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ttiger2k7
Feb3-09, 09:16 PM
1. The problem statement, all variables and given/known data

A gas has a Henry's law constant of 0.150 M/atm. How much water would be needed to completely dissolve 1.49 L of the gas at a pressure of 720 torr and temp of 14 C.


2. Relevant equations

m = amount solute (in mol)/mass solvent(in kg)
M = amount solute (in mol)/volume solution (in L)
Henry's Law: S_{gas} = k_{H}*P_{gas}

3. The attempt at a solution

Convert torr to atm:

S_{gas} = k_{H}*P_{gas}
S_{gas} = .150 M/atm*(720/760) atm
S_{gas} = .1421 M

.1420 \frac{mol}{L} = \frac{?}{1.49 L}

1.49 L of this gas contains .2117 moles.

That's where I get stuck. Not sure about the dissolving portion of this problem. Any help would be appreciated.

negatifzeo
Feb3-09, 11:03 PM
Im stuck on an almost identical problem. If I no one posts a solution I will go in tomorrow and get help and post the solution here.

Borek
Feb4-09, 02:57 AM
You know the molar concentration of the gas in solution, you know number of moles of the gas, just use definition of molar concentration.

negatifzeo
Feb4-09, 07:18 AM
I don't understand. It seems straightforward enough, .1421 moles of gas can be in 1 liter of water. there are .2117 moles. So about 1.48 L of water should disolve the gas. I use this approach but the answer isn't right...

Borek
Feb4-09, 07:56 AM
Approach is correct, but it looks to me like you have confused volumes.

1.49 L of gas at 14 deg C and 720 torr is NOT 0.2117 mole.

negatifzeo
Feb4-09, 07:59 AM
PV=NRT! Thanks!

ttiger2k7
Feb4-09, 10:34 AM
Ah, got it.