7: Abnormal Molecular Mass

Correct answer: A. 0.70

Explanation: For association, the van’t Hoff factor is found by counting the final groups relative to the original particles. The single fraction is 1 − 0.25 − 0.30 = 0.45. Dimerising particles contribute 0.25/2 = 0.125 groups, and tetramerising particles contribute 0.30/4 = 0.075 groups. Hence i = 0.45 + 0.125 + 0.075 = 0.70, so option A is correct. Dividing by the group sizes is essential.

Correct answer: C. 90%

Explanation: MgCl₂ produces three ions, Mg²⁺ and two Cl⁻ ions, on complete dissociation. For a substance yielding n ions, i = 1 + α(n − 1); therefore here i = 1 + 2α. Substitution gives 2.8 = 1 + 2α, so α = 1.8/2 = 0.90, or 90%. Option C is correct. A value of i = 3 would represent complete dissociation, so 2.8 indicates incomplete but extensive dissociation.

Correct answer: C. 0.168 K

Explanation: Dimer association reduces the number of solute particles, so its van’t Hoff factor is i = 1 − α/2. With α = 0.60, i = 1 − 0.60/2 = 0.70. Colligative elevation is multiplied by i, giving observed ΔTb = 0.70 × 0.24 K = 0.168 K. Therefore option C is correct. Association cannot increase the elevation here; the unchanged value would incorrectly ignore particle reduction.

Correct answer: B. 1.8

Explanation: When one formula unit gives n particles on complete dissociation, the van’t Hoff factor is i = 1 + α(n − 1). Here α = 0.20 and n = 5, so i = 1 + 0.20(5 − 1) = 1 + 0.80 = 1.80. Therefore option B is correct. The value 5 would apply only to complete dissociation, whereas the stated 20% dissociation produces only a partial increase in particle number.

Correct answer: B. 60%

Explanation: The particle ratio is i = 160/100 = 1.60. AB produces two ions, so i = 1 + α. Therefore 1.60 = 1 + α, giving α = 0.60 or 60%. Equivalently, each dissociated unit adds one extra particle, and 60 extra particles indicate 60% dissociation. Option B is correct.

Correct answer: C. 84%

Explanation: For dimerisation, two original solute particles combine to form one particle, so the van’t Hoff factor is i = 1 − α/2. Using i = 0.58 gives 0.58 = 1 − α/2; hence α/2 = 0.42 and α = 0.84. The degree of dimer association is therefore 84%, making option C correct. The 42% value is only α/2, not the final association percentage.

Correct answer: B. 50%

Explanation: First calculate the elevation without dissociation: ΔTb,ideal = Kb m = 0.50 × 0.30 = 0.150 K. The van’t Hoff factor is therefore i = 0.225/0.150 = 1.50. For dissociation into two ions, i = 1 + α, so α = 1.50 − 1 = 0.50, or 50%. Thus option B is correct; the increased elevation reflects the larger number of particles after dissociation.

Correct answer: C. 80%

Explanation: K₂SO₄ dissociates as K₂SO₄ ⇌ 2K⁺ + SO₄²⁻, so complete dissociation would produce three particles. For partial dissociation, i = 1 + α(3 − 1) = 1 + 2α. Substituting i = 2.6 gives 2α = 1.6 and α = 0.80. Converting the fraction into a percentage gives 80%, so option C is correct; 100% dissociation would instead give i = 3.

Correct answer: D. 90%

Explanation: For a colligative property, the ratio of the observed value to the value expected for no association is the van’t Hoff factor, so i = 0.55. When monomers form dimers, i = 1 − α/2 because two original particles become one particle. Thus 0.55 = 1 − α/2, giving α/2 = 0.45 and α = 0.90, or 90%. Hence option D is correct.

Correct answer: C. 75%

Explanation: BaCl₂ dissociates according to BaCl₂ ⇌ Ba²⁺ + 2Cl⁻, so complete dissociation produces three ions. For a degree of dissociation α, the van’t Hoff factor is i = 1 + α(3 − 1) = 1 + 2α. Using i = 2.5 gives 2α = 1.5 and α = 0.75, or 75%. Thus option C is correct; complete dissociation would require i = 3.

Correct answer: C. 72%

Explanation: For dimer association, two monomer particles combine into one dimer, so the van’t Hoff factor is i = 1 − α/2. Substituting i = 0.64 gives 0.64 = 1 − α/2; therefore α/2 = 0.36 and α = 0.72. The degree of association is consequently 72%, making option C correct. The value is also above the complete-dimer limit i = 0.50.

Correct answer: C. 400

Explanation: Out of 250 formula units, 20%, or 50 units, dissociate. The remaining 200 units stay undissociated and count as 200 particles. Each dissociated AX₃ unit produces four ions, so 50 units produce 200 ions. Therefore, the total effective particles are 200 + 200 = 400. The other options do not correctly count both undissociated units and ions.

