JEE Main 2022 · 25 July · evening · Single correct · +4 / −1

Two solutions A and B are prepared by dissolving 1 g of non-volatile solutes X and Y, respectively in 1 kg of water. The ratio of depression in freezing points for A and B is found to be 1 : 4. The ratio of molar masses of X and Y is

  1. (a) 1 : 4
  2. (b) 1 : 0.25
  3. (c) 1 : 0.20
  4. (d) 1 : 5
Answer and solution

Answer (b) 1 : 0.25

Solution

Chemistry ma’am · 4 steps

Same mass, same water, so the freezing point drop depends only on how many moles you get.

  1. Link to molar mass. For a non-volatile solute, Δ T_f = K_f × m. The molality is m = w/(M × W), where w is the solute mass in g and W is the water mass in kg.

    Δ T_f = K_f × w/(M × W)

  2. Cancel what is common. Both solutions use water, so K_f is the same. Both have w = 1 g and W = 1 kg. Only M is left, so Δ T_f ∝ 1/M.
  3. Take the ratio. (Δ T_A)/(Δ T_B) = M_Y/M_X = 1/4. Flip it to get M_X/M_Y = 4/1.
  4. So. M_X : M_Y = 4 : 1. Divide both sides by 4 to get 1 : 0.25, which is option (b).

The idea to keep. Same mass of two solutes in the same solvent: every colligative property is inversely proportional to molar mass. A heavier solute gives fewer particles and a smaller effect.

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

Study Solutions & Colligative Properties

Asked about 150 times in JEE Main since 2020. This question is from 2022.