# Chmistry

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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

• Chmistry - ,

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

• Chmistry - ,

Shouldn"t it be P*V=P*V instead of V/P=V/P.

• Chemistry - ,

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

• Chmistry - ,

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.

• Chmistry - ,

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).