Question:

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

Updated On: Nov 13, 2025
  • P3 > P1 > P2

  • P2 > P1 > P3

  • P1 > P2 > P3

  • P2 > P3 > P1

Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation

To determine the decreasing order of osmotic pressures of the given solutions, we need to consider the concept of osmotic pressure, which is a colligative property. Colligative properties depend on the number of solute particles present in a solution, not on the identity or nature of the solute. 

  1. The osmotic pressure (\(\pi\)) of a solution is given by the formula: \(\pi = i \cdot C \cdot R \cdot T\), where:
    • \(i\) = Van't Hoff factor (number of particles the solute splits into),
    • \(C\) = Concentration of the solution in moles per liter,
    • \(R\) = Universal gas constant, and
    • \(T\) = Temperature in Kelvin.
  2. Let's calculate the Van't Hoff factor and concentration for each solution:
    • Glucose (\(C_6H_{12}O_6\)):
      • Does not ionize in solution (i = 1).
      • Molar mass of glucose = 180 g/mol.
      • Moles = \(\frac{10}{180} \approx 0.0556\) mol.
      • Concentration = \(\frac{0.0556}{0.25} \approx 0.2224\) M.
    • Urea (\(CH_4N_2O\)):
      • Does not ionize in solution (i = 1).
      • Molar mass of urea = 60 g/mol.
      • Moles = \(\frac{10}{60} \approx 0.1667\) mol.
      • Concentration = \(\frac{0.1667}{0.25} \approx 0.6668\) M.
    • Sucrose (\(C_{12}H_{22}O_{11}\)):
      • Does not ionize in solution (i = 1).
      • Molar mass of sucrose = 342 g/mol.
      • Moles = \(\frac{10}{342} \approx 0.0292\) mol.
      • Concentration = \(\frac{0.0292}{0.25} \approx 0.1168\) M.
  3. The osmotic pressure is proportional to the concentration and Van't Hoff factor:
    • For glucose (P1): 0.2224 M × 1 = 0.2224
    • For urea (P2): 0.6668 M × 1 = 0.6668
    • For sucrose (P3): 0.1168 M × 1 = 0.1168
  4. Order of osmotic pressures (highest to lowest):
    • P2 > P1 > P3

Therefore, the correct answer is P2 > P1 > P3, which indicates that urea has the highest osmotic pressure due to its higher concentration, followed by glucose, and then sucrose.

Was this answer helpful?
0
0

Concepts Used:

Colligative Properties

Colligative Property of any substance is entirely dependent on the ratio of the number of solute particles to the total number of solvent particles but does not depend on the nature of particles. There are four colligative properties: vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.

Examples of Colligative Properties

We can notice the colligative properties of arrangements by going through the accompanying examples:

  • On the off chance that we add a spot of salt to a glass full of water, its freezing temperature is brought down impressively than its normal temperature. On the other hand, the boiling temperature is likewise increased and the arrangement will have a lower vapor pressure. There are also changes observed in its osmotic pressure.
  • In the same way, if we add alcohol to water, the solution’s freezing point goes down below the normal temperature that is usually observed for either pure alcohol or water.

Types of Colligative Properties

  1. Freezing point depression: ΔTf =1000 x kf x m2 /(M2 x m1)
  2. Boiling point elevation: ΔTb = kb m
  3. Osmotic pressure: π = (n2/V) RT
  4. Relative lowering of vapor pressure: (Po - Ps)/Po