25 mL of an aqueous solution of KCl was found to require 20 mL of 1 M AgNO₃ solution when titrated using K₂CrO₄ as an indicator. What is the depression in freezing point of KCl solution of the given concentration? _________ (Nearest integer).
(Given : K_f=2.0 K kg mol⁻¹)
Assume 1) 100% ionization and 2) density of the aqueous solution as 1 g mL⁻¹

Solution

Chemistry ma’am · 3 steps

The titration hands you the moles of KCl. Then count both ions, because one KCl gives two particles.

1. Moles from titration. Each Cl⁻ takes one Ag⁺ (Ag⁺ + Cl⁻ → AgCl), so moles of KCl equal moles of AgNO₃: 20 mL × 1 M = 0.02 mol in 25 mL. That is a molarity of 0.02/0.025 = 0.8 M.
2. Convert to molality. Take 1 L of solution, which weighs 1000 g. It holds 0.8 mol KCl, which is 0.8 × 74.5 = 59.6 g, so the water is 1000 - 59.6 = 940.4 g. Molality is 0.8/0.9404 = 0.851 mol kg⁻¹.

   Δ T_f = i K_f m = 2 × 2.0 × 0.851 = 3.40 K

3. So. KCl ionises fully into K⁺ and Cl⁻, so i = 2. The depression in freezing point is 3.40 K, which rounds to 3.

The common slip: 2. Leaving out i = 2 for KCl gives 2.0 × 0.851 = 1.7, which rounds to 2, not 3.

The idea to keep. Colligative properties count particles, not formula units. Find the moles, convert to molality using the solvent mass, then multiply by the van 't Hoff factor i.

Answer from the NTA final answer key. Solution by PrepWiser.