Gen Chem: Solubility Homework - Dissolving 1.49 L of Gas

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To determine how much water is needed to dissolve 1.49 L of gas at 720 torr and 14°C, the Henry's law constant of 0.150 M/atm is applied. The gas concentration is calculated as 0.1421 M after converting pressure from torr to atm. The initial attempt at finding moles resulted in confusion, as the correct number of moles should be calculated using the ideal gas law (PV=NRT). After clarification, it was noted that the volume of water needed to dissolve the gas is approximately 1.48 L, but this was initially miscalculated. The discussion emphasizes the importance of correctly applying gas laws in solubility problems.
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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.
 
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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.
 
You know the molar concentration of the gas in solution, you know number of moles of the gas, just use definition of molar concentration.
 
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...
 
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.
 
PV=NRT! Thanks!
 
Ah, got it.
 
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