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Homework Help: Freezing Point Depression and Ethylene glycol

  1. Oct 18, 2007 #1
    1. The problem statement, all variables and given/known data

    Ethylene glycol, the primary ingredient in antifreeze, has the chemical formula C_2H_6O_2. The radiator fluid used in most cars is a half-and-half mixture of water and antifreeze.

    What is the freezing point of radiator fluid that is 50% antifreeze by mass? K_f for water is 1.86 degrees Celsius/m.

    2. Relevant equations

    [tex]\Delta[/tex]T_f = K_f * m

    m (molality) = # of moles of solute/mass of solution (kg)

    3. The attempt at a solution

    Okay, so I assumed 100 g of solution. And the molar mass of ethylene glycol is 62.08 g/mol. So you find the moles of ethylene glycol by multiplying 50 g by the molar mass which gives you 0.805 mol. Then you assume 100 g of water (which is 0.1 kg). And you solve for molality (m) and then you just sub it into the first equation? Is this correct?
  2. jcsd
  3. Oct 19, 2007 #2


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    Not quite. Remember this:

    do not "...assume 100 g of water (which is 0.1 kg)."
  4. Apr 7, 2010 #3
    m (molality) = moles of solute/ mass of SOLVENT (KG) "NOTTTTTTTTTT SOLUTION"
  5. Apr 8, 2010 #4


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    Staff: Mentor

    You not only necroposted in the thread that is over two years old, but you also missed the fact that chemisttree already addressed the problem.

  6. Apr 13, 2010 #5


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    Boy! Those were the good old days. eh?
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