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Chemistry = Freezing Point & Molality

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

    You have a 10.4g mixture of sugar (C12H22O11) and table salt (NaCl). When this mixture is dissolved in 150g of water the freezing point is found to be –2.24 degrees centigrade. Calculate the percent by mass of sugar in the original mixture.

    2. Relevant equations

    delta T = (i) (Kf solvent)(m)
    Mass % = mass of substance 1 / total mass of all substances

    3. The attempt at a solution

    First I tried to find the molality of the salt
    delta T = 2.24 = 2 (1.86) (m)
    m=.602151= mols salt/kg water

    so .602151m = X mols salt/.15 kg water

    X mols of salt = .090323
    grams of salt = 5.27872 grams of salt.

    So....grams of sugar in the mixture would be 5.12128 grams sugar.

    The percent mass by sugar is then 5.12128/10.40....but that's 49.2% and our teacher (the exam key) said that the answer was 53.8%.

    Does anyone know what I am doing wrong? Thanks so much!!! =)
     
  2. jcsd
  3. Apr 27, 2009 #2

    Borek

    User Avatar

    Staff: Mentor

    Sugar doesn't change the freezing point?
     
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