1. Chapter at a Glance
A solution is a homogeneous mixture of two or more substances. This chapter covers ways of expressing concentration, the solubility of solids and gases (Henry's law), vapour pressure and Raoult's law, ideal and non-ideal solutions, and colligative properties — properties that depend only on the number of solute particles. NEET asks one or two questions every year, most often numericals on colligative properties and the van't Hoff factor.
- Henry's law and the solubility of gases.
- Raoult's law; vapour pressure of mixtures; vapour composition.
- Positive and negative deviations; azeotropes.
- Relative lowering of vapour pressure, ΔTb, ΔTf and osmotic pressure.
- Van't Hoff factor for dissociation and association; molar mass determination.
2. Types of Solutions and Concentration
- The component present in the larger amount is the solvent; the others are solutes. Solutions can be gaseous (air), liquid (sugar in water, O₂ in water) or solid (alloys such as brass).
- Concentration terms: mass %, volume %, ppm, mole fraction, molarity (M), molality (m). Mole fraction and molality do not depend on temperature.
- Solubility of solids in liquids generally depends on temperature (Le Chatelier). Dissolving is exothermic for some salts and endothermic for others. Pressure has little effect on it.
3. Solubility of Gases: Henry's Law
- A higher KH means a lower solubility at the same pressure.
- Solubility of gases increases with pressure and decreases with temperature (KH rises as T rises). Fish prefer cold water because it holds more oxygen.
- Applications: soft drinks are bottled under high CO₂ pressure; scuba divers breathe air diluted with helium to avoid "the bends" (N₂ dissolving in blood at depth); at high altitudes low O₂ pressure causes anoxia.
4. Vapour Pressure and Raoult's Law
- For a mixture of two volatile liquids, the partial vapour pressure of each is proportional to its mole fraction.
- The vapour is richer in the more volatile component (the one with the higher p°).
- For a non-volatile solute, only the solvent contributes: p = xsolventp°. So adding a solute lowers the vapour pressure.
- Raoult's law is a special case of Henry's law with KH = p°.
5. Ideal and Non-ideal Solutions
| Property | Ideal | Positive deviation | Negative deviation |
|---|---|---|---|
| A–B attraction vs A–A, B–B | Equal | Weaker | Stronger |
| Vapour pressure | Obeys Raoult | Higher than expected | Lower than expected |
| ΔmixH | 0 | Positive (endothermic) | Negative (exothermic) |
| ΔmixV | 0 | Positive (expands) | Negative (contracts) |
| Azeotrope | None | Minimum-boiling | Maximum-boiling |
| Examples | n-hexane + n-heptane; benzene + toluene | Ethanol + water; acetone + CS₂ | Chloroform + acetone; HNO₃ + water |
Azeotropes boil at a constant temperature with the same composition in liquid and vapour, so they cannot be separated by fractional distillation. Ethanol–water forms a minimum-boiling azeotrope at about 95% ethanol.
6. Colligative Properties
| Property | Formula | NEET Note |
|---|---|---|
| Relative lowering of vapour pressure | (p° − p)/p° = xsolute | Equals mole fraction of solute |
| Elevation of boiling point | ΔTb = iKbm | Kb(water) = 0.52 K kg mol⁻¹ |
| Depression of freezing point | ΔTf = iKfm | Kf(water) = 1.86 K kg mol⁻¹ |
| Osmotic pressure | π = iCRT | Best for macromolecules (proteins, polymers) |
- Osmosis: solvent flows through a semipermeable membrane from a dilute to a concentrated solution. Osmotic pressure is the pressure needed to stop it.
- Isotonic solutions have the same osmotic pressure (0.9% NaCl, "normal saline", is isotonic with blood). Red blood cells swell in a hypotonic solution and shrink in a hypertonic one.
- Reverse osmosis: applying pressure greater than π pushes solvent from the solution into pure solvent; used to desalinate sea water.
- Molar mass from a colligative property: M = (K × w × 1000)/(ΔT × W), where W is the mass of solvent in grams.
