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Posted by on Tuesday, January 29, 2013 at 9:22pm.

The boiling point of an aqueous solution is 102.48 °C. What is the freezing point?
I know that formula for freezing point is delta Tf = Kf*m but what do I plug in for each?

  • Chemistry - , Tuesday, January 29, 2013 at 10:11pm

    delta T = Kb*m
    Substitute and solve for m

    Then plug m and Kf into your delta T = Kf*m formula and solve for delta T, then freezing point.

  • Chemistry - , Tuesday, January 29, 2013 at 10:37pm

    Is the freezing point 372 degree C?
    Kb = 0.512
    Kf = 1.86

  • Chemistry - , Tuesday, January 29, 2013 at 11:19pm

    When you do this you should show your work. I can tell at a glance where your trouble is.
    For boiling point data:
    delta T =Kb*m
    (102.48-100) = 0.512*m
    m = about 4.84m

    Then delta T = Kf*m
    dT = 1.86*4.84
    dT = about9.00 = about 9.00 degrees lower than the normal freezing point.
    0-9.00 = -9.00 C.

  • Chemistry - , Friday, May 22, 2015 at 1:39pm

    7.62 'C

    2.1/0.512=4.10
    4.10*1.86=7.62

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