MadSci Network: Chemistry Query:

### Re: The freeze point depression for benzene is -5.12 C/m.

Date: Wed Apr 15 01:08:23 1998
Posted By: Jeremy Starr, Grad Student, Chemistry, California Institute of Technology
Area of science: Chemistry
ID: 892393877.Ch
Message:
```
Hi Nick,

This sounds like a homework question so I'm not going to answer it. Instead
I will write more generally about freezing point depression.

Freezing point depression is the change in freezing temperature of a
mixture as the concentrations of the components of that mixture change. It
refers specifically to the difference between the freezing point of a
dilute solution of something in a "solvent" and the freezing point of the
pure solvent. The equation which describes this behavior for a two
component system is:

dT = [(R*T^2)*M*m]/dH

dT = change in freezing point (K)
T  = freezing point of pure solvent (K)
M  = molecular weight of solvent (in Kg)
m  = molal concentration of solute in solvent (mol/Kg)
dH = heat of fusion of solvent (J)
R  = gas constant: 8.31431 (J / mol-K)

A simplified version of the same equation can be formulated by combining
all of the solvent dependent variables into one symbol (Kf):

dT = Kf*m

Kf = "freezing point depression constant" (K/molal)
Kf = 5.12 for benzene
= 1.855 for water
= 40.0 for camphor

m = molal concentration of solute in solvent (mol/Kg)

This version is easy to use if you know the freezing point depression
constant of your solvent. Just calculate the molal concentration of the
solute in the solvent then multiply it times the Kf of your solvent to get
the value of the freezing point depression of that solution.

I was careful to specify "dilute" solutions above because the equation for
freezing point depression is based on two approximations: (1) That dH (heat
of fusion) is temperature independent over small differences in freezing
temperature and (2) That dT (the actual freezing point depression) is small
compared to the freezing point of pure solvent. These approximations were
incorporated in order to reduce freezing point problem to a relatively
simple algebraic equation.

One interesting application of freezing point depression is the
determination of the molecular weight of an unknown pure substance. In the
past (>50 yrs ago) synthetic chemists relied on this method to determine
the molecular weights of new synthetic intermediates or other unknowns.
Camphor was typically used as "solvent" because it has a relatively high
freezing point depression constant, allowing more accurate measurements.
The weighed amount of unknown would be dissolved in a known amount of
camphor then the freezing point would be measured. The molality of the
solution would then be calculated from the observed freezing point
depression and used to determine how many moles of unknown had been used to
make the solution. Dividing the weight of unknown added by the calculated
number of moles gave the molecular weight of the unknown.

I hope this is enough info about freezing point depression to help you

Good luck,

Jeremy.

```

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