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3 years ago

10

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

e air is 0.2

1 answer:

pochemuha3 years ago

7 0

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.

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