7: Abnormal Molecular Mass

A₂B produces three particles, namely two A-type ions and one B-type ion, on complete dissociation. For dissociation, the van’t Hoff factor is i = 1 + α(n − 1), where α is the fractional dissociation and n is the total number of particles formed. Thus, i = 1 + 0.60(3 − 1) = 2.20. Therefore, option A is correct; 1.6 would incorrectly treat the solute as producing only two particles.

The relation between true molar mass M, observed molar mass M_obs, and van’t Hoff factor is M_obs = M/i. Substituting the given values gives M_obs = 150/1.5 = 100. Since i is greater than 1, dissociation has increased the particle number and consequently makes the experimentally inferred molar mass smaller.

Use i = M_true/M_observed = 80/160 = 0.5. A van’t Hoff factor below 1 means that the effective number of solute particles has decreased. This occurs when separate solute molecules combine into larger species, such as dimers. Therefore association is the suitable explanation; dissociation would make i greater than 1.

Calculate the factor as i = M_true/M_observed = 56/28 = 2. Thus the solution contains twice the effective particle number expected from undissociated molecules. An increase in particle number is characteristic of dissociation, where one solute unit produces multiple particles. Association would instead give i less than 1 and a larger observed molar mass.

Complete dimerisation combines two original solute molecules into one effective particle. Hence the particle number becomes half and i = 1/2. Since the apparent molar mass is M_app = M/i, substitution gives M_app = M/(1/2) = 2M. Association therefore makes the experimentally calculated molar mass larger than the true monomer mass.

AB produces two ions on complete dissociation, so n = 2. For partial dissociation, i = 1 + α(n − 1), where α = 60/100 = 0.60. Substitution gives i = 1 + 0.60(2 − 1) = 1.60. The factor is greater than 1 because dissociation increases the number of particles. Option B is only α, not i, and 2.60 overestimates the increase.

The relation is M_app = M_true/i, so i = M_true/M_app. Substituting gives i = 120/80 = 1.50. Because the factor is greater than one, the solute has produced more effective particles than its undissociated form would produce, usually because of dissociation. The value 0.67 is the reciprocal and is not i.

For MX₂, complete dissociation produces three ions, so n = 3. The fractional dissociation is α = 0.75. Using i = 1 + α(n − 1), we obtain i = 1 + 0.75(3 − 1) = 1 + 1.50 = 2.50. The value 3.00 would correspond to complete dissociation, not 75% dissociation. Thus, option B correctly accounts for both undissociated and dissociated salt.

If M_app = 2M_true, then i = M_true/M_app = 1/2. A factor of one-half means that two original solute particles behave as one effective particle. Complete dimerisation produces exactly this change. Complete dissociation would increase i and make the apparent molar mass smaller, while trimerisation would give i = 1/3 and M_app = 3M.

For dimer association, two original solute molecules combine to form one dimer. If α is the fraction associated, the particle factor is i = 1 − α/2. Substituting i = 0.75 gives 0.75 = 1 − α/2, so α/2 = 0.25 and α = 0.50. Therefore, the degree of association is 50%. The value is not 75%; that is the measured factor, not the association fraction.

For a solute that dissociates into two ions, n = 2 and i = 1 + α(n − 1) becomes i = 1 + α. Therefore 1.25 = 1 + α, so α = 0.25. Multiplying by 100 gives a degree of dissociation of 25%. The shortcut α = i − 1 works here only because exactly two particles are formed.

In trimerisation, three original molecules combine to form one trimer. If α = 0.60 is the fraction of original molecules involved, the particle factor is i = 1 − α + α/3. Therefore, i = 1 − 0.60 + 0.60/3 = 0.40 + 0.20 = 0.60. The factor is below 1 because association reduces particle number. Using 1 − α alone would incorrectly ignore the trimers formed.

The normal freezing-point depression without abnormality is ΔT_f° = K_fm = 2.0 × 0.20 = 0.40 K. The modified relation is ΔT_f = iK_fm, so i = 0.60/0.40 = 1.50. Since i exceeds one, the solute has more effective particles than an undissociated solute at the same concentration.

