7: Abnormal Molecular Mass

Carboxylic acids can associate, often as hydrogen-bonded dimers, in a non-polar solvent such as benzene. Association reduces the number of independent solute particles, so the van’t Hoff factor becomes less than one. The aqueous ionic salts generally dissociate into ions and therefore tend to give i greater than one, although real values can depend on ion interactions.

FeCl₃ dissociates according to FeCl₃ → Fe³⁺ + 3Cl⁻, so complete dissociation would produce four particles. For partial dissociation, i = 1 + α(n − 1) = 1 + 3α. Substituting i = 3.1 gives 3.1 = 1 + 3α, hence 3α = 2.1 and α = 0.70, or 70%. Therefore option C is correct. The coefficient is 3, not 4, because the increase over the original one particle is four minus one.

The relation between apparent mass and true mass is M_app = M/i. Hence i = 125/50 = 2.5. AB₃ produces four ions on complete dissociation, so i = 1 + 3α. Therefore 2.5 = 1 + 3α, giving α = 0.50, or 50%. The apparent mass is smaller because dissociation increases the number of solute particles.

Apparent molar mass is related to true molar mass by Mapparent = M/i. A higher apparent mass means i < 1, so the number of effective solute particles has decreased. If both independent colligative measurements show this trend, association is the consistent explanation. Thus option A is best.

For dimerisation, if α is the fraction of original particles that associates, the particle factor is i = 1 − α/2. Thus 0.875 = 1 − α/2, giving α/2 = 0.125 and α = 0.25. Therefore, 25% of 200 particles participate: 0.25 × 200 = 50 particles. The number 50 refers to original particles, not the number of dimers formed.

A₂B dissociates as A₂B → 2A⁺ + B²⁻, giving three particles when dissociation is complete. For a substance producing n particles, i = 1 + α(n − 1). Here α = 0.65 and n = 3, so i = 1 + 0.65(3 − 1) = 1 + 1.30 = 2.30. Therefore option B is correct. The value 3.00 would require complete dissociation, whereas 1.65 incorrectly adds α only once.

The ratio of observed to normal osmotic pressure is the van’t Hoff factor, so i = 0.80. For dimer association, i = 1 − α/2. Therefore, 0.80 = 1 − α/2, giving α/2 = 0.20 and α = 0.40, or 40%. The reduced osmotic pressure indicates fewer effective particles because two original particles form one dimer.

Because M_app = M/i, the given ratio gives i = 1/0.625 = 1.60. XY₂ produces three ions, so i = 1 + 2α. Substitution gives 1.60 = 1 + 2α, hence α = 0.30 or 30%. The apparent mass is lower than the true mass because dissociation increases the number of particles.

For tetramerisation, four original solute particles combine to form one tetramer. If α is the fraction associated, the particle factor is i = (1 − α) + α/4 = 1 − 3α/4. With i = 0.70, 0.70 = 1 − 3α/4, so 3α/4 = 0.30 and α = 0.40. Thus the degree of association is 40%, making option C correct. The smaller choices result from using an incorrect particle-change factor.

Freezing-point depression follows ΔTf = iKf m. Without dissociation, Kf m = 1.5 × 0.20 = 0.30 K. The observed value is 0.45 K, so i = 0.45/0.30 = 1.50. For dissociation into two ions, i = 1 + α because each dissociated formula unit adds one particle. Hence α = 1.50 − 1 = 0.50, or 50%. Option B is correct; the larger depression reflects the increased particle count.

First calculate i from the mass relation: i = M/M_app = 90/150 = 0.60. For dimer association, i = 1 − α/2. Thus 0.60 = 1 − α/2, so α/2 = 0.40 and α = 0.80, or 80%. The increased apparent molar mass is consistent with association, which reduces the number of effective particles.

MX₃ dissociates into M³⁺ and three X⁻ ions, so complete dissociation would give four particles. For 70% dissociation, i = 1 + α(4 − 1) = 1 + 0.70 × 3 = 3.10. Since apparent molar mass is related by M_app = M/i, M_app/M = 1/i = 1/3.10. Therefore option B, 1/3.1, is correct. The value 1/4 would apply only to complete dissociation.

A 45% increase means the van’t Hoff factor is i = 1.45. AB₂ gives three ions on complete dissociation, so i = 1 + 2α. Thus 1.45 = 1 + 2α, giving α = 0.225 or 22.5%. The 45% increase is not itself the dissociation percentage because each dissociated unit adds two particles over the original one.

