6: Molar Mass Determination

Correct answer: D. 4

Explanation: First calculate the theoretical molality using the normal molar mass: moles of solute = 1.6/80 = 0.020 mol, and solvent mass = 0.200 kg, so m = 0.020/0.200 = 0.100 mol kg⁻¹. The normal freezing-point depression is Kf m = 1.86 × 0.100 = 0.186 K. Thus i = observed depression/theoretical depression = 0.744/0.186 = 4. Option D is correct.

Correct answer: C. 0.85

Explanation: Assume one mole of solute initially. Of this, 80% remains as individual molecules and contributes 0.80 effective particles. The remaining 20% associates into tetramers; four original molecules form one particle, so this portion contributes 0.20/4 = 0.05 particles. Therefore i = 0.80 + 0.05 = 0.85, making option C correct.

Correct answer: C. 80 g mol−1

Explanation: For AB2, dissociation produces one A particle and two B particles. At degree of dissociation α = 0.50, the van’t Hoff factor is i = 1 + 2α = 2. Using ΔTf = iKf m, the molality is m = 0.558/(2 × 1.86) = 0.150 mol kg−1. The solvent mass is 250 g = 0.250 kg, so moles of solute = 0.150 × 0.250 = 0.0375 mol. Hence the true molar mass is 3/0.0375 = 80 g mol−1.

Correct answer: C. 200 g mol−1

Explanation: Use the osmotic-pressure equation π = iCRT, where C is the concentration of solute particles in the solution. Thus C = 0.492/(0.5 × 0.082 × 300) = 0.040 mol L−1. The solution volume is 300 mL = 0.300 L, so the number of moles represented by the colligative concentration is 0.040 × 0.300 = 0.012 mol. Therefore the true molar mass of the original solute is 2.4/0.012 = 200 g mol−1. The given i has already been included in calculating C.

Correct answer: B. 160 g mol⁻¹

Explanation: The actual moles in 750 mL are n = M V = 0.02 × 0.750 = 0.015 mol. The molar mass calculated without correcting for dissociation or association is 3.0/0.015 = 200 g mol⁻¹. For abnormal colligative behaviour, observed molar mass M_obs = M_true/i. Hence M_obs = 200/1.25 = 160 g mol⁻¹, so option B is correct.

Correct answer: C. 120 g mol⁻¹

Explanation: For dimerization, two monomer particles become one dimer particle. If the degree of association is α, the van’t Hoff factor is i = 1 − α/2. With α = 0.40, i = 1 − 0.40/2 = 0.80. Since M_obs = M_true/i, the true molar mass is M_true = iM_obs = 0.80 × 150 = 120 g mol⁻¹. Therefore, option C is correct.

Correct answer: B. 112.5 g mol⁻¹

Explanation: For formation of trimers, three monomer particles form one trimer, so the van’t Hoff factor is i = 1 − 2α/3. With α = 0.30, i = 1 − (2 × 0.30)/3 = 0.80. The observed molar mass is related by M_obs = M_true/i. Therefore, M_obs = 90/0.80 = 112.5 g mol⁻¹, making option B correct.

Correct answer: C. 140 g mol⁻¹

Explanation: For AB → A + B, one solute unit produces two particles on complete dissociation. Thus, for degree of dissociation α, i = 1 + α. With α = 0.75, i = 1.75. Since M_obs = M_true/i, the true molar mass is M_true = iM_obs = 1.75 × 80 = 140 g mol⁻¹. Hence option C is correct.

Correct answer: C. 40%

Explanation: Because M_obs = M_true/i, the given ratio M_obs/M_true = 5/9 gives i = 9/5 = 1.8. On dissociation, A₂B produces 2A + B, or three particles in total. Therefore i = 1 + (3 − 1)α = 1 + 2α. Solving 1.8 = 1 + 2α gives α = 0.40, or 40%. Thus option C is correct.

Correct answer: B. 60 g mol⁻¹

Explanation: For AB → A + B with 50% dissociation, i = 1 + α = 1.5. Using ΔT_f = iK_fm, the actual molality is m = 0.651/(1.5 × 1.86) ≈ 0.233 mol kg⁻¹. For 200 g = 0.2 kg water, moles of solute are 0.233 × 0.2 ≈ 0.0467 mol. Thus M = 2.8/0.0467 ≈ 60 g mol⁻¹, so option B is correct.

Correct answer: B. i = 1.6; dissociation

Explanation: For abnormal molar masses, the van’t Hoff factor is related by i = true molar mass / observed molar mass. Therefore, i = 1/0.625 = 1.6. Since i is greater than 1, the number of solute particles has increased. This happens when the solute dissociates into two or more particles in solution, so the behaviour is dissociation.

Correct answer: A. i = 0.8; association

Explanation: The van’t Hoff factor is calculated from i = true molar mass / observed molar mass. Here the observed value is 1.25 times the true value, so i = 1/1.25 = 0.80. A factor below 1 indicates fewer solute particles than expected because individual molecules combine. Thus, the solute shows association, such as dimerization.

Correct answer: B. 80 g mol⁻¹

Explanation: For boiling-point elevation, ΔTb = iKb m, where m is the actual molality of the solute. Thus, the effective molality is 0.104/0.52 = 0.20 mol kg⁻¹. Since i = 2, the actual molality is 0.20/2 = 0.10 mol kg⁻¹. The solvent mass is 150 g = 0.150 kg, so moles of solute = 0.10 × 0.150 = 0.015 mol. Therefore, molar mass = 1.2/0.015 = 80 g mol⁻¹.

