7: Abnormal Molecular Mass

First calculate the van’t Hoff factor: i = Mnormal/Mobserved = 180/72 = 2.5. For dissociation AB₂ → A + 2B, one formula unit produces three particles, so i = 1 + 2α, where α is the degree of dissociation. Thus 2.5 = 1 + 2α, giving α = 0.75 or 75%. Therefore, option B is correct.

FeCl₃ produces four particles on complete dissociation: Fe³⁺ and three Cl⁻ ions. Therefore, i = 1 + α(4 − 1) = 1 + 3α. With α = 0.60, i = 1 + 1.80 = 2.80. Since i = M_true/M_observed, M_true = 2.80 × 81.25 = 227.5 g mol⁻¹. Thus, the corrected answer is option D.

A₂B dissociates into three particles, so for a fractional dissociation α, the van’t Hoff factor is i = 1 + α(3 − 1) = 1 + 2α. With α = 0.40, i = 1 + 0.80 = 1.80. Since i = M_true/M_observed, the true molar mass is 1.80 × 72 = 129.6 g mol⁻¹. Therefore, option C is correct.

For dimerization, two solute molecules combine to form one particle. If α is the fraction dimerized, the van’t Hoff factor is i = 1 − α/2. With α = 0.30, i = 1 − 0.15 = 0.85. The relation i = M_true/M_observed gives M_observed = M_true/i = 180/0.85 = 211.76 g mol⁻¹, approximately 211.8 g mol⁻¹. Hence, option C is correct.

For the same concentration and solvent, i equals the observed colligative effect divided by the normal effect: i = 0.30/0.50 = 0.60. For dimer association, i = 1 − α/2. Therefore 0.60 = 1 − α/2, giving α/2 = 0.40 and α = 0.80, or 80%. The reduced depression is due to fewer particles after association.

AB₃ produces four particles on complete dissociation, so n = 4. With α = 0.40, i = 1 + 0.40(4 − 1) = 2.20. The relation between apparent and true molar mass is M_app = M_true/i, because the increased particle number makes the measured colligative effect look as though the solute has a lower molar mass. Hence M_app/M_true = 1/2.2, not 2.2.

For dilute solutions, freezing-point depression is ΔT_f = iK_fm. Since the solvent, K_f, and molality are the same, the ratio of depressions equals the ratio of van’t Hoff factors. Complete dimerisation gives i = 1/2, because two particles become one. Complete dissociation into two ions gives i = 2. Therefore, ΔT_f(first):ΔT_f(second) = (1/2):2 = 1:4. Option C is correct.

For trimer formation, i = 1 − α + α/3 = 1 − 2α/3. Substituting i = 0.40 gives 0.40 = 1 − 2α/3, so 2α/3 = 0.60 and α = 0.60 × 3/2 = 0.90. Thus, approximately 90% of the original solute molecules are associated. A value of 40% confuses the measured i with α, while 60% does not satisfy the equation.

Osmotic pressure follows π = iCRT. First convert the temperature to kelvin: T = 27 + 273 = 300 K. The ordinary contribution is CRT = 0.01 × 0.082 × 300 = 0.246 atm. Therefore, i = π/(CRT) = 0.369/0.246 = 1.50. The essential step is using kelvin rather than 27 directly; otherwise the calculated factor would be physically incorrect.

The apparent molar mass is related to the true molar mass by M_app = M_true/i. Since M_app = 0.40M_true, we get 0.40 = 1/i, so i = 2.5. For dissociation into three ions, i = 1 + α(3 − 1) = 1 + 2α. Thus, 2.5 = 1 + 2α, giving α = 0.75 or 75%. The key is to calculate i before finding α.

First calculate i = M_true/M_app = 60/96 = 0.625. For dimerisation, i = 1 − α/2. Thus 0.625 = 1 − α/2, giving α/2 = 0.375 and α = 0.75. Therefore approximately 75% of the acid units are associated as dimers. The larger apparent mass is evidence of association rather than dissociation.

Count effective particles rather than original particles. Assume 100 original solute particles: 40 make dimers, producing 20 groups; 30 make trimers, producing 10 groups; and the remaining 30 stay single. Thus the final number of effective particles is 20 + 10 + 30 = 60. Therefore i = 60/100 = 0.60, so option B is correct. Directly adding the percentages would wrongly ignore grouping.

