen: This option is wrong — you substituted 250 without converting nanometers to meters (250 nm = 2.50 × 10⁻⁷ m), then matched the tiny result to rotation.
Feedback if chosen: Correct. E = hc/λ = (6.626 × 10⁻³⁴ × 3.00 × 10⁸)/(2.50 × 10⁻⁷) = 7.95 × 10⁻¹⁹ J, and absorbing an ultraviolet photon promotes one of the molecule's electrons — an electronic transition.
Feedback if chosen: This option is wrong — you multiplied h and c by the wavelength instead of dividing by it. Combining E = hν with c = λν gives E = hc/λ.
Check — multiple choice
A solution of a food dye is measured at the dye's wavelength of maximum absorbance in a 1.50 cm cuvette and reads A = 0.90. At this wavelength the dye's molar absorptivity is ε = 240 L·mol⁻¹·cm⁻¹. What is the concentration of the dye?
Feedback if chosen: This option is wrong — you left the path length out. c = A/(εb) = 0.90/(240 × 1.50), not 0.90/240.
Feedback if chosen: This option is wrong — you divided εb by A, inverting the rearrangement. Making c the subject of A = εbc gives c = A/(εb).
Feedback if chosen: This option is wrong — you multiplied the absorbance by εb instead of dividing. c = A/(εb) = 0.90/360 = 0.0025 M.
Feedback if chosen: Correct. Making c the subject of A = εbc gives c = A/(εb) = 0.90/(240 × 1.50) = 0.90/360 = 0.0025 M.
Check — multiple choice
The calibration line shown was built from four standard solutions of a red food dye, measured at 504 nm in a 1.00 cm cuvette; the reading for the second standard is labeled. An unknown solution of the same dye, measured the same way, reads A = 0.75. What is its concentration?
Feedback if chosen: Correct. The line passes through the origin, so absorbance and concentration stay in one ratio: the labeled standard gives A = 0.50 at 2.0 × 10⁻⁵ M, so A = 0.75 reads (0.75/0.50) × 2.0 × 10⁻⁵ = 3.0 × 10⁻⁵ M — safely below the highest standard.
Feedback if chosen: This option is wrong — you inverted the proportion, scaling by 0.50/0.75. The unknown absorbs more than the labeled standard, so its concentration must be higher, not lower.
Feedback if chosen: This option is wrong — you echoed the labeled standard. That point reads A = 0.50; the unknown reads A = 0.75, so it sits farther up the line.
Feedback if chosen: This option is wrong — you treated the standards' own readings as the only usable points. Reading between the standards is exactly what the line is for; only a reading beyond the highest standard is off-limits.
Check — multiple choice
CH₄ (10 electrons), SiH₄ (18 electrons), and GeH₄ (36 electrons) are all nonpolar, tetrahedral molecules. In the liquid state, which substance has the strongest London dispersion forces between its molecules?
Feedback if chosen: This option is wrong — you chose the smallest electron cloud. A small, tightly held cloud distorts least, so CH₄ is the least polarizable of the three and has the weakest London dispersion forces.
Feedback if chosen: This option is wrong — you stopped partway up the trend. SiH₄'s 18-electron cloud distorts more easily than CH₄'s, but GeH₄'s 36-electron cloud distorts more easily still.
Feedback if chosen: Correct. GeH₄ has the largest, most easily distorted electron cloud, so its temporary dipoles are the largest and its London dispersion forces are the strongest of the three.
Feedback if chosen: This option is wrong — you treated all nonpolar molecules as attracting identically. London dispersion force strength grows with polarizability, and polarizability grows with the size of the electron cloud.
Check — multiple choice
In which pure liquid does hydrogen bonding occur between the molecules?
Feedback if chosen: This option is wrong — you treated the S–H hydrogen as a hydrogen-bond donor. Hydrogen bonding requires an H atom bonded to N, O, or F; sulfur does not qualify.
Feedback if chosen: This option is wrong — you saw F and H in the same molecule, but the H atom is bonded to C. With no N–H, O–H, or F–H, the molecule has no hydrogen-bond donor.
Feedback if chosen: This option is wrong — the O atom could accept a hydrogen bond, but every H atom in CH₃CHO is bonded to C, so pure CH₃CHO has no donor and cannot hydrogen-bond with itself.
Feedback if chosen: Correct. Each N–H hydrogen in N₂H₄ is a donor, and the N atom of a neighboring molecule is an acceptor, so hydrazine molecules hydrogen-bond with one another.
