80 mole percent of $\mathrm{MgCl}_{2}$ is dissociated in aqueous solution. The vapour pressure of 1.

Question

80 mole percent of $\mathrm{MgCl}_{2}$ is dissociated in aqueous solution. The vapour pressure of 1.0 molal aqueous solution of $\mathrm{MgCl}_{2}$ at $38^{\circ} \mathrm{C}$ is $\qquad$ mm Hg .\ (Nearest integer)\ Given : Vapour pressure of water at $38^{\circ} \mathrm{C}$ is 50 mm Hg .\

Professor PC Thomas Classes · Answer

(48) : $\frac{p^{\circ}-p_{s}}{p^{\circ}}=i 	imes x_{B}$\ $\mathrm{MgCl}_{2} ightleftharpoons \mathrm{Mg}^{2+}+2 \mathrm{Cl}^{-}$\ $\begin{array}{ccc}1 & 0 & 0 \ 1-0.8 & 0.8 & 0.8 	imes 2\end{array}$\ $i=1-0.8+0.8+1.6=2.6$\ As the solution is 1 molal thus 1 mole is present in 1000 g water\ Moles of water $=\frac{1000}{18}=55.5$\ $\frac{p^{\circ}-p_{s}}{p^{\circ}}=2.6 	imes \frac{1}{1+55.5}$\ $\frac{50-p_{s}}{50}=2.6 	imes \frac{1}{56.5}$\ $\Rightarrow 50-p_{s}=\frac{50 	imes 2.6}{56.5}=2.36$\ $p_{s}=50-20.36=47.7 \approx 48$\