The air is a mixture of a number of gases.

Solution:

Percentage of oxygen (O2) in air = 20 %

Percentage of nitrogen (N2) in air = 79%

Also, it is given that water is in equilibrium with air at a total pressure of 10 atm, that is, (10 × 760) mm Hg = 7600 mm Hg

Therefore,

Partial pressure of oxygen, $p_{\mathrm{O}_{2}}=\frac{20}{100} \times 7600 \mathrm{~mm} \mathrm{Hg}$

= 1520 mm Hg

Partial pressure of nitrogen, $p_{\mathrm{N}_{2}}=\frac{79}{100} \times 7600 \mathrm{mmHg}$

= 6004 mmHg

Now, according to Henry’s law:

p = KH.x

For oxygen:

$p_{\mathrm{O}_{2}}=K_{\mathrm{H}} \cdot x_{\mathrm{O}_{2}}$

$\Rightarrow x_{\mathrm{O}_{2}}=\frac{p_{\mathrm{O}_{2}}}{K_{\mathrm{H}}}$

$=\frac{1520 \mathrm{~mm} \mathrm{Hg}}{3.30 \times 10^{7} \mathrm{~mm} \mathrm{Hg}} \quad\left(\right.$ Given $\left.K_{\mathrm{H}}=3.30 \times 10^{7} \mathrm{~mm} \mathrm{Hg}\right)$

$=4.61 \times 10^{-5}$

For nitrogen:

$p_{\mathrm{N}_{2}}=K_{\mathrm{H}} \cdot x_{\mathrm{N}_{2}}$

$\Rightarrow x_{\mathrm{N}_{2}}=\frac{p_{\mathrm{N}_{2}}}{K_{\mathrm{H}}}$

$=\frac{6004 \mathrm{~mm} \mathrm{Hg}}{6.51 \times 10^{7} \mathrm{~mm} \mathrm{Hg}}$

$=9.22 \times 10^{-5}$

Hence, the mole fractions of oxygen and nitrogen in water are 4.61 ×10−5and 9.22 × 10−5 respectively.