JEE Main 2026 — Solutions Question with Solution
JEE Main 2026 (24 January Shift 1)
Question
A solution is prepared by dissolving 0.3 g of a non-volatile non-electrolyte solute 'A' of molar mass and 0.9 g of a non-volatile non-electrolyte solute ' ' of molar mass in at . Osmotic pressure of the solution will be
[Given: ]
[Given: ]
Choose an option
Show full solutionCorrect option: B
Correct answer
B2.46 atm
Step-by-step explanation
Calculate osmotic pressure using where n is total moles of solute particles.
Moles of solute A: mol
Moles of solute B: mol
Total moles: mol
Volume of solution: 100 mL = 0.1 L
Temperature: 27°C = 300 K
Osmotic pressure: atm
Moles of solute A: mol
Moles of solute B: mol
Total moles: mol
Volume of solution: 100 mL = 0.1 L
Temperature: 27°C = 300 K
Osmotic pressure: atm
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