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Chemistry Question 55 – JEE-MAIN 2026

Given below are two statements : Chamber 1Semi-permeable membraneChamber 218 g glucose30 g glucosein 100 mLin 250 mLaqueousaqueoussolutionsolution Statement I : H2O molecules move from the chamber 1 to chamber 2. Statement II : The osmotic pressure of a solution prepared by dissolving 50 mg of potassium sulphate (molar mass = 174 g/mol) in 2 L of water (at 27 C) is 0.0107 bar. (Given: R = 0.083 bar K1 mol1 and assume complete dissociation of electrolyte) In the light of the above statements, choose the correct answer from the options given below :

Understand the direction of water movement in osmosis based on solute concentration and how to calculate osmotic pressure for electrolyte solutions.

Step 1: Evaluate Statement I (Osmosis Direction)✦ Active

Calculate the molar concentration of glucose in each chamber. The molar mass of glucose (C6H12O6) is 180 g/mol.

C1=18 g180 g/mol/0.1 L=0.1 mol/0.1 L=1.0 M C2=30 g180 g/mol/0.25 L=1/6 mol1/4 L=46=23 M0.667 M

Since C1>C2, Chamber 1 has a higher solute concentration than Chamber 2. In osmosis, water moves from a region of lower solute concentration to a region of higher solute concentration. Therefore, water will move from Chamber 2 to Chamber 1. Statement I, which claims H2O molecules move from Chamber 1 to Chamber 2, is false.

Step 2: Evaluate Statement II (Osmotic Pressure Calculation)○ Expand

Use the osmotic pressure formula Π=iCRT. For K2SO4, complete dissociation means K2SO42K++SO42, so the van't Hoff factor i=3. Given values are: mass of K2SO4=50 mg=0.050 g, molar mass =174 g/mol, volume V=2 L, temperature T=27C=300 K, and R=0.083 bar K1 mol1.

C=massmolar mass×V=0.050 g174 g/mol×2 L=0.050348 M Π=iCRT=3×0.050348 M×0.083 bar K1 mol1×300 K Π0.01073 bar

The calculated osmotic pressure is approximately 0.0107 bar, which matches the value given in Statement II. Therefore, Statement II is true.

Step 3: Conclusion○ Expand

Based on the analysis, Statement I is false, and Statement II is true. This corresponds to option 4.

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