JEE Main 2024 Chemistry question
A solution containing 10g of an electrolyte AB₂ in 100g 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 = 200g mol⁻¹. (molal boiling point elevation const. of water) = 0.52 K kg mol⁻¹, boiling point of water = 100°C; AB₂ ionises as AB₂ → A²⁺ + 2B⁻]
Answer: 5
Solution
m = (10/200)/0.1 = 0.5 mol kg⁻¹; 0.52 = i × 0.52 × 0.5 ⇒ i = 2 = 1 + 2α ⇒ α = 0.5 = 5 × 10⁻¹.