Check — multiple choice
In liquid H₂CO (a polar molecule in which both H atoms are bonded to the central C atom), which list names every intermolecular force present between the molecules?
Feedback if chosen: This option is wrong — you treated H₂CO as nonpolar. The molecule has a partially negative O end and a partially positive H end, so neighboring molecules also align dipole-to-dipole.
Feedback if chosen: This option is wrong — you dropped the London dispersion forces. They act between all particles, whatever other attractions are also present.
Feedback if chosen: Correct. H₂CO is polar, so London dispersion forces and dipole-dipole attractions are both present; with every H bonded to C there is no N–H, O–H, or F–H donor, so there is no hydrogen bonding.
Feedback if chosen: This option is wrong — you added hydrogen bonding because the molecule contains both H and O. The H atoms are bonded to C, so the molecule has no hydrogen-bond donor.
Check — multiple choice
At 25 °C the vapor pressure of liquid pentane (C₅H₁₂, 42 electrons) is about 80 times the vapor pressure of liquid 1-butanol (C₄H₉OH, 42 electrons). Which statement best explains the difference?
Feedback if chosen: Correct. Escaping 1-butanol means completely overcoming hydrogen bonds as well as London dispersion forces, so at 25 °C far fewer 1-butanol molecules escape, and its vapor pressure is far lower.
Feedback if chosen: This option is wrong — you argued from mass. The two molecules are nearly identical in mass (72 and 74 g/mol), and escape from a liquid is decided by the strength of the attractions between molecules, not by particle mass.
Feedback if chosen: This option is wrong — you had covalent bonds breaking during evaporation. Evaporation overcomes attractions between molecules; every bond within each molecule stays intact.
Feedback if chosen: This option is wrong — stronger attractions would give pentane a lower vapor pressure, not a higher one, and with equal electron counts the two clouds are similarly polarizable. The real difference is 1-butanol's hydrogen bonding.
Check — multiple choice
| Property | Observation |
|---|---|
| Melting point | above 2700 °C |
| Conducts electricity as a solid? | no |
| Conducts electricity when molten? | no |
| Hardness | scratches quartz |
The table shows the properties measured for solid E. Based on this evidence, solid E is best classified as which type of solid?
Feedback if chosen: Correct. No conductivity in either state rules out metallic and ionic solids, and the extreme melting point and hardness show covalent bonds extending throughout the structure — melting solid E means breaking those bonds.
Feedback if chosen: This option is wrong — you matched the high melting point but skipped the conductivity evidence. An ionic solid conducts when molten, because its ions become free to move; solid E does not conduct when molten.
Feedback if chosen: This option is wrong — a metallic solid conducts electricity as a solid through its sea of mobile valence electrons. Solid E conducts in neither state.
Feedback if chosen: This option is wrong — a molecular solid is held together by intermolecular attractions, so it is comparatively soft and melts low. Solid E melts above 2700 °C and scratches quartz.
Check — multiple choice
Xe and HI have the same number of electrons (54 each), and Xe (131.3 g/mol) is slightly more massive than HI (127.9 g/mol). NaI is an ionic solid. Which correctly ranks the three substances by boiling point, lowest to highest?
Feedback if chosen: This option is wrong — you ranked Xe above HI by molar mass. With equal electron clouds their London dispersion forces are similar, and HI's dipole-dipole attractions add to its total, so HI boils higher than Xe.
Feedback if chosen: This option is wrong — you ranked the ionic solid lowest. Boiling NaI requires completely overcoming attractions between full ionic charges, which takes far more energy than overcoming any intermolecular force.
Feedback if chosen: This option is wrong — you placed molecular HI above ionic NaI. HI's dipole-dipole and dispersion attractions are much weaker than the full-charge attractions holding NaI together.
Feedback if chosen: Correct. Compare class first: ionic NaI boils highest (1304 °C). Within the molecular pair, the equal electron counts make the London dispersion forces similar, and HI's dipole-dipole attractions lift it above Xe (−35 °C versus −108 °C).
Check — multiple choice
To harden a titanium component, nitrogen atoms are diffused into the metal near its surface. The treated metal is noticeably less malleable than pure titanium. Which statement explains the change?
Feedback if chosen: Correct. Interstitial atoms lodge in the holes between the metal atoms and block the planes of atoms from sliding past one another, so the treated metal deforms less readily.
