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Mole Fraction problem

  1. Sep 24, 2008 #1
    this problem should be simple, but i must be doing it wrong.

    At 250 degrees Celsius, a certain liquid X has a vapor pressure of 900 mm Hg and a second liquid Y has a pressure of 600 mm Hg. What is the minimum mole fraction of X necessary for the mixture to boil at this temperature (assume atmosphere pressure is 1 atm).

    MY WORK:

    Since I know that the total pressure at this temperature is 1500 mm Hg, and only 760 mm Hg is needed, then only about 51% of the pressures are needed. But when I do (51%)(900) and (49%)(600), it doesn't add up to 760 mm Hg.

    When I do guess and check, i get about 53.3% of compound X. How do you do this problem!
    Thank you.
  2. jcsd
  3. Sep 25, 2008 #2


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    Staff: Mentor

    What is expression that describes total vapor pressure over the mixture?

    Hint: total pressure is not 1500 mmHg.
  4. Sep 25, 2008 #3
    I think I got it, but I just want to confirm the answer.

    Ptot= (Xa)Px + (1-Xa)Py.
    760 = (Xa)900 + 600 - (Xa)600
    760 - 600 = 160 = 300(Xa)

    Xa = 160/300 = 53.33%?
  5. Sep 25, 2008 #4


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    Staff: Mentor

    Looks OK.
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