Thermodynamics: Freezing Point Depression

[theow]

Homework Statement

200 g of ethylene glycol (HOCH₂CH₂OH) is dissolved in 1.00 L of water at 25°C.
The cryoscopic constant of water is 1.86 °C molal^-1.
If the solution is chilled to -25°C, how much ice (in g) will be separated out?

Homework Equations

ΔT=K_fm, where K_f is and cryoscopic constant and m is the molality of nonvolatile impurity

The Attempt at a Solution

Calculated molality = (200/40)/1.00 = 5 molal
Freezing point = -5×1.86 = -9.3°C

But I suppose since the temperature is way lower than the freezing point, all 1000 g of ice would be frozen out?

Thanks a lot:)

 

[Borek]

No, when the water freezes it is only water that is removed from the liquid phase, so concentration of glycol in solution goes up.

 

[theow]

Thanks for your reply, Borek.
But how about the mass of ice? It won't change, will it?

 

[theow]

Maybe I'll put it this way. Here's a simpler version:

20.0 g xylitol (C₅H₁₂O₅) is dissolved in 1.00 L of water at 25°C. The cryoscopic constant for water in 1.86 °C molal^-1
If the solution is chilled to -25°C, how much ice (in g) will separate out?​

I hope it's clear to you. Please help:)

 

[Borek]

I already told you what is happening. Water freezes till concentration of solute grows to the point when rest of the solution is high enough to prevent freezing. Amount of solute doesn't change, amount of solvent does.

 

[theow]

hmm... as solute concentration increases, freezing point will continue to decrease. So why would you say

concentration of solute grows to the point when rest of the solution is high enough to prevent freezing

?

 

[Borek]

Sorry, it should read

concentration of solute grows to the point when it is high enough to prevent freezing of the rest of the solution

or something like that.