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

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To determine the amount of water needed to dissolve 1.48L of gas at 730 torr and 28°C, the Henry's law constant of 0.190 M/atm is applied. Converting 730 torr to atmospheres gives approximately 0.9605 atm. Using the equation S = kH * Pgas, the solubility (S) is calculated as 0.1825 M, which corresponds to about 0.2701 moles of gas. The discussion highlights confusion regarding the conversion from moles of gas to the volume of water required for dissolution. Clarification is needed on how to calculate the liters of water based on the moles of gas dissolved.
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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.
 
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How many moles of the gas? How many dissolved per liter of water?
 
.2701 moles of the gas
 
No. 1.48 L of gas at 730 torr and 28°C is not 0.27 moles.
 
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