A 0.5 percent solution of potassium chloride was found to freeze at -0.24°C. The percentage dissociation of potassium chloride is ______________. (Nearest integer)
(Molal depression constant for water is 1.80 K kg mol⁻¹ and molar mass of KCl is 74.6 g mol⁻¹)

Solution

Chemistry ma’am · 3 steps

Find the van 't Hoff factor from the freezing point first, then read the dissociation straight off it.

1. Molality. A 0.5 percent (by mass) solution has 0.5 g KCl in 100 g of solution, so the water is 100 - 0.5 = 99.5 g. Moles of KCl = 0.5/74.6 = 6.70 × 10⁻³ mol, so m = (6.70 × 10⁻³)/0.0995 = 0.0674 mol kg⁻¹.

   i = (Δ T_f)/(K_f m) = 0.24/(1.80 × 0.0674) = 1.98

2. Link i to α. KCl → K⁺ + Cl⁻ gives 2 ions, so i = 1 + α(2 - 1) = 1 + α. That makes α = 1.98 - 1 = 0.98.
3. So. The percentage dissociation is 0.979 × 100 = 97.9, which rounds to 98.

The common slip: 99. Taking 100 g of water instead of 99.5 g gives i = 1.99 and an answer of 99, not 98.

The idea to keep. For a salt that splits into n ions, i = 1 + (n-1)α. Get i from the colligative property, then solve for α. In a mass percent question, the solvent mass is 100 minus the solute mass.

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