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The air is a mixture of a number of gases. The major components are oxygenand 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 the Henry’s law constants for oxygen and nitrogen are 3.30 × 107mm and 6.51 × 107 mm respectively, calculate the composition of these gases in water.
Asked by Vineeth Last Modified
Answer 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, 
= 1520 mm Hg
Partial pressure of nitrogen, 
= 6004 mmHg
Now, according to Henry’s law:
p = KH.x
For oxygen:
For nitrogen:
Hence, the mole fractions of oxygen and nitrogen in water are 4.61 ×10−5and 9.22 × 10−5 respectively.
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