20 % of acetic acid is dissociated when its 5 g is added to 500 mL of water. The depression in freezing point of such water is _________ × 10⁻³ ° C.
Atomic mass of C, H and O are 12,1 and 16 a.m.u. respectively.
[Given : Molal depression constant and density of water are 1.86 K kg mol⁻¹ and 1 g cm⁻³ respectively.]

Solution

Chemistry ma’am · 3 steps

Acetic acid splits into ions here, so multiply by the van't Hoff factor before you finish.

1. Find the molality. Molar mass of CH₃COOH is 60 g mol⁻¹, so 5 g is 5/60 = 1/12 mol. With density 1 g cm⁻³, 500 mL of water is 0.5 kg, so m = (1/12)/0.5 = 1/6 mol kg⁻¹.
2. Van't Hoff factor. CH₃COOH ⇌ CH₃COO⁻ + H⁺ gives 2 particles per molecule. With α = 0.2, i = 1 + (2 - 1)α = 1.2.

   Δ T_f = i K_f m = 1.2 × 1.86 × 1/6 = 0.372 °C

3. So. Δ T_f = 0.372 °C = 372 × 10⁻³ °C, so the answer is 372.

The common slip: 310. Leaving out the van't Hoff factor (i = 1) gives 0.31 °C, that is 310, not 372.

The idea to keep. Whenever a solute dissociates or associates, colligative properties count particles, not molecules. Find i from α first, then use Δ T_f = iK_f m.

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