2 molal solution of a weak acid HA has a freezing point of 3.885°C. The degree of dissociation of this acid is ___________ × 10⁻³. (Round off to the Nearest Integer).
[Given : Molal depression constant of water = 1.85 K kg mol⁻¹ Freezing point of pure water = 0° C]

Solution

Chemistry ma’am · 4 steps

Find the van't Hoff factor from the freezing-point drop, then read α off i = 1 + α.

1. Size of the drop. A solute always lowers the freezing point of water, so the solution freezes at -3.885 °C. That gives Δ T_f = 3.885 K.

   Δ T_f = i K_f m ⇒ i = 3.885/(1.85 × 2) = 3.885/3.7 = 1.05

2. Link i to α. For HA ⇌ H⁺ + A⁻, each mole of acid gives (1 - α) mol HA, α mol H⁺ and α mol A⁻. The total is 1 + α mol of particles, so i = 1 + α.
3. Solve for α. α = 1.05 - 1 = 0.05.
4. So. α = 0.05 = 50 × 10⁻³, so the answer is 50.

The common slip: 5. Reading α = 0.05 as a percentage gives 5. The blank is in units of 10⁻³, so 0.05 must be written as 50 × 10⁻³.

The idea to keep. For a weak electrolyte, colligative data gives you i first. Then write i = 1 + (n - 1)α, where n is the number of ions per formula unit, and solve for α.

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