When 12.2 g of benzoic acid is dissolved in 100 g of water, the freezing point of solution was found to be -0.93°C (K f (H₂O) = 1.86 K kg mol⁻¹). The number (n) of benzoic acid molecules associated (assuming 100% association) is ___________.

Solution

Chemistry ma’am · 4 steps

Compare the freezing point drop you expect with the one you see; the ratio tells you how many molecules join.

1. Molality first. Benzoic acid, C₆H₅COOH, has molar mass 122 g mol⁻¹, so 12.2 g is 0.1 mol. In 100 g (0.1 kg) of water, the molality is m = 0.1/0.1 = 1 mol kg⁻¹.
2. Find i. The observed depression is 0.93 K, since pure water freezes at 0°C. Put this into Δ T_f = i K_f m and solve for the van't Hoff factor i.

   i = (Δ T_f)/(K_f m) = 0.93/(1.86 × 1) = 0.5

3. Link i to n. When n molecules join into one cluster with 100% association, the number of particles drops to 1/n of the original. So i = 1/n, and 1/n = 0.5 gives n = 2.
4. So. Benzoic acid exists as a hydrogen-bonded dimer, so n = 2.

The common slip: 0.5. Reporting the van't Hoff factor i instead of the association number gives 0.5. The question asks for n, which is 1/i = 2.

The idea to keep. If the observed colligative effect is smaller than expected, the solute is associating. For complete association of n molecules into one, i = 1/n.

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