How do I solve this problem on the freezing point of a solution?

Question

What is the freezing point of a solution containing 27.8 g of calcium chlorine in 250 g of water? Is calcium chloride an ionic or molecular splits and why is this important?

Answer 1

Freezing point depression is equal to the molal freezing point depression constant (Kf) times the molality of the solution times the van't Hoff factor.
deltaTf = Kf x m x i

Since the Kf value was not given, we could do a quick reference search to find that the Kf value of our solvent, water is 1/86 °C/mol.

 

Since calcium chloride is a salt, it dissociates in water to give three particles per mole in solution:

CaCl2 --> Ca²⁺ + 2Cl-

Therefore, its van't Hoff factor is 3. 

 

Then we just need to find the molality of the solution and substitute everything into the equation.

Molality is equal to moles of solute divided by kilograms of solvent. 

m = moles solute/kg solvent

To the find the moles of calcium chloride, we take the mass given and divide by its molar mass:

moles Calcium Chloride = 27.8 g x (1 mole/110.98 g) = 0.250 moles

To find the kg of water we divide the grams given by 1000:

kg water = 250 g x (1 kg/1000g) = 0.250 kg 

So the molality of the solution is 

m = moles solute/kg solvent

    = 0.250 moles/0.250m

    = 1.00 m

 

From there, we substitute everything into the equation for freezing point depression:

deltaTf = Kf x m x i
= (1.86°C/m)(1.00m)(3)
= 5.58°C
Therefore the freezing point depressed by 5.58°C.

 

It was important to know that the solute, calcium chloride was ionic, otherwise our van't Hoff factor would have been 1 for a molecular solute and changed our calculations significantly. More particles, means a bigger van't Hoff factor and greater freezing point depression.