Feedback if chosen: This option is wrong — you had the alloy losing its electron sea. Alloys keep their mobile valence electrons (they still conduct); the added rigidity comes from atoms locked between the metal cores, not from lost bonding.
Feedback if chosen: This option is wrong — you described a substitutional alloy. Small atoms like N slip into the spaces between the much larger metal atoms rather than replacing them: an interstitial alloy.
Feedback if chosen: This option is wrong — the treated metal is still a metallic solid. The N atoms restrict how the Ti atoms can rearrange; they do not convert the metal into a covalent network.
Check — multiple choice
One mole of liquid CCl₄ occupies 97 cm³. Which is closest to the volume occupied by one mole of solid CCl₄, just below its freezing temperature?
Feedback if chosen: This option is wrong — you shrank the volume tenfold. The molecules in a liquid are already in close contact, so freezing barely changes how much space a mole of them occupies.
Feedback if chosen: Correct. The molecules remain in close contact in both the solid and the liquid, so the two phases have similar molar volumes — about 90 cm³ versus 97 cm³.
Feedback if chosen: This option is wrong — you expanded the volume tenfold. Melting and freezing keep the particles in close contact; molar volume changes only slightly between solid and liquid.
Feedback if chosen: This option is wrong — that is a gas-like molar volume. Only in the gas phase are the particles widely separated, making the molar volume enormously larger.
Check — multiple choice
Samples of Kr (83.8 g/mol), F₂ (38.0 g/mol), and NH₃ (17.0 g/mol) are all held at 500 K. Which correctly ranks the gases by average particle speed, slowest to fastest?
Feedback if chosen: This option is wrong — you ranked the most massive gas as the fastest. At one temperature all three samples share the same average kinetic energy, so the least massive particles must move fastest.
Feedback if chosen: This option is wrong — you placed F₂ above NH₃, but F₂ (38.0 g/mol) is more than twice as massive as NH₃ (17.0 g/mol). With equal average kinetic energy, the lighter NH₃ molecules move faster.
Feedback if chosen: This option is wrong — equal temperature fixes the average kinetic energy, not the speed. For ½mv² to be equal across the samples, the lighter particles must move faster.
Feedback if chosen: Correct. Same temperature means same average kinetic energy, so average speed ranks opposite to molar mass: Kr (83.8 g/mol) slowest, then F₂ (38.0), then NH₃ (17.0) fastest.
Check — multiple choice
A student sketches the Maxwell-Boltzmann speed distributions for samples of CH₄ (16.0 g/mol) and CO₂ (44.0 g/mol), both at 350 K, as shown. What is the error in the sketch?
Feedback if chosen: Correct. At one temperature both gases share the same average kinetic energy, so the lighter CH₄ molecules move faster on average — the CH₄ curve belongs at the higher speeds.
Feedback if chosen: This option is wrong — equal temperature means equal average kinetic energy, not equal speed. For ½mv² to match, the lighter CH₄ molecules must move faster, so the two peaks sit at different speeds.
Feedback if chosen: This option is wrong — you tied the area under a curve to particle mass. The area shows the number of particles in the sample, not how heavy they are.
Feedback if chosen: This option is wrong — you accepted a sketch in which the more massive gas peaks at the higher speed. At the same temperature the lighter gas's curve sits at higher speeds.
Check — multiple choice
A sealed, rigid container holds a gas at 2.40 atm and 27 °C. The container is warmed to 127 °C. What is the new pressure of the gas?
Feedback if chosen: This option is wrong — you inverted the temperature ratio. At constant volume, pressure grows in proportion to the Kelvin temperature: P = 2.40 atm × (400 K/300 K), not × (300 K/400 K).
Feedback if chosen: This option is wrong — you kept the pressure constant. A rigid container fixes the volume, not the pressure: the faster-moving particles strike the walls harder and more often, so the pressure rises.
Feedback if chosen: Correct. With volume and amount constant, pressure divided by Kelvin temperature stays constant: P = 2.40 atm × (400 K/300 K) = 3.20 atm.
Feedback if chosen: This option is wrong — you used the Celsius ratio (127/27). Gas proportions hold for Kelvin temperatures: 27 °C is 300 K and 127 °C is 400 K.
Check — multiple choice
What volume does 2.00 mol of an ideal gas occupy at 27 °C and 1.20 atm? (R = 0.08206 L·atm·mol⁻¹·K⁻¹.)
Feedback if chosen: This option is wrong — you substituted 27 into PV = nRT without converting to Kelvin. The temperature must be 27 + 273 = 300 K.
