Solutions
1. Introduction
A solution is a homogeneous mixture of two or more substances. This chapter covers the physical chemistry of solutions — how solute particles affect the properties of the solvent.
2. Concentration Terms
Molarity (M): moles of solute per litre of solution. Varies with temperature. Molality (m): moles of solute per kg of solvent. Independent of temperature. Mole fraction (x): ratio of moles of one component to total moles. Mass percentage: (mass of solute/mass of solution) × 100%.
3. Raoult's Law
For a volatile solute: P_A = x_A P_A⁰, where P_A⁰ is the vapour pressure of pure A.
For a non-volatile solute: P_solution = x_solvent P_solvent⁰.
Lowering of vapour pressure: ΔP = P⁰ - P = x_solute P⁰.
3.1 Ideal and Non-Ideal Solutions
Ideal solutions: Obey Raoult's law at all concentrations. ΔH_mix = 0, ΔV_mix = 0. Example: Benzene + Toluene, n-Hexane + n-Heptane.
Non-ideal solutions: Deviate from Raoult's law. Positive deviation (A-B interactions weaker): ΔH_mix > 0, ΔV_mix > 0. Example: Ethanol + Acetone. Negative deviation (A-B interactions stronger): ΔH_mix < 0, ΔV_mix < 0. Example: Acetone + Chloroform.
4. Colligative Properties
Properties that depend only on the number of solute particles, not their nature.
4.1 Relative Lowering of Vapour Pressure
(P⁰ - P)/P⁰ = x_solute = n/(n + N)
4.2 Elevation of Boiling Point
ΔT_b = K_b × m, where K_b is the ebullioscopic constant.
4.3 Depression of Freezing Point
ΔT_f = K_f × m, where K_f is the cryoscopic constant.
4.4 Osmotic Pressure
π = iCRT (for van't Hoff equation).
Osmosis: net flow of solvent from lower to higher concentration through a semipermeable membrane.
5. Determination of Molar Mass Using Colligative Properties
5.1 From Elevation of Boiling Point
Molar mass M = (K_b × w₂ × 1000)/(ΔT_b × w₁) where w₂ = mass of solute (g), w₁ = mass of solvent (g).
5.2 From Depression of Freezing Point
M = (K_f × w₂ × 1000)/(ΔT_f × w₁)
5.3 From Osmotic Pressure
M = (w₂RT)/(πV)
'Osmotic pressure method is the most accurate for determining molar masses of macromolecules because the osmotic pressure is measurable even for very dilute solutions.'
6. Abnormal Molar Masses
When a solute dissociates or associates in solution, the observed molar mass differs from the expected molar mass.
Dissociation: Observed M < Calculated M (more particles, greater colligative effect). i = Calculated M/Observed M.
Association: Observed M > Calculated M (fewer particles, smaller colligative effect). i = Calculated M/Observed M.
6.1 Degree of Dissociation
For dissociation of AₙBₘ → nA + mB: α = (i - 1)/((n + m) - 1)
6.2 Degree of Association
For association of n molecules of A: α = (1 - i)/(1 - 1/n)
7. Van't Hoff Factor
i = (observed colligative property)/(calculated colligative property).
i = 1 + (n - 1)α, where α is the degree of dissociation/association.
'For dissociation, i > 1. For association, i < 1. For non-electrolytes, i = 1.'
6. Worked Problems
Problem 1: Calculate the boiling point of a solution containing 18 g of glucose (C₆H₁₂O₆) in 100 g of water. K_b for water = 0.52 K kg/mol. Solution: Moles of glucose = 18/180 = 0.1 mol. m = 0.1/0.1 = 1 mol/kg. ΔT_b = 0.52 × 1 = 0.52°C. Boiling point = 100 + 0.52 = 100.52°C.
Problem 2: The freezing point of a 0.1 m solution of NaCl is -0.372°C. K_f for water = 1.86 K kg/mol. Find i. Solution: ΔT_f observed = 0.372. ΔT_f calculated = 1.86 × 0.1 = 0.186. i = 0.372/0.186 = 2. NaCl dissociates into Na⁺ and Cl⁻, giving i ≈ 2.
7. Common Mistakes
'Students often forget that molarity depends on temperature while molality does not. For colligative property calculations, use molality (not molarity).'
8. ISC Exam Focus
| Topic | Theory Marks | Practical Marks |
|---|---|---|
| Concentration terms | 2 | 2 |
| Raoult's law | 3 | 1 |
| Colligative properties | 5 | 3 |
| Van't Hoff factor | 3 | 2 |
9. Self-Test Questions
- Define the term 'colligative property'. Name four colligative properties.
- 10 g of a non-volatile solute dissolved in 200 g of benzene raises its boiling point by 1°C. K_b of benzene = 2.53 K kg/mol. Find molar mass.
- Calculate the osmotic pressure of 0.1 M glucose solution at 27°C (R = 0.0821 L atm/mol K).
- Explain positive and negative deviations from Raoult's law with examples.
- The freezing point of 0.1 m acetic acid solution is -0.188°C. K_f = 1.86. Calculate the degree of dissociation of acetic acid.
