How to use this review
Use this page after the Unit 3 diagnostic and any topics it flagged. Answer all seven questions, each combining two or more topics, before opening any answer.
These are original Marlbridge practice questions, not past exam questions. Liquid L and the salt Na₂J are fictional, with invented data. Each rubric is a suggested Marlbridge rubric, not official scoring, and your total gives no predicted score.
Data. R = 0.08206 L atm mol⁻¹ K⁻¹; K = °C + 273.15; h = 6.626 × 10⁻³⁴ J s; c = 2.998 × 10⁸ m s⁻¹; N_A = 6.022 × 10²³ mol⁻¹. Cl 35.45 g mol⁻¹.
Big ideas of the unit
- Forces between particles are not bonds. Dispersion, dipole–dipole, hydrogen bonding and ion–dipole forces are Coulombic attractions between particles (Topic 3.1).
- Properties follow from the particles and their forces. Boiling point and vapor pressure track force strength; the four kinds of solid differ because their particles differ (Topic 3.2).
- Phases differ in spacing and motion, not in the particles. Solids and liquids touch, so their molar volumes are similar; gas particles are far apart (Topic 3.3).
- An ideal gas obeys PV = nRT, and each gas in a mixture acts alone. Partial pressure is proportional to mole fraction (Topic 3.4).
- Kelvin temperature measures average kinetic energy. At the same temperature lighter particles move faster; pressure comes from collisions (Topic 3.5).
- Real gases fail the model where Topic 3.1 matters. Attractions lower PV/nRT near condensation; particle volume raises it at very high pressure (Topic 3.6).
- Solutions are counted in moles per litre of solution. Ions are counted separately, and dilution conserves moles (Topic 3.7), which a correct particle diagram shows (Topic 3.8).
- Separation and solubility both compare attractions. Chromatography and distillation exploit differences in forces (Topic 3.9); a solute dissolves when the new attractions can replace the ones broken (Topic 3.10).
- A photon is absorbed only if its energy matches a gap. The region tells you the transition (Topic 3.11), E = hν gives the size (Topic 3.12), and absorbance counts the absorbing particles in the beam (Topic 3.13).
Key relationships and methods
| Idea | Relationship or method | Topic |
|---|---|---|
| Comparing forces | more electrons → stronger dispersion; H on N, O or F → hydrogen bonding | 3.1 |
| Boiling and vapor pressure | stronger forces → higher boiling point, lower vapor pressure | 3.2 |
| Molar volume | V_m = M ÷ density; solid ≈ liquid ≪ gas | 3.3 |
| Ideal gas | PV = nRT (T in K); P_A = X_A × P_total; P_total = P_A + P_B + … | 3.4 |
| Kinetic energy | KE = ½mv²; average KE ∝ T; at equal T, v ∝ 1/√M | 3.5 |
| Real gases | PV/nRT < 1: attractions dominate; > 1: particle volume dominates | 3.6 |
| Molarity and dilution | M = n ÷ V (L of solution); n before = n after dilution | 3.7, 3.8 |
| Chromatography | Rf = solute distance ÷ solvent distance; more attracted to stationary phase → lower Rf | 3.9 |
| Photons | c = λν; E = hν; per mole × N_A; photon energy = gap | 3.11, 3.12 |
| Beer–Lambert | A = εbc; error chain: effect on A, then on c | 3.13 |
Question 1 (multiple choice · mixed)
Two identical rigid flasks are at 300 K. One holds 1.00 mol of helium; the other 1.00 mol of sulfur dioxide, SO₂, at moderate pressure. Which comparison is correct?
- (A) Same average kinetic energy, same average speed and exactly the same pressure.
- (B) Same average kinetic energy; He atoms are faster; the SO₂ pressure is slightly lower.
- (C) SO₂ has the higher average kinetic energy because it is heavier, so its pressure is higher.
- (D) Same average kinetic energy; He atoms are faster, so the He pressure is well above the PV = nRT value.
Answer and explanation
Answer: (B). Same Kelvin temperature means same average KE (Topic 3.5), so the lighter He atoms move faster. PV = nRT predicts equal pressures (Topic 3.4), but polar SO₂ molecules attract each other, softening their wall collisions and lowering the pressure a little (Topic 3.6).
- (A) ignores the mass difference and SO₂’s non-ideal behaviour.
