Question

Air contains $O_2$ and $N_2$in the ratio of 1:4. Calculate the ratio of solubilities in terms of mole fraction of $O_2$ and $N_2$ dissolved in water at atmospheric pressure and room temperature at which Henry's constant for $O_2$ and $N_2$ are $3.30\times {10}^{7}torr$ and $6.60\times {10}^{7}torr$ respectively.

Answer 1

Given,

The ratio of the $O_2\&N_2=1:4$

So, $p{O}_2=K_{H}X{O}_2$

$1=K_H^{2}X{O}_2$

$X{O}_2=\frac{1}{K_H^2}$

$p{N}_2=K_{H}X{N}_2$

$1=K{H}^{N_2}X{N}_2$

$X{N}_2=\frac{1}{K_H^{N_2}}$

Thus the ratio is:

$\frac{\frac{1}{K_H^2}}{\frac{1}{K_H^{N_2}}}$

$X{O}_2:X{N}_2=\frac{1}{3.33\times {10}^7}:\frac{1}{6.6\times {10}^7}$

$1:2$