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Freezing Point Problem relating to antifreeze

Submitted by antgoblue on Sun, 05/25/2008 - 20:35

Ok we are reviewing for our final coming up and I cannot remember how to do these types of problems. For example---

  How many grams of ethylene glycol, C2H4(OH)2, must be added to 500 g of water to yield a solution that will freeze at -7.7 oC?

Thanks for the help!!!!!



Thanks Spock but I still don't really get how to start the problem.

The molar mass ethlene glycol is 62.068.
I don't know what to do!!


Ok, will give you a hint.
You start by using the formula
        change in freezing pt = Kf x molality
where Kf is the freezing point depression constant (1.86 C/molal) and molality is moles of solute/kg of solvent.

You know the following:
          change in freezing point = 7.7 C
          Kf = 1.86 C/molal
          mass of solvent =  500 g = .500 kg
          mole mass of solute = 62.1 g
         
You need to find:  (in this order)
            molality
            moles of solute
            mass of solute


ok so
molaity= 7.7/1.86=4.1397
moles solute= 4.1397 m x .500kg = 2.069 moles
mass solute=2.069 moles/1molesx 62.1g = 128.333 g and its correct!!!!!!

Thanks again Spock!