Henry's law constants at 25 °C for N₂ and O₂ in water are $4.34\times {10}^5\mathrm{Torr}/\left({\mathrm{gN}}_2/100 \mathrm{g} {\mathrm{H}}_2\mathrm{O}\right)$$1.93\times {10}^5\text{ Torr }/\left({\mathrm{gO}}_2/100 \mathrm{g} {\mathrm{H}}_2\mathrm{O}\right)$ respectively. If the $p_{{\mathrm{N}}_2}=608$ Torr and $p_{{\mathrm{O}}_2}=152$Torr above the solution, the ratio of mass of O₂ to N₂ in water

We have

For ${\mathrm{N}}_{2}608\text{ Torr }=\left[4.34\times {10}^5\text{ Torr }/\left({\mathrm{gN}}_2/100{\mathrm{gH}}_2\mathrm{O}\right)\right]m_{{\mathrm{N}}_2}$

For ${\mathrm{O}}_{2}152\text{ Torr }=\left[1.93\times {10}^5\mathrm{Torr}/\left({\mathrm{gO}}_2/100{\mathrm{gH}}_2\mathrm{O}\right]m_{{\mathrm{O}}_2}\right.$ 

Hence, m water$\frac{m_{{\mathrm{O}}_2}}{m_{{\mathrm{N}}_2}}=\frac{\left(152/1.93\times {10}^5\right)\mathrm{g}}{\left(608/4.34\times {10}^5\right)\mathrm{g}}=\frac{7.88\times {10}^{-4}\mathrm{g}}{1.40\times {10}^{-3}\mathrm{g}}=0.56$

The corresponding ratio in air $\frac{m_{{\mathrm{O}}_2}}{m_{{\mathrm{N}}_2}}=\frac{152\text{ Torr }}{608\text{ Torr }}=0.25$