Solutions Class 12 Notes - CBSE Chemistry Chapter 2 (Free PDF)

Chapter summary

Solutions covers how to express concentration (molarity, molality, mole fraction, mass percent), how solubility of solids and gases responds to temperature and pressure, and Henry’s and Raoult’s laws for vapour pressure of ideal and non-ideal mixtures. Its heart is the four colligative properties, which depend only on the number of solute particles, plus the van’t Hoff factor that corrects them for dissociation and association. It is a high-yield NEET chapter because its formulas reliably appear as numerical problems and the colligative concepts link directly to molar-mass determination.

Chapter notes
🃏 Flash Cards: Solutions

Class 12 Chemistry · Chapter 1 – swipe through all 10 cards to understand the whole chapter.

🧪Start here1/10

Concentration Terms

Concentration says how much solute sits in a given amount of solvent or solution.

M = n_solute / V_solution(L) · m = n_solute / mass_solvent(kg)

Molality & mole fraction are temperature-independent; molarity & normality change with T.

  • Mole fraction: x_A + x_B = 1 (moles of one / total moles)
  • Mass % = (mass solute / mass solution) × 100; ppm for very dilute
  • Molarity ↔ molality conversion needs the density
🥤Core law2/10

Solubility & Henry’s Law

Solubility is the max solute that dissolves at a given temperature; gases behave opposite to solids.

p = K_H · x

K_H is the Henry’s-law constant (pressure units); higher K_H → lower gas solubility.

  • Most solids: solubility rises with T (endothermic); pressure barely matters
  • Gases: solubility rises with pressure, falls with T (warm soda goes flat)
  • K_H increases with T; diluted-air scuba tanks avoid the ‘bends’
💨Key law3/10

Raoult’s Law

Each volatile component’s partial vapour pressure equals its pure value times its mole fraction.

p_A = p°_A x_A · p_total = p°_A x_A + p°_B x_B

For a non-volatile solute only solvent vaporises: p_solution = p°_solvent · x_solvent.

  • A non-volatile solute always lowers vapour pressure (x_solvent < 1)
  • Ideal solution obeys Raoult at all x: ΔH_mix = 0, ΔV_mix = 0
  • Examples: benzene + toluene, n-hexane + n-heptane
⚖️Deviations4/10

Non-ideal Solutions & Azeotropes

When A-B forces differ from A-A/B-B forces, solutions deviate and form constant-boiling azeotropes.

Positive dev → min-boiling azeotrope · Negative dev → max-boiling azeotrope

Azeotropes boil at constant composition and can’t be split by simple distillation.

  • Positive: weaker A-B forces, VP higher than predicted (ethanol + water)
  • Negative: stronger A-B forces, VP lower than predicted (acetone + chloroform)
  • Positive ΔH > 0 & ΔV > 0; negative ΔH < 0 & ΔV < 0
🌡️Big idea5/10

Colligative Properties

These depend only on the NUMBER of solute particles, not their nature.

(p° − p_s)/p° = x_solute

1 mol glucose and 1 mol urea shift the solution by the same amount.

  • Relative lowering of vapour pressure = solute mole fraction
  • Built on molality because it is temperature-proof
  • They reveal dissociation or association of the solute
♨️Property6/10

Boiling-point Elevation

A non-volatile solute raises the solvent’s boiling point.

ΔT_b = K_b · m

K_b = molal elevation (ebullioscopic) constant; m = molality.

  • ΔT_b = T_b(solution) − T_b(pure solvent)
  • More solute particles → larger ΔT_b
  • K_b is a property of the solvent, not the solute
❄️Property7/10

Freezing-point Depression

A solute lowers the freezing point, which is why salt melts ice on roads.

ΔT_f = K_f · m

K_f = molal depression (cryoscopic) constant; m = molality.

  • ΔT_f = T_f(pure solvent) − T_f(solution)
  • Depends only on particle count, not identity
  • Used to find molar mass of an unknown solute
🧫Property8/10

Osmotic Pressure

The pressure needed to stop solvent flowing across a semipermeable membrane into the solution.