Feedback if chosen: This option is wrong — you left the amount of gas out of the calculation. V = nRT/P needs n = 2.00 mol in the numerator.
Feedback if chosen: Correct. V = nRT/P = (2.00 × 0.08206 × 300)/1.20 = 41.0 L.
Feedback if chosen: This option is wrong — you calculated nRT but did not divide by the pressure. V = nRT/P = 49.2/1.20 = 41.0 L.
Check — multiple choice
A container holds a mixture of C₂H₄(g) and N₂(g). The mole fraction of C₂H₄ is 0.55, and the total pressure is 2.00 atm. What is the partial pressure of C₂H₄?
Feedback if chosen: This option is wrong — you multiplied by the mole fraction of the other gas: 0.45 × 2.00 atm = 0.90 atm is the partial pressure of N₂.
Feedback if chosen: Correct. A component's partial pressure is its mole fraction times the total pressure: 0.55 × 2.00 atm = 1.10 atm.
Feedback if chosen: This option is wrong — you reported the total pressure. C₂H₄ contributes only its share of the collisions with the walls: 0.55 of the total.
Feedback if chosen: This option is wrong — you divided the total pressure by the mole fraction instead of multiplying: 2.00/0.55 = 3.6 atm, which is larger than the total pressure and impossible for one component.
Check — multiple choice
Four gas samples are held at the same low temperature and very high pressure. Which gas is expected to deviate most from ideal behavior?
Feedback if chosen: This option is wrong — CH₄ is nonpolar with a small 10-electron cloud, so its attractions are weak and it stays closer to ideal than a polar, more polarizable gas.
Feedback if chosen: This option is wrong — N₂ is nonpolar, and its 14-electron cloud is far smaller than CH₃OCH₃'s. Weak attractions mean small deviations.
Feedback if chosen: This option is wrong — you picked a gas that deviates least. Ne's tiny, tightly held cloud gives it the weakest attractions of the four, so it behaves most ideally.
Feedback if chosen: Correct. CH₃OCH₃ has the largest electron cloud and the only permanent dipole, so its attractions are the strongest of the four — and at low temperature and high pressure those attractions pull its behavior farthest from ideal.
Check — multiple choice
The figure shows a developed plate on which the three colored components of a dye mixture were separated. Which component is the most polar?
Feedback if chosen: This option is wrong — you chose the component that traveled farthest. With a nonpolar mobile phase, the component carried farthest is the one attracted least to the polar coating: the least polar.
Feedback if chosen: This option is wrong — D2 traveled farther than D3, so D2 is held less strongly by the polar coating. The component held most strongly, the one nearest the start line, is the most polar.
Feedback if chosen: Correct. The stationary coating is polar, so the most polar component grips it hardest and travels least. D3 stopped closest to the start line.
Feedback if chosen: This option is wrong — the components separated precisely because their attractions to the two phases differ. Equally polar components would be held equally and would travel together.
Check — multiple choice
Solid SrCl₂ dissolves readily in water but not in the nonpolar solvent toluene. Which statement explains the difference?
Feedback if chosen: Correct. Each ion that leaves the lattice trades its ion-ion attractions for many ion-dipole attractions to water molecules. Nonpolar toluene offers no such replacement, so the ions stay in the solid.
Feedback if chosen: This option is wrong — dissolving is not a chemical reaction. The ions leave the lattice unchanged, as Sr²⁺ and Cl⁻, held by attractions to the water molecules around them.
Feedback if chosen: This option is wrong — you made molecule size the deciding factor. Toluene fails because it is nonpolar and cannot form strong attractions with the ions, not because of its size.
Feedback if chosen: This option is wrong — you named hydrogen bonding, but the attraction that frees an ion from the lattice is ion-dipole: the ion's full charge attracting the oppositely charged partial end of a polar molecule.
Check — multiple choice
After absorbing radiation, gaseous HBr molecules rotate faster, having moved to higher rotational energy levels. The absorbed radiation is in which spectral region?
Feedback if chosen: This option is wrong — you paired rotation with the infrared region. Infrared photons match the larger gaps between vibrational levels; the gaps between rotational levels are smaller still.
Feedback if chosen: Correct. Rotational energy levels are the most closely spaced, and microwave photons carry the small energies that match those gaps.
Feedback if chosen: This option is wrong — ultraviolet photons carry enough energy to promote electrons between electronic levels, far more than a change in rotation requires.
