Chemistry - Liquid Solution Question with Solution | TestHub

$$\begin{array}{rl}& {\mathrm{n}}_2=1\mathrm{mol}\\ & {\mathrm{n}}_1=\frac{36}{18}=2\mathrm{mol}\end{array}$$ Relative lowering of vapour pressure $$=\frac{{\mathrm{P}}_1^0-{\mathrm{P}}_1}{{\mathrm{P}}_1^0}=x_2=\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1+{\mathrm{n}}_2}$$ $\begin{array}{ll}\therefore & \frac{32\mathrm{mm}\mathrm{Hg}-{\mathrm{P}}_1}{32\mathrm{mm}\mathrm{Hg}}=\frac{1}{3}\\ & 96\mathrm{mm}\mathrm{Hg}-3{\mathrm{P}}_1=32\mathrm{mm}\mathrm{Hg}\\ & 64\mathrm{mm}\mathrm{Hg}=3{\mathrm{P}}_1\\ \therefore & {\mathrm{P}}_1=21.33\mathrm{mm}\mathrm{Hg}\approx 21.44\mathrm{mm}\mathrm{Hg}\end{array}$