6: Molar Mass Determination

Correct answer: A. 18 g mol⁻¹

Explanation: For a dilute solution containing a non-volatile solute, relative lowering of vapour pressure equals the mole fraction of the solute: x₂ = n₂/(n₁+n₂). Water moles are 45/18 = 2.5 mol. Thus 0.10 = n₂/(2.5+n₂), giving n₂ = 0.278 mol approximately. Therefore the solute molar mass is 5/0.278 ≈ 18 g mol⁻¹, so option A is correct.

Correct answer: B. 46.2 g mol⁻¹

Explanation: For AB₂ dissociation, one formula unit produces three particles, so the van’t Hoff factor is i = 1 + 2α. With α = 0.80, i = 1 + 2(0.80) = 2.6. The observed molar mass is lower because dissociation increases the number of particles: M observed = M true/i = 120/2.6 ≈ 46.2 g mol⁻¹. Hence B is correct.

Correct answer: B. 180 g mol⁻¹

Explanation: For association into trimers, three original particles become one particle. If α is the fraction associated, the van’t Hoff factor is i = 1 − 2α/3. With α = 0.25, i = 1 − (2 × 0.25)/3 = 0.8333 approximately. Therefore M observed = M true/i = 150/0.8333 ≈ 180 g mol⁻¹. Association lowers i and raises the observed molar mass, so B is correct.

Correct answer: A. 225 g mol⁻¹

Explanation: For 50% dimerisation, the van’t Hoff factor is i = 1 − α/2 = 1 − 0.50/2 = 0.75, because two solute molecules combine to form one particle. Using π = iCRT, C = 0.164/(0.75 × 0.082 × 300) = 0.00889 mol L⁻¹. In 1 L, the amount is 0.00889 mol. Therefore, the true molar mass is 2.0/0.00889 ≈ 225 g mol⁻¹, so option A is correct.

Correct answer: B. 100 g mol⁻¹

Explanation: For AB₂, dissociation produces three ions, so at degree of dissociation α = 0.50, i = 1 + 2α = 2. The freezing-point relation is ΔTf = iKf m; hence m = 0.744/(2 × 1.86) = 0.200 mol kg⁻¹. Since the solvent mass is 0.200 kg, moles of solute = 0.200 × 0.200 = 0.0400 mol. Therefore, molar mass = 4.0/0.0400 = 100 g mol⁻¹, so B is correct.

Correct answer: C. 300 g mol⁻¹

Explanation: Use the osmotic-pressure equation π = iCRT. The molar concentration is C = 0.738/(0.75 × 0.082 × 300) = 0.0400 mol L⁻¹. The volume is 250 mL = 0.250 L, so the amount of solute is n = 0.0400 × 0.250 = 0.0100 mol. Thus, the true molar mass is M = 3.0/0.0100 = 300 g mol⁻¹. Therefore, option C is correct.

Correct answer: B. 50%

Explanation: The van’t Hoff factor is the ratio of true to observed molar mass: i = Mtrue/Mobserved = 95/47.5 = 2. For MgCl₂, dissociation into Mg²⁺ and 2Cl⁻ gives three particles at complete dissociation, so i = 1 + 2α. Substituting i = 2 gives 2 = 1 + 2α, hence α = 0.50 or 50%. Therefore, option B is correct.

Correct answer: C. 40%

Explanation: For abnormal molar mass, i = Mtrue/Mobserved = 120/150 = 0.80. When monomer molecules associate to form dimers, the number of particles decreases and i = 1 − α/2, where α is the fraction associated. Thus, 0.80 = 1 − α/2, so α/2 = 0.20 and α = 0.40. The degree of association is therefore 40%, making option C correct.

Correct answer: C. 137.5 g mol⁻¹

Explanation: For AB₃, dissociation produces four particles from one formula unit, so i = 1 + 3α. With α = 0.40, i = 1 + 3(0.40) = 2.20. The relation between true and observed molar masses is Mtrue = i × Mobserved. Therefore, Mtrue = 2.20 × 62.5 = 137.5 g mol⁻¹. Thus, option C is correct.

Correct answer: C. 100 g mol⁻¹

Explanation: From π = iCRT, the van’t Hoff factor is i = 1.476/(0.04 × 0.082 × 300) = 1.50. Since i is greater than 1, dissociation has increased the number of particles, causing the observed molar mass to be smaller than the true value. The relation is Mobserved = Mtrue/i = 150/1.50 = 100 g mol⁻¹. Therefore, option C is correct.

Correct answer: C. 60 g mol⁻¹

Explanation: Use the colligative-property equation ΔTb = iKb m. Thus, the molality is m = 0.156/(1.5 × 0.52) = 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 = 2.4/0.040 = 60 g mol⁻¹, so option C is correct.

Correct answer: B. 60 g mol⁻¹

Explanation: For freezing-point depression, ΔTf = iKf m. Taking Kf for water as 1.86 K kg mol⁻¹, the effective molality is 0.279/1.86 = 0.15 mol kg⁻¹. Since i = 0.75, the actual molality is 0.15/0.75 = 0.20 mol kg⁻¹. In 0.150 kg water, moles are 0.030; hence M = 1.8/0.030 = 60 g mol⁻¹. Option B is correct.

Correct answer: B. 19 g mol⁻¹

Explanation: The relative lowering of vapour pressure is (p⁰ − p)/p⁰ = (120 − 114)/120 = 0.05. For a dilute solution, this equals the solute mole fraction: x₂ = n₂/(n₁ + n₂). Water moles are 54/18 = 3. Hence 0.05 = n₂/(3 + n₂), giving n₂ = 0.1579 mol. The molar mass is 3/0.1579 ≈ 19 g mol⁻¹, so option B is correct.