Correct answer: B. 45 g mol⁻¹

Explanation: For freezing-point depression, ΔTf = iKf m. Hence the true molality is m = 0.279/(0.75 × 1.86) = 0.20 mol kg⁻¹. Since 100 g water equals 0.100 kg, the amount of solute is 0.20 × 0.100 = 0.020 mol. Its true molar mass is therefore 0.9/0.020 = 45 g mol⁻¹.

Correct answer: C. 50%

Explanation: For one formula unit of A₂B₃, complete dissociation produces 2 + 3 = 5 ions or particles. For a compound producing ν particles, the relation is i = 1 + α(ν − 1), where α is the degree of dissociation. Substituting i = 3 and ν = 5 gives 3 = 1 + 4α. Hence α = 2/4 = 0.5, or 50%. Therefore, option C is correct.

Correct answer: C. 111 g mol⁻¹

Explanation: CaCl₂ dissociates as CaCl₂ → Ca²⁺ + 2Cl⁻, producing three particles when completely dissociated. Therefore, i = 1 + α(3 − 1) = 1 + 2α. With α = 0.75, i = 1 + 1.5 = 2.5. The observed molar mass is related to the true molar mass by Mobserved = Mtrue/i. Hence Mtrue = i × Mobserved = 2.5 × 44.4 = 111 g mol⁻¹.

Correct answer: A. 24 g mol⁻¹

Explanation: For a dilute solution, relative lowering of vapour pressure is approximately the mole fraction of solute: (p⁰ − p)/p⁰ = 4/100 = 0.04. Water contributes 36/18 = 2 mol. Using x₂ = n₂/(2 + n₂), 0.04 = n₂/(2 + n₂), giving n₂ ≈ 0.0833 mol. Thus the molar mass is 2/0.0833 ≈ 24 g mol⁻¹.

Correct answer: B. 96 g mol⁻¹

Explanation: An AB₂ electrolyte produces three particles on complete dissociation: AB₂ → A²⁺ + 2B⁻. Thus, for degree of dissociation α, i = 1 + α(3 − 1) = 1 + 2α. With α = 0.25, i = 1 + 0.50 = 1.50. Since dissociation makes the observed molar mass smaller, Mobserved = Mtrue/i. Therefore, Mtrue = 1.50 × 64 = 96 g mol⁻¹.

Correct answer: C. 222.2 g mol⁻¹

Explanation: For dimerization, two original solute molecules combine to form one particle. Therefore, when the fraction dimerized is α, i = 1 − α/2. With α = 0.20, i = 1 − 0.10 = 0.90. Because i = true molar mass/observed molar mass, the observed molar mass is 200/0.90 ≈ 222.2 g mol⁻¹.

Correct answer: B. 150 g mol⁻¹

Explanation: Osmotic pressure is given by π = iCRT, so the molar concentration is C = π/(iRT). Using R = 0.082 L atm mol⁻¹ K⁻¹, C = 0.615/(0.75 × 0.082 × 300) ≈ 0.0333 mol L⁻¹. The solution volume is 600 mL = 0.600 L, so moles of solute = 0.0333 × 0.600 ≈ 0.0200 mol. Therefore, true molar mass = 3.0/0.0200 = 150 g mol⁻¹.

Correct answer: B. 62.5 g mol⁻¹

Explanation: For freezing-point depression, ΔTf = iKf m. Thus the actual molality is m = 0.465 ÷ (1.25 × 1.86) = 0.20 mol kg⁻¹. The solvent mass is 200 g = 0.200 kg, so moles of solute = 0.20 × 0.200 = 0.040 mol. Therefore, molar mass = 2.5 g ÷ 0.040 mol = 62.5 g mol⁻¹. Hence option B is correct.

Correct answer: B. 60 g mol⁻¹

Explanation: The elevation in boiling point is given by ΔTb = iKb m. Hence the true molality is m = 0.117 ÷ (1.5 × 0.52) = 0.150 mol kg⁻¹. The solvent mass is 400 g = 0.400 kg, so the amount of solute is 0.150 × 0.400 = 0.060 mol. Therefore, molar mass = 3.6 ÷ 0.060 = 60 g mol⁻¹, making option B correct.

Correct answer: A. 100 g mol⁻¹

Explanation: Using ΔTf = iKf m, the true molality is m = 0.186 ÷ (0.5 × 1.86) = 0.20 mol kg⁻¹. The solvent mass is 100 g = 0.100 kg, so the moles of solute are 0.20 × 0.100 = 0.020 mol. Hence the molar mass is 2.0 ÷ 0.020 = 100 g mol⁻¹. Therefore, option A is correct.

Correct answer: C. 360 g mol⁻¹

Explanation: For osmotic pressure, π = iCRT. Thus the molar concentration is C = 0.738 ÷ (1.2 × 0.082 × 300) ≈ 0.025 mol L⁻¹. The solution volume is 0.500 L, so moles of solute are 0.025 × 0.500 = 0.0125 mol. Therefore, molar mass = 4.5 ÷ 0.0125 = 360 g mol⁻¹. Option C is correct.

Correct answer: C. 60 g mol⁻¹

Explanation: Using ΔTf = iKf m, the true molality is m = 0.558 ÷ (1.2 × 1.86) = 0.25 mol kg⁻¹. The solvent mass is 200 g = 0.200 kg, so moles of solute are 0.25 × 0.200 = 0.050 mol. Therefore, the molar mass is 3.0 g ÷ 0.050 mol = 60 g mol⁻¹. Hence option C is correct.