AB 2 is 10 % dissociated in water to A 2 + and B - . The boiling point of 10 . 0 molal aqueous solution of…

AB2 is 10% dissociated in water to A2+ and B-. The boiling point of 10.0 molal aqueous solution of AB2 is-- C. (Round off to the Nearest Integer).

[Given : Molal elevation constant of water Kb=0.5 K kg mol-1 boiling point of pure water=100°C]

Solution

AB2A2++2 B-t=0a00t=0a-2

nT=a-aα+aα+2aα

=a1+2α

so i=1+2α

Now ΔTb=i×m×Kb

ΔTb=1+2α×m×Kb

α=0.1   m=10   Kb=0.5

ΔTb=1.2×10×0.5

=6

So boiling point =106

Asked in: JEE Main 2021 (16 Mar Shift 1)

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