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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.

1 Answer

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 by sreemathi.v
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