- (C) Average KE depends only on temperature.
- (D) PV = nRT already accounts for speed; helium is nearly ideal.
Question 2 (multiple choice · mixed)
A blue complex ion has an electronic energy gap of 3.31 × 10⁻¹⁹ J, and ε = 5.00 × 10³ M⁻¹ cm⁻¹ at the matching wavelength. A solution in a 1.00 cm cuvette gives A = 0.450 there. Which wavelength should the spectrophotometer use, and what is the concentration?
- (A) 450 nm; 9.00 × 10⁻⁵ M
- (B) 600 nm; 1.11 × 10⁴ M
- (C) 300 nm; 9.00 × 10⁻⁵ M
- (D) 600 nm; 9.00 × 10⁻⁵ M
Answer and explanation
Answer: (D). The absorbed photon’s energy equals the gap (Topics 3.11, 3.12): λ = hc/E = (6.626 × 10⁻³⁴ × 2.998 × 10⁸) ÷ 3.31 × 10⁻¹⁹ = 6.00 × 10⁻⁷ m = 600 nm, visible, as an electronic transition should be. Then c = A ÷ (εb) = 0.450 ÷ 5.00 × 10³ = 9.00 × 10⁻⁵ M (Topic 3.13).
- (A) is the blue colour you see. A blue solution absorbs orange light; the absorbed wavelength matches the gap.
- (B) inverts the rearrangement (ε ÷ A).
- (C) doubles the energy; 300 nm photons do not match this gap.
Question 3 (multiple choice · mixed)
Which liquid is miscible with water and has a lower vapor pressure than water at 25 °C?
- (A) Hexane, C₆H₁₄
- (B) Ethanol, CH₃CH₂OH
- (C) Ethane-1,2-diol, HOCH₂CH₂OH
- (D) Octane, C₈H₁₈
Answer and explanation
Answer: (C). Two O–H groups let it hydrogen bond strongly with water, so it mixes in all proportions (Topic 3.10). Its molecules form more hydrogen bonds than ethanol’s and have more electrons than water’s, so fewer escape: lower vapor pressure (Topics 3.1, 3.2).
- (A) Nonpolar hexane does not mix with water, and its weak forces give a high vapor pressure.
- (B) Ethanol mixes, but its vapor pressure is higher than water’s.
- (D) Octane does have a lower vapor pressure, but it is nonpolar and does not mix.
Question 4 (constructed response · mixed)
A student finds the molar mass of a volatile liquid L (invented data), which boils at 92 °C. She heats a little L in a 0.250 L flask in boiling water at 100.0 °C until the flask holds only vapor at 1.00 atm. After cooling, the condensed L has a mass of 0.505 g.
(a) Calculate the amount of L, assuming the vapor is ideal. (b) Calculate the molar mass of L. (c) The true molar mass is 60.10 g mol⁻¹. Calculate the percentage error. (d) Explain, at the particle level, why the vapor is not ideal here and why this makes the result too high. (e) Predict whether repeating the experiment at a higher temperature (still 1.00 atm) gives a result closer to 60.10 g mol⁻¹. Justify with the kinetic molecular theory.
Worked solution and suggested Marlbridge rubric
(a) T = 373.15 K. n = PV/RT = (1.00 × 0.250) ÷ (0.08206 × 373.15) = 8.16 × 10⁻³ mol.
(b) M = 0.505 ÷ 0.008164 = 61.9 g mol⁻¹.
(c) (61.85 − 60.10) ÷ 60.10 × 100 = 2.9% too high.
(d) Only 8 °C above its boiling point, the vapor is close to condensing, and attractions between the slow molecules soften their wall collisions. To reach 1.00 atm the flask must hold more molecules than PV/RT predicts (truly 0.505 ÷ 60.10 = 8.40 × 10⁻³ mol). Their whole mass is divided by the smaller ideal amount, so M is too high.
(e) Closer. Faster molecules (greater average KE) are less affected by brief attractions, so the vapor behaves more ideally.
| Point | What earns it |
|---|---|
| 1 | (a) Kelvin temperature and 8.16 × 10⁻³ mol |
| 1 | (b) 61.9 g mol⁻¹ (carry forward) |
| 1 | (c) 2.9% too high |
| 1 | (d) Attractions between molecules near condensation named as the cause |
| 1 | (d) More molecules (more mass) than PV/RT predicts, so M too high |
| 1 | (e) Closer, because greater average KE makes attractions matter less |
Total: 6 points. Using 100.0 instead of 373.15 K gives an absurd 16.6 g mol⁻¹.
