Freezing point depression. Help!

[blackhole123]

How many liters of antifreeze ethylene glycol (C2H6O2, density 1.11 g/ml) would you add to a car radiator containing 6.5L of water if the coldest winter temperature in your area is -20C?

 

So I figured I can use the freezing point depression equation to figure out what the molality of the resulting solution must be if the freezing point of water is depressed 20 degrees C. I got 10.75m. Now I don't know how to figure out how much of the solution to add to the 6.5L of water to make it this molality.

 

Some sort of ratio? I'm lost.

[insane_alien]

well the molality is the mol per kilogram. you have 6.5L so thats roughly 10.75mol(and a bit) ethylene glycol is 62g/mol so over 620g of glycol.

 

if i had a calculator i'd have done it properly. probably nearer 700g

[blackhole123]

Don't you have to take into account the added volume of the solution of the antifreeze? You aren't adding pure ethylene glycol, you are adding a solution of it, so if you act calculate it without taking that into account your calculated value will be lower than what you actually need to add. I believe the antifreeze comes out to be 17.833m using the density.

[insane_alien]

mm forgot about that, should be easy enough to tack on for someone who isn't doped up at the moment(me).

[blackhole123]

Right, thats my problem, I can't figure out how to tack that on. I'm sure its not difficult once I get it, I just don't know how to get it.