First calculate i from M_app = M/i: i = 122/200 = 0.61. For dimer association, i = 1 − α/2. Therefore 0.61 = 1 − α/2, so α/2 = 0.39 and α = 0.78, or about 78%. The larger apparent mass is evidence of association, which lowers the effective particle count.

Use a convenient basis of 100 original particles. The 20 particles forming dimers become 20/2 = 10 groups, while the 40 particles forming tetramers become 40/4 = 10 groups. The remaining 40 particles stay single and contribute 40 groups. Hence the total effective groups are 10 + 10 + 40 = 60, so i = 60/100 = 0.60. Option B is correct; association lowers i below one.

Out of 100 original particles, 40 remain single. The dimer-forming 10 particles give 5 groups, the trimer-forming 30 give 10 groups, and the tetramer-forming 20 give 5 groups. Total groups = 40 + 5 + 10 + 5 = 60, so i = 0.60. Each aggregate type must be divided by its size.

For association in groups of m molecules, the van’t Hoff factor is i = 1 − α + α/m. Here the degree of association is α = 0.60 and trimerisation means m = 3. Therefore, i = 1 − 0.60 + 0.60/3 = 0.40 + 0.20 = 0.60. Association lowers the effective particle count, so i is less than one.

One formula unit of AB₄ produces five ions on complete dissociation: one A ion and four B ions. For partial dissociation into n particles, i = 1 + α(n − 1). Here α = 0.20 and n = 5, so i = 1 + 0.20(5 − 1) = 1 + 0.80 = 1.80. The value is not 5 because complete dissociation has not occurred.

The ratio of observed osmotic pressure to the ideal expected osmotic pressure is the van’t Hoff factor, so i = 0.75. The apparent or observed molar mass is related to the true molar mass by Mobserved = Mtrue/i. Therefore, Mobserved = Mtrue/0.75 = 4Mtrue/3. A reduced particle count makes the measured molar mass appear larger.

Benzene is a non-polar solvent, so carboxylic acids are not extensively ionised in it. Instead, two acid molecules can associate through two hydrogen bonds to form a cyclic dimer. Two original particles then behave as one effective particle, giving i less than one. Consequently, colligative measurements calculate a molar mass higher than the true molecular mass.

For a solute that produces three ions on complete dissociation, the partial-dissociation equation is i = 1 + α(n − 1) = 1 + 2α. Substituting i = 2.8 gives 2.8 = 1 + 2α, so 2α = 1.8 and α = 0.90. Converting the fraction to a percentage gives 90%, so option C is correct.

During dimerisation, two original solute particles combine to form one particle. If α is the fraction dimerised, the van’t Hoff factor is i = 1 − α/2. Substituting i = 0.6 gives 0.6 = 1 − α/2, so α/2 = 0.4 and α = 0.8. Therefore, the degree of dimerisation is 80%, making option C correct. The value of i below unity is consistent with association because the total number of particles decreases.

For m-fold association, the van’t Hoff factor is i = 1 − α + α/m, where α is the fraction associated and m is the number of molecules forming one aggregate. For tetramerisation, α = 0.30 and m = 4. Thus i = 1 − 0.30 + 0.30/4 = 0.70 + 0.075 = 0.775. Association lowers the number of particles, so a value below one is chemically reasonable. Therefore, option A is correct.

The van’t Hoff factor is obtained from the molar-mass relation i = Mtrue/Mobserved. Therefore, i = 120/180 = 2/3. For dimerisation, two molecules form one particle and i = 1 − α/2. Substituting gives 2/3 = 1 − α/2, so α/2 = 1/3 and α = 2/3. Converting to a percentage gives approximately 66.7%. Hence option C is correct.

Because i = Mtrue/Mobserved and Mobserved = (5/8)Mtrue, the van’t Hoff factor is i = 8/5 = 1.6. A 1:2 electrolyte forms three ions on complete dissociation, so i = 1 + α(3 − 1) = 1 + 2α. Hence 1.6 = 1 + 2α, giving 2α = 0.6 and α = 0.30. Therefore, the degree of dissociation is 30%, or option B.

In complete m-fold association, every group of m original molecules produces one associated particle. Consequently, the van’t Hoff factor is i = 1/m. Using i = 0.4 gives 0.4 = 1/m, so m = 1/0.4 = 2.5. Mathematically, the stated idealised model therefore gives 2.5, which is option B. Since a real single association process usually has an integral m, a non-integral result could indicate that the simplified model does not represent a real single aggregate perfectly.