Class 12 Chemistry · NCERT Chapter 1

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Solutions Class 12 Chemistry Notes

Solutions is the study of homogeneous mixtures and their properties. This chapter covers concentration terms, Henry's law, Raoult's law, ideal and non-ideal solutions, colligative properties, osmotic pressure and the van't Hoff factor — with the board-style numericals that dominate the CBSE paper.

Class:12Subject:ChemistryCovers:Colligative properties · Raoult's lawExams:CBSE · CUET · NEET

8 Key Formulas2 Solved Examples40 Practice MCQs

DWritten byDeep Narayan

UpdatedOctober 2026

Key Concept Summary

What are the four colligative properties?

Relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure. All four depend only on the number of solute particles in a fixed amount of solvent — not on their identity — and follow from Raoult's law.

Types of Solutions and Concentration Terms

A solution is a homogeneous mixture of two or more substances. The component in larger proportion is the solvent and the others are solutes. Solutions may be gaseous, liquid or solid — a 9-type grid is built from the states of solute and solvent.

Gaseous solutions: gas in gas (air), vapour in gas (camphor), gas in liquid (oxygen in water).
Liquid solutions: gas in liquid (CO₂ in soda), liquid in liquid (ethanol in water), solid in liquid (sugar in water).
Solid solutions: gas in solid (H₂ in palladium), liquid in solid (mercury in zinc amalgam), solid in solid (brass = Cu–Zn).

Definition

Concentration terms

Mass percentage, mole fraction (x), molarity (M, mol L⁻¹) and molality (m, mol kg⁻¹) all express how much solute is present. Mole fraction and molality are temperature-independent; molarity is not.

Mole fraction of solute B in a binary solution

x_{B} = \frac{n _{B}}{n _{A} + n _{B}}, x_{A} + x_{B} = 1

MCQ Questions

Which concentration term is independent of temperature?

Brass is an example of a:

Solubility and Henry's Law

The maximum amount of solute that can dissolve in a given amount of solvent at a given temperature fixes the solubility. Solid solutes in liquids generally become more soluble as temperature rises; gases in liquids become less soluble.

Definition

Henry's law

At a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas over the solution: p = KH x. KH is solvent- and gas-specific and increases with temperature.

Raises KH: application in carbonated beverages (CO₂ under pressure) and soft-drink bottling.
Scuba diving: nitrogen dissolves under pressure; rapid ascent causes decompression sickness ('bends').
Inverse use: dissolved oxygen in water falls on heating — anoxic 'dead zones'.

Note

Henry's law describes the equilibrium of a gas over a dilute solution; it is exact for ideal gas–liquid equilibria and fails at high concentrations.

MCQ Questions

The Henry's law constant for a gas in a liquid increases with:

According to Henry's law, the solubility of a gas is proportional to its:

Vapour Pressure and Raoult's Law

Every volatile liquid exerts a characteristic vapour pressure at a given temperature. When a non-volatile solute is added, the vapour pressure of the solvent is lowered because fewer solvent molecules reach the surface.

Definition

Raoult's law (volatile liquids)

For a solution of two volatile liquids A and B, the partial vapour pressure of each component is proportional to its mole fraction: pA = pA° xA and pB = pB° xB. The total pressure p = pA + pB.

Definition

Relative lowering of vapour pressure

For a non-volatile, non-electrolyte solute: (p° − p)/p° = xB. This is a colligative property — it depends only on the mole fraction of the solute.

Exam trap

Relative lowering of vapour pressure = mole fraction of solute, and equals nB/(nA + nB). Do not multiply by 100 unless the question asks for a percentage.

MCQ Questions

The vapour pressure of a solution of a non-volatile solute is always:

The relative lowering of vapour pressure equals the:

Ideal and Non-Ideal Solutions

Definition

Ideal solution

A solution that obeys Raoult's law at all concentrations: ΔHmix = 0, ΔVmix = 0 and A–B interactions equal A–A and B–B interactions. Examples: benzene + toluene, n-hexane + n-heptane, chloroform + benzene.

Definition

Non-ideal solution

A solution that deviates from Raoult's law: A–B forces are weaker (positive deviation, ΔH > 0, ΔV > 0) or stronger (negative deviation, ΔH < 0, ΔV < 0) than the pure-component forces.

Positive deviation: ethanol + water, acetone + ethanol, chloroform + acetone mixtures that form minimum-boiling azeotropes.
Negative deviation: HNO₃ + water, chloroform + acetone — maximum-boiling azeotropes.
Azeotropes boil at constant composition and cannot be separated by distillation.