Feedback if chosen: This option is wrong — visible photons drive electronic transitions. A change in rotation involves a much smaller energy than promoting an electron.
Check — multiple choice
What is the energy of one photon of ultraviolet radiation with a wavelength of 300 nm? (c = 3.00 × 10⁸ m/s; h = 6.626 × 10⁻³⁴ J·s.)
Feedback if chosen: This option is wrong — you did not convert the wavelength to meters: 300 nm = 3.00 × 10⁻⁷ m, which gives ν = 1.00 × 10¹⁵ s⁻¹, not 1.00 × 10⁶ s⁻¹.
Feedback if chosen: Correct. ν = c/λ = (3.00 × 10⁸ m/s)/(3.00 × 10⁻⁷ m) = 1.00 × 10¹⁵ s⁻¹, and E = hν = (6.626 × 10⁻³⁴ J·s)(1.00 × 10¹⁵ s⁻¹) = 6.63 × 10⁻¹⁹ J.
Feedback if chosen: This option is wrong — you calculated the frequency (1.00 × 10¹⁵ s⁻¹) and reported it as an energy. Multiply the frequency by Planck's constant to get the photon energy.
Feedback if chosen: This option is wrong — you divided the frequency by Planck's constant instead of multiplying by it. E = hν.
Free-response question
Source note (not shown as part of the question): authored to archetypes A14 (IMF comparative property explanation) + A19 (ideal gas law), with A23 (particulate representation) and A24 (experimental error analysis)
A student is given a mixture of two colorless liquids: dichloromethane (CH₂Cl₂, 42 electrons per molecule, molar mass 84.93 g/mol) and 2-propanol ((CH₃)₂CHOH, 34 electrons per molecule, molar mass 60.10 g/mol). Both molecules are polar. In 2-propanol the O–H group is bonded to the central C atom; in dichloromethane both H atoms are bonded to the C atom. The student is asked to separate the mixture by distillation and to identify the liquid that collects first.
(a) Identify every intermolecular force present between the molecules in each pure liquid. [1 point]
(b) Draw a particle-level representation of two 2-propanol molecules in the liquid, showing the strongest attraction between the two molecules as a dashed line. Your drawing must make clear which atom of each molecule the dashed line connects, showing all atoms and bonds in each molecule. [1 point]
(c) The student heats the mixture in a distillation apparatus. Predict which liquid collects first in the receiving flask. Justify your prediction in terms of the intermolecular forces between the molecules of each liquid. [2 points]
(d) The figure above represents the particles in a small sample of the vapor rising from the boiling mixture early in the distillation. The total pressure of this vapor sample is 0.500 atm. Calculate the partial pressure of CH₂Cl₂ in the sample. [1 point]
(e) The student collects the liquid that condenses first. To identify it, the student pours a small sample of it into a rigid 0.250 L flask, covers the mouth of the flask with foil pierced by a pinhole, and heats the flask in a boiling-water bath at 100.0 °C until the liquid has just vaporized completely, so that the flask holds only vapor at 100.0 °C and 1.00 atm. Calculate the number of moles of vapor in the flask. (R = 0.08206 L·atm·mol⁻¹·K⁻¹.) [1 point]
(f) The student removes the flask from the bath and lets it cool, and the vapor condenses back to a liquid. The mass of the liquid in the flask is 0.694 g. Calculate the molar mass of the collected liquid, and identify the liquid. [1 point]
(g) To confirm the identification, the student shakes a few milliliters of the collected liquid with water: the collected liquid does not dissolve, and two layers form. A few milliliters of 2-propanol shaken with water dissolve completely. Explain both observations in terms of the intermolecular forces between each liquid's molecules and water molecules. [1 point]
(h) A second student repeats the procedure of parts (e) and (f) but removes the flask from the boiling-water bath while some of the liquid has not yet vaporized. The second student cools the flask, measures the mass of the liquid in it, and calculates the molar mass using the same pressure, volume, and temperature values as in part (e). Will the second student's calculated molar mass be greater than, less than, or equal to the molar mass calculated in part (f)? Justify your answer. [2 points]
✓ Model response (the key)
(a)
Dichloromethane: London dispersion forces and dipole-dipole attractions.
2-Propanol: London dispersion forces, dipole-dipole attractions, and hydrogen bonding (each O–H hydrogen is a donor, and the O atom of a neighboring molecule is an acceptor).
(b)
The drawing shows two complete 2-propanol molecules with every covalent bond intact.
