Two volatile liquids ‘A’ and ‘B’ are mixed in the mole ratio 1 : 3. Vapour pressure of pure liquid ‘A’ is 200 mmHg and pure liquid ‘B’ is 600 mmHg. Mole fraction of ‘B’ in vapour phase will be 

Given

$$\begin{gathered} {\mathrm{n}}_{\mathrm{A}}:{\mathrm{n}}_{\mathrm{B}}=1 : 3 \\ {\mathrm{X}}_{\mathrm{A}}=\frac{1}{4}; \\ {\mathrm{X}}_{\mathrm{B}}=\frac{3}{4}; \\ \mathrm{Applying} \mathrm{Raoult}\text{'}\mathrm{s} \mathrm{law} \\ {\mathrm{p}}_{\mathrm{A}}={\mathrm{p}}_{\mathrm{A}}^0{\mathrm{X}}_{\mathrm{A}};{\mathrm{p}}_{\mathrm{B}}={\mathrm{p}}_{\mathrm{B}}^0{\mathrm{X}}_{\mathrm{B}} \\ {\mathrm{p}}_{\mathrm{A}}=200\left(\frac{1}{4}\right)=50 \mathrm{mmHg} \\ {\mathrm{p}}_{\mathrm{B}}=600\left(\frac{3}{4}\right)=450 \mathrm{mmHg} \\ {\mathrm{p}}_{\mathrm{total}}={\mathrm{p}}_{\mathrm{A}}+{\mathrm{p}}_{\mathrm{B}}=500 \mathrm{mmHg} \\ \mathrm{Applying} \mathrm{Dalton}\text{'}\mathrm{s} \mathrm{law} \mathrm{in} \mathrm{vapour} \mathrm{phase} \\ {\mathrm{p}}_{\mathrm{B}}={\mathrm{p}}_{\mathrm{total}}{\mathrm{y}}_{\mathrm{B}}\left({\mathrm{y}}_{\mathrm{B}}=\mathrm{M}.\mathrm{F}. \mathrm{of} \text{'}\mathrm{B}\text{'} \mathrm{in} \mathrm{vapour} \mathrm{phase}\right) \\ {\mathrm{y}}_{\mathrm{B}}=\frac{450}{500}=0.9 \end{gathered}$$