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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.

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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 107 mm and 6.51 x 107 mm respectively, calculate the composition of these gases in water.

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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, pO2 = \(\frac{20}{100}\) x 7600 mm Hg

= 1520 mm Hg

Partial pressure of nitrogen, pN2 = \(\frac{79}{100}\) x 7600 mm Hg

= 6004 mmHg

Now, according to Henry's law:

p = KH.x

For oxygen:

pO2 = KH.xO2

⇒ xO2 = \(\frac{p_{O_2}}{K_H}\)

= \(\frac{1520mmHg}{3.30 \times 10^7 mm Hg}\) (Given KH = 3.30 x 107 mm Hg)

= 4.61 x 10-5

pN2 = KH.xN2

⇒ xN2 = \(\frac{p_{N_2}}{K_H}\)

= \(\frac{6004 \ mmHg}{6.51 \times 10^7 mm Hg}\)

= 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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