Henry's Law: Solving for Water Needed to Dissolve 1.48L of Gas

[LisaNels2196]

Homework Statement

How much water would be needed to completely dissolve 1.48L of the gas at a pressure of 730torrand a temperature of 28∘C? A gas has a Henry's law constant of 0.190M/atm

Homework Equations

S=kH * Pgas

The Attempt at a Solution

730 torr= .960526316 atm

S=(.190M/atm)(.960526316 atm)
S= .1825 M
.1825 mol = X = .2701 moles
1 L 1.48L

After solving that part I don't understand how to get liters of water. Maybe PV=nRT... but I tried that and I got the answer wrong.

 

[Borek]

How many moles of the gas? How many dissolved per liter of water?

 

[LisaNels2196]

.2701 moles of the gas

 

[Borek]

No. 1.48 L of gas at 730 torr and 28°C is not 0.27 moles.