Three solutions were prepared by dissolving 10 g of glucose ( P_1 ), 10 g of urea ( P_2 ), and 10 g of sucrose ( P_3 ) in 250 ml of water. The right o - Chemistry MCQ NEET 2021 2026 | ExamTest.live

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Three solutions were prepared by dissolving 10 g of glucose (P_{1}), 10 g of urea (P_{2}), and 10 g of sucrose (P_{3}) in 250 ml of water. The right option for the decreasing order of osmotic pressure of these solutions is:

Solution & Step-by-step Explanation

Osmotic pressure π = CRT = \frac{n}{V} RT = \frac{W}{M \times V} RT .Since W (10 g), V (250 ml), R, and T are constant, π \propto \frac{1}{M} (molar mass).Molar mass of Urea (P_{2}) = 60 g/mol Molar mass of Glucose (P_{1}) = 180 g/mol Molar mass of Sucrose (P_{3}) = 342 g/mol Smallest molar mass gives highest osmotic pressure.Order: P_{2} > P_{1} > P_{3} .

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Three solutions were prepared by dissolving 10 g of glucose (P_{1}), 10 g of urea (P_{2}), and 10 g of sucrose (P_{3}) in 250 ml of water. The right option for the decreasing order of osmotic pressure of these solutions is:

P_{2} > P_{3} > P_{1}

P_{3} > P_{1} > P_{2}

P_{2} > P_{1} > P_{3}

P_{1} > P_{2} > P_{3}

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