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 :

The correct option is P₂ > P₁ > P₃.

Step-by-step Explanation:

Osmotic pressure ($\pi$ or $P$) is a colligative property, which means it depends on the concentration of solute particles in the solution. The formula for osmotic pressure is given by:
$P=i·C·R·T$
where:
• $i$ is the van 't Hoff factor (which is equal to 1 for non-electrolytes like glucose, urea, and sucrose because they do not dissociate in water).
• $C$ is the molar concentration (molarity) of the solution.
• $R$ is the universal gas constant.
• $T$ is the absolute temperature.

Since the mass of each solute (10 g), the volume of water (250 ml), and the temperature are the same for all three solutions, the osmotic pressure depends directly on the number of moles of solute dissolved:
$P\propto n$
where $n$ is the number of moles, calculated as:
$n=\frac{\mathrm{mass}}{\mathrm{Molar}\mathrm{Mass}\left(M\right)}$
Thus, osmotic pressure is inversely proportional to the molar mass of the solute:
$P\propto \frac{1}{M}$

Let us find the molar masses of the three solutes:
1. Urea (CH₄N₂O) for solution P₂:

2. Glucose (C₆H₁₂O₆) for solution P₁:
$M=(6·12)+(12·1)+(6·16)=180g/\mathrm{mol}$
3. Sucrose (C₁₂H₂₂O₁₁) for solution P₃:
$M=(12·12)+(22·1)+(11·16)=342g/\mathrm{mol}$

Comparing the molar masses:
$M_{\mathrm{urea}}<M_{\mathrm{glucose}}<M_{\mathrm{sucrose}}$
$60g/\mathrm{mol}<180g/\mathrm{mol}<342g/\mathrm{mol}$

Since osmotic pressure is inversely proportional to the molar mass, the decreasing order of osmotic pressure is:
$P_{\mathrm{urea}}>P_{\mathrm{glucose}}>P_{\mathrm{sucrose}}$
Which is:
$P_2>P_1>P_3$