A solution is made with 23.50 g of calcium chloride, CaCI2 and 250.0 mL of water. The water is 25.0°C. What is the freezing point of this solution?
Assume the CaCI2 dissolves completely in the water.. The density of water at 25.0°C is 0.997 g/L, and the KF of H20 is 1.858°C kg/mol.

-4741 °C

Something is strange, because this is a weird number.

Explanation:

ΔT = Kf . m. i

That's the colligative property of freezing point depression.

Kf = Cyroscopic constant

m = molality (moles of solute in 1kg of solvent)

i = Van't Hoff factor (numbers of ions dissolved)

We assume 100% dissociation:

CaCl₂ → Ca²⁺ + 2Cl⁻   i = 3

ΔT = Freezing point of pure solvent - Freezing point of solution

Let's determine molality

Solute = CaCl₂

Moles of solute = 23.5 g . 1 mol/ 110.98 g = 0.212 moles

We determine the mass of solvent by density. Density's data is in g/L. We need to convert the volume from mL to L

250 mL . 1L / 1000 mL = 0.250 L

0.997 g/L = mass of water / volume of water → 0.997 g/L . volume of water = mass of water

0.997 g/L . 0.250L = 0.249 g

Now, we convert the mass of water from g to kg

0.249 g . 1 kg / 1000 g = 2.49×10⁻⁴ kg

Molality = mol/kg → 0.212 mol / 2.49×10⁻⁴ kg = 850.5 m

We replace data:

0°C - Freezing point of solution = 1.858 °C . kg /mol . 850.5 mol/kg . 3

Freezing point of solution = -4741 °C

The freezing point of the solution is -4.74 °C

Explanation:

Step 1: data given

Mass of calcium chloride CaCl2 = 23.50 grams

Step 2: Calculate moles CaCl2

Moles CaCl2 = mass CaCl2 / molar mass CaCl2

Moles CaCl2 = 23.50 grams / 110.98 g/mol

Moles CaCl2 = 0.212 moles

Step 3: Calculate mass H2O

Density = mass / volume

Mass = density * volume

Mass H2O = 997 g/L * 0.250 L

MAss H2O = 249.25 grams

Step 3: calculate molality of the solution

Molality = moles CaCl2 / mass H2O

Molality = 0.212 moles / 0.24925 kg

Molality = 0.851 molal

ΔT = i*Kf*m  

⇒with ΔT = the freezing point depression = TO BE DETERMNED

⇒ with i = the van't Hoff factor = 3

⇒with Kf = the freezing point depression constant of water = 1.858 °C /m

⇒with m = the molality = moles CaCl2 / mass water = 0.851 molal

ΔT = 3 * 1.858 * 0.851

ΔT =   4.74 °C

Step 4: Calculate the freezing point of the solution

ΔT = T (pure solvent) − T (solution)

ΔT  = 0°C - 4.74 °C

The freezing point of the solution is -4.74 °C

NOTE: the density of water = 0.997 kg/L or 997 g/L

the answer is learned