Question

The total vapour pressure of a $4$ mole solution of $N{H}_3$ in water at $293K$ is $50.0torr$, the vapour pressure of pure water is $17.0torr$ at this temperature. Applying Henry's and Raoult's laws, calculate the total vapour pressure for a $5$ mole solution.

• A. $58.25torr$
• B. $33torr$
• C. $42.1torr$
• D. $52.25torr$

Answer 1

The correct option is A $58.25torr$

The relationship between the relative lowering of vapour pressure of solvent and the molecular weight of solute and solvent is as shown below.
$\frac{\Delta P}{P}=\frac{W}{M}\times \frac{m}{w}$
4 mole% ammonia means 68 g ammonia in 1800 g water.
$\frac{17-P}{17}=\frac{68}{17}\times \frac{18}{1800}$
Vapour pressure of water $P=17-0.68=16.32$
Vapour pressure of ammonia $=50.0-16.32=33.68torr$
From henry's law, $S=KP$
Hence, $K=\frac{S}{P}=\frac{4}{33.68}$
5 mole% ammonia means 85 g ammonia in 1800 g water.
$\frac{17-P}{17}=\frac{85}{17}\times \frac{18}{1800}$
Vapour pressure of water $P=17-0.85=16.15$
From henry's law, $S=KP$
$5=\frac{4}{33.68}\times P$
Vapour pressure of ammonia $P=42.1$
Total vapour pressure $=16.15+42.1=58.25torr$