In just about every problem I've encountered earlier, all such questions describe the process for calculating the depression in freezing point for a solution wherein there is only one solute. However, on a take-home exam, we were asked to solve the following problem:

What is the freezing point of a solution containing $\pu{0.2 mol}$ glucose and $\pu{0.2 mol}$ $\ce{KCl}$ in $\pu{250 g}$ $\ce{H2O}$?

My thinking is to use the general formula

$$\Delta T_\mathrm{f} = imK_\mathrm{f}.$$

What I'm concerned about is how to reconcile the multiple molalities, as both glucose and $\ce{KCl}$ have the same molality, but ought we add them together? Multiply them? Just a little confused.

Also, he gave us a set of “useful equations”, but on the sheet he did not specify a van 't Hoff factor for $\ce{KCl}.$ This is confusing to me because I'm pretty sure that $\ce{KCl}$ is a strong electrolyte and ought to have one. Should we just assume $i = 2$ since there are two species $(\ce{K+}$ and $\ce{Cl-})$ we are sure to get afterwards?

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    $\begingroup$ Use your logic. You need a value with certain dimension to plug into that formula. If you multiply two molalities, what will happen to their dimension? (As for the second question, yes, it's 2.) $\endgroup$ Feb 18, 2016 at 20:17

1 Answer 1


Freezing point depression is a colligative property, well defined by this Wikipedia article (emphasis mine):

In chemistry, colligative properties are properties of solutions that depend on the ratio of the number of solute particles to the number of solvent molecules in a solution, and not on the nature of the chemical species present.

This means the problem is the same regardless of the number of different solvent species. You simply add them up to get the total solute concentration and calculate the freezing point depression in the same way you would if they were all the same species. As you mentioned, you do need to keep track of dissociating species like $\ce{KCl},$ i.e. $\pu{1 mol}$ $\ce{KCl}$ dissociates into $\pu{2 mol}$ of total solutes.


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