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# If $N_2$ gas is bubbled through water at 293 K, how many millimoles of $N_2$ gas would dissolve in 1 litre of water. Assume that $N_2$ exerts a partial pressure of 0.987 bar. Given that Henry’s law constant for $N_2$ at 293 K is 76.84 Kbar.

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The solubility of gas is related to its mole fraction in the aqueous solution. The mole fraction of the gas in the solution is calculated by applying Henry’s law.
Thus, $X_{N_2}=\large\frac{P_{N_2}}{K_{H}}=\frac{0.987\;bar}{76480\;bar}$$=1.29\times 10^{-5} As 1 litre water contains 55.5 mol of it, therefore, if n represents number of moles of N_2 in solution, X_{N_2}=\large\frac{n\;mol}{n\;mol+55.5\;mol}=\frac{n}{55.5}$$=1.29\times 10^{-5}$
Thus, $n = 1.29 \times 10^{−5} \times 55.5 mol = 7.16 \times 10^{−4 }$mol
$\Rightarrow 7.16\times 10^{-4}mol\times \large\frac{1000m\;mol}{1\;mol}$$=0.716m\;mol$
Hence (A) is the correct answer.
answered Jun 9, 2014