Correct answer: B. 240 g mol⁻¹

Explanation: For osmotic pressure, π = iCRT, where C is the molar concentration. Therefore, C = π/(iRT) = 0.492/(0.8 × 0.082 × 300) = 0.025 mol L⁻¹. The solution volume is 0.400 L, so moles of solute = 0.025 × 0.400 = 0.010 mol. Thus, molar mass = 2.4/0.010 = 240 g mol⁻¹. Option B is correct.

Correct answer: B. 72 g mol⁻¹

Explanation: For water, Kf = 1.86 K kg mol⁻¹. The observed depression gives effective molality = ΔTf/Kf = 0.465/1.86 = 0.25 mol kg⁻¹. Since ΔTf = iKf m, the actual molality is 0.25/1.5 = 0.1667 mol kg⁻¹. With 0.250 kg solvent, moles of solute = 0.1667 × 0.250 = 0.04167 mol. Therefore, molar mass = 3/0.04167 ≈ 72 g mol⁻¹, so option B is correct.

Correct answer: A. 100 g mol⁻¹

Explanation: Use the elevation formula ΔTb = iKb m. Thus, the actual molality is m = 0.078/(1.5 × 0.52) = 0.100 mol kg⁻¹. The solvent mass is 500 g = 0.500 kg, so the amount of solute is 0.100 × 0.500 = 0.050 mol. Hence, molar mass = mass/moles = 5/0.050 = 100 g mol⁻¹. Therefore, option A is the only correct answer.

Correct answer: B. 25 g mol⁻¹

Explanation: Effective molality is i times the true molality, so mtrue = meffective/i = 0.10/0.5 = 0.20 mol kg⁻¹. For 0.4 kg of solvent, moles of solute = 0.20 × 0.4 = 0.08 mol. The molar mass is therefore mass divided by amount: M = 2 g/0.08 mol = 25 g mol⁻¹. Thus, option B is correct.

Correct answer: B. i = 1.5, dissociation

Explanation: For abnormal molar mass, the relation is Mobserved = Mtrue/i, or i = Mtrue/Mobserved. Therefore, i = 240/160 = 1.5. A value of i greater than one means that the number of solute particles has increased, which generally occurs through dissociation into ions or smaller particles. Hence, option B is correct.

Correct answer: A. 112.5 g mol⁻¹

Explanation: For water, Kf = 1.86 K kg mol⁻¹. The effective molality is 0.186/1.86 = 0.100 mol kg⁻¹. Since i = 0.75, the true molality is 0.100/0.75 = 0.1333 mol kg⁻¹. In 0.100 kg water, moles of solute = 0.1333 × 0.100 = 0.01333 mol. Hence M = 1.5/0.01333 ≈ 112.5 g mol⁻¹, so option A is correct.

Correct answer: D. 360 g mol⁻¹

Explanation: For osmotic pressure, π = iCRT, so C = π/(iRT) = 0.492/(1.2 × 0.082 × 300) = 0.01667 mol L⁻¹. The solution volume is 300 mL = 0.300 L, giving moles = 0.01667 × 0.300 = 0.00500 mol. Therefore, the true molar mass is 1.8/0.00500 = 360 g mol⁻¹. Thus, option D is correct.

Correct answer: A. 50 g mol⁻¹

Explanation: For water, Kf = 1.86 K kg mol⁻¹. The effective molality is ΔTf/Kf = 0.837/1.86 = 0.450 mol kg⁻¹. Correcting for i, the true molality is 0.450/1.5 = 0.300 mol kg⁻¹. Since 300 g water = 0.300 kg, moles of solute = 0.300 × 0.300 = 0.090 mol. Thus M = 4.5/0.090 = 50 g mol⁻¹, so option A is correct.

Correct answer: A. 23 g mol⁻¹

Explanation: For a dilute solution containing a non-volatile solute, the relative lowering of vapour pressure equals the mole fraction of the solute: x₂ = n₂/(n₁+n₂). Water moles are 36/18 = 2 mol. Thus 0.08 = n₂/(2+n₂), giving n₂ = 0.16/0.92 ≈ 0.174 mol. Therefore, molar mass = 4/0.174 ≈ 23 g mol⁻¹. Hence option A is correct.

Correct answer: B. 144 g mol⁻¹

Explanation: The molar mass obtained from a colligative property is affected by the van’t Hoff factor. The relation is M_observed = M_true/i, because dissociation increases the number of particles and makes the observed molar mass appear smaller. Rearranging gives M_true = i × M_observed = 2.25 × 64 = 144 g mol⁻¹. Therefore, option B is correct.

Correct answer: B. 75 g mol⁻¹

Explanation: For AB₂, one formula unit produces three ions on complete dissociation. At degree of dissociation α, the van’t Hoff factor is i = 1 + (3 − 1)α = 1 + 2α. With α = 0.70, i = 1 + 1.40 = 2.40. The observed molar mass is M_true/i = 180/2.40 = 75 g mol⁻¹. Hence option B is correct.

Correct answer: C. 300 g mol⁻¹

Explanation: For association into trimers, three original solute molecules combine to form one particle. If α is the fraction associated, the particle factor is i = 1 − α(1 − 1/3) = 1 − 2α/3. With α = 0.60, i = 1 − 0.40 = 0.60. Therefore M_observed = M_true/i = 180/0.60 = 300 g mol⁻¹, so option C is correct.