Gen Chem: Solubility Homework - Dissolving 1.49 L of Gas

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Discussion Overview

The discussion revolves around a chemistry homework problem involving the solubility of a gas in water, specifically applying Henry's law to determine how much water is needed to dissolve a given volume of gas at specified conditions of pressure and temperature. The focus is on the calculations and reasoning related to molar concentration and the application of relevant equations.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant presents the problem statement and attempts to apply Henry's law to find the solubility of the gas, converting pressure from torr to atm.
  • Another participant expresses difficulty with a similar problem and indicates a willingness to seek help if no solutions are provided in the thread.
  • A third participant suggests using the definition of molar concentration to find the mass of solvent needed based on the number of moles of gas.
  • One participant calculates that approximately 1.48 L of water should dissolve the gas based on their interpretation of the molar concentration but doubts the correctness of their answer.
  • Another participant challenges the calculation of moles, suggesting that the initial assumption about the number of moles derived from the volume of gas may be incorrect.
  • A later reply references the ideal gas law (PV=NRT) as a potential method to clarify the calculations involved.

Areas of Agreement / Disagreement

Participants express differing views on the correct interpretation of the problem and the calculations involved, indicating that there is no consensus on the solution or the approach to take.

Contextual Notes

There are unresolved issues regarding the conversion of gas volume to moles and the subsequent calculations for the amount of water needed, with some participants questioning the accuracy of the initial assumptions and calculations.

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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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