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 (integer)
48
Solution
$$
\begin{array}{ccc}
\mathrm{MgCl}_2 & \rightleftharpoons ~\mathrm{Mg}^{2+}+2 \mathrm{Cl}^{-} \\
1 & 0 & 0 \\
1-0.8 & 0.8 & 1.6
\end{array}
$$<br/><br/>
Hence overall molality will be equal to $=2.6$<br/><br/>
$\frac{p^{\circ}-p}{p^{\circ}}=\frac{2.6}{\frac{1000}{18}+2.6}$<br/><br/>
For dil solution<br/><br/>
$$
\begin{aligned}
& \frac{p^{\circ}-p}{p^{\circ}}=\frac{2.6}{\frac{1000}{18}} \\\\
& p=47.66 \simeq 48 \mathrm{~mm} \mathrm{Hg}
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: States of Matter · Topic: Gas Laws
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