What is vapour pressure of solution containing 0.1 mole solute dissolved in \(1.8 \times 10^{-2} \mathrm{~kg} \mathrm{H}_2 \mathrm{O}\) ?
\(
\left(\mathrm{P}_1^0=24 \mathrm{~mm} \mathrm{Hg}\right)
\)

\(\begin{aligned} & \frac{\mathrm{P}^{\circ}-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}^{\circ}}=\mathrm{X}_{\text {solute }}=\frac{\mathrm{n}_{\text {solute }}}{\mathrm{n}_{\text {solute }}+\mathrm{n}_{\text {solvent }}} \\ & \text { Mass of water }=1.8 \times 10^{-2} \mathrm{~kg} \\ & =1.8 \times 10^{-2} \times 1000 \mathrm{~g}=18 \mathrm{~g} \\ & \text { Moles of water (solvent) }=\frac{18}{18}=1 \\ & \frac{24-\mathrm{P}_{\mathrm{s}}}{24}=\frac{0.1}{0.1+1}=\frac{0.1}{1.1} \\ & \frac{24-\mathrm{P}_{\mathrm{s}}}{24}=\frac{1}{11} \\ & \mathrm{P}_{\mathrm{s}}=24-\frac{24}{11}=21.8 \mathrm{~mm} \mathrm{Hg} .\end{aligned}\)