7: Abnormal Molecular Mass

On complete dissociation, one formula unit of A₂B produces three particles: two A⁺ ions and one B²⁻ ion. For dissociation into n particles, the van’t Hoff factor is i = 1 + α(n − 1). Here n = 3, so i = 1 + 2α. Substituting i = 2.25 gives 2.25 = 1 + 2α, hence α = 0.625. Converting this fraction into a percentage gives 62.5%, so option B is correct.

In a non-polar solvent, carboxylic acid molecules commonly associate in pairs through two hydrogen bonds, forming cyclic dimers. Association decreases the number of independent solute particles, so the van’t Hoff factor becomes less than 1. Since the apparent molar mass is related by M observed = M true divided by i, division by a value below 1 produces an observed molar mass higher than the true molar mass.

In the reaction A → B + C + D, one original particle becomes three particles after complete dissociation. For a dissociation fraction α, the van’t Hoff factor is i = 1 + α(n − 1), where n is the total number of particles formed. Here α = 25/100 = 0.25 and n = 3. Therefore, i = 1 + 0.25(3 − 1) = 1 + 0.50 = 1.50. The value 3.00 would apply only to complete dissociation.

On complete dissociation, one AlCl₃ formula unit produces four ions: one Al³⁺ and three Cl⁻ ions. For a compound producing n particles and dissociating to the extent α, i = 1 + α(n − 1). Here α = 0.30 and n = 4, so i = 1 + 0.30(4 − 1) = 1 + 0.90 = 1.90. Therefore, option C is correct. The value 4 would occur only for complete dissociation.

For equal molality and solvent, the ratio of freezing-point depressions equals the van’t Hoff factor, so i = 2.4. An AB₂ formula unit gives three ions on complete dissociation. For dissociation fraction α, i = 1 + α(3 − 1) = 1 + 2α. Solving 2.4 = 1 + 2α gives α = 0.70, or 70%. Therefore the correct choice is option B.

One A₂B₃ formula unit produces five ions on complete dissociation: two A ions and three B ions. For a dissociation fraction α, the van’t Hoff relation is i = 1 + α(n − 1), where n is the number of ions formed. Substituting α = 0.75 and n = 5 gives i = 1 + 0.75(4) = 4.0. Therefore option C is correct; 5.0 would represent complete dissociation.

For abnormal molar mass, the relation is Mobserved = Mtrue/i. Substituting i = 0.625 gives Mobserved/Mtrue = 1/0.625 = 1.6. Thus the measured molar mass is 1.6 times the true value. A factor below one indicates fewer particles because of association, so the apparent molar mass becomes larger.

In trimerisation, three original monomer particles combine to form one trimer particle. If α is the fraction trimerised, the effective particle factor is i = (1 − α) + α/3 = 1 − 2α/3. With α = 0.25, i = 1 − 2(0.25)/3 = 1 − 0.1667 = 0.8333, approximately 0.833. Thus option A is correct; association lowers i below one.

For an AB4-type salt, complete dissociation produces five ions, so the particle-number formula is i = 1 + alpha(5 − 1) = 1 + 4alpha. Substituting i = 3.0 gives 3 = 1 + 4alpha, hence alpha = 2/4 = 0.50. Therefore the degree of dissociation is 50%, so option C is correct. A 75% value would give i = 4, not 3.

During dimerisation, two solute molecules combine to form one particle, so the van’t Hoff factor is i = 1 − alpha/2. With alpha = 0.70, i = 1 − 0.70/2 = 0.65. The observed molar mass is Mobs = Mtrue/i = Mtrue/0.65 = 1.538Mtrue, approximately 1.54 times the true value. Hence A is correct; 0.65 is i itself, not the observed-mass ratio.

The governing idea is that dissociation increases the number of effective particles. One X unit gives four particles when fully dissociated, so i = 1 + alpha(n − 1). Using alpha = 0.20 and n = 4, i = 1 + 0.20(4 − 1) = 1 + 0.60 = 1.60. Therefore option C is correct. The value 1.2 would count too few additional particles, while 1.8 and 1.4 do not result from the stated equation and extent.

Al2(SO4)3 produces two Al3+ ions and three sulfate ions, five particles in total, when completely dissociated. Therefore i = 1 + alpha(n − 1) = 1 + 0.40(5 − 1) = 1 + 1.60 = 2.60. Option B is correct. The value 5.0 is the complete-dissociation limit, whereas 3.0 and 2.2 do not account correctly for the four additional particles and the 40% extent.

For dimerisation, two original particles become one, so the van’t Hoff factor is i = 1 − alpha/2. Substituting i = 0.70 gives 0.70 = 1 − alpha/2, hence alpha/2 = 0.30 and alpha = 0.60. Converting the fraction to a percentage gives 60%, so option C is correct. The common error is to report 30%, which is only half the association fraction.

