JEE Main 2022 · 27 July · morning · Single correct · +4 / −1

250 g solution of D-glucose in water contains 10.8 % of carbon by weight. The molality of the solution is nearest to
(Given: Atomic Weights are, H, 1 u ; C, 12 u ; O, 16 u)

  1. (a) 1.03
  2. (b) 2.06
  3. (c) 3.09
  4. (d) 5.40
Answer and solution

Answer (b) 2.06

Solution

Chemistry ma’am · 4 steps

Use the carbon to find the glucose, then remember molality divides by the water alone.

  1. Mass of carbon. Carbon is 10.8% of the solution, so its mass is 0.108 × 250 = 27 g. All of this carbon comes from glucose, because water has none.
  2. Mass of glucose. In C₆H₁₂O₆ (molar mass 180 g mol⁻¹), carbon makes up 72 g of every 180 g. So the glucose mass is 27 × 180/72 = 67.5 g, which is 67.5/180 = 0.375 mol.
  3. Mass of water. The solvent is the solution minus the solute: 250 - 67.5 = 182.5 g = 0.1825 kg.

    m = (0.375 mol)/(0.1825 kg) ≈ 2.05 mol kg⁻¹

  4. So. The molality is about 2.05, and the nearest option is 2.06, option (b).

The idea to keep. When a percentage of one element is given, trace it back to the compound that contains it. For molality, always subtract the solute to get the solvent mass in kg.

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

Study Mole Concept & Stoichiometry

Asked about 210 times in JEE Main since 2020. This question is from 2022.