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

May 7, 2016

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 $\text{NaCl}$ per liter, the other with 50 grams CaCl"_2 per liter.

For the $\text{NaCl}$ solution: The molecular weight for one formula unit is about $22.99 + 35.45 = 58.44 \text{g/mol}$. Divide that into 50 grams and remember that each mole of $\text{NaCl}$ 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 ${\text{CaCl}}_{2}$ solution. The molecular weight for one formula unit of ${\text{CaCl}}_{2}$ is about $40.08 + \left(2 \setminus \times 35.45\right) = 110.90 \text{ g/mol}$, and each mole of ${\text{CaCl}}_{2}$ 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 ($1.353 \text{ mol ions/l vs. 1.711 mol ions/l}$). 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 ${\text{CaCl}}_{2}$ than with $\text{NaCl}$. So the maximum possible freezing point depression is greater with ${\text{CaCl}}_{2}$.