# How do you calculating freezing point from molality?

Aug 7, 2017

One of the colligative properties of solutions is freezing-point depression.

This phenomenon helps explain why adding salt to an icy path melts the ice, or why seawater doesn't freeze at the normal freezing point of $0$ $\text{^"o""C}$, or why the radiator fluid in automobiles don't freeze in the winter, among other things.

The equation for freezing point depression is given by

$\underline{\overline{| \stackrel{\text{ ")(" "DeltaT_f = imK_f" }}{|}}}$

where

• $\Delta {T}_{f}$ represents the change in freezing point of the solution

• $i$ is called the van't Hoff factor, which is essentially the number of dissolved particles per unit of solute (for example, $i = 3$ for calcium chloride, because there is $1$ ${\text{Ca}}^{2 +} + 2$ ${\text{Cl}}^{-}$).

• $m$ is the molality of the solution, the number of moles of solute dissolved per kilogram of solvent:

$\text{molality" = "mol solute"/"kg solvent}$

• ${K}_{f}$ is the molal freezing-point depression constant for the solvent, which the following table lists some values for certain solvents:

(the far-right column shows the ${K}_{f}$)

Once you've calculated the change in freezing point, to find the new freezing point, you subtract the $\Delta {T}_{f}$ quantity from the normal freezing point of the solvent:

ul("new f.p." = "normal f.p." - DeltaT_f

(I'd like to point out that depending on how you're taught, the $\Delta {T}_{f}$ quantity may be negative (possibly because the constant ${K}_{f}$ was negative). Just know that the magnitude of the $\Delta {T}_{f}$ quantity (regardless of sign) represents by how much the freezing point is lowered.)