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Van't Hoff Factor

The freezing ...

Question

The freezing point of a 0.08 molar aqueous solution of $N a H S O_{4}$ is $- {0.372}^{\circ} C$ . The dissociation constant for the following reaction is:
( $K_{f}$ for water= 1.86 $K k g m o l^{- 1}$ )
$H S O_{4}^{-} \to H^{+} + S O_{4}^{- 2}$ .
(Assume molarity and molality to be same)

0.04

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0.02

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0.01

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0.2

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Solution

The correct option is A 0.04

$0.08 M N a H S O_{4} = 0.08 M N a^{+} + 0.08 M H S O_{4}^{-}$

$H S O_{4}^{-} \to H^{+} S O_{4}^{- 2} 0.08 M (1 - α) 0.08 M α 0.08 M α$

Effective molality = $[N a^{+}] + i [H S O_{4}^{-}] = 0.08 m + 0.08 m (1 + α)$

$Δ T_{f} = K_{f} i m \Rightarrow 0.372 = 1.86 \times 0.08 (2 + α)$

$α = 0.5$
So, $K_{a} = \frac{c α^{2}}{1 - α} = \frac{0.08 \times (0.5)^{2}}{1 - 0.5} = \frac{0.02}{0.5} = 0.04$

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