In a $0.2$ molar aqueous solution of a weak acid HX the degree of ionization is $0.3$ The freezing point of the solution will be nearest to .

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Solution

$H X ⇌ H^{+} + I^{-}$

$i = 1 + (n - 1) α$
$\Rightarrow i = 1 + 0.3 = 1.3$

$K_{f} = 1.85 K k g m o l^{- 1}$

$Δ T_{f} = i K_{f} m = 1.3 \times 1.85 \times 0.2 = 0.481 K$

$\Rightarrow {T_{f}}_{s o l v e n t} - {T_{f}}_{s o l u t i o n} = 0.481 K$

$\Rightarrow 273 K - {T_{f}}_{s o l u t i o n} = 0.481 K$

$\Rightarrow {T_{f}}_{s o l u t i o n} = 272.519 K$

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