Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Gen. Chem: Solubility

  1. Feb 3, 2009 #1
    1. The problem statement, all variables and given/known data

    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.


    2. Relevant equations

    m = amount solute (in mol)/mass solvent(in kg)
    M = amount solute (in mol)/volume solution (in L)
    Henry's Law: [tex]S_{gas} = k_{H}*P_{gas}[/tex]

    3. The attempt at a solution

    Convert torr to atm:

    [tex]S_{gas} = k_{H}*P_{gas}[/tex]
    [tex]S_{gas} = .150 M/atm*(720/760) atm[/tex]
    [tex]S_{gas} = .1421 M[/tex]

    [tex].1420 \frac{mol}{L} = \frac{?}{1.49 L}[/tex]

    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.
     
    Last edited: Feb 3, 2009
  2. jcsd
  3. Feb 3, 2009 #2
    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.
     
  4. Feb 4, 2009 #3

    Borek

    User Avatar

    Staff: Mentor

    You know the molar concentration of the gas in solution, you know number of moles of the gas, just use definition of molar concentration.
     
  5. Feb 4, 2009 #4
    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...
     
  6. Feb 4, 2009 #5

    Borek

    User Avatar

    Staff: Mentor

    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.
     
  7. Feb 4, 2009 #6
    PV=NRT! Thanks!
     
  8. Feb 4, 2009 #7
    Ah, got it.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook