When 18 g of ethylene glycol C2H6O2 is dissolved in 150 g of pure water, the freezing point of the solution is

Question

_C . (The freezing point depression constant for water is 1.86C kg)

what i tried was 18/62 to find the moles. then i got .29/.150 to solve for the molality. then i got (.9350)(1.86) to solve for the temperature which is 3.6. i keep getting it wrong im i doing it right? please help. thanks

Answer 1

delta T = Kb*m
m = 18/62 = 0.29 mols
molality = 0.29/0.15 = 1.93 m
delta T = 1.86 x 1.93 = 3.6 degrees.
The normal freezing point is zero, so the new freezing point must be 0.0 - 3.6 = -3.6. I suspect that's your problem. You are stopping at delta T which IS 3.6 but the freezing point is -3.6 C.

Answer 2

thanks sooo much