Solutions - Previous Year Questions Solved Examples (Class 12 Chemistry)
Solutions are homogeneous mixtures studied through colligative properties and solution chemistry. These questions test understanding of solubility, concent
TL;DR: Solutions are homogeneous mixtures studied through colligative properties and solution chemistry. These questions test understanding of solubility, co…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
Solutions are homogeneous mixtures studied through colligative properties and solution chemistry. These questions test understanding of solubility, concent
Solutions - Previous Year Questions — Solved Numerical Examples (Step by Step)
Example 1: Define molarity, molality, and mole fraction. How are they related? [3 marks]
Solution: Molarity (M) = moles of solute / liters of solution. Molality (m) = moles of solute / kilograms of solvent. Mole fraction (x) = moles of component / total moles. Relationship: If density is d g/mL and molecular weight of solvent is M, then approximately m = 1000M / (1000 + m × M). At same dilute concentrations, molarity < molality. Mole fraction is independent of temperature, unlike molarity and molality.
Example 2: What are colligative properties? Name four colligative properties of solutions. [3 marks]
Solution: Colligative properties are properties that depend only on the number of solute particles, not on their nature. Four examples: (1) Vapor pressure lowering: Presence of solute decreases vapor pressure of solvent. (2) Boiling point elevation: Solution boils at higher temperature than pure solvent. (3) Freezing point depression: Solution freezes at lower temperature than pure solvent. (4) Osmotic pressure: Tendency of solvent to move into solution by osmosis. All depend on number of moles of solute particles and are independent of solute identity.
Example 3: Calculate the boiling point elevation when 10 g of NaCl (M = 58.5 g/mol) is dissolved in 100 g of water. Kb for water = 0.52 K·kg/mol. [3 marks]
Solution: Molality m = moles of solute / kg of solvent. Moles of NaCl = 10 / 58.5 = 0.171 mol. Molality = 0.171 / 0.1 = 1.71 m. NaCl dissociates into 2 ions (i = 2). ΔTb = i × Kb × m = 2 × 0.52 × 1.71 = 1.78 K. Boiling point of solution = 100 + 1.78 = 101.78°C.
Example 4: Explain Henry's Law and its limitations. Give an example. [3 marks]
Solution: Henry's Law: The solubility of a gas in a liquid is directly proportional to its partial pressure above the liquid. S = k_H × P, where S is solubility, P is partial pressure, and k_H is Henry's law constant. Example: CO2 is more soluble in water at higher pressure (carbonated drinks). Limitations: (1) Applicable only to ideal gases at low pressures, (2) Does not apply when gas reacts with solvent (HCl in water), (3) Fails at very high pressures, (4) Temperature dependence not accounted for.
Example 5: What is osmosis? Explain its biological significance. [3 marks]
Solution: Osmosis is the spontaneous flow of solvent molecules from dilute solution to concentrated solution through a semipermeable membrane. Osmotic pressure π = i × M × R × T (van't Hoff equation). Biological significance: (1) Water movement in plant cells causes turgor, maintaining rigidity, (2) Red blood cells shrink in hypertonic solution (crenation) or swell in hypotonic solution (hemolysis), (3) Kidney tubules reabsorb water based on osmotic pressure, (4) Nutrient absorption in small intestine depends on osmosis, (5) Plant nutrient uptake from soil.
Example 6: Derive Raoult's Law and explain its applications. [4 marks]
Solution: Raoult's Law: For an ideal solution, partial vapor pressure of a component equals its mole fraction multiplied by its pure vapor pressure. Pi = xi × Pi°. For volatile solvent over non-volatile solute: P = x_solvent × P°_solvent. Vapor pressure lowering: ΔP = x_solute × P°_solvent. Applications: (1) Calculating vapor pressure of solutions, (2) Deriving boiling point elevation formula, (3) Deriving freezing point depression formula, (4) Determining relative molecular mass from vapor pressure data.
Example 7: A 0.2 M aqueous solution of a non-electrolyte froze at -0.186°C. Kf for water = 1.86 K·kg/mol. Verify if the solute behaves ideally. [3 marks]
Solution: ΔTf = Kf × m. 0.186 = 1.86 × m. m = 0.1 mol/kg. For 0.2 M solution in dilute form, assuming density ≈ 1 g/mL, molarity ≈ molality, so m ≈ 0.2. But observed m = 0.1, which is half the expected value. This suggests the solute may be associating or not fully dissolving. For ideal behavior, observed freezing point should be -0.372°C, not -0.186°C.
Tips
- Remember conversion: 1 M ≈ 1 m for dilute aqueous solutions when density ≈ 1 g/mL.
- Always multiply by 'i' (van't Hoff factor) for electrolytes: i = 1 for non-electrolytes, 2 for binary electrolytes.
- Colligative properties depend on particle count, not particle nature; this distinguishes them.
- In thermodynamic calculations, use Kelvin temperatures and SI units consistently.
Frequently Asked Questions
Why does salt dissolve better in hot water?
For most salts, solubility increases with temperature because dissolution is an endothermic process. Higher thermal energy provides the energy needed.
Why are colligative properties more pronounced for molecular than ionic solutes?
At same molarity, ionic solutes produce more particles due to dissociation (i > 1), resulting in stronger colligative effects.
More Chemistry Solved Examples
- Mole Concept and Stoichiometry
- Atomic Structure (Numericals)
- Electrochemistry (Numericals)
- Chemical Equations and Balancing
- Thermodynamics
- Mole Concept and Stoichiometry
🤖 Stuck on any of these? Ask Syllab's free AI Tutor to explain step by step →