7: Abnormal Molecular Mass

Of the 100 initial AB units, 60 remain undissociated and contribute 60 particles. The 40 dissociated units each split into two ions, contributing 40 × 2 = 80 particles. Therefore, the total effective number is 60 + 80 = 140. Equivalently, i = 140/100 = 1.40. The increase occurs because each dissociated AB unit changes from one particle into two particles.

AB₂ produces three ions on complete dissociation, so the partial-dissociation relation is i = 1 + α(3 − 1) = 1 + 2α. Using the measured value, 2.7 = 1 + 2α, so 2α = 1.7 and α = 0.85. Converting the fraction to a percentage gives 85%. The result is below complete dissociation because the maximum factor for AB₂ would be 3.

The apparent molar mass is related to the true mass by M_app = M_true/i. Therefore M_app = M/1.75 = 0.5714M, approximately 0.57M. Because i is greater than one, dissociation has increased the particle number and lowered the apparent molar mass. The fact that two ions are formed is consistent with, but not needed for, this final conversion.

Use M_app = M_true/i. Since M_app = 0.5M_true, we have 0.5 = 1/i, giving i = 2. M₂X dissociates into two M-type ions and one X-type ion, so n = 3. Hence i = 1 + α(3 − 1) = 1 + 2α. Substitution gives 2 = 1 + 2α, so α = 0.50 or 50%. Complete dissociation would instead give i = 3.

A 25% decrease means the observed colligative property is 75% of the normal value, so i = 0.75. For dimer association, i = 1 − α/2. Hence 0.75 = 1 − α/2, giving α/2 = 0.25 and α = 0.50. Thus 50% of the original solute units are associated into dimers.

A 30% increase means the observed colligative property is 1.30 times the normal value, so i = 1.30. For dissociation into two ions, i = 1 + α because each dissociated unit changes from one particle to two. Therefore α = 1.30 − 1 = 0.30, or 30%. The 130% value is i expressed as a percentage, not the dissociation fraction.

Out of 200 units, 45%, or 90 units, dissociate and 110 remain unchanged. Each dissociated AX₂ unit produces three ions, so 90 × 3 = 270 particles arise. The total effective particle count is therefore 110 + 270 = 380. The other choices result from miscounting the ions or ignoring the undissociated units.

The relation between apparent and true molar mass is M_app = M_true/i. Given M_app = 2.5M_true, we obtain 2.5M_true = M_true/i, so i = 1/2.5 = 0.40. Since i is below one, the number of solute particles has decreased, which indicates association, such as dimer formation. Hence option A is correct; dissociation would produce i greater than one.

An AB₂ formula unit produces three ions on complete dissociation, so for α = 0.40 the van’t Hoff factor is i = 1 + α(n − 1) = 1 + 0.40(3 − 1) = 1.80. Apparent molar mass is related by M_app = M_true/i. Thus M_app = 120/1.80 = 66.7 g mol⁻¹, making option B correct. Dissociation increases particle number and therefore lowers the apparent molar mass.

For association into tetramers, i = 1 − α + α/4, because α original particles become α/4 groups. With α = 0.80, i = 1 − 0.80 + 0.80/4 = 0.40. The observed colligative effect equals i times the normal effect, so ΔTf = 0.40 × 0.50 = 0.20 K. Association lowers the depression.

The governing concept is the van’t Hoff factor for abnormal colligative behaviour. At constant concentration and temperature, i = observed osmotic pressure/normal osmotic pressure = 0.90/1.20 = 0.75. For dimerisation, two solute particles become one effective particle, so i = 1 − α/2. Hence 0.75 = 1 − α/2, giving α = 0.50, or 50%. Therefore option B is correct; 25% is only the pressure reduction, while 75% and 100% are incompatible with the calculation.

For abnormal molar mass, Mapp = M/i, where M is the true molar mass. With the actual i = 2.5, Mapp = M/2.5 = 0.40M. Thus the experimentally inferred apparent mass is 40% of the true mass. Assuming i = 1 would incorrectly ignore ion formation, but the comparison requested is still 0.4 times.

For tetramerisation, the van’t Hoff factor is i = 1 − α + α/4, because the associated fraction changes from α original particles to α/4 tetramer groups. With α = 0.30, i = 1 − 0.30 + 0.075 = 0.775. The apparent molar mass is M_app = M/i = M/0.775 ≈ 1.29M. Therefore option A is correct; association lowers i and consequently raises the apparent molar mass.

