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📌 Illustrated Diagram: Colligative Properties

Solutions – Complete CBSE Class 12 Chemistry Notes

Unit 1 | Solutions | CBSE 2026–27

A solution is a homogeneous mixture. In a binary solution, the solvent is the component present in larger amount and the solute is the component dissolved in it.

1. Concentration of Solutions

Worked Example 1

5.85 g NaCl (molar mass 58.5 g mol⁻¹) is present in 500 mL solution. Moles = 5.85/58.5 = 0.10 mol. Therefore M = 0.10/0.500 = 0.20 M.

2. Solubility and Henry’s Law

The solubility of a gas in a liquid generally increases with pressure. Henry’s law is written as p = K_Hx. This explains why carbonated drinks are bottled under pressure and why divers can experience problems when dissolved gases come out of solution rapidly.

3. Vapour Pressure and Raoult’s Law

For an ideal solution, the partial vapour pressure of a volatile component is proportional to its mole fraction: pᵢ = xᵢpᵢ°. For a solution containing a non-volatile solute, the vapour pressure of the solvent decreases.

4. Ideal vs Non-Ideal Solutions

5. Colligative Properties

These depend primarily on the number of solute particles: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure.

ΔT_b = iK_bm   ΔT_f = iK_fm   π = iCRT

Worked Example 2

For a non-electrolyte solution with m = 0.50 mol kg⁻¹ and K_f = 1.86 K kg mol⁻¹, ΔT_f = 1.86 × 0.50 = 0.93 K.

6. van’t Hoff Factor

Association decreases the number of particles, while dissociation increases it. The factor i accounts for this abnormal particle count. For example, ideal NaCl dissociation gives approximately two ions per formula unit, so i may approach 2 in the appropriate dilute limit.

🧠 Visual Concept Map

📝 Competency Practice

1. A student calculates molarity before and after heating a solution. Explain why molarity can change even when the amount of solute remains constant.
2. Why are soft drinks stored under pressure?
3. A solute associates in solution. Predict the direction of change in observed molar mass.
4. Explain why osmotic pressure is useful for determining molar mass of large molecules.

🎯 Board-Style Numericals

1. Calculate molality when 18 g glucose (180 g mol⁻¹) is dissolved in 500 g water.
2. Calculate ΔT_f for a specified molality and K_f.
3. Use osmotic pressure data to calculate molar mass.
4. Determine the van’t Hoff factor from observed freezing-point depression.

MCQs – Chapter Specific

1. Which concentration is based on mass of solvent? Molality.
2. Which law relates gas pressure and mole fraction in solution? Henry’s law.
3. A non-volatile solute generally lowers the solvent’s vapour pressure. True.
4. If a solute dissociates into more particles, the van’t Hoff factor tends to be greater than 1. True.

Deep-Dive: Core Concepts & Numericals

Concentration Terms

Molarity is moles of solute per litre of solution, while molality is moles of solute per kilogram of solvent. Mole fraction is the ratio of moles of a component to total moles. Always identify whether the question gives solution volume or solvent mass before choosing the formula.

Worked Numerical

For 0.5 mol solute dissolved to make 2 L solution, M=0.5/2=0.25 mol L⁻¹.

Colligative Properties

Relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure depend on the number of solute particles. For dilute solutions, ΔT_b=K_bm and ΔT_f=K_fm.

Visual Map

Common Mistakes

• Confusing molarity with molality.
• Using mass of solution instead of mass of solvent for molality.
• Forgetting van’t Hoff factor for association/dissociation.

Practice

1. Calculate molarity and molality from given masses.
2. Apply Raoult’s law to a binary solution.
3. Calculate elevation/depression of temperature.
4. Determine molar mass using osmotic pressure.

Concept Builder: Solutions

Concentration Terms

Use molarity, molality, mole fraction and mass percentage carefully. Molarity depends on volume of solution, while molality depends on mass of solvent and is therefore less affected by temperature.

Raoult’s Law — Visual

Colligative Properties

For dilute solutions, elevation in boiling point and depression in freezing point depend on the number of solute particles. For association or dissociation, the van’t Hoff factor i must be considered.

Worked Numerical

For a solution with molality 0.50 m and K_f=1.86 K kg mol⁻¹, assuming i=1, ΔT_f=K_fm=1.86×0.50=0.93 K.

Exam Traps

• Do not confuse molarity with molality.
• Use the van’t Hoff factor when solute particles associate or dissociate.
• Check whether the question gives mass of solvent or solution.

CBSE Application Practice

1. Calculate molality from masses of solute and solvent.
2. Determine molar mass from elevation/depression data.
3. Explain abnormal molar mass using i.
4. Interpret a vapour-pressure graph.