Question

calculate the solubility of oxygen in water at 25C when the total external pressure is 1 at and the mole fractionof oxygen in the air is 0.2

Answer 1

The given question is incomplete. The complete question is as follows.

The value of Henry's law constant $k_{H}$ for oxygen in water at $25^{o}C$ is $1.66 \times 10^{-6}$ M/torr.

Calculate the solubility of oxygen in water at $25^{o}C$ when the total external pressure is 1 atm and the mole fraction of oxygen in the air is 0.20 atm.

Explanation:

Formula to calculate partial pressure of a gas is as follows.

   Partial pressure of oxygen = mole fraction of oxygen x total pressure

Putting the given values into the above equation as follows.

       = $0.20 \times 760$ = 152 torr

Therefore, solubilty (concentration) of oxygen in water  will be calculated as follows.

        Solubility = Henry's law constant x partial pressure of oxygen

                       = $1.66 \times 10^{-6} M/torr \times 152 torr$

                       = $2.52 \times 10^{-4}$ M

Thus, we can conclude that solubility of given oxygen is $2.52 \times 10^{-4}$ M.