7: Abnormal Molecular Mass

The van’t Hoff factor is defined as the effective moles of particles divided by the initial moles of solute. Therefore i = 0.024/0.010 = 2.4. This means that, on average, each initial mole of electrolyte contributes 2.4 moles of effective particles. The answer is not 0.24 because the decimal division must be performed carefully, and 24 is ten times too large.

Relative lowering of vapour pressure depends on the effective number of solute particles. Dissociation increases this number, making the observed lowering larger than the value expected from intact molecules. When the formula for a normal solute is used, a larger effect corresponds to a smaller calculated molar mass. Dimerisation would instead give a larger apparent mass.

For dimerisation, i = 1 − α/2 because each dimerisation event reduces two particles to one. With α = 0.75, i = 1 − 0.75/2 = 1 − 0.375 = 0.625. The value is substantially below one because association is extensive. A value above one would indicate dissociation, not dimerisation, and 0.75 would incorrectly ignore the factor of one-half.

A two-ion electrolyte has i = 1 + α, with a theoretical maximum of 2 when dissociation is complete. Since the measured value 1.9 is very close to 2, α = 1.9 − 1 = 0.9, or 90% dissociation. Thus the substance is nearly completely dissociated. Association would give i below one, and no dissociation would give i = 1, so those choices contradict the observation.

Colligative properties depend primarily on the number of dissolved particles, not on the chemical identity of each particle. Dissociation increases the count, whereas association decreases it; consequently the measured effect differs from the normal prediction and the calculated molar mass becomes abnormal. Colour, vessel shape and the other listed factors do not govern this principle.

One formula unit of X produces three particles when it dissociates: two particles of Y and one particle of Z. Thus, n = 3 and n − 1 = 2. For dissociation, i = 1 + α(n − 1). With α = 40/100 = 0.40, i = 1 + 0.40 × 2 = 1.80. Therefore, option B is correct. The value is not 3 because complete dissociation would be required for that.

The correct relation is i = Mtrue/Mobs, obtained from Mobs = Mtrue/i. Reversing the ratio replaces i by its reciprocal. A true value above one would then appear below one, and a value below one would appear above one; consequently dissociation could be labelled association and vice versa. The error changes the conclusion, not merely the unit.

Dissociation changes one solute unit into two or more particles, so the effective particle number rises and i becomes greater than one. Association has the opposite effect, giving i below one. Normal behaviour gives i = 1, not zero, and dimerisation generally gives i between 0.5 and 1 depending on its extent, not i = 2.

For dimerisation, i = 1 − α/2. Substituting i = 0.75 gives 0.75 = 1 − α/2. Rearranging, α/2 = 0.25 and α = 0.50. Therefore the degree of dimerisation is 50%. The result is physically reasonable: zero association would give i = 1, while complete dimerisation would give i = 0.5, so 0.75 lies midway between these limits.

CaCl₂ dissociates into one Ca²⁺ ion and two Cl⁻ ions, so the total number of ions on complete dissociation is n = 3. For α = 0.70, i = 1 + (n − 1)α = 1 + 2(0.70) = 2.40. The value is below the complete-dissociation limit of 3 because only 70% dissociation occurs. Treating CaCl₂ as a two-ion electrolyte would undercount the particles.

K₂SO₄ dissociates as K₂SO₄ → 2K⁺ + SO₄²⁻, so one formula unit gives three ions on complete dissociation. Therefore, n = 3. For a dissociation fraction α = 0.50, i = 1 + α(n − 1) = 1 + 0.50(3 − 1) = 1 + 1 = 2.0. Hence option B is correct. The value 3.0 would correspond to complete, not 50%, dissociation.

For a given concentration and temperature, osmotic pressure is proportional to the effective number of solute particles. The van't Hoff factor is therefore i = πobserved/πnormal, where the normal pressure is calculated for unchanged solute particles. The other listed ratios do not directly represent the particle-number correction in the osmotic-pressure equation.

