Why is it that I keep getting this freezing point depression problem wrong? Could someone check it?

Question

The question is as follows:

What volume of ethylene glycol (C2H6O2), a nonelectrolyte, must be added to 12.0 L of water to produce an antifreeze solution with a freezing point of -24.0°C? (The density of ethylene glycol is 1.11 g/cm3, and the density of water is 1.00 g/cm3.)

I keep getting an answer of 10.8 L, and WebAssign keeps telling me I am wrong.

Also, there is a part II to the Question, asking what the boiling point is to the solution. If anyone could help me with that, it'd be great.

I am not neccesarily looking for answers (though they'd be appreciated), just an explanation for how to find them. Thanks!

Answer 1

Kf for water is 1.858 K·kg/mol, or, 1.858°C/m.
To reduce the water freezing point 24.0°C, the C2H6O2 concentration must be: (24.0°C)/(1.858°C/m) = 12.92m.
For water, 12.0L is of 12.0kg. Hence the required C2H6O2 is: 12.92m*12.0kg = 155 mol.
The molar mass of ethylene glycol is 62.068 g/mol. Thus the required C2H6O2 is: 155 mol*62.068 g/mol = 9.62 kg.
Since the density of ethylene glycol is 1.11 g/cm3 = 1.11 kg/L, the required C2H6O2 is 9.62 kg/(1.11 kg/L) = 8.67 L.
Your answer of 10.8L is wrong.

For Part II, Kb for water is 0.52°C/m. We also know the C2H6O2 concentration must be 12.92m. Thus the boiling point elevation is: (0.52°C/m)*(12.92m) = 6.7°C.
Thus the boiling point is 106.7°C.