Henry’s law constants at ${25}^{0}C$ for ${\mathrm{N}}_2$ and ${\mathrm{O}}_2$ in water are $4.34\times {10}^5$Torr$\left(\mathrm{g}{\mathrm{N}}_2/100\mathrm{g}{\mathrm{H}}_2\mathrm{O}\right)$and $1.93\times {10}^5$ Torr/$\left(\mathrm{g}{\mathrm{O}}_2/100\mathrm{g}{\mathrm{H}}_2\mathrm{O}\right)$, respectively . If the ${\mathrm{\rho }}_{{\mathrm{N}}_2}=608$ Torr and ${\mathrm{\rho }}_{{\mathrm{O}}_2}=152$Torr above the solution, the ratio of mass of ${\mathrm{O}}_2$ to ${\mathrm{N}}_2$ in water

We have

For  $N_2$ $602Torr=\left[4.34\times {10}^{5}Torr(g{N}_2/100g{H}_{2}O\right]m_{N_2}$

For  $O_{2}152Torr=\left[1.93\times {10}^{5}Torr(g{O}_2/100g{H}_{2}O\right]m_{O_2}$

Hence in water  $\frac{m_{O_2}}{m_{N_2}}=\frac{\left(152/1.93\times {10}^5\right)g}{\left(608/4.34\times {10}^5\right)g}$ $=\frac{7.88\times {10}^{-4}g}{1.40\times {10}^{-3}g}=0.56$

The corresponding ratio in air   $\frac{m_{O_2}}{m_{N_2}}=\frac{152Torr}{608Torr}=0.25$