Solutions Chemistry Class 12 Notes for CBSE, JEE and NEET

Solutions is Unit 1 of 10 in the current NCERT Class 12 Chemistry book (it was Chapter 2 in older books), and it carries 7 marks in the CBSE theory paper for 2026-27. It is a Physical Chemistry unit with both concept and numerical questions, and it uses the mole concept from Class 11. This page is about the chapter named Solutions. For the textbook exercise answers, see eSaral’s NCERT Solutions for Class 12.

Solutions Chemistry Class 12 Notes: Chapter Overview and What Each Exam Needs

The CBSE unit covers: types of solutions, expression of concentration of solutions, solubility, vapour pressure of liquid solutions, ideal and non-ideal solutions, colligative properties and determination of molar mass, and abnormal molecular masses.

The number of questions varies every year, so check previous years’ papers and the latest syllabus of your exam.

Tip from the eSaral, IIT Bombay Faculty: In numericals, write the given data with units first, convert all the masses to the units that the formula needs, and then substitute. Most mistakes come from a mixed-up unit (g and kg, or mL and L).

Types of Solutions and Concentration Terms in Solutions Chemistry Class 12 Notes

A solution is a homogeneous mixture of two or more components. The component present in the larger amount is the solvent, and the other is the solute.

Ways of Expressing Concentration

Molarity changes with temperature, because the volume changes. Molality does not change with temperature, because mass does not change. Use molality in colligative-property calculations. Normality is not part of the current NCERT chapter. For worked conversions between these terms, see eSaral’s mole concept notes for Class 11.

Solubility and Henry’s Law in Solutions Chemistry Class 12 Notes

The solubility of a gas in a liquid increases with pressure and decreases with temperature. Henry’s law says that the partial pressure of a gas in the vapour phase is proportional to its mole fraction in the solution:

p = K_H × x, where K_H is Henry’s law constant. A higher K_H means a lower solubility at the same pressure.

Vapour Pressure and Raoult’s Law in Solutions Chemistry Class 12 Notes

The vapour pressure of a liquid is the pressure of its vapour in equilibrium with the liquid at a given temperature, and it increases with temperature.

• Raoult’s law for a non-volatile solute: p₁ = x₁ p₁°, where x₁ is the mole fraction of the solvent and p₁° is the vapour pressure of the pure solvent.
• Relative lowering of vapour pressure: (p₁° − p₁)/p₁° = x₂, the mole fraction of the solute. For dilute solutions, it is approximately n₂/n₁ = w₂M₁/(M₂w₁).
• Two volatile liquids: p_total = x₁p₁° + x₂p₂°. The mole fraction of a component in the vapour is its partial pressure divided by the total pressure.

Ideal and Non-Ideal Solutions in Solutions Chemistry Class 12 Notes

Azeotropes are liquid mixtures that boil at a constant temperature, and their vapour has the same composition as the liquid. Positive-deviation mixtures form minimum-boiling azeotropes (ethanol and water), and negative-deviation mixtures form maximum-boiling azeotropes (nitric acid and water).

India's Best Exam Preparation for Class 12th - Download Now

Solutions | Chemistry Class 12 Notes | IIT-JEE & NEET

India's Best Exam Preparation for Class 12th - Download Now

India's Best Exam Preparation for Class 12th - Download Now

India's Best Exam Preparation for Class 12th - Download Now 

Colligative Properties in Solutions Chemistry Class 12 Notes

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

Elevation of Boiling Point in Solutions Chemistry Class 12 Notes

ΔT_b = K_b × m, and the molar mass is M₂ = (K_b × w₂ × 1000)/(ΔT_b × w₁), with w₁ and w₂ in grams. K_b depends only on the solvent: water 0.52 K kg/mol and benzene 2.53 K kg/mol.

Depression of Freezing Point in Solutions Chemistry Class 12 Notes

ΔT_f = K_f × m, and M₂ = (K_f × w₂ × 1000)/(ΔT_f × w₁). K_f for water is 1.86 K kg/mol and for benzene it is 5.12 K kg/mol. Antifreeze in a car radiator and salt on icy roads both lower the freezing point.

