Solution 9:
\begin{tabular}{l}
\hline $\mathrm{P} \square=23.8 \mathrm{~mm} \mathrm{Hg}$ \\
$\mathrm{W} 2=50 \mathrm{~g}, \mathrm{M}_{2}($ urea $)=60 \mathrm{~g} \mathrm{~mol}^{-1}$ \\
$\mathrm{w} 1=850 \mathrm{~g}, \mathrm{M}_{1}($ water $)=18 \mathrm{~g} \mathrm{~mol}^{-1}$ \\
To find: $\mathrm{Ps}$ and $\left(\mathrm{P}^{\circ}-\mathrm{P}^{\mathrm{s}}\right) / \mathrm{P}^{\mathrm{o}}$
\end{tabular}
To find: Ps and $\left(\mathrm{P}^{\mathrm{o}}-\mathrm{P}^{\mathrm{s}}\right) / \mathrm{P}^{\mathrm{o}}$ Solution: Applying Raoult’s law,
$$
\begin{array}{l}
\frac{P^{o}-P_{s}}{P^{o}}=\frac{n_{2}}{n_{1}+n_{2}} \frac{w_{2} / M_{2}}{w_{1} / M_{1}+w_{2} / M_{2}} \\
\therefore \frac{P^{o}-P_{s}}{P^{o}}=\frac{50 / 60}{850 / 18+50 / 60} \\
=\frac{0.83}{47.22+0.83}=0.017
\end{array}
$$
Putting $\mathrm{P}^{\circ}=23.8 \mathrm{~mm}$, we have
$23.8 \mathrm{~mm}$, we have
$$
\frac{23.8-P_{s}}{P_{s}}=0.017
$$
$23.8-\mathrm{P}_{\mathrm{s}}=0.017 \mathrm{P}_{\mathrm{s}}$
or $1.017 \mathrm{P}_{\mathrm{s}}=23.8$
or $\mathrm{P}_{\mathrm{s}}=23.4 \mathrm{~mm}$