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 * 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.

Step-by-step 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 * 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 * 10^(-6) M/torr * 152 torr`

=
`2.52 * 10^(-4)` M

Thus, we can conclude that solubility of given oxygen is
`2.52 * 10^(-4)` M.