When solute–solvent interactions are weaker than solvent–solvent ones, escaping tendency rises (positive deviation, higher vapour pressure). When they are stronger, vapour pressure falls (negative deviation).

MCQ Questions

An ideal solution shows:

Ethanol and water form a minimum-boiling azeotrope because the solution:

Colligative Properties

Colligative properties depend only on the number of solute particles, not on their nature. The four are: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure.

The three measurable colligative laws (m = molality, C = molarity, R = gas constant)

Δ T_{b} = K_{b} m, Δ T_{f} = K_{f} m, π = C R T

Boiling point elevation: adding solute raises the boiling point (salt on roads, boiling vegetables).
Freezing point depression: antifreeze in radiators, ice-cream salt freezing mixtures.
Osmotic pressure: reverse osmosis for water purification, isotonic saline in drips.
Isotonic, hypertonic and hypotonic solutions describe equal, higher and lower osmotic pressure.

Why are they useful?

All four are colligative, so a single measurement — say ΔTf — reveals the moles (and hence molar mass) of a dissolved solute.

MCQ Questions

Which is NOT a colligative property?

Q10

A 0.1 molal aqueous solution of a weak acid HX is 30% ionised. Its freezing point will be:

van't Hoff Factor and Abnormal Molar Masses

Definition

van't Hoff factor i

The ratio of observed colligative effect to the value expected for a non-electrolyte: i = observed molar mass / normal molar mass. i > 1 when the solute dissociates into more particles, i < 1 when it associates.

Dissociation (i > 1): NaCl → 2 ions, i = 2 for complete dissociation; K₄[Fe(CN)₆] → 5 ions.
Association (i < 1): acetic acid dimerises in benzene, i ≈ 0.5 at high concentration.
Corrected formulas: ΔTf = iKf m, ΔTb = iKb m, π = iCRT.

Definition of the van't Hoff factor

i = \frac{normal molar mass}{observed (abnormal) molar mass} = \frac{particles in solution}{particles dissolved}

Exam trap

When solving numericals, ALWAYS multiply the colligative formula by i. For electrolytes, i = 1 + (n − 1)α where α is the degree of dissociation of a solute giving n ions.

MCQ Questions

Q11

For which solute is i > 1?

Q12

The osmotic pressure of a solution is π = iCRT. Here i:

NCERT Exercise Questions

From the textbook

NCERT questions for this chapter, in textbook order. Answers are given at a glance below each set.

Exercises (NCERT Q1–Q4)

1
Define the term solution. How many types of solutions are formed? Briefly describe each type with an example.
2
Express the concentration of a solution containing 2.5 g of sodium chloride dissolved in 50 g of water in terms of mass percentage.
3
Calculate the molarity of a solution prepared by dissolving 5.85 g of NaCl in enough water to make 500 mL of solution. (Molar mass of NaCl = 58.5 g/mol)
4
State Henry's law and mention its two important applications.

Answers at a glance

1
A solution is a homogeneous mixture of two or more substances. Based on the solvent, there are three main types: gaseous (e.g., air), liquid (e.g., sugar water), and solid (e.g., brass).
2
Mass percentage of NaCl = 4.76%
3
Molarity of NaCl solution = 0.2 M
4
Henry's law: solubility of a gas ∝ its partial pressure. Applications: carbonated beverages and scuba diving.

CBSE Pattern Questions

How CBSE asks this chapter

CBSE typically asks 2-3 questions from Solutions, focusing on numerical problems of concentration terms, Henry's law, Raoult's law, colligative properties and van't Hoff factor.

MCQ Questions

Q13

Which of the following concentration terms is independent of temperature?

Q14

The Henry's law constant for a gas in a liquid increases with:

Q15

A 0.1 molal aqueous solution of a weak acid HX is 30% ionised. The freezing point of the solution will be:

Q16

The osmotic pressure of a solution is given by π = iCRT. For which of the following does i > 1?

1 Mark Questions

1
Define molality.
2
State Henry's law.
3
What is van't Hoff factor?

2 Mark Questions

1
Calculate the molarity of a solution containing 5.85 g of NaCl in 500 mL of solution.
2
Explain why aqueous KCl solution freezes at a lower temperature than pure water.

3 Mark Questions

1
State and explain Raoult's law for volatile liquids. What are ideal and non‑ideal solutions?
2
Describe how the molecular mass of a non‑volatile solute can be determined from osmotic pressure measurements.

