# What is the freezing point of a solution prepared by dissolving 11.3 g of Ca(NO_3)_2 (formula weight = 164 g/mol) in 115 g of water?

## The molal freezing point depression constant for water is 1.86° C/m.

Jan 24, 2016

$- {3.34}^{\circ} \text{C}$

#### Explanation:

In order to find the freezing point of this solution, you must first find the freezing-point depression associated with dissolving that much calcium nitrate, "Ca"("NO"_3)_2, in that much water.

Freezing-point depression is a colligative property, which means that it depends on the number of particles of solute present in solution, and not on the nature of those particles.

Calcium nitrate is a soluble ionic compound that dissociates completely in aqueous solution to form

${\text{Ca"("NO"_3)_text(2(s]) -> "Ca"_text((aq])^(2+) + 2"NO}}_{\textrm{3 \left(a q\right]}}^{-}$

As you an see, every mole of calcium nitrate will produce three moles of ions in solution, one mole of calcium cations and two moles of nitrate anions.

Keep this in mind.

Now, the equation that describes freezing-point depression looks like this

$\textcolor{b l u e}{\Delta {T}_{f} = i \cdot {K}_{f} \cdot b} \text{ }$, where

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

In your case, the cryoscopic constant of water is said to be equal to

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

Now, the key here is the value of the van't Hoff factor, which tells you how many moles of particles are produced in solution per mole of dissolved solute.

Since we've established that one mole of calcium nitrate produces three moles of ions in aqueous solution, the van't Hoff factor will be

$i = 3$

To find the molality of the solution, you need to know two things

• the number of moles of solute
• the mass of the solvent expressed in kilograms

The number of moles of solute can be determined by using the given formula weight

11.3 color(red)(cancel(color(black)("g"))) * ("1 mole Ca"("NO"_3)_2)/(164color(red)(cancel(color(black)("g")))) = "0.0689 moles Ca"("NO"_3)_2

The molality of the solution will thus be

$b = {\text{0.0689 moles"/(115 * 10^(-3)"kg") = "0.599 mol kg}}^{- 1}$

Now plug all this into the equation for freezing-point depression and get the value of $\Delta {T}_{f}$

$\Delta {T}_{f} = 3 \cdot {1.86}^{\circ} \text{C" color(red)(cancel(color(black)("kg")))color(red)(cancel(color(black)("mol"^(-1)))) * 0.599color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1)))) = 3.34^@"C}$

The freezing point of the solution will thus be

$\textcolor{b l u e}{\Delta {T}_{f} = {T}_{f}^{0} - {T}_{f}} \text{ " implies" } {T}_{f} = {T}_{f}^{\circ} - \Delta {T}_{f}$

Here ${T}_{f}^{\circ}$ is the freezing point of the pure solvent*.

In your case, you will have

T_f = 0^@"C" - 3.34^@"C" = color(green)(-3.34^@"C")

The answer is rounded to three sig figs.