If the degree of dissociation of aqueous solution of weak monobasic acid is determined to be 0.3, then the observed freezing point will be ___________% higher than the expected/theoretical freezing point. (Nearest integer)

Solution

Chemistry ma’am · 3 steps

Find the van't Hoff factor first; how far i sits above 1 is how much bigger the effect is.

1. Count the particles. A weak monobasic acid splits as HA ⇌ H⁺ + A⁻. Start with 1 mol of HA. At equilibrium you have 1 - α of HA, plus α of H⁺ and α of A⁻, so 1 + α mol of particles in total.

   i = (Δ T_f (observed))/(Δ T_f (expected)) = 1 + α = 1 + 0.3 = 1.3

2. Read the wording. The quantity that changes by a percentage here is the depression Δ T_f, because the freezing point itself goes down. The observed depression is 1.3 times the expected one, so it is (1.3 - 1) × 100 = 30% larger.
3. So. The observed freezing point depression is 30% higher than the theoretical value.

The common slip: 130. Reporting the observed depression as a percentage of the expected one gives 130, but the question asks how much higher it is, which is 130 - 100 = 30.

The idea to keep. For any dissociating solute, i = 1 + (n-1)α. The percentage by which a colligative property exceeds its expected value is (i - 1) × 100.

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