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Molality Question

  1. Aug 27, 2006 #1

    How many grams of antifreeze must be dissolved in 300 g of water in order to reduce the freezing point of water to –13.02o C under 1 atm pressure (Kf: 1.86 o C/molal and C3H5(OH)3: 92.09 g/mol).

    I have not learned molality yet and I am completely confused as to where to start. I only know that molality = moles of solute / solvent. If someone could please take a look at this question and perhaps explain to me the steps to solving it, I would really appreciate it. Thanks.
  2. jcsd
  3. Aug 27, 2006 #2


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    Molality (m) = Moles of solute / kg of solvent

    To find the freezing point depression, use this formula,
    (delta T) = k_f * m
    where (delta T) is the change in the solution's freezing point from the pure solution, k_f is the freezing point depression constant, and m is the concentration of the solution in terms of molality.
    From here is just becomes a "plug n' chug" type of problem.

    Molality is just another way of representing the concentration of a solution, just like molarity, mass %, mole fraction, ....

    The freezing point of an aqeous anti-freeze solution will be lower than the freezing point of pure water. The amount the freezing point will drop depends on the concentration of the solution and the freezing point depression constant.

    If you know you want to lower the freezing point by 13.02 degrees C (0 degrees - -13.02 degrees C), simply plug this into the equation as (delta T) and solve for the concentration since you know the k_f value.

    molality = (delta T) / k_f
    Once you find the concentration, use the unit of molality to help you calculate the number of moles of ethylene glycol, and from that, the number of grams.
  4. Aug 28, 2006 #3


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