A 0.2 molal aqueous solution of a weak acid (HX) is 20% ionised. The freezing point of this solution is (Given Kf=1.86oC for water)

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−0.31oC

−0.45oC

−0.53oC

−0.90oC

answer is B.

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Detailed Solution

ΔTf=molality×Kf×(1+α)α=0.2, Molality=0.2, Kf=1.86ΔTf=0.2×1.2×1.86=0.4464oFreezing point = -0.45oC

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