Chmistry
posted by Anonymous on .
28.31 mL of O2(g) at 1.00 atm pressure dissolve in 1.00 L of water at 25.0 °C. If the pressure of O2(g) is raised to 3.86 atm what will the new concentration of oxygen be expressed in mol/L

The LONG way of doing this (but you want to understand what you're doing) follows.
p = KcC with p in atm and C in M = mols/L.
We have 28.31 mL O2 in 1.00 L. That's
28.31/22,400 = ?mols/L = M = about 0.00126
Kc = p/C = 1.00 atm/0.00127 = about 790 but you need to do that more accurately.
Then p = KcC
C = p/Kc = 3.86/790 = about 0.005 M
The short way is
(28.31 mL/1.00 atm) = (x mL/3.86)
x mL = 28.31 x 3.86 = about 109 mL
and convert that to mols = about 109/22,400 = about 0.005M 
Shouldn"t it be P*V=P*V instead of V/P=V/P.

And wouldn't dividing the V by the molar constant result in moles(n) not M(mol/L)?

I don't think so.
If 28.31 dissolves at 1.00 atm, then
28.31 x (3.86/1.00) will dissolve at 3.86 atm
That's V1 x p2/p1 = v2 and that is
v1p2 = p1v2 and that is
v1/p1 = p1/p2
pv = pv won't give the same answer as the long way either. 
Two items.
1. Yes, you are correct that dividing volume by molar constant does result in moles (n) BUT that is mols in 1.00 L (from the problem) and that is M.
2. I did make an error, however, in that 22,400 is not the molar constant at 25C (298K) but at 273K. Therefore, I should have corrected for that by
22,400 x 298/273 = 24,451
Thus the molarity of that 109.3 mL will be 109.3/22,451 = 0.00447 moles and that is in 1.00 L which is 0.00447 M.(as opposed to 0.00488M using the 22,400 value at 273K).