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In just about every problem both in my book and that I find online, all freezing point depression problems describe the process for calculating freezing point depression for a solution wherein there is only one solute. However, on a take-home exam (and yes, we are allowed to use online sources) I was given, we are asked to solve the problem:

"What is the freezing point of a solution of a solution of .2 moles glucose and .2 moles $\ce{KCl}$ in 250 g $\ce{H2O}$?"

My thinking is to use the general formula:

$\mathrm{\Delta}$$T_f=iK_fm$

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, or should we just assume that it's 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.) $\endgroup$ – Ivan Neretin Feb 18 '16 at 20:17
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Freezing point depression is a colligative property, well defined by this Wikipedia article:

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. (All emphasis mine).

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. 1 mol $\ce{KCl}$ dissociates into 2 mol of total solutes).

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