Question

Henry’s law constant for the solubility of $N_2$ gas in water at 298K is $1.0\times {10}^5$ atm. The molefraction of $N_2$ in air is 0.6. The number of moles of $N_2$ from air dissolved in 10 moles of water at 298K and 5 atm pressure is

• A. $3.0\times {10}^{-4}$
• B. $4.0\times {10}^{-5}$
• C. $5.0\times {10}^{-4}$
• D. $6.0\times {10}^{-6}$

Answer 1

The correct option is A $3.0\times {10}^{-4}$

Partial pressure of $N_2$ in air $\left(P_{N_2}\right)=P_{Total}\times X_{N_2}$ (in air)
= 5 $\times$ 0.6
$P_{N_2}\left(inair\right)=K_H\times X_{N_2}\left(in{H}_{2}O\right)$
$5\times 0.6=1\times {10}^5\times X_{N_2}\left(in{H}_{2}O\right)$
$X_{N_2}$ in 10 moles of water $=\frac{5\times 0.6}{1\times {10}^5}=3\times {10}^{-5}$
$X_{N_2}=\frac{n{N}_2}{n{N}_2+n_{H_2}o}$
$3\times {10}^{-5}=\frac{n{N}_2}{n{N}_2+10}\Rightarrow n{N}_2\times 3\times {10}^{-5}+3\times {10}^{-5}\times 10=n_2$
$3\times {10}^{-4}=n{N}_2(1-3\times {10}^{-5})$
$n{N}_2=3\times {10}^{-4}$