0.5 molal aqueous solution of a weak acid (^H X) is 20% ionised. If K_f for water is 1.86 K kg m o 1^−1, the lowering in freezing point of the solution is

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Correct answer: B. (b) H X^−1−α−H^++X^−α α Total =1+α i=1+α=1+0.2=1.2 ∆T_f=i×K_f×m=1.2×1.8 6×0.5=1.1 1 6 K=1.1 2 K

$0.5 molal aqueous solution of a weak acid (^{H X}) is 20% ionised. If K_{f} for water is 1.86 K kg m o 1^{- 1}, the lowering in freezing point of the solution is$

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