chemistry
posted by david on .
Under an O2 pressure of 1.00 atm, 28.31 mL of O2 dissolves in 1.00 L H2O at 25 degrees.
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 20.80 \%.
(Have no idea how to go through with this problem.)

See my response below.

Okay, so this is how I think I should set it up? can you confirm it at least. Anyways I would try to find moles of O2 using ideal gas law (1.00 atm)(0.283)/(0.0821)(298) = n moles O2.
then I would multiply 0.208 by the 1 L of water??
Then divide the moles of O2 / by L of water to get molarity? 
sorry would i multiply the moles of O2 by the 1(0.208) to get the concentration which is proportional to the molarity?

The first part, if you wish to work in literatm, appears to be ok. But I think that first part is to allow you to calculate k
partial pressure O2 = kc
So you have partial pressure O2 in atm and c in moles/L, that gets k.
Go from there. Using k from above and the new pressure, calculate moles under the new conditions with the same formula of p = kc.