7: Abnormal Molecular Mass

For trimer association, three original particles combine to form one particle. If α is the fraction associated, the van’t Hoff factor is i = (1 − α) + α/3 = 1 − 2α/3. With i = 0.70, 0.30 = 2α/3. Multiplying by 3/2 gives α = 0.45, or 45%. Thus option B is correct. The reduction factor 2/3 represents the net loss of two particles for every three that associate.

A 40% decrease means the observed colligative effect is 60% of the normal value, so i = 0.60. Since M_app = M/i, M_app = M/0.60 = 1.666...M, approximately 1.67M. The increased apparent mass is consistent with association, which lowers the particle count and the boiling-point elevation.

The governing concept is the van’t Hoff correction for association in a colligative property. When three solute particles associate to form one trimer, the effective particle factor for an associated fraction α is i = 1 − α + α/3. For 50% association, α = 0.50, so i = 1 − 0.50 + 0.50/3 = 0.6667, approximately 2/3. The observed boiling-point elevation is therefore ΔTb(observed) = i × ΔTb(normal) = (2/3) × 0.32 K = 0.2133 K, which rounds to 0.213 K. Hence option C is correct. Option B incorrectly multiplies by 0.50 directly and ignores that each trimer still represents one particle, while option D ignores association. Option A is also inconsistent with the correct factor and calculation.

The particle factor is the ratio of effective to initial particles: i = 130/100 = 1.30. AB dissociates into two ions, so i = 1 + α. Therefore 1.30 = 1 + α, giving α = 0.30, or 30%. Equivalently, each dissociated unit adds one extra particle, so 30 additional particles mean 30% dissociation.

First calculate the elevation without dissociation: ΔTb° = Kb m = 0.50 × 0.40 = 0.20 K. The van’t Hoff factor is the ratio of observed to normal elevation, so i = 0.30/0.20 = 1.50. For dissociation into two ions, i = 1 + α; hence α = 1.50 − 1 = 0.50, or 50%. Therefore option B is correct.

The ideal maximum van’t Hoff factor is obtained when every formula unit of an electrolyte dissociates completely into the maximum possible number of ions. If the measured i is lower, some formula units have remained undissociated or ion formation is incomplete. Association could lower i in a particular system, but incomplete dissociation is the best general explanation for an electrolyte below its ideal maximum.

KCl is an ionic electrolyte. When dissolved in water, its formula units can separate into K⁺ and Cl⁻ ions, increasing the effective number of solute particles and making the van’t Hoff factor greater than one. Glucose, urea and sucrose are molecular nonelectrolytes under ordinary conditions; they dissolve mainly as intact molecules and do not show dissociation-based abnormal molar mass.

For an abnormal solution, the van’t Hoff factor relates the true and observed molar masses as i = Mtrue/Mobserved. Hence Mobserved = Mtrue/i = 100/1.25 = 80 g mol⁻¹. Since i is greater than one, the solute produces more particles than expected, as in dissociation; consequently, a colligative-property measurement gives an apparent molar mass smaller than the true molar mass. Thus option A is correct.

A 25% decrease means that the observed osmotic pressure is 75% of the ideal pressure. Therefore, the van’t Hoff factor is i = 0.75. Since osmotic-pressure calculations give Mobserved = Mtrue/i, the ratio is Mobserved/Mtrue = 1/0.75 = 4/3. Thus the observed molar mass is four-thirds of the true molar mass. The increase is associated with a decrease in the number of solute particles, as in association.

AB dissociates into two ions, so for a fraction α dissociated, i = 1 + α because the number of particles increases from one to two. With α = 0.25, i = 1.25. The observed molar mass is Mobserved = Mtrue/i, so Mobserved/Mtrue = 1/1.25 = 0.80 = 4/5. Dissociation therefore makes the observed molar mass smaller than the true value. Hence option A is correct.

One formula unit of CaCl₂ can produce three ions: one Ca²⁺ and two Cl⁻ ions. For partial dissociation into three particles, i = 1 + α(3 − 1) = 1 + 2α. Substituting i = 2.5 gives 2.5 = 1 + 2α, so 2α = 1.5 and α = 0.75. Therefore, the degree of dissociation is 75%, making option C correct. Complete dissociation would give i = 3.

