# Question 9413c

Sep 26, 2015

$- 5.401 \text{^@"C}$

#### Explanation:

!! LONG ANSWER !!

What you need to do here is find the mole fraction of calcium chloride in the solution, then use a sample of this solution to try and find its freezing-point depression.

So, you know that you're dealing with a solution of calcium chlorate, ${\text{CaCl}}_{2}$, a soluble ionic compound that dissociates in aqueous solution to give

${\text{CaCl"""_text(2(aq]) -> "Ca"_text((aq])^(2+) + 2"Cl}}_{\textrm{\left(a q\right]}}^{-}$

Now, you can use the vapor pressure of the pure water and the vapor pressure of the solution to find the mole fraction of water.

To do that, use Raoult's Law, which tells you that the vapor pressure of the solution will depend on the mole ratio and the pure vapor pressure of water

${P}_{\text{sol" = chi_"water" * P_"water}}^{0}$

This means that the mole fraction of water is

chi_"water" = P_"sol"/P_"water"^0 = (0.02970color(red)(cancel(color(black)("atm"))))/(0.03126color(red)(cancel(color(black)("atm")))) = 0.9501

Since your solution only has two components, you can say that

${\chi}_{\text{water" + chi_"solute}} = 1$

${\chi}_{\text{solute" = 1 - chi_"water}} = 1 - 0.9501 = 0.04990$

So, calcium chloride has a mole fraction of $0.04990$ in this solution.

Now, in order to find the freezing-point depression, you first need the molality of the solution. To make the calculations easier, I will pick a sample of solution that contains $\text{1 kg}$ of water.

You know what the mole fraction of particles other than water, which I'll call solute ions, is in this sample, but you don't know exactly how many moles of the solute you have.

Use water's molar mass to find how many moles of water you get in $\text{1 kg}$ of water

1color(red)(cancel(color(black)("kg"))) * (1000color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * "1 mole"/(18.02color(red)(cancel(color(black)("g")))) = "55.494 moles"

The mole fraction of solute ions will be

${n}_{\text{solute ions")/(n_"solute ions" + n_"water}} = 0.04990$

This means that the solution that contains $\text{1 kg}$ of water will contain

${n}_{\text{solute ions" - 0.04990 * n_"solute ions}} = 55.494 \cdot 0.04990$

${n}_{\text{solute ions" = "2.9146 moles of ions}}$

Now, you need to take into account the fact every mole of calcium chloride will produce 3 moles of ions in solution, one mole of calcium ions, ${\text{Ca}}^{2 +}$, and two moles of chloride ions, ${\text{Cl}}^{-}$.

This means that the actual number of moles of calcium chloride added to the water is three time smaller

${n}_{C a C {l}_{2}} = {n}_{\text{solute ions" /3 = "2.9146 moles"/3 = "0.97153 moles}}$

The molality of the sample will thus be

$b = {n}_{C a C {l}_{2}} / {m}_{\text{water" = "0.97153 moles"/"1 kg" = "0.97153 molal}}$

SIDE NOTE You will need to multiply this value by 3 again to get the freezing-point depression, but I wanted to show you how many moles of calcium chloride were added to the solution.

The equation for freezing-point depression looks like this

$\Delta {T}_{f} = i \cdot {K}_{f} \cdot b \text{ }$, 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 van't Hoff factor will be equal to $3$, since every formula unit of calcium chloride produces three ions in solution.

The cryoscopic constant of water is $1.853 {\text{^@"C kg mol}}^{- 1}$, and so the freezing-point depression will be

DeltaT_f = 3 * 1.853 ""^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.97153 color(red)(cancel(color(black)("mol")))/color(red)(cancel(color(black)("kg")))

$\Delta {T}_{f} = 5.401 \text{^@"C}$

The freezing-point depression is defined as

$\Delta {T}_{f} = {T}_{\text{f"^0 - T_"f sol}}$

The freezing point of the solution will thus be

${T}_{\text{f sol}} = {T}_{f}^{0} - \Delta {T}_{f}$

T_"f sol" = 0""^@"C" - 5.401""^@"C" = color(green)(-5.401""^@"C")#

I'll leave the answer rounded to four sig figs, the number of sig figs you gave for the two vapor pressures.