Vapour pressure of pure water at $$298\,K$$  is $$23.8\,\,mm\,\,Hg.$$   $$50\,g$$  of urea is dissolved in $$850\,g$$  of water. The vapour pressure of water for this solution and its relative lowering are respectively

Solution :
$$\eqalign{ & {\text{Given,}}\,\,\,{p^ \circ } = 23.8\,\,mm\,\,Hg \cr & {w_2} = 50\,g,\,{M_2}{\text{(urea)}} = 60\,\,g\,\,mo{l^{ - 1}},{p_s} = ?,\frac{{{p^ \circ } - {p_s}}}{{{p^ \circ }}} = ? \cr & {w_1} = 850\,g,\,{M_1}{\text{(water ) = 18}}\,g\,mo{l^{ - 1}} \cr & \therefore \,\,{n_2} = \frac{{50}}{{60}} = 0.83,\,\,{n_1} = \frac{{850}}{{18}} = 47.22 \cr & {\text{Applying Raoult's 1aw,}}\,\,\frac{{{p^ \circ } - {p_s}}}{{{p^ \circ }}} = \frac{{{n_2}}}{{{n_1} + {n_2}}} \cr & {\text{or,}}\,\,\,\frac{{{p^ \circ } - {p_s}}}{{{p^ \circ }}} \cr & = \frac{{0.83}}{{47.22 + 0.83}} \cr & = \frac{{0.83}}{{48.05}} \cr & = 0.017 \cr & {\text{Thus, relative lowering of vapour pressure = 0}}{\text{.017}} \cr & {\text{Again}},\,\,\frac{{\Delta p}}{{{p^ \circ }}} = 0.017 \cr & \therefore \,\,\Delta p = 0.017 \times 23.8;\,{p^ \circ } - {p_s} = 0.017 \times 23.8 \cr & {p_s} = 23.8 - 0.017 \times 23.8;{p_s} = 23.4\,\,mm\,\,Hg \cr & {\text{Thus, vapour pressure of water in the solution}} = 23.4\,\,mm\,\,Hg \cr} $$