8. The vapour pressure of pure liquids A and B are 450 and 700 mm Hg at 350 K respectively. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that}: \\ {\mathrm{p}}_{\mathrm{A}}^{°} = 450\mathrm{mm} \mathrm{of} \mathrm{Hg} \\ {\mathrm{p}}_{\mathrm{B}}^0 = 700\mathrm{mm} \mathrm{of} \mathrm{Hg} \\ {\mathrm{p}}_{\mathrm{total}} = 600\mathrm{mm} \mathrm{of} \mathrm{Hg} \\ \mathrm{from} \mathrm{Raoult}\text{'}\mathrm{s} \mathrm{law} , \mathrm{we} \mathrm{have}: \\ {\mathrm{p}}_{\mathrm{A}} = {\mathrm{p}}_{\mathrm{A}}^{°}{\mathrm{x}}_{\mathrm{A}} \\ {\mathrm{p}}_{\mathrm{B}} ={\mathrm{p}}_{\mathrm{B}}^0 {\mathrm{x}}_{\mathbf{B}} = {\mathrm{p}}_{\mathrm{B}}^{°}(1-{\mathrm{x}}_{\mathrm{A}}) \mathrm{therefore} , \mathrm{total} \mathrm{pressure}, {\mathrm{p}}_{\mathrm{total}} = {\mathrm{p}}_{\mathrm{A}} + {\mathrm{p}}_{\mathrm{B}} \\ \Rightarrow {\mathrm{p}}_{\mathrm{total}}= {\mathrm{p}}_{\mathrm{A}}^{°}{\mathrm{x}}_{\mathrm{A}} + {\mathrm{p}}_{\mathrm{B}}^{°}(1-{\mathrm{x}}_{\mathrm{A}}) \\ \Rightarrow {\mathrm{p}}_{\mathrm{total}} = {\mathrm{p}}_{\mathrm{A}}^{°}{\mathrm{x}}_{\mathrm{A}} + {\mathrm{p}}_{\mathrm{B}}^{°}-{\mathrm{p}}_{\mathrm{B}}^0 {\mathrm{x}}_{\mathrm{A}} \\ \Rightarrow {\mathrm{p}}_{\mathrm{total}} = ( {\mathrm{p}}_{\mathrm{A}}^{°} - {\mathrm{p}}_{\mathrm{B}}^{°}){\mathrm{x}}_{\mathrm{A}} + {\mathrm{p}}_{\mathrm{B}}^{°} \\ \Rightarrow 600= (450-750){\mathrm{x}}_{\mathrm{A}} + 700 \\ \Rightarrow -100 = -250{\mathrm{x}}_{\mathrm{A}} \\ \Rightarrow {\mathrm{x}}_{\mathrm{A}} = 0.4 \\ \mathrm{therefore} , \\ {\mathrm{x}}_{\mathbf{B}} = 1-{\mathrm{x}}_{\mathrm{A}} =1-0.4 \\ =0.6 \\ \mathrm{Now}, {\mathrm{p}}_{\mathrm{A}} = {\mathrm{p}}_{\mathrm{A}}^{°}{\mathrm{x}}_{\mathrm{A}} \\ =450\times 0.4 \\ =180 \mathrm{mm} \mathrm{of} \mathrm{Hg} \\ {\mathrm{p}}_{\mathrm{B}} = {\mathrm{p}}_{\mathrm{B}}^0 {\mathrm{x}}_{\mathbf{B}} \\ = 700\times 0.6 \\ = 420\mathrm{mm} \mathrm{of} \mathrm{Hg} \\ \mathrm{Now}, \mathrm{fraction} \mathrm{of} \mathrm{liquid} \mathrm{A} = \frac{{\mathrm{p}}_{\mathrm{A}}}{{\mathrm{p}}_{\mathrm{A}+}{\mathrm{p}}_{\mathrm{B}}} \\ = \frac{180}{180+420} = \frac{180}{600} = 0.30 \end{gathered}$$

mole fraction of liquid b = 1-0.30 = 0.70