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 :

Osmotic pressure (Π) of a dilute solution is given by the van’t Hoff equation:

$\Pi \; =\; i M R T$

For the solutes considered—glucose, urea and sucrose—none dissociate in water, so the van’t Hoff factor i = 1. At a fixed temperature T and with the same gas constant R, Π is directly proportional to the molarity M (moles of solute per litre of solution). Therefore we only need to compare their molar concentrations.

First compute the number of moles of each solute from the given mass (10 g) and their molar masses:

• Glucose (C₆H₁₂O₆) M_r = 6·12 + 12·1 + 6·16 = 180 g mol⁻¹  n_glucose = 10 g / 180 g mol⁻¹ = 0.0556 mol

• Urea (CH₄N₂O) M_r = 12 + 4·1 + 2·14 + 16 = 60 g mol⁻¹  n_urea = 10 g / 60 g mol⁻¹ = 0.1667 mol

• Sucrose (C₁₂H₂₂O₁₁) M_r = 12·12 + 22·1 + 11·16 = 342 g mol⁻¹  n_sucrose = 10 g / 342 g mol⁻¹ = 0.0292 mol

All solutions are prepared in the same volume, V = 250 mL = 0.250 L. Their molarities are:

M_glucose = 0.0556 mol / 0.250 L = 0.222 M

M_urea = 0.1667 mol / 0.250 L = 0.667 M

M_sucrose = 0.0292 mol / 0.250 L = 0.117 M

Since Π ∝ M, the osmotic pressures follow the same order as the molarities:

Π_urea > Π_glucose > Π_sucrose

Translating back to the notation used in the question:

P₂ > P₁ > P₃