The following solutions were prepared by dissolving 10 g of glucose ( C₆H₁₂O₆ ) in 250 ml of water ( P₁ ), 10 g of urea ( CH₄N₂O ) in 250 ml of water ( P₂ ) and 10 g of sucrose ( C₁₂H₂₂O₁₁ ) in 250 ml of water ( P₃ ). The right option for the decreasing order of osmotic pressure of these solutions is : 1. P₂ > P₁ > P₃ 2. P₁ > P₂ > P₃ 3. P₂ > P₃ > P₁ 4. P₃ > P₁ > P₂

Topic: Colligative properties· Solutions

The following solutions were prepared by dissolving 10 g of glucose (C_6H_12O_6) in 250 ml of water (P_1), 10 g of urea (CH_4N_2O) in 250 ml of water (P_2) and 10 g of sucrose (C_12H_22O_11) in 250 ml of water (P_3). The right option for the decreasi…

The following solutions were prepared by dissolving 10 g of glucose (\(C_6H_{12}O_6\)) in 250 ml of water (\(P_1\)), 10 g of urea (\(CH_4N_2O\)) in 250 ml of water (\(P_2\)) and 10 g of sucrose (\(C_{12}H_{22}O_{11}\)) in 250 ml of water (\(P_3\)). The right option for the decreasing order of osmotic pressure of these solutions is :

1. \(P_2 > P_1 > P_3\)

2. \(P_1 > P_2 > P_3\)

3. \(P_2 > P_3 > P_1\)

4. \(P_3 > P_1 > P_2\)

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Answer: 1. \(P_2 > P_1 > P_3\)

Key idea: for equal mass in equal volume, osmotic pressure \(\pi=CRT\) is proportional to moles, i.e. inversely proportional to molar mass (all three are non-electrolytes).

Molar masses: urea 60, glucose 180, sucrose 342.

Moles in 10 g: urea 0.167 > glucose 0.056 > sucrose 0.029.

So \(\pi\): \(P_2>P_1>P_3\).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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The following solutions were prepared by dissolving 10 g of glucose (C_6H_12O_6) in 250 ml of water (P_1), 10 g of urea (CH_4N_2O) in 250 ml of water (P_2) and 10 g of sucrose (C_12H_22O_11) in 250 ml of water (P_3). The right option for the decreasi…

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