150 g of acetic acid was contaminated with 10.2 g ascorbic acid (C₆ H₈ O₆) to lower down its freezing point by (x × 10⁻¹)° C. The value of x is ___________. (Nearest integer)
[Given K_f=3.9 K kg mol⁻¹; molar mass of ascorbic acid =176 g mol⁻¹]

Solution

Chemistry ma’am · 3 steps

Find the molality of ascorbic acid in acetic acid, multiply by K_f, then read off x from the tenths.

1. Moles of solute. Ascorbic acid is the solute: n = 10.2/176 = 0.05795 mol. It is a non-electrolyte here, so i = 1.
2. Molality. The solvent is acetic acid, 150 g = 0.150 kg. So m = 0.05795/0.150 = 0.3864 mol kg⁻¹.

   Δ T_f = K_f m = 3.9 × 0.3864 = 1.507 °C

3. So. 1.507 °C = 15.07 × 10⁻¹ °C, so x is 15 to the nearest integer.

The common slip: 2. Reporting Δ T_f itself, 1.5, and rounding it gives 2. The question asks for x in x × 10⁻¹, so multiply by 10 first.

The idea to keep. For freezing point depression, use Δ T_f = iK_f m, where molality is moles of solute per kg of solvent. Always convert the solvent mass to kg first.

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