The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportions of $20 \%$ and $79 \%$ by volume at $298 \ K$. The water is in equilibrium with air at a total pressure of $10 \ atm$. At $298 \ K$,if the Henry's law constants for oxygen and nitrogen are $3.30 \times 10^{7} \ mm \ Hg$ and $6.51 \times 10^{7} \ mm \ Hg$ respectively,calculate the mole fractions of these gases in water.

(N/A) Given:
Total pressure of air,$P_{total} = 10 \ atm = 10 \times 760 \ mm \ Hg = 7600 \ mm \ Hg$
Percentage of $O_{2} = 20 \%$,Percentage of $N_{2} = 79 \%$
Partial pressure of $O_{2}$,$P_{O_{2}} = \frac{20}{100} \times 7600 \ mm \ Hg = 1520 \ mm \ Hg$
Partial pressure of $N_{2}$,$P_{N_{2}} = \frac{79}{100} \times 7600 \ mm \ Hg = 6004 \ mm \ Hg$
According to Henry's law,$P = K_{H} \cdot x$,where $x$ is the mole fraction.
For $O_{2}$:
$x_{O_{2}} = \frac{P_{O_{2}}}{K_{H(O_{2})}} = \frac{1520 \ mm \ Hg}{3.30 \times 10^{7} \ mm \ Hg} = 4.61 \times 10^{-5}$
For $N_{2}$:
$x_{N_{2}} = \frac{P_{N_{2}}}{K_{H(N_{2})}} = \frac{6004 \ mm \ Hg}{6.51 \times 10^{7} \ mm \ Hg} = 9.22 \times 10^{-5}$
Thus,the mole fractions of $O_{2}$ and $N_{2}$ in water are $4.61 \times 10^{-5}$ and $9.22 \times 10^{-5}$ respectively.