A weak electrolyte, $\mathrm{AB}$, is $5 \%$ dissociated in aqueous solution. What is the freezing point of a $0.100$ molal aqueous solution of $\mathrm{AB}$ ? $\mathrm{K}_{\mathrm{f}}$ for water is $1.86 \mathrm{~deg}/\mathrm{molal}$ .

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Aicharya · Accepted Answer

Degree of dissociation, $\alpha, A B=$ $\frac{5}{100}=0.05$
No. of moles dissolved $\begin{array}{c}\mathrm{AB} & ightarrow \mathrm{A}^{+}+\mathrm{B}^{-} \\ \mathrm{m} & 0 & 0\end{array}$
No. of moles after dissociation $\mathrm{m}(1-\alpha) \mathrm{m}\alpha \quad \mathrm{m} \alpha$
$\begin{array}{cc}\text { After dissociation } & \end{array} \begin{array}{lll}0.1(1-0.05) & 0.1 \times 0.05 & 0.1 \times 0.05\end{array}$
Total moles $=$ Molality
$=0.1(1-0.05)+0.1 \times 0.05+0.1 \times 0.05$
$=0.095+0.005+0.005=0.105 \mathrm{~m}$
$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \mathrm{m}$ or $\mathrm{T}_{\mathrm{f}}^{0}-\mathrm{T}_{\mathrm{f}}=1.86 \mathrm{~K} / \mathrm{m} \times 0.105 \mathrm{~m}$
$=0.1953$ $\mathrm{deg}$
$\mathrm{T}_{\mathrm{f}}=0^{\circ} \mathrm{C}-0.1953=-0.1953^{\circ} \mathrm{C}$

A weak electrolyte, $\mathrm{AB}$, is $5 \%$ dissociated in aqueous solution. What is the freezing point of…

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