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Posted by on Monday, February 18, 2013 at 10:57pm.

A solution prepared by dissolving 3.00 grams of ascorbic acid (vitamin C, C6H8O6), in
50.0 grams of acetic acid has a freezing point that is depressed by T = 1.33 oC below that of
pure acetic acid. What is the value of the molar freezing point-depression constant for acetic acid?

  • chem - , Monday, February 18, 2013 at 11:49pm

    mols ascorbic acid = grams/molar mass
    m = mols/kg solvent

    Then 1.33 = Kf*m
    Substitut for m and solve for Kf.

  • chem - , Monday, February 18, 2013 at 11:52pm

    ^ but what about the i?

  • chem - , Tuesday, February 19, 2013 at 12:04am

    I assume the problem expects you to use i = 1.

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