Correct answer: A. i = 0.8, association

Explanation: For abnormal molar mass, the relation is M_app = M_true/i. If M_app = 1.25M_true, then 1.25 = 1/i, giving i = 0.80. A van’t Hoff factor below one means that the effective number of solute particles has decreased. This is characteristic of association, such as dimer formation, rather than dissociation.

Correct answer: C. 91.4 g mol⁻¹

Explanation: AB₃ produces four ions on complete dissociation, so for degree of dissociation α, i = 1 + (4 − 1)α. With α = 0.25, i = 1 + 3(0.25) = 1.75. The apparent molar mass is M_app = M_true/i = 160/1.75 = 91.43 g mol⁻¹, approximately 91.4 g mol⁻¹. Dissociation lowers the apparent mass.

Correct answer: B. 30%

Explanation: For dissociation into n ions, i = 1 + (n − 1)α. Here n = 3, so i = 1 + 2α. Substituting i = 1.6 gives 1.6 = 1 + 2α, hence 2α = 0.6 and α = 0.30. Converting the fraction to a percentage gives 30%. The factor of two is essential because one original particle produces three, an increase of two particles.

Correct answer: C. 60%

Explanation: For fixed concentration and temperature, the ratio of actual to normal osmotic pressure gives i. Thus i = 3.20/2.00 = 1.60. A solute producing two ions follows i = 1 + α, because complete dissociation would increase the particle count from one to two. Therefore 1.60 = 1 + α, so α = 0.60 or 60%. The increased pressure confirms dissociation.

Correct answer: A. Two-thirds

Explanation: AB produces two ions on dissociation, so n = 2. With α = 0.50, i = 1 + (2 − 1)α = 1.50. Since M_obs = M_true/i, the observed mass is M_true/1.5 = (2/3)M_true. Thus it is two-thirds of the true molar mass. Dissociation increases particle number and therefore makes the experimentally calculated molar mass smaller.

Correct answer: C. 66.7%

Explanation: For a solute that forms n particles on complete dissociation, the van’t Hoff factor is i = 1 + α(n − 1), where α is the degree of dissociation. AlCl₃ forms four ions, so n = 4. Substituting i = 3 gives 3 = 1 + 3α, hence α = 2/3 = 0.667. Therefore, the degree of dissociation is 66.7%. The other percentages do not satisfy this equation.

Correct answer: A. Trimerisation

Explanation: Because M_obs = M_true/i, an observed mass equal to three times the true mass gives i = 1/3. For complete association of m molecules into one particle, i = 1/m. Therefore i = 1/3 corresponds to three molecules combining to form one particle, called trimerisation. Dimerisation would give i = 1/2, while dissociation would produce i greater than one and a smaller observed mass.

Correct answer: B. 50%

Explanation: AB₃ gives four ions on complete dissociation, so n = 4. Using i = 1 + (n − 1)α gives 2.5 = 1 + 3α. Therefore 3α = 1.5 and α = 0.50. Expressed as a percentage, the degree of dissociation is 50%. Complete dissociation would give i = 4, while 25% would give only i = 1.75.

Correct answer: B. 25%

Explanation: First use M_obs = M_true/i. Since M_obs = 0.8M_true, i = 1/0.8 = 1.25. A 1:1 electrolyte gives two ions, so i = 1 + α. Hence α = 1.25 − 1 = 0.25, or 25%. The observed mass is lower than the true mass because dissociation increases the number of particles and strengthens the colligative effect.

Correct answer: A. 4/3 times

Explanation: For 50% dimerisation, α = 0.50 and the factor is i = 1 − α/2 = 1 − 0.25 = 0.75. The apparent molar mass is related by M_obs = M_true/i. Therefore M_obs/M_true = 1/0.75 = 4/3. Association makes the observed molar mass larger than the true value, so 3/4 is the factor for i, not for the observed mass.

Correct answer: B. 50%

Explanation: The relation between observed molar mass and the van’t Hoff factor is M observed = M true/i. Since the observed molar mass is half the true value, i = 2. On complete dissociation, AB₂ gives three ions, so n = 3. Using i = 1 + α(n − 1), we get 2 = 1 + 2α, giving α = 0.5. Thus, the degree of dissociation is 50%; 100% dissociation would give i = 3.

Correct answer: A. It is nearly 80% dissociated

Explanation: Glucose is a non-electrolyte, so its van't Hoff factor is approximately one. At equal molality, the ratio of freezing-point depressions therefore gives iNaCl ≈ 1.8. NaCl produces two ions, for which i = 1 + α. Hence α = 1.8 − 1 = 0.8, or about 80% dissociation. Association would make the ratio below one.

Correct answer: B. 2.8

Explanation: One formula unit of FeCl₃ gives four ions on complete dissociation: one Fe³⁺ ion and three Cl⁻ ions. For a dissociation fraction α into n particles, i = 1 + α(n − 1). Therefore, i = 1 + 0.60(4 − 1) = 1 + 1.80 = 2.80. The result is less than 4, the value for complete dissociation, because only 60% of FeCl₃ is dissociated.