# Question #9413c

##### 1 Answer

#### Answer:

#### 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*, *soluble* ionic compound that dissociates in aqueous solution to give

#"CaCl"""_text(2(aq]) -> "Ca"_text((aq])^(2+) + 2"Cl"_text((aq])^(-)#

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_"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_"water" + chi_"solute" = 1#

#chi_"solute" = 1 - chi_"water" = 1 - 0.9501 = 0.04990#

So, calcium chloride has a mole fraction of

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

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

#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_("solute ions")/(n_"solute ions" + n_"water") = 0.04990#

This means that the solution that contains

#n_"solute ions" - 0.04990 * n_"solute ions" = 55.494 * 0.04990#

#n_"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,

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

#n_(CaCl_2) = n_"solute ions" /3 = "2.9146 moles"/3 = "0.97153 moles"#

The molality of the sample will thus be

#b = n_(CaCl_2)/m_"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

#DeltaT_f = i * K_f * b " "# , where

*van't Hoff factor*;

**cryoscopic constant** of the solvent;

In your case, the van't Hoff factor will be equal to **three ions** in solution.

The *cryoscopic constant* of water is

#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")))#

#DeltaT_f = 5.401 ""^@"C"#

The *freezing-point depression* is defined as

#DeltaT_f = T_"f"^0 - T_"f sol"#

The freezing point of the solution will thus be

#T_"f sol" = T_f^0 - DeltaT_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.