π = C R T = (n/V) R T

R = 0.0821 L atm mol⁻1 K⁻1, T in kelvin; equal-π solutions are isotonic.

  • Most accurate colligative property, best for macromolecules
  • Molar mass: M = w R T / (π V)
  • Isotonic = same π; hyper/hypotonic shrink or swell cells
🔢Correction9/10

Van’t Hoff Factor

Factor i corrects colligative formulas when a solute dissociates or associates.

i = observed property / normal property = normal M / observed M

Modified: ΔT_b = i K_b m, ΔT_f = i K_f m, π = i C R T.

  • Dissociation i > 1: NaCl → 2 (i ≈ 2), BaCl2 → 3, K4[Fe(CN)6] → 5
  • Association i < 1: benzoic acid dimerises in benzene (i ≈ 0.5)
  • No change i = 1: glucose, urea, sucrose
📐Advanced10/10

Degree of Dissociation/Association

From i you can back out how completely a solute splits or pairs up.

α_diss = (i − 1)/(n − 1) · α_assoc = (1 − i)/(1 − 1/n)

n = number of ions (dissociation) or molecules combining (association).

  • Higher i → larger colligative effect → lower observed molar mass
  • Dissociation raises particle count; association lowers it
  • Abnormal molar mass arises whenever i ≠ 1
Swipe Click a card to focus 10 cards
📝 Practice Solutions — 10 NEET PYQs
Real previous-year questions · with answers & solutions
Start →Close ✕
Tap an option to check your answer and see the worked solution. Every question is a real NEET previous-year question.
Q1NEET 2021
At 45 degC the vapour pressure of benzene is 280 mmHg and that of octane is 420 mmHg. Assuming an ideal solution, the total vapour pressure of a solution containing benzene and octane in molar ratio 3 : 2 is:
Correct answer: C. Mole fractions: benzene = 3/5, octane = 2/5. pₜₒₜₐₗ = 280 × (3/5) + 420 × (2/5) = 168 + 168 = 336 mmHg.
🔎 See the full step-by-step solution in the app →
Q2NEET 2021
10 g of glucose (C₆H₁₂O₆), 10 g of urea (CH₄N₂O) and 10 g of sucrose (C₁₂H₂₂O₁₁) are each dissolved in 250 mL of water giving osmotic pressures p₁, p₂ and p₃ respectively. The correct order of decreasing osmotic pressure is:
Correct answer: A. All three are non-electrolytes (i = 1), so osmotic pressure depends on moles. Moles: glucose = 10/180 = 0.056; urea = 10/60 = 0.167; sucrose = 10/342 = 0.029. Order urea > glucose > sucrose, i.e. p₂ > p₁ > p₃.
🔎 See the full step-by-step solution in the app →
Q3NEET 2020
A cylinder contains a mixture of 7 g of N₂ and 8 g of Ar. If the total pressure of the mixture is 27 bar, the partial pressure of N₂ is (atomic masses: N = 14, Ar = 40):
Correct answer: B. Moles: N₂ = 7/28 = 0.25; Ar = 8/40 = 0.20. Mole fraction of N₂ = 0.25/0.45 = 5/9. By Dalton’s law (Raoult/partial pressure), p_(N₂) = x_(N₂) × pₜₒₜₐₗ = (5/9) × 27 = 15 bar.
🔎 See the full step-by-step solution in the app →
Q4NEET 2020
Isotonic solutions have the same:
Correct answer: C. Isotonic solutions are defined as having the same osmotic pressure at a given temperature (π = CRT), which means they have the same molar concentration of particles.
🔎 See the full step-by-step solution in the app →
Q5NEET 2017
Which of the following modes of expressing concentration is dependent on temperature?
Correct answer: B. Molarity = moles of solute / volume of solution (in L). Volume expands or contracts with temperature, so molarity is temperature dependent. Molality, mole fraction and weight percentage are all based on mass, which is temperature independent.
🔎 See the full step-by-step solution in the app →
Q6NEET 2016
The van’t Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide, Ba(OH)₂, is:
Correct answer: D. Strong electrolytes dissociate completely. Ba(OH)₂ arrow Ba²⁺ + 2OH⁻ gives 3 ions per formula unit, so i = 3.
🔎 See the full step-by-step solution in the app →
Q7NEET 2015
A gas such as carbon monoxide would be most likely to obey the ideal gas law at:
Correct answer: A. Real gases approach ideal behaviour when intermolecular forces and molecular volume are negligible. This occurs at high temperature (high kinetic energy overcomes attractions) and low pressure (molecules far apart).
🔎 See the full step-by-step solution in the app →
Q8NEET 2015
The boiling point of a 0.2 molal solution of X in water is greater than that of an equimolal solution of Y in water. Which statement is true?
Correct answer: A. Δ T_b = i K_b m. At equal molality, a larger Δ T_b means a larger van’t Hoff factor i, i.e. X gives more particles in solution. This indicates X is dissociating (ionising) in water, raising its effective particle count.
🔎 See the full step-by-step solution in the app →
Q9NEET 2010
25.3 g of sodium carbonate (Na₂CO₃, molar mass 106 g/mol) is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely, the molar concentrations of Na⁺ and CO₃²⁻ ions respectively are:
Correct answer: B. Moles of Na₂CO₃ = 25.3/106 = 0.2387 mol; molarity = 0.2387/0.250 = 0.955 M. Each formula unit gives 2 Na⁺ and 1 CO₃²⁻, so [Na⁺] = 2 × 0.955 = 1.910 M and [CO₃²⁻] = 0.955 M.
🔎 See the full step-by-step solution in the app →
Q10NEET 2007
Concentrated aqueous sulphuric acid is 98% H₂SO₄ by mass and has a density of 1.80 g/mL. The volume of this acid required to make one litre of 0.1 M H₂SO₄ solution is:
Correct answer: D. Moles of H₂SO₄ needed = 0.1 × 1 = 0.1 mol = 0.1 × 98 = 9.8 g. Concentrated acid: 1 mL has mass 1.80 g, of which 98% is H₂SO₄ = 1.764 g/mL. Volume = 9.8/1.764 = 5.55 mL.
🔎 See the full step-by-step solution in the app →
View 20+ more practice questions, gamified →
Free · no signup · works in your browser
Studying this chapter? Track it - saved on this device, no login.