Use Mobserved = Mtrue/i. Since Mobserved = (5/9)Mtrue, i = Mtrue/Mobserved = 9/5 = 1.8. AB2 gives three particles on complete dissociation, so i = 1 + 2alpha. Hence 1.8 = 1 + 2alpha, giving alpha = 0.4 or 40%. Therefore option B is correct. The other percentages result from using the mass ratio directly or miscounting the three particles.

Tetramerisation means that four original solute particles combine to form one associated particle. If alpha is the fraction undergoing association, the effective particle factor is i = (1 − alpha) + alpha/4 = 1 − 3alpha/4. Substituting alpha = 0.40 gives i = 1 − 0.30 = 0.70. Thus option A is correct; association lowers i below one, unlike dissociation.

In dimerisation, two original solute particles combine to form one A2 particle. If alpha is the associated fraction, the effective factor is i = (1 − alpha) + alpha/2 = 1 − alpha/2. For alpha = 0.90, i = 1 − 0.45 = 0.55. Thus option B is correct. Complete dimerisation would give the limiting value 0.50, so 0.55 is consistent with 90% association.

For abnormal molar mass, the relation is Mobserved = Mtrue/i. Since the observed value is half the true value, 1/i = 1/2 and therefore i = 2. An AB3 salt gives four ions on complete dissociation, so i = 1 + alpha(4 − 1) = 1 + 3alpha. Solving 2 = 1 + 3alpha gives alpha = 1/3 = 33.3%. Hence option B is correct.

For tetramerisation, four particles become one associated particle. The van’t Hoff factor is i = 1 − 3alpha/4. At alpha = 0.50, i = 1 − 0.375 = 0.625. Since Mobserved = Mtrue/i, the ratio Mobserved/Mtrue = 1/0.625 = 1.60. Therefore option A is correct. Association lowers i and consequently makes the observed molar mass appear larger than the true value.

During trimerisation, three original particles combine into one particle. If alpha is the fraction trimerised, i = (1 − alpha) + alpha/3 = 1 − 2alpha/3. Substituting i = 0.75 gives 0.75 = 1 − 2alpha/3, so 2alpha/3 = 0.25 and alpha = 0.375. Thus the degree of trimerisation is 37.5%, making option B correct.

The van’t Hoff factor for dissociation into three ions is i = 1 + α(n − 1), where α is the fraction dissociated and n = 3. Therefore, 2.2 = 1 + α(3 − 1) = 1 + 2α. Hence 2α = 1.2 and α = 0.60. Converting the fraction into a percentage gives 60% dissociation, so option B is correct. The other values do not satisfy the equation.

For dimerisation, two solute molecules combine to form one particle, so the van’t Hoff relation is i = 1 − α/2. Substituting i = 0.8 gives 0.8 = 1 − α/2, hence α/2 = 0.2 and α = 0.4. Thus 40% of the original molecules participate in dimerisation. The normal, unassociated fraction is 1 − 0.4 = 0.6, or 60%, so B is correct.

One formula unit of A3B produces four ions on complete dissociation: three A-containing ions and one B-containing ion. For a compound producing n particles, i = 1 + α(n − 1). Here α = 0.30 and n = 4, so i = 1 + 0.30(4 − 1) = 1 + 0.90 = 1.90. Therefore option B is correct. The value 4.0 would apply only to complete dissociation.

BaCl2 dissociates as BaCl2 → Ba2+ + 2Cl−, producing three particles when dissociation is complete. For partial dissociation, i = 1 + α(n − 1). With α = 0.50 and n = 3, i = 1 + 0.50(3 − 1) = 1 + 1 = 2.0. Thus option B is correct. A value of 3.0 would incorrectly assume complete dissociation.

The governing concept is the van’t Hoff factor for partial electrolyte dissociation. An AB₂ salt dissociates as AB₂ → A + 2B, so one formula unit produces three ions on complete dissociation. For a substance producing ν particles, i = 1 + α(ν − 1), where α is the fraction dissociated. Here ν = 3, so i = 1 + 2α. Using i = 2.0 gives 2.0 = 1 + 2α, hence 2α = 1 and α = 0.50. The percentage dissociation is therefore 0.50 × 100 = 50%, making option B correct. If dissociation were 25%, i would be 1.5; at 75% it would be 2.5; and complete dissociation would give i = 3. These checks also confirm that the alternatives are not compatible with the stated factor.

If one formula unit dissociates into five ions, the van’t Hoff factor is given by i = 1 + (n − 1)α, where n is the number of ions and α is the degree of dissociation. Substituting i = 3.25 and n = 5 gives 3.25 = 1 + 4α, so α = 2.25/4 = 0.5625. Converting this fraction to a percentage gives 56.25%, so option B is correct. The other values do not satisfy the equation.