The total vapour pressure of a 4 mole% solution of NH₃ in water at 293k is 50.0torr. The vapour pressure of pure water is 17.0 torr at this temperature. Applying Henry’a law and Raoult’s laws, calculate the total vapour pressure for 5mole% solution

Given, P^o(water)=17.0 torr; 

P_total(4 mole % solution) =${\mathrm{P}}_{{\mathrm{NH}}_3}+{\mathrm{P}}_{\mathrm{water} }=50.0 \mathrm{torr}$

${\mathrm{X}}_{{\mathrm{NH}}_3}= 0.04 \mathrm{and} {\mathrm{X}}_{\mathrm{water}}= 0.96$

Now according to Raoult's law;

 $$\begin{gathered} {\mathrm{P}}_{\mathrm{water}}={\mathrm{X}}_{\mathrm{water}} {\mathrm{P}}^{\mathrm{o}}\left(\mathrm{water}\right)=0.96\times 17=16.32\mathrm{torr} \\ {\mathrm{P}}_{{\mathrm{NH}}_3}=50-16.32=33.68\mathrm{torr} \\ {\mathrm{K}}_{\mathrm{H}}\left({\mathrm{NH}}_3\right)=\frac{\mathrm{P}}{\mathrm{X}}=\frac{33.68}{0.04}=842\mathrm{torr} \end{gathered}$$

For 5 mole % solution,

$$\begin{gathered} {\mathrm{X}}_{{ }_{{\mathrm{NH}}_3}}=0.05 ; {\mathrm{X}}_{\mathrm{water}}=0.95 \\ {\mathrm{P}}_{{\mathrm{NH}}_3}={\mathrm{K}}_{\mathrm{H}}\left({\mathrm{NH}}_3\right){\mathrm{X}}_{{\mathrm{NH}}_3}=842\left(0.05\right)=42.1\mathrm{torr} \\ {\mathrm{P}}_{\mathrm{water}}={\mathrm{P}}^{\mathrm{o}}\mathrm{X}=17(0.95)=16.15\mathrm{torr} \\ {\mathrm{P}}_{\mathrm{total}}=42.1+16.15=58.25 \end{gathered}$$