Respuesta :
The given question is incomplete. The complete question is as follows.
The value of Henry's law constant [tex]k_{H}[/tex] for oxygen in water at [tex]25^{o}C[/tex] is [tex]1.66 \times 10^{-6}[/tex] M/torr.
Calculate the solubility of oxygen in water at [tex]25^{o}C[/tex] 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.
= [tex]0.20 \times 760[/tex] = 152 torr
Therefore, solubilty (concentration) of oxygen in water will be calculated as follows.
Solubility = Henry's law constant x partial pressure of oxygen
= [tex]1.66 \times 10^{-6} M/torr \times 152 torr[/tex]
= [tex]2.52 \times 10^{-4}[/tex] M
Thus, we can conclude that solubility of given oxygen is [tex]2.52 \times 10^{-4}[/tex] M.