During dimerisation, two solute molecules combine to form one associated particle. If α is the fraction of molecules dimerised, the van’t Hoff factor is i = 1 − α/2. Substituting i = 0.65 gives 0.65 = 1 − α/2, so α/2 = 0.35 and α = 0.70. Therefore, the degree of dimerisation is 70%. The value of i below 1 confirms association.

One formula unit of A₃B₂ produces five ions on complete dissociation: three A ions and two B ions. For 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. Thus, the correct answer is 1.8. The factor exceeds one because dissociation increases the number of solute particles.

The relation between observed and true molar mass is M_observed = M_true/i. Since i = 1.75 = 7/4, the ratio is M_observed/M_true = 1/(7/4) = 4/7. Therefore, the observed molar mass is 4/7 times the true molar mass. Because i is greater than 1, dissociation has increased the particle count and caused the apparent molar mass to decrease.

For dimerisation, two original solute particles combine into one particle. If α is the fraction dimerised, i = 1 − α/2. On substituting i = 0.875, we get 0.875 = 1 − α/2. Thus α/2 = 0.125 and α = 0.25. Expressed as a percentage, the degree of dimerisation is 25%. The value i < 1 is consistent with association.

A 25% higher observed molar mass means M_observed = 1.25M_true. The van’t Hoff factor is defined by i = M_true/M_observed. Hence i = M_true/(1.25M_true) = 1/1.25 = 0.80. Since i is less than one, the solute particles have associated, reducing the effective number of particles. The value 1.25 is only the mass ratio, not i.

The observed molar mass is calculated using M_observed = M_true/i. Substituting the given values gives M_observed = 60/0.75 = 80 g mol⁻¹. Because i is below one, association has reduced the number of effective solute particles. Consequently, the apparent molar mass calculated from colligative properties is larger than the true molar mass. Therefore, option C is correct.

On complete dissociation, one formula unit of AB₂ produces three ions: one A ion and two B ions. For dissociation fraction α, i = 1 + α(n − 1), so here i = 1 + 2α. Substituting i = 2.6 gives 2.6 = 1 + 2α, hence 2α = 1.6 and α = 0.80. Therefore, the percentage dissociation is 80%. Complete dissociation would correspond to i = 3.

Take one mole of A initially. The 0.50 mol that remains monomeric contributes 0.50 mol of particles. The other 0.50 mol undergoes dimerisation, and two molecules form one dimer, so it contributes 0.50 divided by 2, or 0.25 mol of particles. The total is 0.50 + 0.25 = 0.75 mol particles. Therefore, i = 0.75.

For a substance that produces two ions, the van’t Hoff factor is i = 1 + α, where α is the fraction dissociated. Substitution gives 1.2 = 1 + α, so α = 0.2, or 20% dissociation. This is a small or partial dissociation. Complete dissociation would give i = 2, while association would make i less than 1.

In trimerisation, three original molecules combine to form one associated particle. If α = 0.20 of the molecules undergo trimerisation, the particle factor is i = 1 − α + α/3. Therefore, i = 1 − 0.20 + 0.20/3 = 0.80 + 0.0667 = 0.8667, which rounds to 0.867. The decrease from 1 reflects association.

For dimerisation, two solute molecules combine to form one particle, so the van’t Hoff factor is i = 1 − α/2. Here α = 0.80, hence i = 1 − 0.80/2 = 0.60. The relation between true and observed molar mass is M observed = M true/i. Therefore, M observed = 100/0.60 = 166.7 g mol⁻¹ approximately. Association decreases the number of solute particles and consequently increases the apparent molar mass.

The relation between true and observed molar mass is M observed = M true/i. If the observed molar mass is twice the true value, then 2M true = M true/i, so i = 1/2 = 0.5. A van’t Hoff factor less than unity indicates that the number of effective particles has decreased. This occurs through association, such as two molecules joining to form a dimer. Dissociation would produce i greater than one.