If m molecules associate completely to form one particle, the van’t Hoff factor is i = 1/m. Here i = 0.25 = 1/4, so m = 4. Therefore four molecules combine to form one associated particle. Complete dimerisation would give i = 1/2, while one molecule forming two particles is dissociation and would give i = 2. The value 0.25 specifically indicates fourfold association.

The relation is M_obs = M_true/i, so i = M_true/M_obs. Given M_obs = 0.4M_true, i = M_true/(0.4M_true) = 1/0.4 = 2.5. Since the observed mass is smaller than the true mass, i must be greater than one, which is consistent with dissociation or another particle-increasing process. The reciprocal calculation rules out 0.4.

One X₂ particle produces two X particles on complete dissociation, so n = 2. For α = 0.30, i = 1 + (n − 1)α = 1 + (2 − 1)(0.30) = 1.30. The factor is greater than one because particle splitting increases the effective number of particles. A value of 2 would require complete dissociation, while 0.70 would incorrectly describe a decrease in particles.

The van't Hoff factor is the ratio of actual effective particles to the particles expected from the original solute. i = 1 means these numbers are equal under the experimental conditions. It does not prove that no microscopic association or dissociation occurs, because opposing effects could sometimes cancel. Therefore only normal effective particle count is a safe conclusion.

A van’t Hoff factor of 0.5 means that the effective number of particles is half the number expected for an undissociated solute. Complete dimerisation converts two original molecules into one associated particle, giving i = 1/2 = 0.5. Dissociation would increase particle number and give i greater than 1, while a normal solute would have i close to 1.

The van’t Hoff factor is the ratio of effective particles after association to the original number of solute particles. Consider one mole initially. The 40% remaining normal contributes 0.40 mol particles. The 60% dimerised portion forms dimers, so it contributes 0.60/2 = 0.30 mol particles. Total particles = 0.40 + 0.30 = 0.70 mol; hence i = 0.70 and option B is correct.

The factor i compares the actual effective particle count with the count expected without association or dissociation. If i < 1, fewer particles are present, commonly because molecules associate into dimers or larger groups. Such association generally makes the experimentally calculated molar mass higher, not lower; dissociation would give i > 1.

First calculate the van’t Hoff factor from the particle amounts: i = effective particle moles/original solute moles = 0.075/0.05 = 1.5. For AB, one formula unit becomes two ions on complete dissociation, so i = 1 + α. Therefore α = 1.5 − 1 = 0.50. Converting the fraction to a percentage gives 50% dissociation, so option B is correct.

The van’t Hoff factor measures the number of effective particles formed per original solute unit. Complete dissociation of AB produces two ions, A⁺ and B⁻, from one formula unit; therefore i = 2. Complete dissociation of AB₂ would produce three particles, giving i = 3. Dimerisation and trimerisation combine particles and give factors below one. Hence option A is the simplest explanation.

This expression describes a substance whose one original unit remains unchanged in the undissociated fraction and produces n particles in the dissociated fraction. Here α is the degree of dissociation and n is the number of particles from complete dissociation. Association, such as dimerisation, requires a different particle-balance expression.

The relation between the true and observed molar masses is i = Mtrue/Mobserved. Therefore, i = 150/100 = 1.5. Since i is greater than one, the number of solute particles has increased, which is consistent with dissociation rather than association. The osmotic-pressure measurement therefore gives the lower apparent molar mass.

For an AB electrolyte, complete dissociation changes one solute unit into two ions. Thus, for dissociation fraction α, the van’t Hoff factor is i = 1 + α(2 − 1) = 1 + α. Here α = 35/100 = 0.35, so i = 1 + 0.35 = 1.35. A value below one would indicate association rather than dissociation, while 1.70 uses an incorrect multiplier. Option A is correct.

Abnormal molar mass is related to the van’t Hoff factor by Mobserved = Mtrue/i. Given Mobserved = 1.6Mtrue, rearrangement gives i = Mtrue/(1.6Mtrue) = 1/1.6 = 0.625. Thus option A is correct. A value below one indicates association, because association reduces the number of effective particles. Option B reverses the relation, while C and D do not follow from the stated ratio.