5 Mark Questions

1
A solution contains 10 g of urea (M = 60 g mol−1) per litre. Calculate its osmotic pressure at 300 K. (R = 0.0821 L atm K−1 mol−1)
2
The vapour pressure of pure benzene at a certain temperature is 0.850 bar. When 0.5 g of a non‑volatile solute is added to 39.0 g of benzene, the vapour pressure of the solution becomes 0.845 bar. Calculate the molecular mass of the solute.

PDF: Assignments & Tests

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Assignment 2 — Topic-wise Practice Set B

Topic-wise practice covering the second half of the chapter, with answer key.

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CBSE Pattern Test 1

Full CBSE-pattern test (90 minutes, 40 marks) with answer key.

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CBSE Pattern Test 2

Second full CBSE-pattern test (90 minutes, 40 marks) with answer key.

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Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Molarity

M = \frac{moles of solute}{volume of solution in L}

Depends on temperature because volume changes with temperature.

Molality

m = \frac{moles of solute}{kg of solvent}

Temperature independent.

Henry's law

p = K_{H} x

Solubility of a gas ∝ its partial pressure over the liquid.

Raoult's law

p_{A} = p_{A}^{\circ} x_{A}

Partial pressure of a component ∝ its mole fraction.

Elevation of boiling point

Δ T_{b} = K_{b} m

ebullioscopic constant K_b depends on the solvent.

Depression of freezing point

Δ T_{f} = K_{f} m

cryoscopic constant K_f is solvent-specific.

Osmotic pressure

π = C R T

π = iCRT when solute dissociates or associates.

van't Hoff factor

i = \frac{normal molar mass}{abnormal molar mass} = \frac{particles in solution}{particles dissolved}

i > 1 for dissociation, i < 1 for association.

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

Concentration numericals (mass %, mole fraction, molarity, molality) open the chapter and are near-certain 2-mark questions.
Henry's law constant KH rises with temperature; this trend is a favourite MCQ.
State and apply Raoult's law; distinguish ideal from non-ideal solutions and give one azeotrope of each kind.
Depression of freezing point and boiling-point elevation numericals using Kf and Kb recur in 3 and 5-mark questions.
Osmotic pressure (π = iCRT) is the preferred method to determine molar mass of macromolecules.
The van't Hoff factor and abnormal molar masses — i for NaCl, K4[Fe(CN)6], acetic acid in benzene — are standard.
Isotonic, hypertonic and hypotonic solutions and osmotic pressure relation with concentration.

Practice Numericals

Step-by-step worked examples

Standard exam-level problems solved with clear conceptual breakdowns.

Molarity of a sodium chloride solution

Problem StatementCalculate the molarity of a solution prepared by dissolving 5.85 g of NaCl in enough water to make 500 mL of solution. (Molar mass of NaCl = 58.5 g mol⁻¹)

Step-by-Step Solution

1Moles of NaCl = 5.85 g / 58.5 g mol⁻¹ = 0.1 mol.
2Volume = 500 mL = 0.5 L.
3Molarity = 0.1 mol / 0.5 L = 0.2 M.

Final Answer

Molarity = 0.2 M

Osmotic pressure of urea

Problem StatementA solution contains 10 g of urea (M = 60 g mol⁻¹) per litre. Calculate its osmotic pressure at 300 K (R = 0.0821 L atm K⁻¹ mol⁻¹).

Step-by-Step Solution

1Concentration C = 10 g / 60 g mol⁻¹ per litre = 0.1667 mol L⁻¹.
2π = CRT = 0.1667 × 0.0821 × 300.

Final Answer

π ≈ 4.11 atm

FAQ

Frequently asked questions

Which concentration term is independent of temperature and why?

Molality and mole fraction, because they are expressed per unit mass or per mole fraction of solvent, which do not change with temperature. Molarity changes because it is per litre of solution, and solution volume expands with temperature.

Why does the solubility of gases decrease with rising temperature?

Dissolution of a gas in a liquid is exothermic. By Le Chatelier's principle, increased temperature shifts the equilibrium towards the gas phase, and Henry's constant KH rises with temperature, so solubility falls.

What is an azeotrope?

An azeotrope is a constant-boiling mixture whose composition in the liquid and vapour phases is identical at its boiling point, so it cannot be separated into pure components by simple distillation. Positive-deviation mixtures (ethanol–water) give minimum-boiling azeotropes; negative-deviation mixtures (HNO₃–water) give maximum-boiling azeotropes.

Which colligative property is best for large molecules?

Osmotic pressure. It is measurable even at very low concentrations (where ΔTf and ΔTb become too small to measure accurately), so it is used to determine molar masses of polymers and proteins.

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