Question

The major components of air are $O_2$ and $N_2$ with approximate proportion of 21% and 79% by volume at $298K$. The water is equilibrium with air at a pressure of $10atm$. At $298K$, $K_H\left(N_2\right)=6.51\times {10}^{7}mm$
What will be the moles of $N_2$ in $1kg{H}_{2}O?$

• A. (i) - r; (ii) - p; (iii) - s; (iv) - q
• B. (i) - q; (ii) - p; (iii) - r; (iv) - s
• C. (i) - p; (ii) - q; (iii) - s; (iv) - r
• D. (i) - s; (ii) - r; (iii) - q; (iv) - p

Answer 1

The correct option is A (i) - r; (ii) - p; (iii) - s; (iv) - q

$K_H$ is in pressure unit.
Since $N_2$ is 79% by volume. According to dalton's law:
$p_{N_2}=(0.79\times 10)atm=(10\times 0.79\times 760)mm$
According to Henry's Law:

$p=K_H\chi$
${\chi }_{N_2}=\frac{p_{N_2}}{K_H}=\frac{10\times 0.79\times 760mm}{6.51\times {10}^{7}mm}=9.22\times {10}^{-5}$

Also, $n_{\left(H_{2}O\right)}=\frac{1000}{18}=55.55$
${\chi }_{N_2}=\frac{n_{N_2}}{n_{N_2+n_{H_{2}O}}}\approx \frac{n_{N_2}}{n_{H_{2}O}}(n_{N_2}\text{being very small)}$
$\therefore n_{N_2}={\chi }_{N_2}\times n_{H_{2}O}=9.22\times {10}^{-5}\times 55.55=5.117\times {10}^{-3}$