Solute A associates in water. When 0.7 g of solute A is dissolved in 42.0 g of water, it depresses the freezing point by 0.2°C. The percentage association of solute A in water, is :
[Given : Molar mass of A = 93 g mol⁻¹. Molal depression constant of water is 1.86 K kg mol⁻¹.]

Solution

Chemistry ma’am · 4 steps

Find the expected depression first, compare it with 0.2, and the van 't Hoff factor gives you the association.

1. Expected depression. Moles of A = 0.7/93 = 7.53 × 10⁻³ mol, so molality = (7.53 × 10⁻³)/0.042 = 0.179 mol kg⁻¹. Without association, Δ T_f = 1.86 × 0.179 = 0.333 K, which is 1/3 K.
2. Van 't Hoff factor. The observed depression is only 0.2 K, so i = 0.2/(1/3) = 0.6. The value i < 1 confirms fewer particles, as association requires.

   2A ⇌ A₂: i = 1 - α + α/2 = 1 - α/2

3. Solve for alpha. Take A to form dimers. Then 1 - α/2 = 0.6, so α/2 = 0.4 and α = 0.8.
4. So. The percentage association of A is 80%, option (d).

The idea to keep. For association, i is below 1. For dimers, i = 1 - α/2. Get i as observed Δ T_f divided by calculated Δ T_f, then solve for α.

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