6: Molar Mass Determination

Correct answer: B. i = 2, dissociation

Explanation: For a dilute solution, osmotic pressure is given by π = iCRT, where i is the van’t Hoff factor. Substituting the data gives i = 0.984 ÷ (0.02 × 0.082 × 300) = 0.984 ÷ 0.492 = 2. Since i is greater than 1, the solute produces more particles in solution than its original formula units, indicating dissociation. Thus, option B is correct.

Correct answer: B. 400 g mol⁻¹

Explanation: Because i = 1, the osmotic-pressure equation is π = CRT. The molar concentration is C = π/(RT) = 0.123/(0.082 × 300) = 0.005 mol L⁻¹. The solution volume is 150 mL = 0.150 L, so solute moles are 0.005 × 0.150 = 0.00075 mol. Hence M = 0.3/0.00075 = 400 g mol⁻¹, option B.

Correct answer: A. 120 g mol⁻¹, 25%

Explanation: The true molar mass is obtained from M_true = iM_observed = 1.25 × 96 = 120 g mol⁻¹. For AB dissociating into A and B, two particles form from one particle, so i = 1 + α. Therefore α = i − 1 = 0.25, or 25%. Both required values are given in option A, which is correct.

Correct answer: D. 200 g mol⁻¹

Explanation: For molar mass determined from a colligative property, M_observed = M_true/i. Substitution gives M_observed = 80/0.4 = 200 g mol⁻¹. Since i is less than one, association has reduced the number of particles in solution, weakened the colligative effect, and made the experimentally observed molar mass greater than the true molar mass. Thus option D is correct.

Correct answer: C. 200 g mol⁻¹

Explanation: Freezing-point depression is given by ΔTf = iKf m. Since i = 1, the molality is m = 0.0186/1.86 = 0.010 mol kg⁻¹. The solvent mass is 250 g = 0.250 kg, so solute moles are 0.010 × 0.250 = 0.00250 mol. Therefore molar mass = 0.5/0.00250 = 200 g mol⁻¹, making option C correct.

Correct answer: A. 60 g mol⁻¹, 80%

Explanation: The observed molar mass is related to the true molar mass by M observed = M true/i. Therefore, M observed = 156/2.6 = 60 g mol⁻¹. For AB₂ dissociating as AB₂ → A + 2B, one mole produces three particles when completely dissociated, so i = 1 + 2α. Hence 2.6 = 1 + 2α, giving α = 0.8 or 80%. Option A is correct.

Correct answer: B. 210 g mol⁻¹

Explanation: For dimerization, two solute molecules combine to form one particle, so the van’t Hoff factor is i = 1 − α/2 = 1 − 0.25/2 = 0.875. Using π = iCRT, C = 0.369/(0.875 × 0.0821 × 300) ≈ 0.01714 mol L⁻¹. Since the solution volume is 1 L, it contains about 0.01714 mol of solute units. Therefore, the true molar mass is M = 3.6/0.01714 ≈ 210 g mol⁻¹. Hence, option B is correct.

Correct answer: A. Include the van’t Hoff factor with i > 1

Explanation: For boiling-point elevation, the general relation is ΔTb = iKbm. Dissociation produces more particles than expected from the original solute units, so i becomes greater than 1 and the observed elevation is larger. If this effect is ignored, the calculation attributes the large elevation to too many solute moles and therefore gives an apparent molar mass that is too low. Including i > 1 corrects the particle-count effect and gives a value closer to the true molar mass.

Correct answer: A. Interparticle association or dissociation effects are decreasing

Explanation: An apparent molar mass differs from the true value when the solute associates into larger units or dissociates into more particles, changing the van’t Hoff factor. Dilution generally reduces interactions between solute particles, so association or other non-ideal effects become weaker. As the solution approaches ideal behaviour, the van’t Hoff factor approaches unity and the apparent molar mass approaches the true molar mass.

Correct answer: A. πV = i nRT

Explanation: For an ideal dilute solution, the osmotic-pressure equation is πV = nRT. If the solute dissociates or otherwise produces more effective particles, the van’t Hoff factor i accounts for that change. The effective number of particles becomes in, so the corrected equation is πV = i nRT. Thus i multiplies the colligative effect; it does not divide the right-hand side.

Correct answer: A. When the solute undergoes association

Explanation: Association joins two or more solute molecules into larger aggregates, reducing the number of independent particles in solution. Consequently, the observed osmotic pressure π becomes smaller than the ideal value. Since π appears in the denominator of M = wRT/(πV), a smaller π produces a calculated molar mass greater than the true molar mass. Therefore, association is the correct condition.