Wednesday

November 25, 2015
Posted by **Anonymous** on Saturday, December 1, 2012 at 5:02pm.

- Chmistry -
**DrBob222**, Saturday, December 1, 2012 at 5:18pmThe 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

- Chmistry -
**Anonymous**, Saturday, December 1, 2012 at 5:45pmShouldn"t it be P*V=P*V instead of V/P=V/P.

- Chemistry -
**Anonymous**, Saturday, December 1, 2012 at 5:51pmAnd wouldn't dividing the V by the molar constant result in moles(n) not M(mol/L)?

- Chmistry -
**DrBob222**, Saturday, December 1, 2012 at 5:55pmI 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 -
**DrBob222**, Saturday, December 1, 2012 at 7:11pmTwo 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).