Osmotic Pressure in Solutions Chemistry Class 12 Notes

Osmosis is the flow of the solvent through a semipermeable membrane from the dilute solution to the concentrated solution. The osmotic pressure is π = CRT = (n₂/V)RT, with R = 0.0821 L atm mol⁻¹ K⁻¹. The molar mass is M₂ = w₂RT/(πV).

• Isotonic solutions have the same osmotic pressure. Hypertonic means a higher osmotic pressure than the reference, and hypotonic means a lower one.
• Osmotic pressure is used for the molar mass of large molecules such as proteins and polymers, because it can measure very dilute solutions at room temperature.
• Reverse osmosis applies a pressure greater than the osmotic pressure to the solution side, so the solvent flows out of the solution. It is used for the desalination of sea water.

Abnormal Molar Mass and van’t Hoff Factor in Solutions Chemistry Class 12 Notes

If the solute dissociates or associates in the solution, the number of particles changes, and the calculated molar mass appears abnormal. The van’t Hoff factor corrects for this:

i = observed colligative property ÷ calculated colligative property = normal molar mass ÷ observed molar mass

The modified formulas are ΔT_b = i K_b m, ΔT_f = i K_f m, π = i CRT and relative lowering = i x₂.

• Degree of dissociation (a solute that gives n ions): i = 1 + α(n − 1), so α = (i − 1)/(n − 1).
• Degree of association (n molecules forming one): i = 1 − α(1 − 1/n).

Key Formulas in Solutions Chemistry Class 12 Notes

Solved Examples from Solutions Chemistry Class 12 Notes

Example 1 (concentration): 18 g of glucose (M = 180) is dissolved in 200 g of water. Find the molality and the mole fraction of glucose.

n = 18/180 = 0.1 mol. Molality = 0.1/0.2 = 0.5 m. Moles of water = 200/18 = 11.11, so x = 0.1/(0.1 + 11.11) ≈ 0.009.

Example 2 (relative lowering of vapour pressure): Find the relative lowering of vapour pressure in Example 1.

The relative lowering equals the mole fraction of the solute, so it is ≈ 0.009. If the vapour pressure of pure water is 23.8 mm Hg, the lowering is about 0.21 mm Hg.

Example 3 (boiling point): 6 g of urea (M = 60) is dissolved in 200 g of water. Find the boiling point (K_b = 0.52 K kg/mol).

m = (6/60)/0.2 = 0.5 mol/kg. ΔT_b = 0.52 × 0.5 = 0.26 K. The boiling point is about 100.26 °C.

Example 4 (freezing point with an electrolyte): Find the freezing point of a 0.1 m NaCl solution, assuming complete dissociation (K_f = 1.86 K kg/mol).

i = 2, so ΔT_f = 2 × 1.86 × 0.1 = 0.372 K. The freezing point is about −0.37 °C.

Example 5 (molar mass from osmotic pressure): 2 g of a polymer in 500 mL of solution gives an osmotic pressure of 0.0123 atm at 300 K. Find its molar mass.

M₂ = w₂RT/(πV) = (2 × 0.0821 × 300)/(0.0123 × 0.5) ≈ 8.0 × 10³ g/mol.

Example 6 (degree of dissociation): A 0.1 m solution of a weak electrolyte HA shows ΔT_f = 0.2 K (K_f = 1.86 K kg/mol). Find the degree of dissociation.

i = ΔT_f/(K_f × m) = 0.2/0.186 ≈ 1.075. α = (i − 1)/(n − 1) = 0.075/1 = 0.075, or 7.5%.

Common Mistakes in Solutions Chemistry Class 12 Notes Problems

How to Study Solutions Chemistry Class 12 Notes

1. Read the NCERT chapter first, and learn the concentration terms. Download the book from eSaral’s NCERT books for Class 12 page.
2. Practise numericals for each colligative property, first without i and then with i.
3. Learn the deviations table and the azeotrope examples as short facts for the boards.
4. Revise the chapter with eSaral’s Class 12 Chemistry notes, and then solve previous year questions.