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Molar Mass From Freezing Point Depression

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  1. Oct 27, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the molar mass of a nonpolar molecular compound if 5.52 grams dissolved in 36.0 grams of benzene begins to freeze at -1.87 C? The freezing point of pure benzene is 5.50 C. The freezing point depression constant is -5.12.

    2. Relevant equations

    (Change of temperature)=(FP constant)(molality)
    MM=moles/gram
    m=moles of solute/kg of solvent


    3. The attempt at a solution

    The change in temperature is 5.50-(-1.87) which is 7.37.
    7.37=5.12x --- x will equal .07. This equates to .07 moles of solute per solvent.

    I'm not sure what to do next to get the molar mass of the compound. Would I divide 5.52 by .07 moles?

    Thank you!
     
  2. jcsd
  3. Oct 27, 2009 #2
    I haven't checked your numbers, but

    Molar mass = grams/moles. Divide grams by moles.
     
  4. Oct 27, 2009 #3
    n=m/M then n/1=m/M then m=n x m
     
  5. Oct 27, 2009 #4
    I know molar mass is grams/mole. I just am not sure if it's OK to divide the 5.16 grams by the .07 moles because it's actually .07 moles per Kg. (molality).
     
  6. Oct 28, 2009 #5

    Borek

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    Staff: Mentor

    You doubts are not unfounded. You don't have 0.07 moles of substance - that you would have in 1000 g of bezene. But there is only 36 g of benzene, so obviously number of moles is much smaller.

    --
    methods
     
  7. Oct 28, 2009 #6
    OK, here's what I've done on the second try.

    The problem setup with the freezing point depression equation is 7.37=5.12x. x will equal to 1.44 which means 1.44 moles of solute per kg of solvent. I don't have that much solvent, just 36g. After changing kg to grams and then multiplying that number of moles by 36, I have .05184 moles.

    I know there is 5.52 g of solvent in the solution. That's 5.52 g per .05184 moles. That translations to 106.48 g per mole...which is the molar mass and the answer.

    Am I doing this correctly?
     
  8. Oct 28, 2009 #7

    Borek

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    Staff: Mentor

    Looks OK to me.

    Beware - you have solved the same equation twice (7.37=5.12x) getting two different results - 0.07 and 1.44. that's not good.

    --
    methods
     
  9. Oct 28, 2009 #8
    Yes I noticed that in my first post. 1.44 should be the right answer. I divided the wrong way last night. I was tired.
     
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