7: Abnormal Molecular Mass

Correct answer: C. 300 g mol⁻¹

Explanation: For association into trimers, three monomer particles become one particle. If α is the fraction associated, the van’t Hoff factor is i = 1 − 2α/3. For α = 0.45, i = 1 − (2 × 0.45)/3 = 1 − 0.30 = 0.70. The observed molar mass is Mobserved = Mtrue/i = 210/0.70 = 300 g mol⁻¹. Therefore, the approximate observed molar mass is given by option C.

Correct answer: C. 70%

Explanation: Because Mobserved = Mtrue/i, the statement Mobserved = Mtrue/2.4 gives i = 2.4. On dissociation, A₂B produces 2A and B, that is, three particles from one formula unit. Therefore, i = 1 + (3 − 1)α = 1 + 2α. Substituting i = 2.4 gives 2.4 = 1 + 2α, so α = 0.70, or 70%. Thus, option C is correct.

Correct answer: C. i = 2.5; dissociation

Explanation: For abnormal molar mass, the van’t Hoff factor is related to true and observed molar masses by i = M_true/M_observed. Since M_observed = 0.4 M_true, i = 1/0.4 = 2.5. A value of i greater than 1 means that the number of solute particles has increased, which indicates dissociation. Therefore, option C is correct.

Correct answer: A. i = 0.625; association

Explanation: The van’t Hoff factor for abnormal molar mass is i = M_true/M_observed. Here M_observed = 1.6 M_true, so i = 1/1.6 = 0.625. Because i is less than 1, the solute produces fewer independent particles than expected. This reduction is generally caused by association, such as dimer formation. Hence, option A is correct.

Correct answer: A. The effective number of solute particles differs under the two experimental conditions

Explanation: The true molar mass of a substance does not change with the measuring method, but its apparent value can change when the number of dissolved particles is abnormal. Association decreases the particle number and gives i < 1, whereas dissociation increases it and gives i > 1. Different solvents, concentrations, or temperatures can produce different association or dissociation, so the two colligative measurements may yield different apparent molar masses. Thus, option A is the scientifically valid explanation.

Correct answer: A. Nearly complete but not complete

Explanation: For a solute that produces two ions on complete dissociation, the maximum van’t Hoff factor is i = 2, assuming ideal behaviour. The observed value i = 1.8 is greater than 1, so dissociation has occurred, but it is less than 2, so dissociation is not complete. The result therefore indicates high or nearly complete, yet incomplete, dissociation rather than association.

Correct answer: A. i = 1 − α/2

Explanation: In dimerization, two original solute molecules combine to form one associated particle: 2A → A₂. If a fraction α of the molecules participates, the loss in particle number is α/2 relative to the original count. Therefore the van’t Hoff factor is i = 1 − α/2. The minus sign reflects fewer particles; a plus sign would describe dissociation instead.

Correct answer: D. Complete dimer association

Explanation: For dissociation into two ions, i = 1 + α, so complete dissociation gives i = 2. For three ions at α = 0.50, i = 1 + 2(0.50) = 2. For four ions at α = 1/3, i = 1 + 3(1/3) = 2. Complete dimer association instead reduces particles and gives i = 1/2, so it cannot produce i = 2.

Correct answer: A. 1.36M

Explanation: For trimerisation, 3α original particles become α particles of trimer groups, while 1 − α remain single. Thus i = 1 − α + α/3. With α = 0.40, i = 1 − 0.40 + 0.1333 = 0.7333. Since M_app = M/i, M_app = M/0.7333 ≈ 1.36M. Association lowers i and therefore raises apparent molar mass.

Correct answer: B. 61.67

Explanation: Thirty particles remain single and contribute 30 particles. The 50 particles assigned to dimers form 50/2 = 25 groups. The 20 particles assigned to trimers form 20/3 = 6.667 groups. Therefore, the total effective particle count is 30 + 25 + 6.667 = 61.667, approximately 61.67. The calculation counts groups, not original particles.

Correct answer: B. 0.38 K

Explanation: In tetramerisation, four original particles combine to form one particle. If α is the fraction associated, the van’t Hoff factor is i = 1 − α + α/4. For α = 0.70, i = 1 − 0.70 + 0.70/4 = 0.475. Since the observed colligative effect equals i times the normal effect, ΔT_f = 0.475 × 0.80 = 0.38 K.

Correct answer: B. 0.70

Explanation: In trimerisation, three original molecules form one associated molecule. If α is the fraction of molecules undergoing association, the van’t Hoff factor is i = 1 − α + α/3, or i = 1 − 2α/3. For α = 0.45, i = 1 − 0.45 + 0.45/3 = 0.55 + 0.15 = 0.70. Thus option B is correct. The value is below one because association reduces the number of particles.