Question 5 (constructed response · mixed)
A fictional salt, Na₂J (250.0 g mol⁻¹), dissolves completely to give Na⁺ and a coloured ion, J²⁻. Only J²⁻ absorbs light.
(a) Calculate the mass of Na₂J needed for 100.0 mL of 4.00 × 10⁻³ M stock, and [Na⁺] in the stock. (b) Calculate the volume of stock needed for 50.00 mL of a 2.00 × 10⁻⁴ M standard. (c) Standards measured at λmax in a 1.00 cm cuvette give:
| c (× 10⁻⁴ M) | 1.00 | 2.00 | 3.00 | 4.00 |
|---|---|---|---|---|
| A | 0.158 | 0.317 | 0.473 | 0.633 |
Find ε for J²⁻, and the concentration of an unknown with A = 0.410. (d) A particle diagram of the stock portion used in (b) shows 4 J²⁻ ions. How many of each ion should a diagram of the whole 50.00 mL standard show? Describe how water molecules surround each ion. (e) Some Na₂J was left undissolved in the stock flask. Predict the effect on the unknown’s concentration.
Worked solution and suggested Marlbridge rubric
(a) n = 4.00 × 10⁻³ × 0.1000 = 4.00 × 10⁻⁴ mol, so mass = 0.100 g. Two Na⁺ per formula unit: [Na⁺] = 8.00 × 10⁻³ M.
(b) Moles are conserved: V₁ = (2.00 × 10⁻⁴ × 50.00 mL) ÷ 4.00 × 10⁻³ = 2.50 mL, made up to 50.00 mL.
(c) A ÷ c is nearly constant (1577 to 1585), so ε = 1.58 × 10³ M⁻¹ cm⁻¹. Unknown: c = 0.410 ÷ 1.58 × 10³ = 2.59 × 10⁻⁴ M.
(d) Dilution adds only water: still 4 J²⁻ and 8 Na⁺, spread through a larger volume. Water points its O atoms (δ−) towards Na⁺ and its H atoms (δ+) towards J²⁻.
(e) Too high. Every standard is more dilute than its label, so each reads low, the slope (εb) is too small, and c = A ÷ slope is too large.
| Point | What earns it |
|---|---|
| 1 | (a) 0.100 g and 8.00 × 10⁻³ M |
| 1 | (b) 2.50 mL |
| 1 | (c) ε = 1.58 × 10³ M⁻¹ cm⁻¹ |
| 1 | (c) 2.59 × 10⁻⁴ M |
| 1 | (d) 4 J²⁻ and 8 Na⁺, unchanged by dilution |
| 1 | (d) Correct water orientation around both ions |
| 1 | (e) Too high, with the chain standards → slope → c |
Total: 7 points.
Question 6 (constructed response · mixed)
Butan-1-ol, CH₃CH₂CH₂CH₂OH, and diethyl ether, CH₃CH₂OCH₂CH₃, are isomers (C₄H₁₀O, 42 electrons). At 1 atm butan-1-ol boils at about 118 °C and diethyl ether at about 35 °C.
(a) Identify all the forces between molecules in each pure liquid. (b) Predict which has the higher vapor pressure at 20 °C, and explain the large difference in boiling points. (c) Diethyl ether is heated from just below to just above its boiling point. State one feature of a particle diagram that changes and one that stays the same. (d) A mixture of the two is a clear solution. Explain why filtering cannot separate it, and which liquid is concentrated in the first distillate. (e) Explain how diethyl ether hydrogen bonds with water although it cannot hydrogen bond with itself.
Worked solution and suggested Marlbridge rubric
(a) Butan-1-ol: dispersion, dipole–dipole and hydrogen bonding. Diethyl ether: dispersion and dipole–dipole only (no H on O).
(b) Diethyl ether. Equal electron counts give similar dispersion forces, so the difference is hydrogen bonding between butan-1-ol molecules, which boiling must also overcome.
(c) Changes: molecules go from touching to far apart, filling the whole box. Same: every molecule stays intact; no covalent bonds break.
