# Air contain O2 and N2 in the ratio of  1 : 4. Calculate the ratio of solubilities in term of mole fraction of O2 and N2 dissolved in water at atmospheric pressure and room temperature at which Henry’s constant for O2 and N2 are 3.30 × 10^7 torr and 6.60 × 10^7 torr respectively.

Air contain O2 and N2 in the ratio of  1 : 4. Calculate the ratio of solubilities in term of mole fraction of O2 and N2 dissolved in water at atmospheric pressure and room temperature at which Henry’s constant for O2 and N2 are 3.30 × 107 torr and 6.60 × 107 torr respectively.

Given Values

KH for O2 = 3.30 × 107 torr

KH for N2 = 6.60 × 107 torr

$$P_{O_{2}} : P_{N_{2}} = 1 : 4$$

As We Know:

$$P_{O_{2}} = K_{H~(O_{2})} ~.~ \chi _{O_{2}}$$

$$P_{N_{2}} = K_{H~(N_{2})} ~.~ \chi _{N_{2}}$$

Calculate

$$\frac{\chi _{O_{2}}}{\chi _{N_{2}}} = \frac{P_{O_{2}}}{K_{H~(O_{2})}} \times \frac{K_{H~(N_{2})}}{P_{N_{2}}}$$

$$\frac{\chi _{O_{2}}}{\chi _{N_{2}}} = \frac{1}{3.30 \times 10^{7}} \times \frac{6.60 \times 10^{7}}{4}$$

$$\frac{\chi _{O_{2}}}{\chi _{N_{2}}} = \frac{2}{4} = \frac{1}{2}$$

$$\chi _{O_{2}} : \chi _{N_{2}} = 1 : 2$$