The vapour pressure, at a given temperature, of an ideal solution containing $$0.2$$ $$mole$$  of a non-volatile solute and $$0.8$$ $$mole$$  of solvent is $$60$$ $$mm$$  of $$Hg.$$  The vapour pressure of the pure solvent at the same temperature is

Solution :
We know that, according to Raoult’s law
$$\eqalign{ & \,\,\,\frac{{{p^ \circ } - p}}{{{p^ \circ }}} = {\chi _B} \cr & \frac{{{p^ \circ } - 60}}{{{p^ \circ }}} = \frac{{{n_B}}}{{{n_A} + {n_B}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{0.2}}{{0.2 + 0.8}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{0.2}}{{1.0}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{2}{{10}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{5} \cr & {p^ \circ } - 60 = \frac{{{p^ \circ }}}{5} \cr & \Rightarrow \,\,\,{p^ \circ } - \frac{{{p^ \circ }}}{5} = 60 \cr & \,\,\,\,\,\frac{{5{p^ \circ } - {p^ \circ }}}{5} = 60 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4{p^ \circ } = 60 \times 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{p^ \circ } = \frac{{60 \times 5}}{4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{300}}{4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 75\,mm\,{\text{of}}\,Hg \cr} $$