7: Abnormal Molecular Mass

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.