Solutions & Colligative Properties

Solutions are homogeneous mixtures, and their behaviour underlies many NEET numericals. Henry's law governs gas solubility: p = $K_{H}$ × x, where p is the partial pressure of the gas and x is its mole fraction in solution. $K_{H}$ increases with temperature (gases become less soluble in hot water). Applications: carbonated beverages release CO₂ when opened (pressure drops), and deep-sea divers face "the bends" when dissolved N₂ forms bubbles during rapid ascent.

Raoult's law for volatile liquid mixtures: $P_{A}$ = $x_{A}$ · P°_A and $P_{total}$ = $x_{A}$ · P°_A + $x_{B}$ · P°_B.
For a non-volatile solute in a volatile solvent: P = $x_{solvent}$ · P°_solvent.

Ideal solutions obey Raoult's law at all compositions (Δ$H_{mix}$ = 0, Δ$V_{mix}$ = 0); examples: benzene + toluene, n-hexane + n-heptane. Non-ideal solutions deviate:

• Positive deviation: $P_{observed}$ > $P_{Raoult}$ (weaker A-B interactions, Δ$H_{mix}$ > 0, Δ$V_{mix}$ > 0) → form minimum boiling azeotropes. Examples: ethanol + water (SMILES: CCO + O), acetone + CS₂.
• Negative deviation: $P_{observed}$ < $P_{Raoult}$ (stronger A-B interactions, Δ$H_{mix}$ < 0, Δ$V_{mix}$ < 0) → form maximum boiling azeotropes.
  Examples: chloroform + acetone (SMILES: ClC(Cl)Cl + CC(=O)C), HCl + water.

Colligative properties depend on the NUMBER of solute particles, not their identity:

1. Relative lowering of vapour pressure: ΔP/P° = $x_{solute}$
2. Elevation of boiling point: ΔTb = Kb · m (water Kb = 0.52 K·kg/mol)
3. Depression of freezing point: ΔTf = Kf · m (water Kf = 1.86 K·kg/mol)
4. Osmotic pressure: π = CRT (C in mol/L, R = 0.0821 L·atm/(mol·K))

Molar mass determination: M₂ = (Kb × w₂ × 1000) / (ΔTb × w₁)

Solved Example 1: 18 g glucose (C₆H₁₂O₆, M = 180 g/mol) in 500 g water. Find boiling point. Molality = ($\frac{18}{180}$) / ($\frac{500}{1000}$) = 0.$\frac{1}{0}$.5 = 0.2 m ΔTb = Kb × m = 0.52 × 0.2 = 0.104 K Boiling point = 100.104°C

Solved Example 2: Osmotic pressure of 0.1 M NaCl at 27°C (i = 1.85). π = iCRT = 1.85 × 0.1 × 0.0821 × 300 Dimensional analysis: (unitless) × (mol/L) × L·atm/(mol·K) × K = atm ✓ π = 4.56 atm

Solved Example 3: ΔTf of 0.1 m acetic acid in benzene = 0.256 K (Kf = 5.12 K·kg/mol). Find van't Hoff factor and degree of association. Expected ΔTf = 5.12 × 0.1 = 0.512 K i = observed/expected = 0.$\frac{256}{0}$.512 = 0.5 For dimerization (n=2): Derivation: If α = degree of association, initial moles = 1, after association moles = (1 − α) + α/2 = 1 − α/2. So i = 1 − α/2. 0.5 = 1 − α/2 → α/2 = 0.5 → α = 1 (100% association)

Van't Hoff factor i = observed colligative property / calculated colligative property.

• Dissociation (i > 1): i = 1 + (n−1)α, where n = number of ions. NaCl: n=2, i ≈ 2.
• Association (i < 1): i = 1 + (1/n − 1)α, where n = number of monomers per aggregate.

Reverse osmosis applies pressure exceeding osmotic pressure to force solvent through a semipermeable membrane (water purification).

The key testable concept is colligative property calculations with the van't Hoff factor for electrolyte and association systems.