# Question f5ef6

Jul 10, 2016

Here's what I got.

#### Explanation:

First of all, you're dealing with freezing-point depression, not freezing-point elevation.

Simply put, the freezing point of a solution will be lower than the freezing point of the pure solvent. The freezing-point depression tells you how low the freezing point of the solution will actually be compared with that of the pure solvent.

Now, the equation that allows you to calculate freezing-point depression looks like this

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \Delta {T}_{f} = i \cdot {K}_{f} \cdot b \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

$\Delta {T}_{f}$ - the freezing-point depression;
$i$ - the van't Hoff factor
${K}_{f}$ - the cryoscopic constant of the solvent;
$b$ - the molality of the solution

As you can see, the freezing-point depression cannot be negative because the van't Hoff factor, the cryoscopic constant, and the molality of the solution are all positive values.

In your case, aluminium nitrate, "Al"("NO"_3)_3, is soluble in aqueous solution, which means that it dissociates completely to form aluminium cations, ${\text{Al}}^{3 +}$, and nitrate anions, ${\text{NO}}_{3}^{-}$

${\text{Al"("NO"_ 3)_ (3(aq)) -> "Al"_ ((aq))^(3+) + 3"NO}}_{3 \left(a q\right)}^{-}$

Since one mole of solute produces $4$ moles of particles of solute in solution, the van't Hoff factor is equal to $4$.

The cryoscopic constant of water is equal to

${K}_{f} = {1.86}^{\circ} {\text{C kg mol}}^{- 1}$

The freezing-point depression will thus be

DeltaT_f = 4 * 1.86^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.5 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))

$\Delta {T}_{f} = {3.72}^{\circ} \text{C}$

This is the correct value for the freezing-point depression of a $\text{0.5 molal}$ aluminium nitrate solution for which water is the solvent.

Since freezing-point depression tells you by how many degrees the freezing point of the solution decreased compared to that of the pure solvent, which is ${0}^{\circ} \text{C}$ for water, you can say that the freezing point of the solution will be

${T}_{\text{f sol" = 0^@"C" - 3.72^@"C" = -3.72^@"C}}$

Now, the question asks you to find the freezing point of a solution in which ethanol is the solvent, not water.

The cryoscopic constant for ethanol is

${K}_{f} = {1.99}^{\circ} {\text{C kg mol}}^{- 1}$

http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf

Assuming that you have complete dissociation, the freezing-point depression will be

DeltaT_f = 4 * 1.99^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.5 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))#

$\Delta {T}_{f} = {3.98}^{\circ} \text{C}$

This means that the freezing point of the solution will be ${3.98}^{\circ} \text{C}$ lower than the freezing point of the pure solvent, which for ethanol is $- {18}^{\circ} \text{C}$

${T}_{\text{f sol" = -18^@"C" - 3.98^@"C" = -21.98^@"C}}$

So remember, the freezing-point depression is always positive and it must be subtracted from the freezing point of the pure solvent in order to get the freezing point of the solution.