JEE MAIN - Chemistry (2023 - 12th April Morning Shift - No. 15)
80 mole percent of $$\mathrm{MgCl}_{2}$$ is dissociated in aqueous solution. The vapour pressure of $$1.0 ~\mathrm{molal}$$ aqueous solution of $$\mathrm{MgCl}_{2}$$ at $$38^{\circ} \mathrm{C}$$ is ____________ $$\mathrm{mm} ~\mathrm{Hg.} ~\mathrm{(Nearest} ~\mathrm{integer)}$$
Given : Vapour pressure of water at $$38^{\circ} \mathrm{C}$$ is $$50 \mathrm{~mm} ~\mathrm{Hg}$$
Answer
48
Explanation
$$
\begin{array}{ccc}
\mathrm{MgCl}_2 & \rightleftharpoons ~\mathrm{Mg}^{2+}+2 \mathrm{Cl}^{-} \\
1 & 0 & 0 \\
1-0.8 & 0.8 & 1.6
\end{array}
$$
Hence overall molality will be equal to $=2.6$
$$ \frac{p^{\circ}-p}{p^{\circ}}=\frac{2.6}{\frac{1000}{18}+2.6} $$
For dil solution
$$ \begin{aligned} & \frac{p^{\circ}-p}{p^{\circ}}=\frac{2.6}{\frac{1000}{18}} \\\\ & p=47.66 \simeq 48 \mathrm{~mm} \mathrm{Hg} \end{aligned} $$
Hence overall molality will be equal to $=2.6$
$$ \frac{p^{\circ}-p}{p^{\circ}}=\frac{2.6}{\frac{1000}{18}+2.6} $$
For dil solution
$$ \begin{aligned} & \frac{p^{\circ}-p}{p^{\circ}}=\frac{2.6}{\frac{1000}{18}} \\\\ & p=47.66 \simeq 48 \mathrm{~mm} \mathrm{Hg} \end{aligned} $$
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