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Vapor pressure of pure water

  1. Oct 18, 2005 #1
    [tex]X_1=\frac{n_1}{n_1+n_2}\implies n_2=\frac{n_1-X_1n_1}{X_1}[/tex]

    where n1 is the moles of solvent and n2 is the moles of solute.

    [tex]22.98=X_1\left(23.76\right)\implies X_1=.9672[/tex]

    and 1000 g of water is equal to 55.49 mol (n1), so plugging this all in gives:

    [tex]n_2=1.88\text{mol}[/tex]

    which would be the same as the molarity.

    However, my textbook says that the molarity is .920 m. Where did I go wrong? Thanks a lot.
     
  2. jcsd
  3. Oct 19, 2005 #2

    Borek

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    Staff: Mentor

    NaCl is dissolved.
     
  4. Oct 23, 2005 #3
    Right, Na++Cl-->NaCl.

    Thanks for the help.
     
  5. Sep 30, 2009 #4
    Re: Molality

    Can someone please explain this to me, because I'm not quite sure how to solve this problem. I only reached the part where I got the mole fraction of water and NaCl, but that's just it. I don't know what to do next. Help would be greatly appreciated. :(
     
  6. Sep 30, 2009 #5

    Borek

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    Staff: Mentor

    Re: Molality

    You mean you have no idea how to convert molar fraction to molality?

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  7. Oct 1, 2009 #6
    Re: Molality

    Okay, I found out how to convert it molality, but I keep getting 1.88 m not .920 m. How does the strong electrolyte/complete ionization of NaCl make a difference?
     
    Last edited: Oct 1, 2009
  8. Oct 1, 2009 #7

    Borek

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    Staff: Mentor

    Re: Molality

    Check what Van't Hoff factor is.

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