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K_H of N₂ is 1x10⁵ atm, the moles of N₂ dissolved in 10 moles of water is
(given pressure of air is 5 atm, mole fraction of N₂ in air is 0.8)

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$3 \times 10^{- 2}$

$2 \times 10^{- 8}$

$4 \times 10^{- 4}$

$5 \times 10^{- 3}$

answer is A.

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Detailed Solution

$P_{N_{2}} = \underset{(N_{2})}{K_{H}} \frac{n N_{2}}{n H_{2} O}$
$\begin{array}{l} 4 = 10^{5} \times \frac{n_{N_{2}}}{N_{H_{2} O}} \\ n_{N_{2}} = 4 \times 10^{- 4} \end{array}$

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