The normal boiling-point elevation is K_bm = 0.52 × 0.50 = 0.26 K. Since the observed value equals iK_bm, i = 0.39/0.26 = 1.50. The value above one indicates that the solute produces additional particles, as expected for partial dissociation or ionisation. Association would instead make the observed elevation smaller than 0.26 K.

AB₂ produces three ions on complete dissociation, so n = 3. For partial dissociation, i = 1 + α(n − 1) = 1 + 2α. Substituting i = 2.2 gives 2.2 = 1 + 2α, hence 2α = 1.2 and α = 0.60. Therefore, the degree of dissociation is 60%. The formula must use three particles, not two, because the two B ions are separate particles.

On complete dissociation, XY₄ gives one X-type ion and four Y-type ions, so n = 5. The partial-dissociation equation is i = 1 + α(n − 1) = 1 + 4α. With i = 3.4, 3.4 = 1 + 4α, so 4α = 2.4 and α = 0.60. Hence the degree of dissociation is 60%. Counting only four particles would produce an incorrect result because the X ion must also be included.

Tetramerisation means that four original solute particles combine to form one particle. For association into n-mers, i = 1 − α + α/n. Here α = 0.50 and n = 4, so i = 1 − 0.50 + 0.50/4 = 0.50 + 0.125 = 0.625. The associated part must be counted as one-fourth as many tetramer particles, not removed entirely; that is why the answer is not 0.50.

For dimerisation, two original molecules become one dimer. If α = 0.30 is the fraction of molecules associated, the van’t Hoff factor is i = 1 − α/2. Therefore, i = 1 − 0.30/2 = 1 − 0.15 = 0.85. The factor is less than 1 because association decreases the number of particles. The value 0.30 is only the associated fraction, not the final particle ratio.

CaCl₂ dissociates as CaCl₂ → Ca²⁺ + 2Cl⁻, giving three particles on complete dissociation. With α = 0.80, the factor is i = 1 + α(n − 1) = 1 + 0.80(3 − 1) = 1 + 1.60 = 2.60. The value 3.0 would require complete dissociation, while 2.2 would correspond to a lower degree of dissociation. Both chloride ions must be counted separately.

AlCl₃ gives Al³⁺ and three Cl⁻ ions on complete dissociation, so n = 4. For α = 0.50, use i = 1 + α(n − 1). Thus, i = 1 + 0.50(4 − 1) = 1 + 1.50 = 2.50. The value 4.0 represents complete dissociation, whereas 2.0 would not reflect the three additional particles produced when dissociation occurs. Therefore, option B is correct.

At fixed solvent, molality, and K_f, the observed freezing-point depression is proportional to i. Therefore, i = observed/normal = 0.24/0.48 = 0.50. For dimer association, i = 1 − α/2. Substitution gives 0.50 = 1 − α/2, so α/2 = 0.50 and α = 1.00, or 100%. Thus, complete dimerisation is required; partial values would give i greater than 0.50.

Formation of A₄ is tetramer association, so four original particles become one particle. For an associated fraction α and tetramer size n = 4, i = 1 − α + α/4. With α = 0.25, i = 1 − 0.25 + 0.25/4 = 0.75 + 0.0625 = 0.8125. The associated 25 particles out of 100 form 6.25 tetramers, while 75 remain single, giving 81.25 effective particles.

The apparent molar mass is M_app = M_true/i, so it is smallest when i is largest. Complete dissociation of AlCl₃ gives four particles per formula unit, making i = 4 under the ideal assumption and M_app = M/4. Dimerisation lowers i below one, while glucose has i near one; neither can produce as small an apparent mass.

Since M_app = M_true/i, the largest apparent mass corresponds to the smallest i. Complete tetramerisation changes four monomer particles into one effective particle, so i = 1/4 and M_app = 4M_true. An undissociated solute has i = 1, while dissociation gives i greater than 1 and therefore lowers the apparent molar mass.

Out of 100 original particles, 80 remain single and therefore contribute 80 particles. The remaining 20 particles associate in pairs; 20 particles make 20/2 = 10 dimers. The total effective particle count is therefore 80 + 10 = 90. The answer is not 80 because the dimers still count as particles, and it is not 100 because association reduces the number of separate particles.