A solution containing 10 g of an electrolyte AB₂ in 100 g of water boils at 100.52° C. The degree of ionization of the electrolyte (α) is _________ × 10⁻¹. (nearest integer)
[Given : Molar mass of AB₂=200 g mol⁻¹, K_b (molal boiling point elevation const. of water) =0.52 K kg mol⁻¹, boiling point of water =100° C ; AB₂ ionises as AB₂ → A²⁺+2 B⁻]

Solution

Chemistry ma’am · 4 steps

Find the van 't Hoff factor from the boiling point rise, then read α off i = 1 + 2α.

1. Molality first. Moles of AB₂ are 10/200 = 0.05 mol in 0.1 kg of water. So m = 0.05/0.1 = 0.5 mol kg⁻¹.

   Δ T_b = i K_b m ⇒ 0.52 = i × 0.52 × 0.5

2. Van 't Hoff factor. The rise is 100.52 - 100 = 0.52 K, so i = 0.52/(0.52 × 0.5) = 2.
3. Link i to alpha. One AB₂ gives A²⁺ and 2B⁻, so n = 3 particles. Then i = 1 + (n-1)α = 1 + 2α = 2, which gives α = 0.5.
4. So. α = 0.5 = 5 × 10⁻¹, so the answer is 5.

The common slip: 10. Counting only 2 ions per formula unit, so that i = 1 + α, gives α = 1, which is reported as 10. AB₂ gives 3 ions, so n - 1 = 2.

The idea to keep. For partial ionization, i = 1 + (n-1)α, where n counts every ion from one formula unit. Get i from the colligative property first, then solve for α.

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