** change in freezing point (ΔT_{f})** is given by:

$\overline{){{\mathbf{\Delta T}}}_{{\mathbf{f}}}{\mathbf{=}}{{\mathbf{T}}}_{\mathbf{f}\mathbf{,}\mathbf{}\mathbf{pure}\mathbf{}\mathbf{solvent}}{\mathbf{-}}{{\mathbf{T}}}_{\mathbf{f}\mathbf{,}\mathbf{}\mathbf{solution}}}$

The ** change in freezing point** is also related to the molality of the solution:

$\overline{){{\mathbf{\Delta T}}}_{{\mathbf{f}}}{\mathbf{=}}{{\mathbf{imK}}}_{{\mathbf{f}}}}$

where:

**i** = van’t Hoff factor

**m** = molality of the solution (in m or mol/kg)

**K _{f}** = freezing point depression constant (in ˚C/m) of the solvent

The problem states that we have aqueous solutions which means the solvent is water. This means the value of K_{f} is the same for all the given solutions.

Therefore, ΔT_{f} will just depend on the osmolality of each solution, given by:

$\overline{){\mathbf{Osmolality}}{\mathbf{=}}{\mathbf{i}}{\mathbf{\times}}{\mathbf{m}}}$

**A ****higher osmolality or concentration will result in a lower freezing point****.**

**I. 0.40 m ****C _{2}H_{6}O**

C_{2}H_{6}O_{2} is a *non-electrolyte*, which means **i = 1**.

Solving for osmolality:

**Osmolality = (1)(0.40) = 0.40**

**II. 0.20 m Na _{3}PO_{4}**

Na_{3}PO_{4} is an *electrolyte*, dissociating into 3 Na^{+} and PO_{4}^{3}^{–}, which means **i = 4**.

Solving for osmolality:

Rank these aqueous solutions from lowest freezing point to highest freezing point.

I. 0.40 m C_{2}H_{6}O_{2}

II. 0.20 m Na_{3}PO_{4}

III. 0.30 m KNO_{3}

IV. 0.20 m C_{6}H_{12}O_{6}

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Freezing Point Depression concept. If you need more Freezing Point Depression practice, you can also practice Freezing Point Depression practice problems.