JEE Main 2024 · 9 April · morning · Single correct · +4 / −1
0.05 M CuSO₄ when treated with 0.01 M K₂ Cr₂ O₇ gives green colour solution of Cu₂ Cr₂ O₇. The two solutions are separated as shown below : [SPM : Semi Permeable Membrane]
Due to osmosis :
- (a) Molarity of K₂ Cr₂ O₇ solution is lowered.
- (b) Green colour formation observed on side Y.
- (c) Molarity of CuSO₄ solution is lowered.
- (d) Green colour formation observed on side X.
Answer and solution
Answer (c) Molarity of CuSO₄ solution is lowered.
Solution
Chemistry ma’am · 3 steps
Only water crosses the membrane, so compare particle counts and see which way the water moves.
- What crosses the SPM. A semipermeable membrane lets solvent through but stops solute ions. So Cu²⁺ and Cr₂O₇²⁻ never meet, and no green Cu₂Cr₂O₇ forms on either side.
π = iCRT
Side Solution Ions per formula unit, i Particle concentration iC X 0.01 M K₂Cr₂O₇ 3 (2K⁺ + Cr₂O₇²⁻) 0.03 M, lower π Y 0.05 M CuSO₄ 2 (Cu²⁺ + SO₄²⁻) 0.10 M, higher π - Direction of flow. Water moves from the side with fewer solute particles to the side with more. Here that is from X into Y.
- So. Side Y gains water, so the CuSO₄ solution is diluted and its molarity is lowered. The answer is (c).
The idea to keep. In osmosis, the solute stays put and the solvent moves towards the higher iC. Always multiply molarity by the van 't Hoff factor before you compare the two sides.
Answer from the NTA final answer key. Solution by PrepWiser.