The total vapour pressure of a $$4\,mole\,\% $$   solution of $$N{H_3}$$  in water at $$293\,K$$  is $$50.0\,torr.$$  The vapor pressure of pure water is $$17.0\,torr$$   at this temperature. Applying Henry’s and Raoult’s laws, the total vapour pressure for a $$5\,mole\,\% $$   solution is

Solution :
$$\eqalign{ & {\text{The given data are}} \cr & {P_{water}} = 17.0\,torr; \cr & {P_{total}}\left( {4\,mole\,\% \,{\text{solution}}} \right) \cr & = {P_{N{H_3}}} + {P_{water}} = 50.0\,torr \cr & {X_{N{H_3}}} = 0.04\,\,{\text{and}}\,\,{X_{water}} = 0.96 \cr & {\text{Now according to Raoult's}}\,{\text{law;}} \cr & {P_{water}} = {X_{water}}\,P_{water}^ \circ \cr & = 0.96 \times 17.0\,torr = 16.32\,torr \cr & {\text{Now Henry's}}\,{\text{law constant for ammonia is}} \cr & {K_H}\left( {N{H_3}} \right) = \frac{{{P_{N{H_3}}}}}{{{X_{N{H_3}}}}} = \frac{{33.68\,torr}}{{0.04}} = 842\,torr \cr & {\text{Hence, for}}\,\,5\,mole\,\% \,\,{\text{solution, we have}} \cr & {P_{N{H_3}}} = {K_H}\left( {N{H_3}} \right){X_{N{H_3}}} \cr & = \left( {842\,torr} \right)\left( {0.05} \right) = 16.15\,torr \cr & {\text{Thus,}}\,{P_{total}}\left( {5\,mole\,\% \,{\text{solution}}} \right) \cr & = {P_{N{H_3}}} + {P_{water}} \cr & = 42.1 + 16.15 \cr & = 58.25\,torr \cr} $$