AB₃ dissociates into A³⁺ and three B⁻ ions, producing four particles when fully dissociated. For partial dissociation, i = 1 + α(4 − 1) = 1 + 3α. Substituting i = 1.9 gives 1.9 = 1 + 3α, so 3α = 0.9 and α = 0.30. Thus the degree of dissociation is 30%, so option B is correct. Dividing 0.9 by four would be wrong because the increase is n − 1 = 3.

For a colligative property, the ratio of the observed value to the value for an undissociated solute is the van’t Hoff factor, i. Thus i = 1.6. Dissociation of one solute particle into two ions gives i = 1 + α, where α is the degree of dissociation. Therefore α = 1.6 − 1 = 0.60, or 60%. Hence option C is correct; 40%, 50%, and 80% do not satisfy the equation.

Formation of A₃ is association into trimers. If α = 0.36 is the fraction associated, the unassociated fraction contributes 1 − α particles, while each three original particles produce one trimer, contributing α/3. Therefore i = 1 − α + α/3 = 1 − 0.36 + 0.12 = 0.76. Option B is correct. The factor is below one because association reduces the number of solute particles.

The osmotic-pressure equation for an electrolyte is π = iCRT. First calculate the value without the van’t Hoff factor: CRT = 0.05 × 0.082 × 300 = 1.23 atm. Hence i = π/(CRT) = 2.46/1.23 = 2.0. Therefore option C is correct. A value of 2 means twice the effective particles expected for an undissociated solute at the same concentration.

For a nonelectrolyte, the normal boiling-point elevation is ΔTb = Kb m = 0.80 × 0.25 = 0.20 K. The van’t Hoff factor is the observed elevation divided by this normal value: i = 0.10/0.20 = 0.50. Since i is less than one, particles have combined and association has occurred. Thus option A is correct; dissociation would give i > 1 and normal behaviour i = 1.

Of the 100 formula units, 75 remain undissociated and contribute 75 particles. Each of the 25 dissociated AB₃ units produces four ions, contributing 25 × 4 = 100 particles. Thus the total effective particle count is 75 + 100 = 175. Counting 25 as four ions is essential; counting only the dissociated units would underestimate the total.

The governing concept is the van’t Hoff factor for association. In tetramerisation, four original solute particles combine to form one associated particle. If α is the fraction of particles undergoing association, the unassociated fraction contributes 1 − α particles, while the associated fraction contributes α/4 particles per original particle. Therefore i = 1 − α + α/4 = 1 − 3α/4. Substituting the given value, 0.625 = 1 − 3α/4. Hence 3α/4 = 0.375, so α = 0.375 × 4/3 = 0.50. Converting this fraction into a percentage gives 50%. Thus option B is correct. A value of 25% would give i = 0.8125, 75% would give 0.4375, and complete association would give i = 0.25, so none of those alternatives satisfies the data.

For dimerisation, i = 1 − α/2 because α original particles become α/2 dimer groups. With α = 0.75, i = 1 − 0.375 = 0.625. Colligative properties are proportional to the effective particle concentration, so the observed value is 0.625 of normal, or 62.5%. Association cannot increase the property above normal.

For AB₂, dissociation produces three ions from one formula unit, so i = 1 + α(3 − 1) = 1 + 2α. At α = 0.75, i = 1 + 2(0.75) = 2.50, exactly the stated value. By contrast, 75% dimer association gives i = 0.625, 50% dissociation of AB gives i = 1.5, and complete trimer association gives i = 1/3. Therefore option A is correct.

Na₂SO₄ dissociates according to Na₂SO₄ → 2Na⁺ + SO₄²⁻, producing three ions. For partial dissociation, i = 1 + α(n − 1) = 1 + 2α. With i = 2.2, 2.2 = 1 + 2α, so α = 1.2/2 = 0.60, or 60%. Hence option C is correct. Complete dissociation would produce i = 3, so the given electrolyte is only partially dissociated.

The ratio of the observed colligative property to its normal value is i, so i = 0.40. For trimer association, i = 1 − α + α/3 = 1 − 2α/3. Therefore 0.40 = 1 − 2α/3, giving 2α/3 = 0.60 and α = 0.90, or 90%. Option C is correct. The value is above the complete-trimer limit of 1/3, so the association is incomplete.

Complete dissociation of AB₄ produces one A particle and four B particles, so n = 5. For a fraction α dissociated, the van’t Hoff factor is i = 1 + α(n − 1). With α = 0.30, i = 1 + 0.30(5 − 1) = 1 + 1.20 = 2.20. Therefore option C is correct. The increase is 1.20 because each dissociated unit creates four additional particles.