AB 2 is 10 % dissociated in water to A 2 + and B - . The boiling point of 10 . 0 molal aqueous solution of AB 2 is-- C ∘ . (Round off to the Nearest Integer). [Given : Molal elevation constant of water K b = 0 . 5 K kg mol - 1 boiling point of pure water = 100 ° C ]

Step-by-step solution

AB 2 → A 2 + + 2   B - t = 0 a 0 0 t = 0 a - aα aα 2 aα n T = a - a α + a α + 2 a α = a 1 + 2 α so i = 1 + 2 α Now ΔT b = i × m × K b ΔT b = 1 + 2 α × m × K b α = 0 . 1      m = 10      K b = 0 . 5 ΔT b = 1 . 2 × 10 × 0 . 5 = 6 So boiling point = 106