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Exhaustive Guide: Colligative Properties | CBSE Class 12 Chemistry

Exhaustive Guide: Colligative Properties | CBSE Class 12 Chemistry | ChemCA

Colligative Properties & Molar Mass

Module 4 | CBSE Class 12 Chemistry | Solutions Chapter

1. What are Colligative Properties?

We have learned that adding a non-volatile solute to a volatile solvent lowers its vapour pressure. Many properties of solutions are directly connected to this lowering of vapour pressure.

Colligative Properties: Properties of a dilute solution that depend only on the number of solute particles (molecules or ions) present, and not on the chemical nature of the solute, are called colligative properties. (Latin: co = together, ligare = to bind).

The four colligative properties are:

  1. Relative lowering of vapour pressure
  2. Elevation of boiling point
  3. Depression of freezing point
  4. Osmotic pressure

Assumption: In this module, we assume the solute is non-volatile and a non-electrolyte (it neither dissociates nor associates in the solvent). Electrolytes will be covered in Module 5 (van't Hoff factor).

2. Relative Lowering of Vapour Pressure (RLVP)

According to Raoult's Law, the vapour pressure of a solvent in a solution (p1) is less than that of the pure solvent (p10). The lowering of vapour pressure is Δp = p10 - p1.

Raoult established that the relative lowering of vapour pressure is equal to the mole fraction of the solute (x2).

(p10 - p1) / p10 = x2

Since x2 = n2 / (n1 + n2) where n2 is moles of solute and n1 is moles of solvent.

For dilute solutions (n2 << n1): We can neglect n2 in the denominator. The equation simplifies to:

(p10 - p1) / p10 = (W2 × M1) / (M2 × W1)

(Where W = Mass, M = Molar Mass. Subscript 1 = solvent, 2 = solute). This formula is highly used in board numericals to find the molar mass of an unknown solute (M2).

3. Elevation of Boiling Point (ΔTb)

Boiling Point: The temperature at which the vapour pressure of a liquid becomes equal to the atmospheric pressure (1 atm or 1.013 bar).

Since the addition of a non-volatile solute lowers the vapour pressure of the solvent, the solution must be heated to a higher temperature to make its vapour pressure equal to atmospheric pressure. Thus, the boiling point of the solution (Tb) is always higher than that of the pure solvent (Tb0).

The elevation of boiling point is given by: ΔTb = Tb - Tb0

Experimentally, for dilute solutions, ΔTb is directly proportional to the molal concentration (molality, m) of the solute.

ΔTb = Kb × m

ΔTb = (Kb × W2 × 1000) / (M2 × W1)

Kb is called the Molal Boiling Point Elevation Constant or Ebullioscopic Constant. Its unit is K kg mol-1.
Definition of Kb: It is the elevation in boiling point when the molality of the solution is unity (1 m).

4. Depression of Freezing Point (ΔTf)

Freezing Point: The temperature at which the vapour pressure of the substance in its liquid phase is equal to its vapour pressure in the solid phase.

When a non-volatile solute is added, the vapour pressure of the solution decreases. It now becomes equal to the vapour pressure of the solid solvent at a lower temperature. Hence, the freezing point of the solution (Tf) is less than that of the pure solvent (Tf0).

The depression of freezing point is given by: ΔTf = Tf0 - Tf

Similar to boiling point, for dilute solutions:

ΔTf = Kf × m

ΔTf = (Kf × W2 × 1000) / (M2 × W1)

Kf is called the Molal Freezing Point Depression Constant or Cryoscopic Constant. Its unit is K kg mol-1.

NCERT Real-World Application (Anti-freeze): Ethylene glycol is added to water in car radiators during winters. It acts as an antifreeze by lowering the freezing point of water, preventing it from freezing in the engine at sub-zero temperatures.

5. Osmosis and Osmotic Pressure (Π)

Certain membranes, like pig's bladder, parchment, or synthetic cellophane, allow only small solvent molecules (like water) to pass through them but block larger solute molecules. These are called Semi-Permeable Membranes (SPM).

Osmosis: The spontaneous flow of solvent molecules through a semi-permeable membrane from a region of lower solute concentration (pure solvent/dilute solution) to a region of higher solute concentration (concentrated solution).
Osmotic Pressure (Π): The excess pressure that must be applied on the solution side to just stop the flow of solvent into the solution through a semi-permeable membrane.

For dilute solutions, osmotic pressure is proportional to the molarity (C) of the solution at a given temperature (T).

Π = C R T

Π = (n2 / V) R T   &implies;   M2 = (W2 R T) / (Π V)

(Where R = Gas constant, V = Volume in Litres, C = Molarity in mol/L)

Why is Osmotic Pressure the preferred colligative property for measuring molar mass of macromolecules (proteins/polymers)?
  1. It uses Molarity instead of Molality.
  2. Measurements are taken at room temperature (macromolecules degrade at high temperatures like boiling points).
  3. The magnitude of Π is large enough to be measured accurately even for very dilute solutions.

