Air contains O₂ and N₂ in the ratio 0.2 : 0.8. If Henry law constant for O₂ and N₂ are 3.3 × 10⁷ torr and 6.6 × 10⁷ torr respectively, then the ratio of mole fractions of O₂ to N₂ dissolved in water at 1 bar pressure is

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2 :1

1 : 3

1 : 2

1 : 1

answer is C.

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Detailed Solution

$From Henry' s law, \frac{P_{O_{2}}}{P_{N_{2}}} = \frac{K_{H} (O_{2}) X_{O_{2}}}{K_{H} (N_{2}) X_{N_{2}}} \frac{X_{O_{2}}}{X_{N_{2}}} = \frac{0.2}{0.8} \times \frac{6.6 \times 10^{7}}{3.3 \times 10^{7}} = \frac{1}{2}$

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