Question

How could the freezing point depression of `CaCl_2` be smaller than `NaCl`?

Answer 1

It depends on whether you are calculating on a molar or mass basis. Calcium chloride gives more freezing point depression per mole than sodium chloride, but less freezing point depression per gram.

Explanation:

Let's see how the mass basis is different. Say you have two solutions, one of them with 50 grams `"NaCl"` per liter, the other with 50 grams `CaCl"_2` per liter.

For the `"NaCl"` solution: The molecular weight for one formula unit is about `22.99+35.45=58.44 "g/mol"`. Divide that into 50 grams and remember that each mole of `"NaCl"` dissociates to make two moles of ions, thus:

`{({50" g NaCl"}/"l")\times(2" mol ions")}/{58.44 " g NaCl"}`
`=1.711 {"mol ions"}/"l"`

Now let's do this with the `"CaCl"_2` solution. The molecular weight for one formula unit of `"CaCl"_2` is about `40.08+(2\times35.45)=110.90" g/mol"`, and each mole of `"CaCl"_2` makes three moles of ions in solution. So:

`{({50" g CaCl_2"}/"l")\times(3" mol ions")}/(110.90 " g CaCl"_2)`
`=1.353 {"mol ions"}/"l"`

The calcium chloride solution, with the same number of grams per liter, makes fewer ions in solution (`1.353" mol ions/l vs. 1.711 mol ions/l"`). So the calcium chloride solution has less freezing point depression at the same mass concentration.

However, you can get a lot more mass concentration into solution with `"CaCl"_2` than with `"NaCl"`. So the maximum possible freezing point depression is greater with `"CaCl"_2`.