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

1 Answer
May 7, 2016

Answer:

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#.