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0.5 molal aqueous solution of a weak acid $(\mathrm{HX})$ is $20 \%$ ionised. If $\mathrm{K}_f$ for water is $1.86 \mathrm{Kg} \mathrm{mol}^{-1}$, the lowering in freezing point of the solution is:
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Verified Answer
The correct answer is:
$1.12 \mathrm{~K}$
$\operatorname{As} \Delta \mathrm{T}_f=i k_f \mathrm{~m}$ For
$\begin{aligned}
& \mathrm{HX} \rightleftharpoons \mathrm{H}^{+}+\mathrm{X}^{-} \\
& t=0 \quad 1 \quad 0 \quad 0 \\
& \begin{array}{llll}
t_{\text {eq }} & (1-0.20) & 0.20 & 0.20
\end{array} \\
&
\end{aligned}$
Total no. of moles $=1-0.20+0.20$
$\begin{aligned}
+0.20 & =1+0.20=1.2 \\
\therefore \Delta \mathrm{T}_f & =1.2 \times 1.86 \times 0.5 \\
& =1.1160
\end{aligned}$
Related Theory
The boiling point and freezing point are colligative properties. Due to the addition of solutes, the boiling point tends to increase, and freezing point tends to decrease.
$\begin{aligned}
& \mathrm{HX} \rightleftharpoons \mathrm{H}^{+}+\mathrm{X}^{-} \\
& t=0 \quad 1 \quad 0 \quad 0 \\
& \begin{array}{llll}
t_{\text {eq }} & (1-0.20) & 0.20 & 0.20
\end{array} \\
&
\end{aligned}$
Total no. of moles $=1-0.20+0.20$
$\begin{aligned}
+0.20 & =1+0.20=1.2 \\
\therefore \Delta \mathrm{T}_f & =1.2 \times 1.86 \times 0.5 \\
& =1.1160
\end{aligned}$
Related Theory
The boiling point and freezing point are colligative properties. Due to the addition of solutes, the boiling point tends to increase, and freezing point tends to decrease.
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