colligative properties

Question

OS Chapter 11 Items9. Calculate the freezing point of a solution of 400.0 g of ethylene glycol (C2H6O2)

dissolved in 600.0 g of water. A) 80.0°C

B) -5.5°C

C) 20.0°C

D) -20.0°C E) 5.5°C

Kf = 1.86°C/m and Kb = 0.512°C/m. (Show work below.)

I got 19.97 but the answer says the answer is B.

i got 19.97 but multipling the molality of the soluntion by kf and the pure water freezing point is 0 for t intial.

i just need help understanding where i went wrong. thanks!

Answer 1

ΔT = imK

ΔT = change in temp = ?

i = vant Hoff factor = 1 for ethylene glycol a non electrolyte

m = molality = mol/kg solvent = 400.0g x1 mol/62g /0.6 kg = 10.8 m

K = freezing constant = 1.86

ΔT = (1)(10.8)(1.86) = 20 degrees

New freezings point is -20 degrees C

You arent wrong. The answer sheet is incorrect. If the question asked for the freezing point DEPRESSION, the B would be the answer. But since it asks for th new freezing point, D is the answer.