7: Abnormal Molecular Mass

The van’t Hoff factor measures the ratio of effective particles to the normal particle number. Since i = 1.75 is greater than one, the effective particle number is 1.75 times the normal number. This commonly suggests dissociation, but i alone does not identify the exact ions or extent. Therefore A is definitely correct.

Starting from M_app = M/i, rearrange to M = iM_app. Substituting the values gives M = 1.5 × 72 = 108 g mol⁻¹. Since i is greater than one, dissociation has increased the particle count and lowered the apparent mass, so the true mass must be greater than 72 g mol⁻¹. This rules out the first two options.

A van’t Hoff factor of four means that each original solute unit produces four effective particles. Complete dissociation into four ions gives i = 4 directly. Association decreases the particle count and therefore gives i below one: complete dimerisation gives 1/2 and complete tetramerisation gives 1/4. An unchanged solute gives i = 1.

The van’t Hoff factor compares the actual number of solute particles with the number expected without association or dissociation. For AB dissociation, one formula unit produces two ions, so i = 1 + α. With α = 0.90, i = 1 + 0.90 = 1.90. AB₂ would give 1 + 2α = 2.80, while association gives i below 1; therefore option A is correct.

AB₄ is assumed to dissociate into one A ion and four B ions, giving n = 5 particles from each completely dissociated formula unit. The van’t Hoff relation is i = 1 + α(n − 1). With α = 0.25, i = 1 + 0.25(5 − 1) = 1 + 1 = 2.00. Therefore option C is correct; the value is between 1 and the complete-dissociation limit of 5.

When one solute unit dissociates into two ions, the particle increase is one additional particle per dissociated unit. Consequently, i = 1 + α. Substituting i = 5/4 gives α = 5/4 − 1 = 1/4. Since 1/4 × 100 = 25, the degree of dissociation is 25%. Therefore option B is correct; the other percentages would produce different i values.

AlCl₃ dissociates as AlCl₃ ⇌ Al³⁺ + 3Cl⁻, so one formula unit can produce four particles. For partial dissociation, i = 1 + α(n − 1). With α = 0.80 and n = 4, i = 1 + 0.80(4 − 1) = 1 + 2.40 = 3.40. Therefore option C is correct. The maximum value, 4, would correspond to complete dissociation.

The van't Hoff factor compares the actual effective particle number with the number expected from undissociated solute. A value just above 1 means that the particle count has increased slightly. Dissociation produces extra particles, so slight dissociation explains i = 1.05. Association would make i less than 1, while complete dimerisation gives about 0.5.

The relation between apparent molar mass and true molar mass is M_app = M_true/i. Since i = 1.25 = 5/4, M_app = M_true × 4/5. Therefore M_app : M_true = 4 : 5. The ratio is below one because i is greater than one, which indicates an increase in effective particles, usually due to dissociation.

The apparent molar mass obtained from a colligative property is Mapp = Mtrue/i. Substituting i = 3.4 gives Mapp = Mtrue/3.4 = 0.294Mtrue, approximately 0.29Mtrue. Ignoring dissociation and taking i as one would therefore make the calculated molar mass much too high compared with the apparent value.

Using Mapp = Mtrue/i, the condition Mapp = Mtrue/2 gives i = 2. Thus the effective number of particles is twice the original number. The simplest matching process is complete dissociation of each solute unit into two particles. Dimerisation or trimerisation would reduce i below one and increase the observed molar mass.

Dimerisation makes two benzoic-acid molecules behave as one effective particle. Therefore the particle number and freezing-point depression are nearly half of the values expected for separate molecules. Since Mapp = Mtrue/i and i is nearly 0.5, the measured molar mass is nearly 2Mtrue. Hence it is nearly double.

A 1:1 electrolyte, such as a compound producing two ions, has n = 2 on complete dissociation. The general relation is i = 1 + (n − 1)α. With α = 80/100 = 0.80, i = 1 + (2 − 1)(0.80) = 1.80. The value is not 2.0 because dissociation is not complete, and it is not 0.8 because i counts total effective particles relative to the original amount.

AB₂ dissociates into three ions, so n = 3. For a degree of dissociation α, i = 1 + (n − 1)α. Substitution gives i = 1 + (3 − 1)(0.60) = 1 + 1.20 = 2.20. The value is below the complete-dissociation limit of 3 because only 60% of the salt units dissociate. Counting AB₂ as a two-ion electrolyte would be incorrect.

A van’t Hoff factor below one indicates association, because several original particles combine and form fewer effective particles. For complete dimerisation, two molecules form one associated particle, giving i = 1/2 = 0.5. Thus the value is the limiting value for nearly complete dimerisation. Dissociation would produce i greater than one, while an inactive or normal solute would have i close to one.

The relation is M_app = M_true/i, so i = M_true/M_app. Substituting the given values gives i = 120/180 = 2/3 = 0.666..., approximately 0.67. Because the observed molar mass is greater than the true value, the factor must be below one, consistent with association. Reversing the ratio would incorrectly give 1.50.

Freezing-point depression is proportional to the number of dissolved particles through ΔTf = iKf m. If the observed depression is larger than the normal value, i is greater than one, meaning that more particles are present than expected. Dissociation of a solute into ions causes this increase. Association would reduce the particle number and give a smaller depression, while the other choices do not explain the abnormality.

For dimerisation, an associated pair contains two original molecules but behaves as one particle. If α is the fraction dimerised, i = 1 − α/2. With α = 0.40, i = 1 − 0.40/2 = 1 − 0.20 = 0.80. The factor is less than one because association decreases the effective particle count; values above one would describe particle-producing dissociation instead.

Association combines two or more solute molecules into larger units, so the number of effective particles becomes smaller than the number calculated from the formula units added. Since osmotic pressure is proportional to particle concentration, it becomes lower than expected. The same reduction would lower boiling-point elevation and freezing-point depression. Higher colligative effects instead suggest dissociation.

For dissociation into two ions, n = 2, so i = 1 + (2 − 1)α = 1 + α. Substituting i = 1.25 gives 1.25 = 1 + α, hence α = 0.25. Expressing this fraction as a percentage gives 25%. A value of 50% would produce i = 1.50 for a two-ion electrolyte, so it is not compatible with the given factor.

The van’t Hoff factor is the ratio of the actual number of solute particles in solution to the number expected without association or dissociation. A normal non-electrolyte dissolves as individual molecules, so the particle number remains unchanged. Hence the ratio is one and i = 1. Values below one indicate association, whereas values above one generally indicate dissociation or another process that increases particle number.

The observed molar mass is lower than the true molar mass. Since Mapp = Mtrue/i, this requires i > 1. A value above one means that the effective number of particles has increased, which occurs when the solute dissociates into ions or smaller particles. Association would give i < 1 and an observed mass above the true value.

Hydrogen bonding joins two acid molecules into one dimer, reducing the number of effective particles. Consequently i becomes less than one. Because Mapp = Mtrue/i, division by a value below one makes the observed molar mass greater than the true monomeric molar mass. Thus both parts of option A are correct.

On complete dissociation, A₂B gives two A ions and one B ion, so n = 3. The van’t Hoff factor is i = 1 + (n − 1)α. With α = 0.25, i = 1 + (3 − 1)(0.25) = 1 + 0.50 = 1.50. The result is between one and the complete-dissociation value three, as expected for partial dissociation.

The relation between observed and true molar mass is M_obs = M_true/i, or equivalently i = M_true/M_obs. If the observed mass is greater than the true mass, the numerator is smaller than the denominator, so i must be less than one. Such a value means fewer effective particles are present, generally because of association. Dissociation would make i greater than one and M_obs smaller.