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.

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

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.

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

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.