Question

The solubility of $N_{2}$ in water at $\text{300} \, \text{K}$ and $\text{500} \, \text{torr}$ partial pressure is $0.01gL^{- 1}.$ The solubility (in $\text{g L}^{- \text{1}}$ ) at $\text{750} \, \text{torr}$ partial pressure is:

• A. $0.0075$
• B. $0.005$
• C. $0.02$
• D. $0.015$

Answer 1

Answer: (D)

$p=K_{H }X$
where $p \rightarrow $ Partial pressure of gas
$X_{2} \rightarrow $ Mole fraction of gas in solution
$K_{H}=$ Henry's law constant
So, $p \propto $ solubility
$\frac{p_{1}}{p_{2}}= \, \frac{s_{1}}{s_{2}} \, \Rightarrow \, \frac{500}{0 . 01}= \, \frac{750}{x}$
$\therefore \, x=0.015 \, g/L$