Two volatile liquids ‘A’ and ‘B’ mixed in the mole ratio 1 : 1 form an ideal solution. Vapour pressure of pure liquid ‘A’ is 200 mm Hg and pure liquid ‘B’ is 600 mm Hg. Mole fraction of ‘B’ in vapour phase will be 

Given

$$\begin{gathered} {\mathrm{n}}_{\mathrm{A}}:{\mathrm{n}}_{\mathrm{B}}=1 : 1 \\ {\mathrm{X}}_{\mathrm{A}}=\frac{1}{2}; \\ {\mathrm{X}}_{\mathrm{B}}=\frac{1}{2}; \\ \mathrm{Applying} \mathrm{Raoult}\text{'}\mathrm{s} \mathrm{law} \\ {\mathrm{p}}_{\mathrm{A}}={\mathrm{p}}_{\mathrm{A}}^0{\mathrm{X}}_{\mathrm{A}}; \\ {\mathrm{p}}_{\mathrm{B}}={\mathrm{p}}_{\mathrm{B}}^0{\mathrm{X}}_{\mathrm{B}}; \\ {\mathrm{p}}_{\mathrm{A}}=200\left(\frac{1}{2}\right)=100 \mathrm{mm} \mathrm{Hg} \\ {\mathrm{p}}_{\mathrm{B}}=600\left(\frac{1}{2}\right)=300 \mathrm{mm} \mathrm{Hg} \\ {\mathrm{p}}_{\mathrm{total}}={\mathrm{p}}_{\mathrm{A}}+{\mathrm{p}}_{\mathrm{B}}=400 \mathrm{mm} \mathrm{Hg} \\ \mathrm{Applying} \mathrm{Dalton}\text{'}\mathrm{s} \mathrm{law} \mathrm{in} \mathrm{vapour} \mathrm{phase} \\ {\mathrm{p}}_{\mathrm{B}}={\mathrm{p}}_{\mathrm{total}}{\mathrm{y}}_{\mathrm{B}} \left({\mathrm{y}}_{\mathrm{B}}=\mathrm{M}.\mathrm{F}. \mathrm{of} \text{'}\mathrm{B}\text{'} \mathrm{in} \mathrm{vapour} \mathrm{phase}\right) \\ {\mathrm{y}}_{\mathrm{B}}=\frac{300}{400}=0.75 \end{gathered}$$