7. Abnormal Molar Masses and the Van't Hoff Factor
- Dissociation (electrolytes) gives i > 1: i = 1 + (n − 1)α, where n = ions per formula unit. For complete dissociation, NaCl i = 2, K₂SO₄ i = 3, Al₂(SO₄)₃ i = 5.
- Association gives i < 1: i = 1 − (1 − 1/n)α. Ethanoic or benzoic acid dimerises in benzene (n = 2), so i ≈ 0.5.
- For the same molality, the solution with the larger i has the larger ΔTb, ΔTf and π, and the lower freezing point.
8. Formula Master Table
| Quantity | Formula |
|---|---|
| Henry's law | p = KHx |
| Raoult's law | p = xAp°A + xBp°B |
| Vapour composition | yA = xAp°A/ptotal |
| Relative lowering | (p° − p)/p° = n₂/(n₁ + n₂) |
| Boiling point elevation | ΔTb = iKbm |
| Freezing point depression | ΔTf = iKfm |
| Osmotic pressure | π = iCRT |
| Molar mass | M = 1000Kw/(ΔT W) |
| Dissociation | i = 1 + (n − 1)α |
| Association | i = 1 − (1 − 1/n)α |
9. Master Summary Table
| Concept | Key idea | NEET Note |
|---|---|---|
| Henry's law | Gas solubility ∝ pressure | Higher KH ⇒ less soluble |
| Raoult's law | p ∝ mole fraction | Vapour rich in volatile part |
| Positive deviation | Weaker A–B forces | Minimum-boiling azeotrope |
| Negative deviation | Stronger A–B forces | Maximum-boiling azeotrope |
| Colligative | Depends on particle number | Not on nature of solute |
| Osmotic pressure | π = CRT | Used for proteins |
| Van't Hoff factor | i > 1 dissociation; i < 1 association | Benzoic acid in benzene |
10. Mind Map
- M, m, x
- ppm
- Henry
- Raoult
- Deviations
- RLVP
- ΔTb, ΔTf
- Osmosis
- Van't Hoff i
- Dissociation
- Association
11. Tips and Tricks
- To compare freezing points of equimolal solutions, compare i: 0.1 m Al₂(SO₄)₃ (i = 5) freezes lowest.
- Isotonic solutions: i₁C₁ = i₂C₂.
- ΔTb/ΔTf for the same solution = Kb/Kf.
- For dilute solutions, relative lowering ≈ n₂/n₁ (moles of solute/moles of solvent).
- Thinking a higher KH means higher solubility.
- Forgetting i for electrolytes like NaCl and CaCl₂.
- Using grams of solution instead of kilograms of solvent for molality.
- Mixing up minimum- and maximum-boiling azeotropes.
12. Important One-liners
- Colligative properties depend only on the number of solute particles.
- The solubility of gases in liquids decreases with rise in temperature.
- 0.9% (w/v) NaCl solution is isotonic with blood.
- Reverse osmosis is used to make drinking water from sea water.
- A mixture of chloroform and acetone shows negative deviation because of hydrogen bonding between them.
- Azeotropes cannot be separated by fractional distillation.
- Osmotic pressure is preferred for finding the molar mass of proteins.
13. Solved Numericals: Average to Advanced
Five average and five advanced numericals. Take Kb(water) = 0.52 and Kf(water) = 1.86 K kg mol⁻¹, and R = 0.0821 L atm K⁻¹ mol⁻¹. Click to reveal each answer and working.
Average level
10 g of glucose (C₆H₁₂O₆) is dissolved in 90 g of water. Find the mole fraction of glucose.
Show answer & solution
Concept: x = nsolute/(nsolute + nsolvent).
- n(glucose) = 10/180 ≈ 0.0556 mol; n(water) = 90/18 = 5 mol.
- x = 0.0556/5.0556 ≈ 0.011.
Answer: about 0.011
Henry's law constant for CO₂ in water at 298 K is 1.67 × 10⁸ Pa. Find the mole fraction of CO₂ dissolved when its partial pressure is 2.5 bar.
