📚 StudyHub

🧪 Chemistry  ·  Class 12  ·  NEET & JEE

Solutions

Learn how substances dissolve and form mixtures. Covers concentration terms, colligative properties (boiling point elevation, freezing point depression, osmotic pressure), and ideal vs non-ideal solution behaviour.

Practice Solutions Quiz — 100% Free →
Reading time~8 min
Revision time~3 min
Last updated2026-07-19
1 Read the chapter ~8 min

🎯 Key Points

  • Colligative properties depend on NUMBER of solute particles, not their identity: vapour pressure lowering, ΔTb, ΔTf, osmotic pressure (π)
  • Molality (not molarity) is used for colligative properties since it's temperature-independent
  • Van't Hoff factor i: dissociation gives i > 1 (e.g. NaCl, i≈2); association gives i < 1 (e.g. benzoic acid dimer, i≈0.5); non-electrolytes i=1
  • Modified equations: ΔTb=iKbm, ΔTf=iKfm, π=iCRT
  • Ideal solutions obey Raoult's law exactly (ΔHmix=0); positive deviation → minimum boiling azeotrope; negative deviation → maximum boiling azeotrope
  • Henry's Law: gas solubility ∝ partial pressure — explains why soda goes flat when opened
Vapour Pressure: Pure Solvent vs SolutionTVapour PPure solventSolution (lower VP)ΔTfTf(soln)Tf(pure)Dissolved solute lowers vapour pressure at every temperature, shifting Tf down and Tb up

Adding a non-volatile solute lowers the solvent's vapour pressure at every temperature, which is the root cause behind all four colligative properties: it shifts the freezing point down and the boiling point up.

Types of Solutions

A solution is a homogeneous mixture. The component in larger amount is the solvent; the smaller is the solute.

Concentration Terms

  • Molarity (M): Moles of solute per litre of solution
  • Molality (m): Moles of solute per kg of solvent (temperature-independent)
  • Mole fraction (X): Moles of component / total moles
  • ppm: Parts per million (for very dilute solutions)

Colligative Properties

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

  • Vapour pressure lowering: ΔP/P° = Xsolute (Raoult's law)
  • Boiling point elevation: ΔTb = Kb × m
  • Freezing point depression: ΔTf = Kf × m
  • Osmotic pressure: π = iMRT

Van't Hoff Factor (i)

  • Electrolytes that dissociate: i > 1
  • Solutes that associate: i < 1
  • Non-electrolytes: i = 1

Henry's Law

Mass of gas dissolved is proportional to its partial pressure: p = KH × X. This explains why carbonated drinks go flat when opened (pressure drops).

Quick Tips

  • Molality is used for colligative properties (not molarity) because it is temperature-independent
  • Antifreeze (ethylene glycol) works by depression of freezing point
  • Osmotic pressure is used to determine molar mass of large molecules like proteins

Raoult's Law for Ideal and Non-Ideal Solutions

  • Raoult's law (general): For a solution of volatile liquids, partial vapour pressure of each component is proportional to its mole fraction: pA = pA° × xA
  • Ideal solutions: Obey Raoult's law at all concentrations; ΔHmixing = 0, ΔVmixing = 0 (e.g., benzene + toluene, n-hexane + n-heptane)
  • Non-ideal solutions with positive deviation: A-B interactions weaker than A-A/B-B; vapour pressure higher than predicted (e.g., ethanol + acetone, water + ethanol); forms a minimum boiling azeotrope
  • Non-ideal solutions with negative deviation: A-B interactions stronger than A-A/B-B; vapour pressure lower than predicted (e.g., chloroform + acetone, HNO₃ + water); forms a maximum boiling azeotrope
Total vapour pressure versus composition: a straight line for ideal Raoult's-law behaviour, a red curve bulging above it (positive deviation, maximum equals positive azeotrope) and a blue curve dipping below it (negative deviation, minimum equals negative azeotrope).

Real solutions deviate from Raoult's law: positive deviation (red) occurs when A–B attractions are weaker than in the pure liquids, giving a higher vapour pressure and a minimum-boiling azeotrope; negative deviation (blue) occurs when A–B attractions are stronger, giving a maximum-boiling azeotrope. Image: Д.Ильин (vectorization), CC0, via Wikimedia Commons.

