Solubility of gases?

Question

Using data from #39, determine the molarity of O2 in an aqueous solution at equilibrium with air at normal atmospheric pressure. The volume percent of O2 in air is 20.95%.

Data in #39: Under an O2(g) pressure of 1 atm, 28.31 mL of O2 dissolves in 1 L H2O at 25 C.

Answer 1

The key concept is that the rates of O2 molecules going into, and coming out of, water is the same at equilibrium. When the pressure of O2 drops down X times, the rate of O2 molecules coliding to the water surface, and thus going into water, would drop X times. This rate is the same as O2 molecules coming out of water. that means the concentration of O2 in the water would also drop X times.
First, let us calculate the molarity of O2 when the O2(g) pressure is 1 atm:
n = PV/RT = 0.02831*1/(0.08206*298.2) = 0.001157 (mole)
For noemal air, the result is:
0.001157mole * 20.95% = 0.0002424 mole