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At $298 \mathrm{~K}$, the vapour pressure of pure water is 25 torr. The vapour pressure of water, when $12 \mathrm{~g}$ of urea (molar mass, $60 \mathrm{~g} \mathrm{~mol}^{-1}$ ) and $36 \mathrm{~g}$ of glucose (molar mass, $180 \mathrm{~g}$ $\mathrm{mol}^{-1}$ ) is dissolved in $100 \mathrm{~g}$ of water at the sametemperature (in torr) is
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23.32
$\mathrm{P}_{\mathrm{H}_2 \mathrm{O}}^{\circ}=25$ torr; $\mathrm{n}_{\mathrm{u}}=12 / 60=0.2$ moles;
$\begin{aligned} & \mathrm{n}_{\mathrm{g}}=36 / 180=0.2 \text { moles; } \mathrm{n}_{\mathrm{H}_2 \mathrm{O}}=100 / 18=5.55 \text { moles } \\ & \therefore \chi_{\mathrm{H}_2 \mathrm{O}}=\frac{5.55}{5.55+0.2+0.2}=\frac{5.55}{5.95} \\ & \therefore \quad \mathrm{P}_{\mathrm{H}_2 \mathrm{O}}=\chi_{\mathrm{H}_2 \mathrm{O}} \mathrm{P}^{\circ}{ }_{\mathrm{H}_2 \mathrm{O}}=\frac{5.55}{5.95} \times 25=23.32\end{aligned}$
$\begin{aligned} & \mathrm{n}_{\mathrm{g}}=36 / 180=0.2 \text { moles; } \mathrm{n}_{\mathrm{H}_2 \mathrm{O}}=100 / 18=5.55 \text { moles } \\ & \therefore \chi_{\mathrm{H}_2 \mathrm{O}}=\frac{5.55}{5.55+0.2+0.2}=\frac{5.55}{5.95} \\ & \therefore \quad \mathrm{P}_{\mathrm{H}_2 \mathrm{O}}=\chi_{\mathrm{H}_2 \mathrm{O}} \mathrm{P}^{\circ}{ }_{\mathrm{H}_2 \mathrm{O}}=\frac{5.55}{5.95} \times 25=23.32\end{aligned}$
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