Question

What is the density of wet air (in $\text{g/L}$) that has the partial pressure of water of $22.5$ torr at $1$ atm and $300\text{K}$?
$($Given: vapour pressure of $H_{2}O$ is $30$ torr and the average molar mass of air is $29\text{g/mol})$

• A. $1.164\text{g/L}$
• B. $2.164\text{g/L}$
• C. $3.164\text{g/L}$
• D. $4.164\text{g/L}$

Answer 1

The correct option is A $1.164\text{g/L}$

According to Dalton's law, $\frac{P_1}{P_2}=\frac{n_1}{n_2}$
where P and n are the partial pressure and the moles of the gases present respectively,
% mole of $H_{2}O$ vapour in air $=\frac{\left(22.5\right)}{760}\times 100=2.96$
molar mass of wet air $=\frac{29\times 97.04+2.96\times 18}{100}$
$=\frac{2814.16+53.28}{100}=28.67$
density of wet air $=\frac{PM}{RT}=\frac{1\times 28.67}{0.0827\times 300}=1.164\text{g/L}$