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Question

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 1.85 oCm1, 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+(n1)α
Where, α = degree of ionisation
So here i=1+(21)0.3=1.3
Again, since the solution is 0.2 molal
Thus,
ΔTf=1.3×1.85×0.2=0.481 oC 
Again,
ΔTf=Tf(solvent)Tf(solution)
0.481=0Tf(solution)
Tf(solution)=0.481 oC

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