Questions & explanations
1. 0.6 g of acetic acid (M = 60 g mol⁻¹) dissolved in 100 g benzene gives ΔT_f = 0.205 K. K_f of benzene is 5.12 K kg mol⁻¹. What is the degree of association (α) of acetic acid in benzene assuming dimerization?
- 0.50
- 0.75
- 1.00
- 0.25
Answer: 0.75
Observed molar mass M_obs = (1000 × K_f × w2) / (ΔT_f × w1) = (1000 × 5.12 × 0.6) / (0.205 × 100) = 149.85 g mol⁻¹. i = M_normal / M_obs = 60 / 149.85 = 0.4004. For dimerization, i = 1 - α/2, so α = 2(1 - i) = 1.1992. Since α cannot exceed 1, complete dimerization (α = 1) is indicated. The closest option is 1.00.
2. In the Ostwald-Walker method, the loss in mass of the solution bulb is proportional to which quantity?
- The vapour pressure of the solution
- The difference in vapour pressure between pure solvent and solution
- The vapour pressure of the pure solvent
- The relative lowering of vapour pressure
Answer: The vapour pressure of the solution
In the Ostwald-Walker method, dry air passes through the solution bulb and gets saturated with solvent vapour. The mass loss of the solution bulb equals the mass of vapour carried away, which is directly proportional to the vapour pressure of the solution (p) according to Dalton's law. Thus, loss in mass ∝ p.
3. What is the molarity of 98% (w/w) H2SO4 solution with density 1.84 g/mL? (Molar mass of H2SO4 = 98 g/mol)
- 36.8 M
- 9.2 M
- 18.4 M
- 1.84 M
Answer: 18.4 M
Molarity = moles of solute per liter of solution. 100 g solution contains 98 g H2SO4 = 1 mol. Volume of 100 g solution = mass/density = 100 g / 1.84 g/mL = 54.35 mL = 0.05435 L. Molarity = 1 mol / 0.05435 L = 18.4 M. Alternatively, formula M = (10 × d × %w/w) / M_solute gives same result.
4. For an ideal binary mixture of acetone (p°=286 torr) and ethyl acetate (p°=90 torr) with liquid mole fraction of acetone 0.5, what is the mole fraction of acetone in the vapour phase?
- 0.50
- 0.76
- 0.24
- 0.86
Answer: 0.76
Using Raoult's law, p_acetone = 286×0.5 = 143 torr, p_ethyl acetate = 90×0.5 = 45 torr, total pressure = 188 torr. Mole fraction in vapour y_acetone = p_acetone / p_total = 143/188 ≈ 0.76. The vapour is richer in the more volatile component (acetone).
5. 4.0 g of a non-volatile solute dissolved in 100 g of water lowers the vapour pressure from 23.76 torr to 23.28 torr at 25°C. What is the molar mass of the solute? (Molar mass of water = 18 g/mol)
- 23.8 g/mol
- 18.0 g/mol
- 71.4 g/mol
- 35.7 g/mol
Answer: 35.7 g/mol
Using Δp/p° = x_solute ≈ n_solute/n_solvent for dilute solution. Δp = 23.76 - 23.28 = 0.48 torr. n_solvent = 100/18 = 5.556 mol. So n_solute = (Δp/p°) × n_solvent = (0.48/23.76) × 5.556 = 0.112 mol. Molar mass = mass/n = 4.0/0.112 = 35.7 g/mol.
6. A 6% w/v aqueous solution of glucose (M = 180 g mol⁻¹) is isotonic with a 3% w/v solution of an unknown non-electrolyte at the same temperature. What is the molar mass of the unknown?
- 90 g mol⁻¹
- 180 g mol⁻¹
- 360 g mol⁻¹
- 720 g mol⁻¹
Answer: 90 g mol⁻¹
Isotonic solutions have equal osmotic pressure, so molar concentrations are equal. For glucose: 6% w/v = 6 g/100 mL = 60 g/L. Molarity = 60/180 = 0.3333 M. For unknown: 3% w/v = 30 g/L. Molarity = 30/M = 0.3333, so M = 30/0.3333 = 90 g/mol.
7. Which of the following is NOT a colligative property?
- Relative lowering of vapour pressure
- Elevation of boiling point
- Elevation of vapour pressure
- Depression of freezing point
Answer: Elevation of vapour pressure
Colligative properties depend on the number of solute particles. The four are: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure. Vapour pressure is lowered, not elevated.
8. At 300 K, pure benzene has vapour pressure 100 mm Hg and pure toluene has 40 mm Hg. What is the total vapour pressure above an ideal solution with mole fraction of benzene 0.6?
- 84 mm Hg
- 60 mm Hg
- 64 mm Hg
- 76 mm Hg
Answer: 76 mm Hg
For an ideal solution, Raoult's law gives total pressure p_total = p_benzene°·x_benzene + p_toluene°·x_toluene. Here p_benzene° = 100, x_benzene = 0.6, p_toluene° = 40, x_toluene = 0.4. So p_total = 100×0.6 + 40×0.4 = 60 + 16 = 76 mm Hg.
9. Which of the following correctly describes osmosis?
- Net flow of solvent from higher solute concentration to lower solute concentration through a semipermeable membrane.
- Net flow of solvent from lower solute concentration to higher solute concentration through a semipermeable membrane.
- Net flow of solute from higher concentration to lower concentration through a semipermeable membrane.
- Net flow of both solute and solvent through a semipermeable membrane.
Answer: Net flow of solvent from lower solute concentration to higher solute concentration through a semipermeable membrane.
Osmosis is defined as the net flow of solvent molecules from a region of lower solute concentration to a region of higher solute concentration through a semipermeable membrane. The solvent moves to dilute the more concentrated solution.
10. 2.0 g benzoic acid (M=122) in 25 g benzene gives ΔT_f=1.62 K. K_f=5.12. What is the degree of association α?
- 0.99
- 0.50
- 0.75
- 1.00
Answer: 0.99
Theoretical molality = (2.0/122)/0.025 = 0.6556 m. Theoretical ΔT_f = 5.12×0.6556 = 3.357 K. Observed i = 1.62/3.357 = 0.4826. For dimerization (n=2), i = 1 - α/2, so α = 2(1 - i) = 2(1-0.4826) = 1.0348 ≈ 0.99 (complete association).
11. Which of the following pairs of liquids is expected to show positive deviation from Raoult's law?
- Ethanol and acetone
- Benzene and toluene
- Acetone and chloroform
- Chlorobenzene and bromobenzene
Answer: Ethanol and acetone
Positive deviation occurs when A-B interactions are weaker than A-A and B-B. Ethanol has strong H-bonding; acetone disrupts it, making ethanol-acetone interactions weaker. Thus vapour pressure is higher than Raoult's law predicts.
12. 98% w/w H2SO4 has density 1.84 g/mL. What volume of this acid is needed to prepare 250 mL of 0.5 M H2SO4?
- 7.6 mL
- 6.8 mL
- 6.2 mL
- 8.4 mL
Answer: 6.8 mL
First, calculate molarity of concentrated acid using M = (10 × d × %w/w) / M_solute = (10 × 1.84 × 98) / 98 = 18.4 M. Then use dilution formula M1V1 = M2V2: 18.4 × V1 = 0.5 × 250, so V1 = (0.5 × 250) / 18.4 = 6.79 mL ≈ 6.8 mL.