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

Phase equilibrium

  1. Jul 17, 2005 #1
    The vapour pressure of a pure liquid solvent B is 75 kPa. When an unknown mass of non volatile solute W was added to 0.2 moles of solvent B,its vapour pressure drops by 20 kPa. If the relative molecular mass of W is 118,how much of W was added to B?

    A 5.07g
    B 8.58g
    C 23.44g
    D 64.9g

    i will show my calculation here,but i don't think its correct because my answer wrong .please state out my mistake.

    P(B) = X(B) x P(pure B)
    20 = (0.2/W+0.2) x 75
    W =0.55

    mass of W = 0.55x118
    = 64.90g
    the actual answer is B,i think my answer is unreasonable too,cause the mass is too big,but from my calculation,i get 64.9g.why??????
     
  2. jcsd
  3. Jul 17, 2005 #2

    GCT

    User Avatar
    Science Advisor
    Homework Helper

    wrong value for the vapor pressure, it has dropped by 20kPa
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Phase equilibrium
  1. Phase rule (Replies: 3)

  2. Gas Equilibriums? (Replies: 5)

  3. Time to equilibrium (Replies: 6)

  4. Chemical equilibrium (Replies: 4)

  5. Equilibrium Constants (Replies: 4)

Loading...