Question

If `k_f=1.86*""^(@)C*mol^-1*kg` for water, what is the freezing point of a solution composed of `255*kg` `CaCl_2`, and `33.8*L` of water?

Answer 1

Well, `DeltaT_"fusion"=k_fxxcxxi` where.................

Explanation:

Where `k_f` is the `"molal freezing point depression constant"`, which you have kindly given, `c="molal concentration"` in `mol*kg^-1`, and `i`, the so-called `"van't Hoff factor"` which reflects the speciation of an ionic solid in a solvent.

Clearly, here, `i=3`. Why? Because in solution, `"calcium chloride"` speciates to give three ions......

`CaCl_2(aq) stackrel(H_2O)rarrCa^(2+) + 2Cl^(-)`

And so...............

`Delta_"fusion"=1.86*""^(@)C*m^-1xx3xx((2.55xx10^3*g)/(110.98*g*mol^-1))/(33.8*kg)=3.80` `""^@C`.

This reflects the depression of the fusion point (the melting point) with RESPECT to PURE SOLVENT. And thus `"fusion point"` `-=` `-3.80` `""^@C`.

In practice, the melting point probably would not be depressed so far. And this is the province of measurement.