Question Details

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 :

Options

A

P₂ > P₁ > P₃

B

P₁ > P₂ > P₃

C

P₂ > P₃ > P₁

D

P₃ > P₁ > P₂

Correct Answer :

P₂ > P₁ > P₃

Solution :

The correct option is P₂ > P₁ > P₃.

To find the decreasing order of osmotic pressure for the given solutions, we can use the formula for osmotic pressure (represented by the symbol π):
π=i·C·R·T
where:
- i is the van 't Hoff factor,
- C is the molar concentration (molarity) of the solute,
- R is the universal gas constant, and
- T is the absolute temperature.

Since glucose, urea, and sucrose are all non-electrolytes, they do not dissociate or associate in water. Therefore, their van 't Hoff factor is the same:
i=1
Additionally, the mass of each solute dissolved is the same (10 g), the volume of water is the same (250 ml), and the temperature T is constant. Thus, the osmotic pressure π is directly proportional to the molar concentration C, which is determined by the number of moles of each solute:
C=nV=mass/Molar MassV
Since the mass and volume V are constant across all three solutions, the osmotic pressure is inversely proportional to the molar mass of the solute:
π1Molar Mass

Let's calculate the molar masses of the three solutes:
1. Urea (CH₄N₂O) in solution P₂:
Molar mass = 12 (C) + 4×1 (H) + 2×14 (N) + 16 (O) = 12 + 4 + 28 + 16 = 60 g/mol.
2. Glucose (C₆H₁₂O₆) in solution P₁:
Molar mass = 6×12 (C) + 12×1 (H) + 6×16 (O) = 72 + 12 + 96 = 180 g/mol.
3. Sucrose (C₁₂H₂₂O₁₁) in solution P₃:
Molar mass = 12×12 (C) + 22×1 (H) + 11×16 (O) = 144 + 22 + 176 = 342 g/mol.

Comparing the molar masses:
Molar Mass (Urea) < Molar Mass (Glucose) < Molar Mass (Sucrose)
60 g/mol < 180 g/mol < 342 g/mol

Since osmotic pressure is inversely proportional to the molar mass of the solute for equal masses dissolved in equal volumes:
πurea>πglucose>πsucrose
which corresponds to:
P₂ > P₁ > P₃

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