1. Chapter at a Glance
A solution is a homogeneous mixture of two or more substances. This chapter explains how much solute dissolves (solubility, Henry's law), how the vapour pressure of a solution depends on its composition (Raoult's law), and how dissolving a solute changes properties such as boiling point and freezing point — the colligative properties, which depend only on the number of solute particles. JEE Main asks one question, usually a numerical on colligative properties or Raoult's law.
- Vapour pressure of ideal solutions (Raoult's law) and relative lowering of vapour pressure.
- Elevation of boiling point and depression of freezing point; molar mass from them.
- Osmotic pressure and isotonic solutions.
- van't Hoff factor for electrolytes; positive and negative deviations.
2. Solubility and Henry's Law
- "Like dissolves like": polar solutes dissolve in polar solvents, non-polar in non-polar.
- Solubility of most solids increases with temperature; solubility of gases in liquids decreases with temperature.
- Henry's law: the partial pressure of a gas above a solution is proportional to its mole fraction in the solution: p = KH x. A larger KH means lower solubility.
- Applications: soda bottles are sealed under high CO₂ pressure; scuba divers risk "bends" when dissolved N₂ comes out of the blood on rapid ascent.
3. Raoult's Law and Ideal Solutions
For a solution of two volatile liquids A and B:
The mole fraction of A in the vapour is yA = pA/Ptotal. The vapour is richer in the more volatile component.
Ideal and non-ideal solutions
| Property | Ideal | Positive deviation | Negative deviation |
|---|---|---|---|
| Interactions | A–B ≈ A–A, B–B | A–B weaker | A–B stronger |
| ΔHmix, ΔVmix | 0, 0 | > 0, > 0 | < 0, < 0 |
| Vapour pressure | Obeys Raoult | Higher than expected | Lower than expected |
| Azeotrope | — | Minimum boiling | Maximum boiling |
| Example | Benzene + toluene | Ethanol + acetone | Chloroform + acetone |
4. Colligative Properties
These depend only on the number of solute particles, not on their nature (for dilute solutions of non-volatile solutes).
| Property | Formula |
|---|---|
| Relative lowering of vapour pressure | (p° − p)/p° = xsolute |
| Elevation of boiling point | ΔTb = Kb m |
| Depression of freezing point | ΔTf = Kf m |
| Osmotic pressure | π = CRT |
Kb (ebullioscopic constant) and Kf (cryoscopic constant) depend only on the solvent. For water, Kb = 0.52 K kg mol−1 and Kf = 1.86 K kg mol−1.
Molar mass from colligative properties
Osmotic pressure is preferred for proteins and polymers because it is measurable at room temperature and gives a large value even for dilute solutions. Isotonic solutions have equal osmotic pressure (0.9% NaCl is isotonic with blood). A solution of higher osmotic pressure is hypertonic; lower is hypotonic.
5. Abnormal Molar Mass and the van't Hoff Factor
Electrolytes dissociate (more particles) and some solutes associate (fewer particles). The van't Hoff factor i corrects the colligative formulas:
- i = (observed colligative property)/(calculated property) = (normal molar mass)/(observed molar mass).
- Dissociation into n ions with degree α: i = 1 + (n − 1)α. NaCl (complete): i = 2; K₂SO₄: i = 3.
- Association of n molecules into one: i = 1 − α(1 − 1/n). Acetic acid dimerises in benzene, so i < 1 (about 0.5).
6. Formula Master Table
| Quantity | Formula |
|---|---|
| Raoult (volatile) | P = xApA° + xBpB° |
| Vapour composition | yA = xApA°/P |
| Henry's law | p = KHx |
| Relative lowering | Δp/p° = x2 ≈ n2/n1 |
| Boiling point elevation | ΔTb = iKbm |
| Freezing point depression | ΔTf = iKfm |
| Osmotic pressure | π = iCRT |
| van't Hoff factor (dissociation) | i = 1 + (n − 1)α |
| Molar mass | M = Kfw21000/(ΔTfw1) |
7. Mind Map
- Like dissolves like
- Henry's law
- Gas solubility ↓ with T
- Raoult's law
- Ideal vs non-ideal
- Azeotropes
- Relative lowering
- ΔTb, ΔTf
- Osmotic pressure
- van't Hoff factor
- Dissociation, association
8. Tips and Tricks
- For comparing colligative properties of solutions, compare i × concentration — the larger product gives the larger effect (and the lower freezing point).
- 0.1 m glucose, 0.05 m NaCl and 0.033 m Na₂SO₄ (fully dissociated) all have about the same freezing point.
