The solution from the following with highest depression in freezing point/lowest freezing point is

Solution

Chemistry ma’am · 4 steps

Count the particles each solute really gives, then check where each solvent starts freezing.

1. Count particles. Δ T_f = i K_f m is a colligative property, so it depends on the number of solute particles, not on their mass. The effective amount of solute is i × moles.
2. Grams to moles. Acetic acid (CH₃COOH, M = 60) gives 180/60 = 3 mol. Benzoic acid (C₆H₅COOH, M = 122) gives 180/122 ≈ 1.48 mol. Glucose (M = 180) gives only 1 mol.
3. Van't Hoff factor. In benzene, carboxylic acids pair up into hydrogen-bonded dimers, so i ≈ 1/2. In water, acetic acid ionises slightly into CH₃COO⁻ and H⁺, so i is a little more than 1. Glucose neither ionises nor associates, so i = 1.

   Solution (1 kg solvent)	Effective moles	K_f of solvent	Freezing point
   (a) acetic acid in benzene	3 × 1/2 = 1.5	5.12	5.5 - 7.68 ≈ -2.2 °C
   (b) acetic acid in water	a little more than 3	1.86	below -5.6 °C
   (c) benzoic acid in benzene	≈ 0.74	5.12	5.5 - 3.8 ≈ 1.7 °C
   (d) glucose in water	1	1.86	-1.86 °C

4. So. Acetic acid in water carries the most solute particles, and water starts freezing at 0 °C, below benzene's 5.5 °C. So it has the lowest freezing point of the four, and the answer is (b).

The idea to keep. Colligative properties count particles, not grams. Convert to moles, apply i (above 1 for ionisation, about 1/2 for acid dimers in benzene), and if solvents differ, compare final freezing points.

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