Solutions — Previous Year Questions (Class 12 Chemistry)
Solutions covers solubility, molarity, molality, colligative properties, and Raoult's law. CBSE tests concentration units, freezing point depression, and b
TL;DR: Solutions covers solubility, molarity, molality, colligative properties, and Raoult's law. CBSE tests concentration units, freezing point depression,…
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Solutions covers solubility, molarity, molality, colligative properties, and Raoult's law. CBSE tests concentration units, freezing point depression, and b
Solutions — Previous Year Questions with Solutions
Q (2023, 2 marks): What is the difference between a saturated, unsaturated, and supersaturated solution?
Answer: Saturated Solution: A solution that contains the maximum amount of dissolved solute at a given temperature. No more solute can dissolve at that temperature. Solute and solvent are in dynamic equilibrium.
Example: 36 g NaCl dissolved in 100 g water at 20°C.
Unsaturated Solution: A solution that contains less dissolved solute than the saturation limit at a given temperature. More solute can still be added and dissolved.
Example: 20 g NaCl in 100 g water (less than 36 g).
Supersaturated Solution: A solution that contains MORE dissolved solute than the saturation limit. This is unstable; it cannot exist in equilibrium and precipitates the excess solute when disturbed or seeded with a crystal.
Example: Sodium acetate solution cooled below its crystallization temperature.
Key difference:
- Saturated: At equilibrium, max solute dissolved.
- Unsaturated: Can dissolve more solute.
- Supersaturated: Unstable, excess solute will precipitate.
Final Answer: Saturated = max solute at equilibrium. Unsaturated = can dissolve more. Supersaturated = unstable, excess will precipitate.
Q (2022, 2 marks): Calculate the molarity of a solution prepared by dissolving 10 g NaOH in water to make 500 mL of solution.
Answer: Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol
Mass of NaOH = 10 g
Moles of NaOH = 10 / 40 = 0.25 mol
Volume of solution = 500 mL = 0.5 L
Molarity M = moles / volume (in L) = 0.25 / 0.5 = 0.5 M
Final Answer: Molarity = 0.5 M
Q (2021, 2 marks): Define molality and calculate the molality of a solution containing 5.85 g NaCl dissolved in 1 kg of water.
Answer: Molality (m): Number of moles of solute per kilogram of solvent.
m = moles of solute / mass of solvent (in kg)
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl = 5.85 g
Moles of NaCl = 5.85 / 58.5 = 0.1 mol
Mass of solvent (water) = 1 kg
Molality = 0.1 / 1 = 0.1 m (or 0.1 mol/kg)
Final Answer: Molality = 0.1 m
Q (2023, 3 marks): If the mole fraction of solute in a solution is 0.02, what is the mass percentage of the solute? (Molar mass of solute = 80 g/mol, molar mass of solvent = 18 g/mol)
Answer: Mole fraction of solute χ_solute = 0.02
Mole fraction of solvent χ_solvent = 1 - 0.02 = 0.98
Let moles of solute = n_s, moles of solvent = n_solv
χ_solute = n_s / (n_s + n_solv) = 0.02
0.02(n_s + n_solv) = n_s
0.02n_s + 0.02n_solv = n_s
0.02n_solv = 0.98n_s
n_solv / n_s = 0.98 / 0.02 = 49
Mass of solute = n_s × 80 = 80n_s
Mass of solvent = 49n_s × 18 = 882n_s
Total mass = 80n_s + 882n_s = 962n_s
Mass percentage = (80n_s / 962n_s) × 100% = (80/962) × 100% ≈ 8.3%
Final Answer: Mass percentage ≈ 8.3%
Q (2022, 2 marks): A solution of urea (molar mass 60 g/mol) in water has a boiling point elevation of 0.3°C. Calculate the molality. (K_b for water = 0.51 K·kg/mol)
Answer: Boiling point elevation: ΔT_b = K_b × m
Where m = molality, K_b = ebullioscopic constant
0.3 = 0.51 × m
m = 0.3 / 0.51 ≈ 0.588 mol/kg
Final Answer: Molality ≈ 0.588 m or 0.59 m
Q (2021, 3 marks): State Raoult's law and explain when it is applicable.
Answer: Raoult's Law: The partial pressure of a component in a solution is equal to the mole fraction of that component multiplied by its vapor pressure in the pure state.
For component i: P_i = χ_i × P_i°
Where P_i = partial pressure of component i in solution, χ_i = mole fraction, P_i° = vapor pressure of pure component i.
For a binary solution (A and B):
P_A = χ_A × P_A° (for component A)
P_B = χ_B × P_B° (for component B)
Total pressure P_total = P_A + P_B = χ_A × P_A° + χ_B × P_B°
Applicability:
1. Valid for ideal solutions (negligible intermolecular interactions between unlike molecules).
2. Valid for volatile components only (components that have measurable vapor pressure).
3. NOT applicable to solutions with strong intermolecular interactions (hydrogen bonding, ionic attraction).
4. Works better at dilute concentrations.
5. For non-volatile solutes, we use the modified form related to freezing point depression and boiling point elevation.
Final Answer: P_i = χ_i × P_i°; applicable to ideal volatile solutions only.
Frequently Asked Questions
Why do colligative properties depend only on the number of particles, not their nature?
Colligative properties (boiling point elevation, freezing point depression, osmotic pressure) depend on the number of solute particles disrupting the solvent's properties, not on the type of solute. Whether the solute is glucose or urea, the effect is the same if the number of particles is the same.
What is the difference between ideal and non-ideal solutions?
Ideal solutions follow Raoult's law perfectly; intermolecular forces between like molecules equal forces between unlike molecules. Non-ideal solutions deviate from Raoult's law due to strong interactions (e.g., hydrogen bonding between H2O and ethanol).
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