Solutions
'A solution is more than the sum of its parts — the interactions between solute and solvent create entirely new properties.'
1. Chapter Overview
Solutions are HOMOGENEOUS mixtures where the components are uniformly distributed at the molecular level. This chapter covers: TYPES of solutions (solid, liquid, gaseous), EXPRESSING CONCENTRATION (mass percent, mole fraction, molarity, molality), RAOULT'S LAW (vapour pressure of solutions), IDEAL and NON-IDEAL solutions, AZEOTROPES, and COLLIGATIVE PROPERTIES (properties that depend ONLY on the NUMBER of solute particles, not their identity) — including relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and OSMOTIC PRESSURE. The van't Hoff factor accounts for dissociation or association of solutes.
2. Concentration Terms
| Term | Symbol | Formula | Unit |
|---|---|---|---|
| Mass percent | w/w% | (mass of solute / mass of solution) × 100 | % |
| Mole fraction | x | moles of component / total moles | NO unit |
| Molarity | M | moles of solute / volume of solution (L) | mol/L |
| Molality | m | moles of solute / mass of solvent (kg) | mol/kg |
- Molarity vs Molality: Molarity depends on TEMPERATURE (volume changes). Molality is TEMPERATURE INDEPENDENT (mass does not change).
3. Solubility
- Solubility of gases: Follows HENRY'S LAW — p = K_H × x (partial pressure of gas = Henry's constant × mole fraction). 'The solubility of a gas INCREASES with pressure and DECREASES with temperature.'
- Applications: Carbonated beverages (CO₂ under pressure), deep-sea diving (N₂ solubility at high pressure causes bends).
4. Raoult's Law
- For a volatile solute: p₁ = p₁° × x₁ (vapour pressure of component = vapour pressure of pure component × its mole fraction).
- For a solution of two volatile components: P_total = p₁°x₁ + p₂°x₂ = p₁°x₁ + p₂°(1−x₁).
- 'Raoult's law says: the vapour pressure of a component in a solution is PROPORTIONAL to its mole fraction.'
Ideal Solutions (Obey Raoult's Law)
- A_mix = 0, V_mix = 0 (no volume change on mixing).
- Similar intermolecular forces. Examples: benzene + toluene, n-hexane + n-heptane.
Non-Ideal Solutions (Deviation from Raoult's Law)
| Type | Deviation | A_mix | V_mix | Example | Explanation |
|---|---|---|---|---|---|
| Positive deviation | P_total > expected | ENDOthermic (+) | Increases (+) | Ethanol + Water, Acetone + CS₂ | A-B forces < A-A and B-B forces |
| Negative deviation | P_total < expected | EXOthermic (−) | Decreases (−) | Acetone + Chloroform, Water + HCl | A-B forces > A-A and B-B forces |
5. Azeotropes
- Minimum boiling azeotrope (positive deviation): Boils at LOWER temperature than either pure component. Example: 95.4% ethanol + 4.6% water (bp 78.2°C).
- Maximum boiling azeotrope (negative deviation): Boils at HIGHER temperature. Example: 68% HNO₃ + 32% water (bp 120.5°C).
- 'An azeotrope is a mixture with a CONSTANT boiling point — the vapour has the SAME composition as the liquid. It CANNOT be separated by simple distillation.'
6. Colligative Properties
6.1 Relative Lowering of Vapour Pressure
- (p° − p)/p° = x₂ (mole fraction of solute).
6.2 Elevation of Boiling Point
- ΔT_b = K_b × m. K_b = ebullioscopic constant.
- T_b(solution) = T_b(solvent) + ΔT_b.
6.3 Depression of Freezing Point
- ΔT_f = K_f × m. K_f = cryoscopic constant.
- T_f(solution) = T_f(solvent) − ΔT_f.
- 'Adding salt to ice LOWERS the freezing point — that is why salt is spread on icy roads.'
6.4 Osmotic Pressure
- π = CRT (van't Hoff equation for dilute solutions).
- π = iCRT (for electrolytes, with van't Hoff factor i).
- 'Osmotic pressure is the pressure required to PREVENT the flow of solvent into a solution through a semipermeable membrane.'