Azeotropes

Mixtures of two liquids that boil at a constant temperature and have the same composition in liquid and vapour phase, so they cannot be separated by simple fractional distillation. Ethanol-water (95.6% ethanol) is a classic minimum boiling azeotrope.

Van't Hoff Factor: Worked Reasoning

  • i = (observed colligative property)/(calculated colligative property) = (normal molar mass)/(observed/abnormal molar mass)
  • NaCl in water: i ≈ 2 (dissociates into Na⁺ and Cl⁻)
  • K₂SO₄ in water: i ≈ 3 (dissociates into 2K⁺ and SO₄²⁻)
  • Benzoic acid in benzene: i ≈ 0.5 (dimerises due to hydrogen bonding, halving the effective number of particles)

Modified Colligative Property Equations (with i)

  • ΔTb = i × Kb × m
  • ΔTf = i × Kf × m
  • π = i × CRT
  • ΔP/P° = i × xsolute

Types of Solutions (by Physical State)

Depending on the state of solute and solvent, nine kinds of solution exist:

SoluteSolventExample
GasGasAir (O₂ in N₂)
GasLiquidCO₂ in soda water
GasSolidH₂ adsorbed in palladium
LiquidLiquidEthanol in water
SolidLiquidSalt/sugar in water
SolidSolidAlloys (brass, bronze)

Solubility: Effect of Temperature and Pressure

  • Solid in liquid: If dissolution is endothermic (most salts, e.g. KNO₃), solubility increases with temperature; if exothermic (e.g. Ce₂(SO₄)₃), it decreases. Pressure has almost no effect on solids/liquids
  • Gas in liquid: Solubility DECREASES with temperature (dissolution of a gas is exothermic) and INCREASES with pressure (Henry's law) — warm soda holds less CO₂ and fizzes more

Osmosis and Types of Solutions

  • Osmosis: Net flow of solvent from a dilute (or pure) solution to a concentrated solution across a semipermeable membrane
  • Osmotic pressure (π): The external pressure that must be applied on the concentrated side to just stop osmosis; π = iCRT
  • Isotonic solutions have the same osmotic pressure (no net flow); a cell in isotonic 0.9% NaCl stays intact
  • Hypertonic (higher π) causes a cell to shrink (exosmosis); hypotonic (lower π) causes it to swell/burst (endosmosis)
  • Reverse osmosis: Applying pressure greater than π forces solvent backward through the membrane — used for desalination of sea water

Abnormal Molar Mass

  • When a solute associates or dissociates, the number of particles differs from what the formula suggests, so the molar mass calculated from a colligative property is "abnormal"
  • Dissociation (more particles) gives an observed molar mass LOWER than the true value (e.g. KCl in water)
  • Association (fewer particles) gives an observed molar mass HIGHER than the true value (e.g. acetic/benzoic acid in benzene, which dimerise)
  • The van't Hoff factor corrects for this: i = normal molar mass / observed molar mass

Applications of Henry's Law

  • Soft-drink bottles are sealed under high CO₂ pressure to increase gas solubility
  • Deep-sea divers face "bends" (nitrogen bubbles in blood) on rapid ascent; diving cylinders use helium-diluted air to reduce dissolved N₂
  • At high altitude the low partial pressure of O₂ lowers blood oxygen, causing anoxia/altitude sickness
  • A higher value of the Henry's constant KH means LOWER solubility of the gas at a given pressure

🚀 JEE Advanced Edge

Degree of dissociation/association from i: For a solute dissociating into n ions with degree of dissociation α: i = 1 + (n−1)α. For a solute associating into clusters of n molecules with degree of association α: i = 1 − α + α/n. Rearranging either equation lets you solve for α directly from an experimentally measured i.

Relative lowering of vapour pressure for molar mass determination: ΔP/P° = (moles solute)/(moles solute + moles solvent) ≈ wB×M_A/(M_B×W_A) for dilute solutions — measuring vapour pressure lowering precisely gives a route to determine the unknown molar mass M_B of a solute.

