Question

At $ 20^{\circ}C $ the solubility of $ N_2 $ gas in water is $ 0.015 \,g/L $ when the partial pressure of $ N_2 $ is $ 580 $ torr. What is the solubility of $ N_2 $ in $ H_2O $ at $ 20^{\circ}C $ when its partial pressure is $ 800 $ torr?

• A. 0.207 g/L
• B. 0.0207 g/L
• C. 0.414 g/L
• D. 0.0414 g/L

Answer 1

Answer: (B)

At $20^{\circ}$ solubility of $N_2$ in water is $0.015 \,g/L$.
When partial pressure is $580$ torr.
According to Henry’s law
$m \propto p$
$ m = K_H p$
$0.015 = K_H \times 580$
$K_H = \frac{580}{0.015}$
For the solubility of $N_2$ in at $20^{\circ}C$
$ = \frac{800}{K_H} = \frac{800}{580} \times 0.015$
$ = 0.0207\, g/L$