Show answer & solution
Concept: x = p/KH.
- p = 2.5 bar = 2.5 × 10⁵ Pa.
- x = 2.5 × 10⁵/1.67 × 10⁸ ≈ 1.5 × 10⁻³.
Answer: about 1.5 × 10⁻³
Liquids A (p° = 100 mm Hg) and B (p° = 40 mm Hg) form an ideal solution with xA = 0.4. Find the total vapour pressure and the mole fraction of A in the vapour.
Show answer & solution
Concept: Raoult's law and Dalton's law.
- pA = 0.4 × 100 = 40 mm; pB = 0.6 × 40 = 24 mm.
- ptotal = 64 mm Hg.
- yA = 40/64 = 0.625 — the vapour is richer in the more volatile A.
Answer: 64 mm Hg; 0.625
18 g of glucose is dissolved in 500 g of water. Find the elevation in boiling point.
Show answer & solution
Concept: ΔTb = Kbm (i = 1 for glucose).
- n = 18/180 = 0.1 mol; m = 0.1/0.5 = 0.2 mol kg⁻¹.
- ΔTb = 0.52 × 0.2 = 0.104 K.
Answer: 0.104 K
Find the osmotic pressure of a 0.1 M solution of a non-electrolyte at 27 °C.
Show answer & solution
Concept: π = CRT.
- T = 300 K.
- π = 0.1 × 0.0821 × 300 = 2.46 atm.
Answer: about 2.46 atm
Advanced level
1.0 g of a non-volatile, non-electrolyte solute dissolved in 50 g of benzene lowers its freezing point by 0.40 K. Kf of benzene is 5.12 K kg mol⁻¹. Find the molar mass of the solute.
Show answer & solution
Concept: M = 1000 Kf w/(ΔTf W).
- M = 1000 × 5.12 × 1.0/(0.40 × 50).
- M = 5120/20 = 256 g mol⁻¹.
Answer: 256 g mol⁻¹
(a) Find the expected freezing-point depression of 0.1 m aqueous K₂SO₄, assuming complete dissociation. (b) If the observed depression is 0.50 K, find the degree of dissociation.
Show answer & solution
Concept: ΔTf = iKfm; i = 1 + (n − 1)α with n = 3.
- (a) i = 3: ΔTf = 3 × 1.86 × 0.1 = 0.558 K.
- (b) i = 0.50/(1.86 × 0.1) ≈ 2.69.
- α = (i − 1)/(n − 1) = 1.69/2 ≈ 0.84 (84%).
Answer: (a) 0.558 K; (b) about 84%
6 g of urea (molar mass 60) is dissolved in 90 g of water at 25 °C, where the vapour pressure of pure water is 23.8 mm Hg. Find the relative lowering of vapour pressure and the vapour pressure of the solution.
Show answer & solution
Concept: (p° − p)/p° = xsolute.
- n(urea) = 0.1 mol; n(water) = 5 mol; x = 0.1/5.1 ≈ 0.0196.
- p = p°(1 − x) = 23.8 × 0.9804 ≈ 23.33 mm Hg.
Answer: about 0.0196; about 23.3 mm Hg
A 5% (w/v) glucose solution is isotonic with a urea solution at the same temperature. Find the concentration of the urea solution in % (w/v).
Show answer & solution
Concept: isotonic ⇒ equal molar concentration (both non-electrolytes).
- Glucose: 50 g per litre ⇒ 50/180 ≈ 0.278 M.
- Urea must also be 0.278 M ⇒ 0.278 × 60 ≈ 16.7 g per litre.
- That is about 1.67% (w/v).
Answer: about 1.67% (w/v)
3 g of benzoic acid (molar mass 122) dissolved in 30 g of benzene lowers its freezing point by 2.05 K (Kf = 4.9 K kg mol⁻¹). Find the observed molar mass, the van't Hoff factor and the percentage of acid that has dimerised.
Show answer & solution
Concept: association: i = 1 − α/2 for dimers.