Worked problem: 0.1 m aqueous solution of an electrolyte AB₂ has ΔTf = 0.6°C (Kf for water = 1.86 K·kg/mol, assume complete dissociation). Find the van't Hoff factor and verify it matches AB₂ → A²⁺ + 2B⁻. Approach: i = ΔTf/(Kf×m) = 0.6/(1.86×0.1) = 3.23 ≈ 3, matching the expected i=3 for complete dissociation into 3 ions (1 A²⁺ + 2 B⁻), confirming near-complete dissociation.

2 Revise ~3 min before the exam

📐 Formula Sheet

  • Concentration: molarity M = mol/L; molality m = mol/kg solvent; mole fraction x
  • Henry's law: p = KH·x (gas solubility ∝ partial pressure)
  • Raoult's law: p = p°·x (ideal solutions of volatile liquids)
  • Relative lowering of vapour pressure: (p° − p)/p° = xsolute
  • Boiling-point elevation: ΔTb = Kb·m  |  Freezing-point depression: ΔTf = Kf·m
  • Osmotic pressure: π = CRT = (n/V)RT
  • Van't Hoff factor i: multiply every colligative property by i (i > 1 for dissociation, < 1 for association)
  • Degree of dissociation: α = (i − 1)/(n − 1)
3 Practice apply it

✍️ Worked Examples

Example 1 — Freezing-point depression
Q: 1.8 g of glucose (M = 180) is dissolved in 100 g of water. Find the depression in freezing point. (Kf = 1.86 K·kg/mol)
Step 1 — Moles of glucose: 1.8/180 = 0.01 mol.
Step 2 — Molality: 0.01 mol / 0.1 kg = 0.1 m.
Step 3 — Apply ΔTf = Kf·m: 1.86 × 0.1 = 0.186 K.
Answer: 0.186 K. Note: glucose does not dissociate, so i = 1 and no correction is needed.

Example 2 — Van't Hoff factor
Q: Why does 0.1 m NaCl depress the freezing point about twice as much as 0.1 m glucose?
Step 1 — NaCl dissociates into Na⁺ and Cl⁻, giving two particles per formula unit, so i ≈ 2.
Step 2 — Colligative properties depend on the number of particles, not their identity.
Step 3 — So ΔTf for NaCl is roughly 2× that of the non-dissociating glucose.
Answer: because NaCl yields twice as many dissolved particles. Key idea: always ask "how many particles?" for colligative problems.

Example 3 — Osmotic pressure
Q: Find the osmotic pressure of 0.2 M glucose at 300 K. (R = 0.0821 L·atm/mol·K)
Step 1 — Use π = CRT.
Step 2 — Substitute: π = 0.2 × 0.0821 × 300.
Step 3 — Compute: ≈ 4.93 atm.
Answer: ≈ 4.93 atm. Note: osmotic pressure is the most sensitive colligative property, which is why it is used to find the molar masses of large molecules like proteins.

Practice Solutions Quiz — 100% Free →

Frequently Asked Questions — Solutions

What are the key concepts in Solutions?
Learn how substances dissolve and form mixtures. Covers concentration terms, colligative properties (boiling point elevation, freezing point depression, osmotic pressure), and ideal vs non-ideal solution behaviour.
Is Solutions important for NEET & JEE?
Yes. Solutions is part of the Chemistry Class 12 NCERT syllabus and is directly tested in NEET and JEE examinations. StudyHub provides structured notes, diagrams, and practice questions covering all exam-level subtopics.
How can I practice Solutions questions on StudyHub?
Open StudyHub and select Chemistry → Solutions. Choose Easy, Medium, or Hard difficulty. Hard-tier questions are at NEET & JEE level with full step-by-step explanations.

References

  1. NCERT Class 12 Chemistry Textbook — Chapter: Solutions
  2. CBSE Curriculum — Chemistry (Class 12)
  3. NTA NEET UG Official Syllabus — subject-wise topic list
  4. NTA JEE Main Official Syllabus — subject-wise topic list