Question

Ratio of $O _{2}$ and $N _{2}$ in the air is $1: 4$. Find out the ratio of their solubilities in terms of mole fractions of $O _{2}$ and $N _{2}$ dissolved in water at atmospheric pressure and room temperature.
$\begin{bmatrix} K_{ H }\left( O _{2}\right)=3.30 \times 10^{7} \text { torr } \\ K_{ H }\left( N _{2}\right)=6.60 \times 10^{7} \text { torr } \end{bmatrix}$

• A. $1: 2$
• B. $2: 1$
• C. $1: 1$
• D. None of these

Answer 1

Answer: (A)

$p_{ O _{2}} =\frac{1}{5} \times 1=0.2\, bar$
$p_{ N _{2}} =\frac{4}{5} \times 1=0.8\, bar$
$p =K_{ H } x$ (from Henry's law)
$\therefore \frac{x_{ O _{2}}}{x_{ N _{2}}} =\frac{p_{ O _{2}}}{K_{ H }\left( O _{2}\right)} \times \frac{K_{( H )}\left( N _{2}\right)}{p_{ N _{2}}}$
$=\frac{0.2 \times 6.60 \times 10^{7}}{0.8 \times 3.30 \times 10^{7}}=\frac{1}{2}$
$x_{ O _{2}}: x_{ N _{2}}=1: 2$