A solute A dimerizes in water. The boiling point of a 2 molal solution of A is 100.52°C. The percentage association of A is __________. (Round off to the Nearest Integer).
[Use : K b for water = 0.52 K kg mol⁻¹ Boiling point of water = 100°C]

Solution

Chemistry ma’am · 4 steps

Find the van 't Hoff factor from the boiling point first, then turn it into the degree of dimerization.

1. Find i. The rise in boiling point is Δ T_b = 100.52 - 100 = 0.52 K. Put it into Δ T_b = i K_b m: 0.52 = i × 0.52 × 2, so i = 0.5.
2. Particles after dimerizing. For 2A ⇌ A₂, start with 1 mol of A and let a fraction α associate. That leaves 1 - α mol of A and makes α/2 mol of A₂.

   i = 1 - α + α/2 = 1 - α/2

3. Solve for alpha. 1 - α/2 = 0.5 gives α = 1. Every molecule of A has paired up.
4. So. The percentage association of A is 100.

The common slip: 50. Taking i = 1 - α and forgetting the α/2 mol of dimer formed gives α = 0.5, that is 50 instead of 100.

The idea to keep. For association into n-mers, i = 1 - α + α/n. For a dimer, i = 0.5 means association is complete.

Answer from the NTA final answer key. Solution by PrepWiser.