7: Abnormal Molecular Mass

Correct answer: C. 0.80

Explanation: Take the initial amount of solute as one mole. The 60% that remains normal contributes 0.60 mole of particles. The remaining 40% forms dimers, and every dimer contains two original molecules, so this part contributes 0.40/2 = 0.20 mole of dimers. The total particle amount is therefore 0.60 + 0.20 = 0.80 mole, giving i = 0.80.

Correct answer: C. 0.034 mol

Explanation: One formula unit of A₂B₃ gives five ions on complete dissociation: 2A and 3B. For 60% dissociation, the van’t Hoff factor is i = 1 + α(n − 1) = 1 + 0.60(5 − 1) = 3.4. Effective moles of particles are the initial moles multiplied by i: 0.01 × 3.4 = 0.034 mol. Thus option C is correct.

Correct answer: A. 4/7

Explanation: AB₃ produces four ions on complete dissociation, so for 25% dissociation the van’t Hoff factor is i = 1 + 0.25(4 − 1) = 1.75 = 7/4. For dissociation, the observed molar mass is related to the true molar mass by M_obs = M_true/i. Hence M_obs/M_true = 1/(7/4) = 4/7. The observed value is smaller because dissociation increases the number of particles.

Correct answer: C. 2.5

Explanation: A 1:3 electrolyte produces 1 + 3 = 4 ions when completely dissociated. For partial dissociation, use i = 1 + α(n − 1), where α = 0.50 and n = 4. Thus i = 1 + 0.50(4 − 1) = 1 + 1.5 = 2.5. The value 4.0 would apply only to complete dissociation, while 1.5 does not account for all three additional ions.