How much water to dissolve 1.49 L of gas using Henry's law?

[ttiger2k7]

Homework Statement

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.

Homework Equations

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

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]

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]

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]

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]

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]

PV=NRT! Thanks!

 

[ttiger2k7]

Ah, got it.