The osmotic-pressure ratio directly gives the van’t Hoff factor, so i = 0.65. For association exclusively into dimers, i = 1 − α/2: each pair formed reduces two original particles to one effective particle. Therefore 0.65 = 1 − α/2, so α/2 = 0.35 and α = 0.70, or 70%. Option C is correct. The 35% choice is only half of the association fraction and does not satisfy the equation.

Use the abnormal-molar-mass relation M_app = M_true/i. Since M_app = 0.40M_true, i = 1/0.40 = 2.50. Dissociation of AX3 produces four ions, so i = 1 + α(4 − 1) = 1 + 3α. Solving 2.50 = 1 + 3α gives α = 0.50, or 50%; therefore option B is correct. The smaller apparent mass reflects the increased number of particles after dissociation.

For tetramer association, the van’t Hoff factor is i = 1 − α + α/4 = 1 − 3α/4. Substituting i = 0.8125 gives 3α/4 = 1 − 0.8125 = 0.1875. Hence α = 0.1875 × 4/3 = 0.25, or 25%. Option B is correct. The result is physically sensible because i lies between 1 for no association and 0.25 for complete tetramerisation.

The governing relation is ΔTf = iKf m. Without dissociation, the normal depression is ΔTf° = Kf m = 2.0 × 0.15 = 0.30 K. Therefore i = observed/normal = 0.54/0.30 = 1.80. For dissociation into two ions, i = 1 + α, so α = 1.80 − 1 = 0.80, or 80%. Hence option C is correct; the larger observed depression indicates increased particle number.

MX4 produces five ions on complete dissociation, so n = 5. For dissociation fraction α = 0.45, the van’t Hoff factor is i = 1 + α(n − 1) = 1 + 0.45(5 − 1) = 2.80. Since apparent molar mass is M_app = M_true/i, M_app/M_true = 1/2.80. Thus option A is correct; 2.80 is i itself, not the requested mass fraction.

The van’t Hoff factor compares the observed colligative property with the value for particles that do not dissociate. A 120% increase means the observed value is 2.20 times the normal value, so i = 2.20. For AB3, one formula unit produces four ions on complete dissociation; therefore i = 1 + 3α. Thus 2.20 = 1 + 3α, giving α = 0.40 or 40%. The increase must not be treated directly as the degree of dissociation.

The governing idea is that the van’t Hoff factor measures the increase in solute particles after dissociation. A2B produces three ions, so for a degree of dissociation α, i = 1 + α(3 − 1) = 1 + 2α. Substituting i = 2.05 gives 2α = 1.05 and α = 0.525. Converting this fraction to a percentage gives 52.5%, so option C is correct. Dividing the excess over one by three would incorrectly ignore the original particle.

For the same concentration and temperature, the ratio of the observed colligative effect to the normal effect is i. Therefore i = 2.4. AB2 produces three ions, so i = 1 + 2α. Solving 2.4 = 1 + 2α gives α = 0.70, or 70%. The increased vapour-pressure lowering is due to the increased number of particles after dissociation.

Formation of A3 is association into trimers, so three original solute particles become one particle. If α = 0.45 of the original particles associates, the particle factor is i = (1 − α) + α/3. Substitution gives i = 0.55 + 0.15 = 0.70. Therefore option B is correct. The value is below one because association reduces the total number of independently moving particles; 0.55 counts only unassociated particles and omits the trimer groups.

For dilute solutions, osmotic pressure follows π = iCRT, where i corrects the concentration for dissociation or association. First calculate the ordinary osmotic pressure: CRT = 0.04 × 0.082 × 300 = 0.984 atm. Hence i = π/(CRT) = 1.476/0.984 = 1.50. Thus option C is correct. A value above one indicates that the solute produces more effective particles than an undissociated nonelectrolyte.

The normal boiling-point elevation for an undissociated solute is ΔTb° = Kb m. Using the data, ΔTb° = 0.60 × 0.20 = 0.12 K. The van’t Hoff factor is the ratio of the observed elevation to this normal elevation, so i = 0.18/0.12 = 1.50. Therefore option C is correct. Since the observed elevation exceeds the normal value, the solute must produce additional effective particles, typically through dissociation.

Particle counting must include both unchanged formula units and the ions formed by dissociation. Of the 100 original AB2 units, 65 remain undissociated and contribute 65 particles. Each of the 35 dissociated units produces three ions, so those units contribute 35 × 3 = 105 particles. The total is therefore 65 + 105 = 170, making option C correct. Counting only the new ions would incorrectly give 105.