Calculate the concentration of nitrogen present in the water. Assuming that the temperature is 25^oC, total pressure is 1 atm and mole fraction of nitrogen is
$0 .78. [K_{H} for nitrogen = 8 .42 \times 10^{- 7} M / mmHg]$

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4.99 M

$4 .99 \times 10^{- 2} M$

None of the above

$4.99 \times 10^{- 4} M$

answer is C.

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Detailed Solution

$p_{N_{2}} = x_{N_{2}} \times p_{total} [∵ 1 atm = 760 mmHg]$
$p_{N_{2}} = 0 .78 \times 1 atm = 0 .78 \times 760 mm = 592 .8 mm$
Concentration in solution
$\begin{array}{l} = K_{H} \times p_{N_{2}} \\ = 8 .42 \times 10^{- 7} M (mm)^{- 1} \times 592 .8 mm \\ = 4 .99 \times 10^{- 4} M \end{array}$