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Ratio of molar mass 2 component ideal solution

  1. Nov 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Component 1 and 2 form an ideal solution. The vapor pressure of pure component 1 is 13.3kPa at 298K, and the corresponding vapor pressure of component 2 is approximately zero. If the addition of 1.00 g of component 2 to 10.0g of component 1 reduces the total vapor pressure to 12.6 kPa, find the ratio of the molar mass of component 2 to that of component 1.

    2. Relevant equations
    x1 = p/p*
    x1 = n1/(n1+n2)
    n2=n1(1/x1-1)

    3. The attempt at a solution
    x1 = p/p* = 12.6/13.3 = 0.947
    Mass of component 1 = m1
    n1 = 10g/m1
    x1 = n1/(n1+n2)
    n1+n2= n1/x1
    n2=n1(1/x1-1)
    =10/m1(1/0.947 - 1) = 0.5597/m1
    m2=1/(0.5597/m1) = m1/0.5597
    m1/m2 = 0.5597
    Can anybody tell me if this looks right? thank you
     
  2. jcsd
  3. Nov 1, 2009 #2

    Borek

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    Staff: Mentor

    Close. Have you used rounded down 0.947 for calculations? You should not round down intermediate results.

    --
    methods
     
  4. Nov 1, 2009 #3
    If I don't round I get m1/m2=0.5556
    Is that right? Thanks!
     
  5. Nov 1, 2009 #4

    Borek

    User Avatar

    Staff: Mentor

    Watch significant digits.

    --
    methods
     
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