5: Colligative Properties

The freezing-point depression is given by ΔT_f = iK_fm. At equal molality and in the same solvent, the solution with the largest van’t Hoff factor has the greatest depression and therefore the lowest freezing point. Urea gives i = 1, NaCl gives i = 2, AlCl₃ gives i = 4, while K₄[Fe(CN)₆] gives four K⁺ ions and one complex ion, so i = 5. Therefore, option D is correct.

Using the normal molar mass, the number of solute moles is 1.0/100 = 0.010 mol. Since the solvent mass is 0.100 kg, the normal molality is 0.010/0.100 = 0.10 m. The normal freezing-point depression would be Kf m = 1.86 × 0.10 = 0.186 K. The observed depression is 0.093 K, exactly half of this value, so i = observed ΔTf/normal ΔTf = 0.093/0.186 = 0.5.

For abnormal molar mass, the van’t Hoff factor is related to the true and observed molar masses by i = Mtrue/Mobserved. Hence i = 133.5/44.5 = 3. AlCl3 dissociates as AlCl3 → Al3+ + 3Cl−, producing four particles on complete dissociation. For a degree of dissociation α, i = 1 + (4 − 1)α = 1 + 3α. Thus 3 = 1 + 3α, so α = 2/3 = 0.667, or approximately 66.7%.

Use π = iCRT to determine the van’t Hoff factor. With π = 1.845 atm, C = 0.05 mol L−1, R = 0.082 L atm K−1 mol−1, and T = 300 K, i = 1.845/(0.05 × 0.082 × 300) = 1.5. The relation between true and observed molar mass is i = Mtrue/Mobserved, so Mobserved = Mtrue/i. Therefore Mobserved = 120/1.5 = 80 g mol−1.