- Mobs = 1000 × 4.9 × 3/(2.05 × 30) ≈ 239 g mol⁻¹.
- i = 122/239 ≈ 0.51.
- 1 − α/2 = 0.51 ⇒ α ≈ 0.98, so about 98% of the acid is dimerised.
Answer: about 239 g mol⁻¹; i ≈ 0.51; about 98%
14. NEET-pattern Questions with Detailed Answers
These questions are original and based on concepts that NEET UG asks again and again.
Which of these is a colligative property?
- Vapour pressure
- Osmotic pressure
- Boiling point
- Density
Show solution
Osmotic pressure depends only on the number of solute particles. Vapour pressure and boiling point are ordinary properties; only their lowering or elevation is colligative. Density is not colligative.
Answer: (B)
The solubility of a gas in a liquid increases with:
- Increase in temperature
- Decrease in pressure
- Increase in pressure
- Increase in KH
Show solution
By Henry's law, solubility is proportional to pressure. It falls when temperature rises, and a higher KH means lower solubility.
Answer: (C)
Which 0.1 m aqueous solution has the lowest freezing point?
- Glucose
- NaCl
- CaCl₂
- Al₂(SO₄)₃
Show solution
ΔTf ∝ i. The values of i are 1, 2, 3 and 5. Al₂(SO₄)₃ gives the most particles, so it lowers the freezing point the most.
Answer: (D)
A mixture of chloroform and acetone shows:
- Positive deviation and forms a minimum-boiling azeotrope
- Negative deviation and forms a maximum-boiling azeotrope
- Ideal behaviour
- Positive deviation and forms a maximum-boiling azeotrope
Show solution
Hydrogen bonding between CHCl₃ and acetone makes the A–B attraction stronger. This lowers the vapour pressure, giving a negative deviation and a maximum-boiling azeotrope. (D) mismatches the type of deviation with the type of azeotrope.
Answer: (B)
Red blood cells placed in 2% NaCl solution will:
- Swell and burst
- Shrink
- Remain unchanged
- First swell, then shrink
Show solution
A 2% solution is hypertonic compared with 0.9% saline, which is isotonic with blood. Water moves out of the cells, so they shrink. In a hypotonic solution they would swell.
Answer: (B)
The van't Hoff factor for acetic acid in benzene is close to:
- 2
- 1
- 0.5
- 3
Show solution
In benzene, acetic acid forms hydrogen-bonded dimers, which halves the number of particles, so i ≈ 0.5. In water it would ionise slightly and i would be a little above 1.
Answer: (C)
A 0.1 M NaCl solution (assume complete dissociation) is isotonic with a glucose solution. The concentration of the glucose solution is:
- 0.05 M
- 0.1 M
- 0.2 M
- 0.3 M
Show solution
Isotonic solutions have equal iC. For NaCl, i = 2, so iC = 2 × 0.1 = 0.2 M of particles. Glucose does not dissociate (i = 1), so it must be 0.2 M. (B) forgets the van't Hoff factor, and (D) would suit a salt that gives 3 ions.
Answer: (C)
Two liquids A and B form an ideal solution. At a certain composition, the vapour pressure is 300 mm, and pure A and pure B have vapour pressures of 450 mm and 150 mm. The mole fraction of A in the liquid is:
- 0.25
- 0.5
- 0.67
- 0.75
Show solution
300 = 450x + 150(1 − x) = 150 + 300x, so x = 0.5. (D) is the mole fraction of A in the vapour: 0.5 × 450/300 = 0.75.
Answer: (B)
15. Quick Revision Summary
- Henry: p = KHx; gases less soluble when hot.
- Raoult: p = Σxip°i; vapour richer in volatile part.
- +ve deviation: weaker A–B, min-boiling azeotrope; −ve: stronger A–B, max-boiling.
- RLVP = xsolute; ΔTb = iKbm; ΔTf = iKfm; π = iCRT.
- i = 1 + (n − 1)α (dissociation); i = 1 − (1 − 1/n)α (association).