How could the freezing point depression of CaCl2 be smaller than NaCl?

1 Answer
Oscar L. · P dilip_k
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 NaCl per liter, the other with 50 grams CaCl2 per liter.

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

(50 g NaCll)×(2 mol ions)58.44 g NaCl
=1.711mol ionsl

Now let's do this with the CaCl2 solution. The molecular weight for one formula unit of CaCl2 is about 40.08+(2×35.45)=110.90 g/mol, and each mole of CaCl2 makes three moles of ions in solution. So:

(50 g CaCl_2l)×(3 mol ions)110.90 g CaCl2
=1.353mol ionsl

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 CaCl2 than with NaCl. So the maximum possible freezing point depression is greater with CaCl2.

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