A dashed line is drawn from the H atom of one molecule's O–H group to the O atom of the other molecule — between the two molecules, not from a C–H hydrogen and not within one molecule.
(c)
Dichloromethane collects first.
2-Propanol molecules hydrogen-bond through their O–H groups; dichloromethane molecules cannot (both H atoms are bonded to C), so even though dichloromethane has the larger electron cloud (42 electrons versus 34), the overall attractions between 2-propanol molecules are stronger.
Vaporizing requires completely overcoming the attractions between molecules, so the more weakly attracted dichloromethane molecules escape the liquid in greater numbers, and dichloromethane vaporizes and reaches the receiving flask first.
(d)
mole fraction of CH₂Cl₂ = 9/(9 + 3) = 0.750
P(CH₂Cl₂) = 0.750 × 0.500 atm = 0.375 atm
(e)
T = 100.0 + 273 = 373 K
n = PV/RT = (1.00 atm × 0.250 L)/(0.08206 L·atm·mol⁻¹·K⁻¹ × 373 K) = 0.00817 mol
(f)
molar mass = 0.694 g ÷ 0.00817 mol = 84.9 g/mol (85.0 g/mol if decimals are carried)
84.9 g/mol matches dichloromethane (84.93 g/mol), not 2-propanol (60.10 g/mol): the liquid that collects first is dichloromethane, consistent with part (c)
(g)
2-Propanol dissolves because its O–H group forms hydrogen bonds with water molecules, and those new attractions between 2-propanol and water replace the hydrogen bonds between water molecules that are given up as the liquids mix.
Dichloromethane cannot hydrogen-bond with water (both of its H atoms are bonded to C), so its dipole-dipole and London dispersion attractions to water are too weak to replace the hydrogen bonds between water molecules — the water molecules stay attracted to one another, and the dichloromethane collects in a separate layer.
(h)
The second student's calculated molar mass will be greater than the molar mass calculated in part (f).
The liquid that never vaporized is weighed together with the condensed vapor, so the measured mass is larger than the mass of the vapor that filled the flask; the number of moles calculated from PV = nRT is unchanged, so mass ÷ moles gives a larger molar mass.
Scoring rubric — 10 points available
Free-response question
Source note (not shown as part of the question): 2021 FRQ Q7
A student investigates gas behavior using a rigid cylinder with a movable piston of negligible mass, as shown in the diagram above. The cylinder contains 0.325 mol of O₂(g).
(a) The cylinder has a volume of 7.95 L at 25 °C and 1.00 atm. Calculate the density of the O₂(g), in g/L, under these conditions. [1 point]
(b) Attempting to change the density of the O₂(g), the student opens the valve on the side of the cylinder, pushes down on the piston to release some of the gas, and closes the valve again. The temperature of the gas remains constant at 25 °C. Will this action change the density of the gas remaining in the cylinder? Justify your answer. [1 point]
(c) The student tries to change the density of the O₂(g) by cooling the cylinder to −55 °C, which causes the volume of the gas to decrease. Using principles of kinetic molecular theory, explain why the volume of the O₂(g) decreases when the temperature decreases to −55 °C. [1 point]
(d) The student further cools the cylinder to −180 °C and observes that the measured volume of the O₂(g) is substantially smaller than the volume that is calculated using the ideal gas law. Assume all equipment is functioning properly. Explain why the measured volume of the O₂(g) is smaller than the calculated volume. (The boiling point of O₂(l) is −183 °C.) [1 point]
✓ Model response (the key)
(a)
mass of O₂ = 0.325 mol × 32.00 g/mol = 10.4 g
D = m ÷ V = 10.4 g ÷ 7.95 L = 1.31 g/L
(b)
No — the density of the remaining gas does not change.
P and T are constant, so the mass and the volume of the O₂ remaining in the cylinder decrease proportionally, and m ÷ V stays the same.
(c)
As the gas cools, the average kinetic energy (speed) of the O₂ molecules decreases, so the molecules rebound with less energy when they collide with the walls of the container.
The volume of the gas must decrease so that the collisions with the walls keep supporting a pressure of 1.00 atm.
(d)
The ideal gas law assumes that gas particles experience no attractions to one another.
At −180 °C, just above the boiling point of O₂(l), the molecules move slowly enough that the attractions between them have a significant effect, drawing the molecules closer together, so the measured volume is smaller than the calculated volume.
Scoring rubric — 4 points available
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