Verified Solution • 2025–26 Session moderate

QJEEM23Chc378: JEE Main 2023 Chemistry - Class 12 JEEM23Chc378 Solution: 80 mole percent of is dissociated in aqueous solution. The…

80 mole percent of MgCl2\mathrm{MgCl}_{2} is dissociated in aqueous solution. The vapour pressure of 1.0 molal1.0 ~\mathrm{molal} aqueous solution of MgCl2\mathrm{MgCl}_{2} at 38∘C38^{\circ} \mathrm{C} is ____________ mm Hg. (Nearest integer)\mathrm{mm} ~\mathrm{Hg.} ~\mathrm{(Nearest} ~\mathrm{integer)}

Given : Vapour pressure of water at 38∘C38^{\circ} \mathrm{C} is 50 mm Hg50 \mathrm{~mm} ~\mathrm{Hg}

Step-by-Step Verified Solution

MgCl2amp;⇌ Mg2++2Cl−1amp;0amp;01−0.8amp;0.8amp;1.6\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=2.6

p∘−pp∘=2.6100018+2.6\frac{p^{\circ}-p}{p^{\circ}}=\frac{2.6}{\frac{1000}{18}+2.6}

For dil solution

amp;p∘−pp∘=2.6100018amp;p=47.66≃48 mmHg\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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