5.1 Isotonic, Hypertonic, and Hypotonic Solutions

Type of Solution Definition Effect on Red Blood Cell (RBC)
Isotonic Two solutions having the same osmotic pressure at a given temperature. No change. RBC fluid is isotonic with 0.9% (mass/volume) NaCl solution.
Hypertonic A solution with higher osmotic pressure (concentration > 0.9% NaCl). Water flows out of the cell. RBC shrinks (Plasmolysis).
Hypotonic A solution with lower osmotic pressure (concentration < 0.9% NaCl). Water flows into the cell. RBC swells and may burst (Hemolysis).

5.2 Reverse Osmosis (RO) and Water Purification

If a pressure larger than the osmotic pressure is applied to the solution side, the direction of osmosis is reversed. Now, the pure solvent flows out of the solution through the semi-permeable membrane. This phenomenon is called Reverse Osmosis.

Desalination of Sea Water: RO is widely used to purify sea water. When pressure greater than the osmotic pressure of sea water is applied, pure water is squeezed out of the sea water through a membrane (typically Cellulose Acetate, which is permeable to water but impermeable to sea salts).

6. NCERT Solved Examples (Step-by-Step)

NCERT Example 2.8: The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. (Kb for benzene = 2.53 K kg mol-1)

Solution:
Step 1: Calculate ΔTb
ΔTb = Tb - Tb0 = 354.11 - 353.23 = 0.88 K

Step 2: Use the boiling point elevation formula
W2 (mass of solute) = 1.80 g
W1 (mass of solvent) = 90 g
Kb = 2.53 K kg mol-1
ΔTb = (Kb × W2 × 1000) / (M2 × W1)

Step 3: Solve for M2
M2 = (2.53 × 1.80 × 1000) / (0.88 × 90)
M2 = 4554 / 79.2 = 58 g mol-1.

NCERT Example 2.11: 200 cm3 of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57 × 10-3 bar. Calculate the molar mass of the protein. (R = 0.083 L bar mol-1 K-1)

Solution:
Volume (V) = 200 cm3 = 0.200 Litres
Temperature (T) = 300 K
Pressure (Π) = 2.57 × 10-3 bar
Mass of solute (W2) = 1.26 g

Using formula: M2 = (W2 R T) / (Π V)
M2 = (1.26 × 0.083 × 300) / (2.57 × 10-3 × 0.200)
M2 = 31.374 / (0.514 × 10-3) = 61,039 g mol-1.

7. Previous Year Questions (PYQs) & Exhaustive Question Bank

Part A: Conceptual (1-2 Marks)

[CBSE 2017, 2020]

Q1. Measurement of osmotic pressure method is preferred for the determination of molar masses of macromolecules such as proteins and polymers. Give two reasons.

Answer: 1. Osmotic pressure is measured at room temperature, whereas boiling point elevation requires high temperatures which could degrade biomolecules like proteins.
2. The magnitude of osmotic pressure is significantly large and easy to measure even for very dilute solutions of polymers that have poor solubility.
[CBSE 2018, 2019]

Q2. What happens when blood cells are placed in pure water? Explain.

Answer: Pure water is hypotonic compared to the fluid inside blood cells (which has an osmotic pressure equivalent to 0.9% NaCl solution). Therefore, due to osmosis, water flows into the blood cells. The cells swell and may eventually burst. This process is called hemolysis.

Part B: Assertion-Reason Type (1 Mark)

[CBSE 2023]

Q3. Assertion (A): Elevation in boiling point is a colligative property.
Reason (R): Elevation in boiling point is directly proportional to the molarity of the solution.

Answer: Assertion is Correct but Reason is Incorrect.
Elevation in boiling point is a colligative property, but it is directly proportional to the molality (m) of the solution, not molarity.
[CBSE Sample Paper 2024]

Q4. Assertion (A): Isotonic solutions must have the same molar concentration at a given temperature.
Reason (R): Osmotic pressure of a solution depends on the concentration of solute particles.

Answer: Both Assertion and Reason are correct, and Reason is the correct explanation for Assertion. Since Π = CRT, if Π and T are identical for two solutions (isotonic), their molar concentrations (C) must also be identical (assuming non-electrolytes).

Part C: Numerical Problems (3 Marks)

[CBSE 2015, 2016]

Q5. 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol-1. Find the molar mass of the solute.

Answer:
Given:
W2 = 1.00 g
W1 = 50 g
ΔTf = 0.40 K
Kf = 5.12 K kg mol-1

Formula: M2 = (Kf × W2 × 1000) / (ΔTf × W1)
M2 = (5.12 × 1.00 × 1000) / (0.40 × 50)
M2 = 5120 / 20 = 256 g mol-1.

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This module is strictly mapped to the latest rationalised NCERT syllabus for Class 12 Chemistry.
Coming up in Module 5: Abnormal Molar Mass and the van't Hoff Factor.

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