7. van't Hoff Factor (i)
- i = (actual number of particles in solution) / (number of formula units dissolved).
- For dissociation: i > 1 (e.g., NaCl → Na⁺ + Cl⁻: i = 2).
- For association: i < 1 (e.g., dimerisation of benzoic acid in benzene: i ≈ 0.5).
- Modified formulas: ΔT_f = iK_f m. ΔT_b = iK_b m. π = iCRT.
| Solute | Type | i (theoretical) | Example |
|---|---|---|---|
| Nonelectrolyte | No dissociation | 1 | Glucose, Urea |
| Binary electrolyte | Dissociates into 2 ions | 2 | NaCl, KCl, NaOH |
| Ternary electrolyte | Dissociates into 3 ions | 3 | CaCl₂, BaCl₂ |
| Associated | Forms dimers, etc. | < 1 | Benzoic acid in benzene |
8. Common Mistakes
- Molality vs Molarity: Molality uses kg of SOLVENT. Molarity uses LITRES of SOLUTION. They are NOT interchangeable.
- Raoult's law for non-volatile solutes: P_total = p° × x_solvent (only the solvent contributes to vapour pressure).
- Van't Hoff factor for weak electrolytes: For weak acids/bases (e.g., CH₃COOH), i is LESS than the theoretical value because dissociation is PARTIAL.
- Osmotic pressure magnitude: Even dilute solutions have SIGNIFICANT osmotic pressure — an impressive demonstration of colligative effects.
9. CBSE Exam Focus
- Expressing concentration — mole fraction, molarity, molality (numerical problems)
- Raoult's law — ideal solutions, vapour pressure calculations
- Non-ideal solutions — positive and negative deviations
- Colligative properties — ΔT_f, ΔT_b, π — numerical problems
- van't Hoff factor — i for dissociation/association, modified formulas
10. Self-Test
Q1: Calculate the molality of a solution containing 10 g of glucose (C₆H₁₂O₆, M = 180) in 250 g of water. A1: Moles of glucose = 10/180 = 0.0556 mol. Molality = 0.0556/0.25 = 0.222 m.
Q2: The vapour pressure of pure water at 25°C is 23.76 mm Hg. What is the vapour pressure of a solution containing 10 g of urea (M=60) in 100 g of water? A2: Moles of urea = 10/60 = 0.167. Moles of water = 100/18 = 5.556. x_water = 5.556/(5.556+0.167) = 0.971. p = p°×x_water = 23.76×0.971 = 23.07 mm Hg.
Q3: 0.5 g of a non-electrolyte dissolved in 50 g of water gives a freezing point of −0.186°C. Find the molecular mass. (K_f of water = 1.86 K·kg/mol) A3: ΔT_f = 0.186 = K_f × m = 1.86 × (0.5/M)/(0.05). 0.186 = 1.86 × 10/M. M = 1.86×10/0.186 = 100 g/mol.
Q4: Calculate the osmotic pressure of 0.1 M glucose solution at 27°C. A4: π = CRT = 0.1 × 0.0821 × 300 = 2.463 atm.
Q5: A 0.1 M aqueous solution of KCl has ΔT_f = 0.335°C. Find the van't Hoff factor. (K_f = 1.86) A5: ΔT_f = iK_f m. 0.335 = i × 1.86 × 0.1. i = 0.335/0.186 = 1.80. (Theoretical i = 2. Observed i < 2 due to ion-pair formation.)
11. Conclusion
Solutions are the MEDIUM of chemistry:
- CONCENTRATION: 'Molarity, molality, mole fraction — each tells you how MUCH solute is present. Choose the right one for your calculation.'
- RAOULT: 'The vapour pressure of a solution depends on the NUMBER of solute particles — the foundation of colligative properties.'
- COLLIGATIVE: 'Freezing point depression, boiling point elevation, osmotic pressure — all depend ONLY on the number of particles.'
- VAN'T HOFF: 'Accounting for dissociation or association — i makes colligative formulas work for ALL solutes.'
'Solutions chemistry connects molecular behaviour to MACROSCOPIC properties — from the salt on our roads to the sugar in our tea.'