Frequently Asked Questions

What is a colligative property?

A colligative property depends only on the number of solute particles in a solution and not on their chemical identity. The four colligative properties are relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure.

What are the key colligative-property formulas in this chapter?

Relative lowering of vapour pressure equals the solute mole fraction; elevation of boiling point is delta T_b = K_b times molality; depression of freezing point is delta T_f = K_f times molality; and osmotic pressure is pi = CRT. When a solute dissociates or associates, each formula is multiplied by the van’t Hoff factor i.

What is the difference between molarity and molality?

Molarity is moles of solute per litre of solution, while molality is moles of solute per kilogram of solvent. Molarity changes with temperature because the solution volume expands, but molality is temperature-independent since it uses mass, which is why colligative-property formulas use molality.

What is the van’t Hoff factor and when is it not equal to 1?

The van’t Hoff factor i is the ratio of the observed colligative effect to the value expected for an undissociated solute. It is greater than 1 for solutes that dissociate, such as NaCl giving about 2 and BaCl2 about 3, and less than 1 for solutes that associate, such as benzoic acid dimerising in benzene; for non-electrolytes like glucose and urea i equals 1.

Is Solutions important for NEET and how does Henry’s law differ from Raoult’s law?

Yes, Solutions is part of the Class 12 NEET chemistry syllabus and is a high-scoring chapter built on direct numerical formulas. Henry’s law relates the partial pressure of a gas above a solution to its mole fraction in the liquid (p = K_H times x), whereas Raoult’s law relates the vapour pressure of a volatile liquid component to its mole fraction (p_A = p_A pure times x_A); Henry’s law is essentially the special case for gas solubility.

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top