Question

The solubility of ${\text{O}}_{\text{2}}\left(\text{g}\right)$ in water exposed to the amosphere when its partial pressure is 506.66 mm is $\text{6}\times \text{1}{\text{0}}^{-4}\text{M}$. Its solubility at 760 mm and at the same temperature is

• A. $4.1\times {10}^{-4}\text{M}$
• B. $6.8\times {10}^{-4}\text{M}$
• C. $9\times {10}^{-4}\text{M}$
• D. $10.8\times {10}^{-4}\text{M}$

Answer 1

The correct option is C $9\times {10}^{-4}\text{M}$

According to Henery's law, the solubility of a gas is directly proportional to its partial pressure at constant temperature.
${\text{K}}_n=\frac{\text{P}}{\text{C}}$
Where, ${\text{K}}_n$ is Henery's constant P is the portial pressure of gas.
C is the solubility of the gas.
Now, $\frac{{\text{P}}_1}{{\text{C}}_1}=\frac{{\text{P}}_2}{{\text{C}}_2}$ or $\frac{{\text{P}}_1}{{\text{P}}_2}=\frac{{\text{C}}_1}{{\text{C}}_2}$
The solubility of ${\text{O}}_2\left(g\right)$ in water is $6\times {10}^{-4}\text{m}$ when partial pressure is 506.66 mm.
The solubility at 760 mm is,
$\frac{{\text{P}}_1}{{\text{P}}_2}=\frac{{\text{C}}_1}{{\text{C}}_2}$
$\frac{506.66}{760}=\frac{6\times {10}^{-4}}{{\text{C}}_2}$
${\text{C}}_2=\frac{6\times {10}^{-4}\times 760}{506.66}=9\times {10}^{-4}\text{m}$
Thus option (C) is the correct option.