The air is a mixture of a number of gases. The major components of air are oxygen and nitrogen with approximate proportion of 20% 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 330 x 10⁷ mm and 6.51 x 10⁷ mm respectively, calculate the composition of these gases in water.

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$N_{2} = 9 .22 \times 10^{- 5}; O_{2} = 4 .6 \times 10^{- 5}$

$N_{2} = 4 .8 \times 10^{- 5}; O_{2} = 2 .3 \times 10^{- 5}$

$N_{2} = 2 .3 \times 10^{- 5}; O_{2} = 4 .8 \times 10^{- 5}$

$N_{2} = 4 .6 \times 10^{- 5}; O_{2} = 9 .22 \times 10^{- 5}$

answer is C.

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Detailed Solution

$p = {xp}_{total}; p_{O_{2}} = 10 \times \frac{20}{100} = 2 atm = 2 \times 760 mm$
$\begin{array}{l} p_{N_{2}} = 10 \times \frac{79}{100} = 7 .9 atm = 7 .9 \times 760 mm \\ x_{O_{2}} = \frac{p_{O_{2}}}{K_{H}} = \frac{2 \times 760 mm}{3 .3 \times 10^{7} mm} = 4 .6 \times 10^{- 5} \\ x_{N_{2}} = \frac{p_{N_{2}}}{K_{H}} = \frac{7 .9 \times 760 mm}{6 .51 \times 10^{7} mm} = 9 .22 \times 10^{- 5} \end{array}$