- Osmotic pressure in atm: π = CRT with R = 0.0821 L atm K−1 mol−1.
- If the solute is non-volatile, the vapour pressure of the solution comes from the solvent only.
- Using molarity instead of molality in ΔTb and ΔTf.
- Forgetting the van't Hoff factor for electrolytes.
- Confusing positive and negative deviation examples — hydrogen bonding between components usually gives negative deviation.
9. Important One-liners
- Colligative properties depend on the number of solute particles, not on their identity.
- Adding a non-volatile solute lowers vapour pressure, raises boiling point and lowers freezing point.
- Reverse osmosis is used for desalination of sea water.
- Salt is spread on icy roads because it lowers the freezing point of water.
- A 0.9% (w/v) NaCl solution is isotonic with blood cells.
- For association, the observed molar mass is greater than the normal molar mass.
10. PYQ-pattern Questions with Solutions
These questions are original and modelled on concepts frequently tested in JEE Main.
Liquids A and B form an ideal solution. pA° = 100 mm Hg and pB° = 60 mm Hg. Find the total vapour pressure (in mm Hg) of a solution with xA = 0.4.
Show solution
P = xApA° + xBpB° = 0.4 × 100 + 0.6 × 60 = 40 + 36 = 76 mm Hg.
Answer: 76
A 0.5 molal aqueous solution of urea is prepared (Kb of water = 0.52 K kg mol−1). The elevation in boiling point is:
- 0.52 K
- 0.26 K
- 1.04 K
- 0.13 K
Show solution
Urea is a non-electrolyte (i = 1). ΔTb = Kbm = 0.52 × 0.5 = 0.26 K.
Answer: (B)
The depression in freezing point of a 0.1 m aqueous NaCl solution (assume complete dissociation, Kf = 1.86 K kg mol−1) is:
- 0.186 K
- 0.372 K
- 0.093 K
- 0.558 K
Show solution
NaCl → Na⁺ + Cl⁻, so i = 2. ΔTf = iKfm = 2 × 1.86 × 0.1 = 0.372 K.
Answer: (B)
The osmotic pressure of a 0.1 M solution of a non-electrolyte at 300 K is (R = 0.0821 L atm K−1 mol−1):
- 0.246 atm
- 2.46 atm
- 24.6 atm
- 1.23 atm
Show solution
π = CRT = 0.1 × 0.0821 × 300 ≈ 2.46 atm.
Answer: (B)
Which of the following mixtures shows negative deviation from Raoult's law?
- Ethanol + acetone
- Benzene + toluene
- Chloroform + acetone
- Carbon disulphide + acetone
Show solution
Chloroform forms a hydrogen bond with the carbonyl oxygen of acetone, so A–B attractions are stronger than A–A and B–B: vapour pressure falls below the ideal value (negative deviation). Benzene + toluene is nearly ideal; the other two show positive deviation.
Answer: (C)
1.0 g of a non-electrolyte dissolved in 50 g of benzene lowers its freezing point by 0.40 K (Kf of benzene = 5.12 K kg mol−1). Find the molar mass (in g mol−1) of the solute.
Show solution
M = Kfw21000/(ΔTfw1) = (5.12 × 1.0 × 1000)/(0.40 × 50) = 5120/20 = 256 g mol−1.
Answer: 256
A salt AB₂ dissociates as AB₂ → A²⁺ + 2B⁻. If its van't Hoff factor in a solution is 2.2, the degree of dissociation is:
- 0.4
- 0.5
- 0.6
- 0.8
Show solution
Each formula unit gives n = 3 ions. i = 1 + (n − 1)α = 1 + 2α = 2.2 ⇒ α = 0.6.
Answer: (C)
Which aqueous solution has the lowest freezing point? (Assume complete dissociation of electrolytes.)
- 0.1 m glucose
- 0.1 m NaCl
- 0.1 m K₂SO₄
- 0.1 m urea
Show solution
Freezing point depression ∝ i × m. Glucose and urea: i = 1; NaCl: i = 2; K₂SO₄: i = 3. The largest depression (lowest freezing point) is for K₂SO₄.
Answer: (C)
11. Quick Revision Summary
- Raoult: p = x p°; ideal solutions give straight-line P–x graphs.
- Henry: p = KHx; gas solubility falls with temperature.
- Δp/p° = x2; ΔTb = iKbm; ΔTf = iKfm; π = iCRT.
- i = 1 + (n − 1)α for dissociation; i < 1 for association.
- Positive deviation: minimum-boiling azeotrope; negative: maximum-boiling azeotrope.