Ncert Exercises & Solutions

2.39 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 the Henry’s law constants for oxygen and nitrogen at 298 K are 3.30 × 107 mm and 6.51 × 107 mm respectively, calculate the composition of these gases in water.

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, $P_{O_{2}} = \frac{20}{100} \times 7600 mm Hg = 1520 mm Hg$

Partial pressure of nitrogen, $P_{N_{2}} = \frac{79}{100} \times 7600 mm Hg = 6004 mm Hg$

Now, according to Henry’s law:

p = K_H.x

For oxygen:

$P_{O_{2}} = K_{H} \cdot x_{O_{2}}$
$\Rightarrow x_{O_{2}} = \frac{P_{O_{2}}}{K_{H}}$
$= \frac{1520 mm Hg}{3.30 \times 10^{7} mm Hg}$
$= 4.61 \times 10^{- 5}$

For nitrogen:

$P_{N_{2}} = K_{H} \cdot x_{N_{2}}$
$\Rightarrow x_{N_{2}} = \frac{P_{N_{2}}}{K_{H}}$
$= \frac{6004 mm Hg}{6.51 \times 10^{7} mm Hg}$
$= 9.22 \times 10^{- 5}$

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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