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$0.5$ molal aqueous solution of a weak acid $(\mathrm{HX})$ is $20 \%$. ionised. If $\mathrm{K}_{\mathrm{f}}$ of water is $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, the lowering in freezing point of solution is
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$1 \cdot 12 \mathrm{~K}$
$\mathrm{HX} \rightleftharpoons \mathrm{H}^{+}+\mathrm{X}^{-}$
For dissociation of an electrolyte,
$\alpha=\frac{i-1}{n-1} \quad \therefore 0.2=\frac{i-1}{2-1}$
$\therefore \mathrm{i}-1=0.2$
$\therefore \mathrm{i}=0.2+1=1.2$
Now, $\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{k}_{\mathrm{f}} \mathrm{m}=1.2 \times 1.86 \times 0.5=1.12 \mathrm{~K}$
For dissociation of an electrolyte,
$\alpha=\frac{i-1}{n-1} \quad \therefore 0.2=\frac{i-1}{2-1}$
$\therefore \mathrm{i}-1=0.2$
$\therefore \mathrm{i}=0.2+1=1.2$
Now, $\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{k}_{\mathrm{f}} \mathrm{m}=1.2 \times 1.86 \times 0.5=1.12 \mathrm{~K}$
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