$0.5$ molal aqueous solution of a weak acid $(\mathrm{HX})$ is $20 \%$ ionised. If $K_{f}$ for water is $1…

$0.5$ molal aqueous solution of a weak acid $(\mathrm{HX})$ is $20 \%$ ionised. If $K_{f}$ for water is $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, the lowering in freezing point of the solution is
  1. $0.56 \mathrm{~K}$
  2. $1.12 \mathrm{~K}$
  3. $-0.56 \mathrm{~K}$
  4. $-1.12 \mathrm{~K}$

Solution

As $\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \cdot \mathrm{m}$
For,
$\begin{array}{lccc} & \mathrm{HX} & ightleftharpoons & \mathrm{H}^{+}+ & \mathrm{X}^{-} \\ \mathrm{t}=0 & 1 & 0 & 0 \\ \mathrm{t}=\mathrm{eq} & (1-0.20) & 0.20 & 0.20\end{array}$
Total no. of moles $=1-0.20+0.20+0.20$
$=1+0.20=1.2$
$\therefore \Delta \mathrm{T}_{\mathrm{f}}=1.2 \times 1.86 \times 0.5=1.1160 \approx 1.12 \mathrm{~K}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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