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

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Discussion Overview

The discussion revolves around calculating the amount of water needed to dissolve 1.48L of a gas using Henry's Law, considering specific conditions of pressure and temperature. Participants are exploring the application of the law and the necessary calculations involved.

Discussion Character

  • Homework-related, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant presents a calculation using Henry's Law to find the solubility of the gas, converting pressure from torr to atm and applying the constant.
  • Another participant questions the calculated number of moles derived from the initial gas volume and conditions, suggesting that the result may be incorrect.
  • There is a mention of using the ideal gas law (PV=nRT) to further the calculations, but uncertainty exists regarding its application in this context.

Areas of Agreement / Disagreement

Participants do not appear to agree on the correctness of the calculated moles of gas, indicating a disagreement on the interpretation or application of the calculations.

Contextual Notes

There are unresolved aspects regarding the assumptions made in the calculations, particularly in converting gas volume to moles and the application of the ideal gas law in this scenario.

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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.
 
Physics news on Phys.org
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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