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Homework Help: Chemistry, colligative properties-freezing point

  1. Dec 13, 2008 #1
    The problem statement, all variables and given/known data

    A solution is made by dissolving 250.0 g of solid potassium chromate in 1.00 kg of water. What will be the freezing point of the new solution?

    molal freezing point-depression constant of water = 1.86 degrees Celsius/molality
    molality = mol/kg

    Relevant equations

    [Delta]T(freezing point)=(Van't Hoff Factor)(molal concentration of solute particles)(molal freezing point-depression constant)

    Van't Hoff Factor = (moles of particles in solution/moles of solute dissolved)


    The attempt at a solution

    (250 g K2CrO4)(1 mol K2CrO4/194.188 K2CrO4) = 1.29 mol of K2CrO4
    (1.29 mol K2CrO4/1.00 kg H2O) = 1.29 mol/kg
    Change in freezing point = (1.29 molality)(1.86 degrees Celsius/molality)
    Change in freezing point = 0 degrees celsius - 2.40 degrees celsius
    Change in freezing point = -2.40 degrees celsius


    Comments

    Answer is -7.18 degrees celsius
    Since ionic compounds rarely dissociate completely the Van't Hoff Factor has to be used. Except I don't know how to use it...
     
    Last edited: Dec 13, 2008
  2. jcsd
  3. Dec 13, 2008 #2

    Borek

    User Avatar

    Staff: Mentor

    This is false statement. Simple ionic salts almost always dissociate 100%.

    What is van't Hoff factor for this salt (assuming 100% dissociation)?
     
  4. Dec 13, 2008 #3
    The Van't Hoff Factor isn't given, which is why I thought that I had to find it myself. But if it does dissociate completely then doesn't that mean that the Van't Hoff Factor would just be 1?
     
  5. Dec 13, 2008 #4

    Borek

    User Avatar

    Staff: Mentor

    No. Write equation of dissociation reaction and use the definition you have already posted.
     
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