Question

What is the freezing point of an aqueous solution that has 25.00 g of calcium iodide dissolved in 1250 g of water?

Answer 1

Answer:

• - 0.380ºC

Explanation:

The lowering of the freezing point of a solvent is a colligative property ruled by the formula:

• $\Delta T_f=K_f\times m\times i$

Where:

• ΔTf is the lowering of the freezing point

• Kf is the molal freezing constant of the solvent: 1.86 °C/m

• m is the molality of the solution

• i is the van't Hoff factor: the number of particles (ions) per unit of ionic compound.

a) molality, m

• m = number of moles of solute/ kg of solvent

• number of moles of CaI₂ = mass in grams/ molar mass

• number of moles of CaI₂ = 25.00g / 293.887 g/mol = 0.0850667mol

• m = 0.0850667mol/1.25 kg = 0.068053m

b) i

• Each unit of CaI₂, ideally, dissociates into 1 Ca⁺ ion and 2 I⁻ ions. Thus, i = 1 + 2 = 3

c) Freezing point lowering

• ΔTf =  1.86 °C/m × 0.068053m × 3 = 0.3797ºC ≈ 0.380ºC

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