What is the freezing point of a solution prepared from 50.0 g ethylene glycol and 85.0 g H2O Kf of water is 1.86 Celsius?

Answer 1

50g C2H6O2(1 mol C2H6O2/62g C2H6O2)=

.81mol C2H6O2

(.81mol C2H6O2/.085kg H2O)=9.53m

Kf=1.86 C/m

delta t=(1.86 C /m)(9.53m)= 17.7 C

t=(0 C-17.7C)=-17.7 C is the freezing point of the solution.

Answer 2

What is the freezing point of a solution made by dissolving 450 g of ethylene glycol (C2H6O2) in 550 g of water? The freezing point of pure water is 0.0ºC and Kf of pure water is -1.86ºC/m.

-24.5 celcius

Answer 3

The freezing point of a water solution can be calculated from ∆T = imK where ∆T is the change in temperature; i is the van't Hoff factor (# of particles); m is the molality of the solute (moles/kg) and K is the freezing point constant for H2O (-1.86ºC/m). Thus, ∆T = (1)(27.5/62.1)/0.65(-1.86) = 1.27 degrees
So, freezing point will be - 1.3 degrees C

Answer 4

The freezing point is approx. -4 oC.

Answer 5

The freezing point is approx. -26 oC.

Answer 6

-30 degrees Celsius

Answer 7

-30 degrees C

Answer 8

-24.5ºC

Answer 9

16.3ºC.

Answer 10

32