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Q.

In a 0.2 molar aqueous solution of a weak acid HX, the degree of ionization is 0.3. taking k_f for water is 1.85, the freezing point of the solution will be nearest to

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a

+0.480⁰C

b

-0.360⁰C

c

-0.260⁰C

d

-0.480⁰C

answer is D.

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Detailed Solution

$\begin{array}{l} α = \frac{i - 1}{n - 1} (or) 0 .3 = \frac{i - 1}{2 - 1} \Rightarrow i = 1 .3 \\ △ T_{f} = i \times k_{f} \times m \\ = 1 .3 \times 1 .85 \times 0 .2 = 0 .480 k \\ freezing point of solution = 0 - 0 .480 = - 0 {.480}^{0} C \end{array}$

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