(d) Dissolved molecules are far smaller than the pores in filter paper. Distillation uses the different vapor pressures: the first distillate is rich in diethyl ether, which has the weaker attractions.
(e) Ether’s O has lone pairs, so it accepts a hydrogen bond from water’s O–H. It has no H on O to donate to another ether molecule.
| Point | What earns it |
|---|---|
| 1 | (a) Correct forces for both, with no hydrogen bonding in pure ether |
| 1 | (b) Ether has the higher vapor pressure |
| 1 | (b) Similar dispersion, so hydrogen bonding explains the difference |
| 1 | (c) One change (spacing) and molecules intact |
| 1 | (d) Filtration fails for a solution |
| 1 | (d) Ether concentrated first, with the reason |
| 1 | (e) Ether’s O accepts; ether has no H on O to donate |
Total: 7 points. “Butan-1-ol has stronger bonds” earns nothing: name the intermolecular force.
Question 7 (constructed response · mixed)
A rigid 1.00 L quartz flask holds chlorine, Cl₂, at 25.0 °C and 0.500 atm. It is lit with 330 nm ultraviolet light, which Cl₂ absorbs, and some molecules split: Cl₂ → 2 Cl. Take the Cl–Cl bond energy as 242 kJ mol⁻¹.
(a) Calculate the amount of Cl₂ before lighting. (b) Calculate the energy of one 330 nm photon and of one mole of them. Name the kind of transition the photon causes. (c) Calculate the longest wavelength that could break one Cl–Cl bond. (d) Back at 25.0 °C, 10.0% of the original Cl₂ has split (no atoms have yet recombined). Calculate the total pressure and P(Cl). (e) Use the kinetic molecular theory to explain the pressure change, and compare the average speeds of Cl and Cl₂.
Worked solution and suggested Marlbridge rubric
(a) n = PV/RT = (0.500 × 1.00) ÷ (0.08206 × 298.15) = 0.0204 mol.
(b) E = hc/λ = (6.626 × 10⁻³⁴ × 2.998 × 10⁸) ÷ 3.30 × 10⁻⁷ = 6.02 × 10⁻¹⁹ J; × N_A = 363 kJ mol⁻¹. UV causes an electronic transition; here the excited molecule falls apart.
(c) One bond: 242 000 ÷ 6.022 × 10²³ = 4.02 × 10⁻¹⁹ J, so λ = hc/E = 494 nm. Longer wavelengths carry too little energy per photon, however bright.
(d) For each 1 mol of Cl₂: 0.900 mol remains and 0.200 mol Cl forms, 1.100 mol in all. At constant V and T, P ∝ n: P = 0.550 atm. P(Cl) = (0.200 ÷ 1.100) × 0.550 = 0.100 atm.
(e) At the same temperature the average KE per particle is unchanged, but more particles make more wall collisions each second. Equal KE with half the mass makes Cl atoms faster by √2 ≈ 1.41.
| Point | What earns it |
|---|---|
| 1 | (a) 0.0204 mol |
| 1 | (b) 6.02 × 10⁻¹⁹ J and 363 kJ mol⁻¹ (accept 362) |
| 1 | (b) Electronic transition |
| 1 | (c) 494 nm, from the energy of one bond |
| 1 | (d) 0.550 atm |
| 1 | (d) P(Cl) = 0.100 atm |
| 1 | (e) More particles at the same average KE; Cl faster by √2 |
Total: 7 points. Forgetting to divide 242 kJ by N_A in (c) gives a wavelength about 6 × 10²³ times too small.
How did you do?
Total: 30 points (3 multiple choice, 27 constructed response). Where you lost points matters more than the total.
- Naming or comparing forces (Questions 3, 6): revisit Topics 3.1, 3.2, 3.9 and 3.10.
- Gas calculations and the particle model (Questions 1, 4, 7): revisit Topics 3.3 to 3.6.
- Solutions and diagrams (Question 5): revisit Topics 3.7 and 3.8.
- Light and absorbance (Questions 2, 5, 7): revisit Topics 3.11 to 3.13.
- Explanations marked down: name the particles, the force and which way it pushes the result.
If several topics were weak, return to the Unit 3 diagnostic. Then tick off each topic checklist: 3.1, 3.2, 3.3, 3.4, 3.5, 3.6, 3.7, 3.8, 3.9, 3.10, 3.11, 3.12, 3.13.