In a 0.2 molal aqueous solution of a weak acid HX, the degree of ionization is 0.3 at room temperature. Taking Kf for water as 1.85oCm−1, the freezing point of the solution will be nearest to
A
−0.481∘C
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B
−0.360∘C
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C
−0.260∘C
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D
−0.150∘C
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Solution
The correct option is A−0.481∘C We know, ΔTf=iKfm Again, we know, i=1+(n−1)α Where, α = degree of ionisation So here i=1+(2−1)0.3=1.3 Again, since the solution is 0.2 molal Thus, ΔTf=1.3×1.85×0.2=0.481oC Again, ΔTf=Tf(solvent)−Tf(solution) ⇒0.481=0−Tf(solution) ⇒Tf(solution)=−0.481oC