help me.....?

Question

can anybody help me with this equation please?


What mass of ethylene glycol (C2H6O2, molecular weight 62.1) must be added to 10.0
L of water to produce a solution for use in a car’s radiator that freezes at -23.3 °C?
(The density of water is 1 kg L-1. The molal freezing point depression constant for
water, Kf, is 1.86 °C kg mol-1)

how the f*ck do you do this?? its driving me crazy! can anyone give me a formula or something? please?! im goin outta my mind!


and before i get the "do your own homework" crap, its not homework, im trying to revise and im asking for help :]

Answer 1

OK...You know all about the equation that relates freezing point depression to the molality of a solution, right? It should look something like:

DTf=Kf m (Where DTf is the freezing point depression and m is the molality of the solution).

Since the freezing point of pure water is 0 C, and you want a solution which freezes at -23.3 C, DTf = -23.3 C
So, you should be able to calculate the molality of the solution to be something like 12.5 molal. Since the definition of molality is moles of solute per kilogram of solvent, your solution will need to have 12.5 moles dissolved in each kg of water, or a total of 125 moles of your solute.

Now, just calculate the mass of 125 moles of ethylene glycol, and you're there.

If something didn't make sense here, message me...

Answer 2

for 10L of water you have 10kg
molal solutions are g solute per 1kg
1.86C/m
you need 23.3 C
23.3/1.86 = 12.52m solution
to make a 12.52m solution you neede 12.52g/1000g
but since you want 10000g of solution you need 125.2g