The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K.
The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if Henry's law constants for oxygen and nitrogen are 3.30 x 10^{7} mm and 6.51 x 10^{7} mm respectively, calculate the composition of these gases in water.
Percentage of oxygen (O_{2}) in air = 20%
Percentage of nitrogen (N_{2}) 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_{O2} = \(\frac{20}{100}\) x 7600 mm Hg
= 1520 mm Hg
Partial pressure of nitrogen, p_{N2} = \(\frac{79}{100}\) x 7600 mm Hg
= 6004 mmHg
Now, according to Henry's law:
p = K_{H}.x
For oxygen:
p_{O2} = K_{H}.x_{O2}
⇒ x_{O2} = \(\frac{p_{O_2}}{K_H}\)
= \(\frac{1520mmHg}{3.30 \times 10^7 mm Hg}\) (Given K_{H} = 3.30 x 10^{7} mm Hg)
= 4.61 x 10^{5}
p_{N2} = K_{H}.x_{N2}
⇒ x_{N2} = \(\frac{p_{N_2}}{K_H}\)
= \(\frac{6004 \ mmHg}{6.51 \times 10^7 mm Hg}\)
= 9.22 x 10^{5}
Hence, the mole fractions of oxygen and nitrogen in water are 4.61 × 10^{5} and 9.22 × 10^{5} respectively.

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