Solutions — Class 12 Chemistry NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Chemistry chapter "Solutions" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Chemistry chapter "Solutions" — 8 important questions with detailed answers for CBSE board exam prepara…
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NCERT Solutions for Class 12 Chemistry — Solutions. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Solutions — All 8 Questions Solved
Q1. Define a solution. What are the main components of a solution?
A solution is a homogeneous mixture of two or more substances where one substance (solute) is uniformly dispersed in another (solvent).
Main components:
1. Solvent: The substance in which solute dissolves (usually present in larger amount). Typically liquid.
2. Solute: The substance being dissolved (present in smaller amount). Can be solid, liquid, or gas.
Example: In salt water, water is the solvent and salt (NaCl) is the solute.
Q2. Calculate the molarity of a solution containing 10.6 g of Na₂CO₃ dissolved in water to make 500 mL of solution.
Given: Mass of Na₂CO₃ = 10.6 g, Volume of solution = 500 mL = 0.5 L
Step 1: Calculate molar mass of Na₂CO₃
Molar mass = (2 × 23) + 12 + (3 × 16) = 46 + 12 + 48 = 106 g/mol
Step 2: Calculate number of moles
Moles = mass / molar mass = 10.6 / 106 = 0.1 mol
Step 3: Calculate molarity
Molarity = moles / volume (in L) = 0.1 / 0.5 = 0.2 M
Answer: Molarity = 0.2 M
Q3. What is the difference between molarity and molality?
Molarity (M):
1. Defined as moles of solute per litre of solution
2. Formula: M = moles / volume of solution (L)
3. Temperature dependent (volume changes with temperature)
4. Used in laboratory work and dilution calculations
Molality (m):
1. Defined as moles of solute per kilogram of solvent
2. Formula: m = moles / mass of solvent (kg)
3. Temperature independent (mass doesn't change with temperature)
4. Used for colligative properties and non-aqueous solutions
Example: A 1 M solution means 1 mole in 1 L of total solution, whereas 1 m solution means 1 mole dissolved in 1 kg of solvent.
Q4. State Henry's Law and explain its application.
Henry's Law states: The solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid, at a constant temperature.
Mathematical form:
p = K_H × x
where p = partial pressure of gas, x = mole fraction of gas in solution, K_H = Henry's law constant
Alternative form: S = K_H × p
where S = solubility of gas, p = partial pressure
Applications:
1. Carbonated beverages: CO₂ remains dissolved under pressure; fizzes when pressure is released
2. Scuba diving: N₂ solubility increases at depth; causes decompression sickness if surfaced too quickly
3. Oxygen in water: Fish obtain dissolved oxygen; reduced by heating (less soluble at higher temperature)
4. Industrial processes: Gas absorption and extraction operations
5. Anesthesia: Gaseous anesthetics in blood follow Henry's Law
Q5. Calculate the mass of solute needed to prepare 500 mL of 0.1 M NaOH solution.
Given: Volume of solution = 500 mL = 0.5 L, Molarity = 0.1 M
Step 1: Calculate molar mass of NaOH
Molar mass = 23 + 16 + 1 = 40 g/mol
Step 2: Use molarity formula to find moles
Molarity = moles / volume (L)
0.1 = moles / 0.5
Moles = 0.1 × 0.5 = 0.05 mol
Step 3: Calculate mass of solute
Mass = moles × molar mass = 0.05 × 40 = 2 g
Answer: Mass of NaOH needed = 2 g
Q6. Explain the process of osmosis and its importance in biological systems.
Osmosis is the spontaneous flow of solvent molecules through a semipermeable membrane from a region of higher solvent concentration (lower solute concentration) to a region of lower solvent concentration (higher solute concentration).
Mechanism:
Solvent molecules move randomly; more pass through membrane in direction of lower solvent concentration, creating net flow of solvent.
Osmotic pressure (π):
π = CRT or π = iMRT
where C = molar concentration, R = gas constant, T = absolute temperature, i = van't Hoff factor
Importance in biological systems:
1. Cell hydration: Cells maintain turgidity through osmotic balance
2. Nutrient absorption: Osmosis aids nutrient transport across cell membranes
3. Kidney function: Osmosis helps filtration and reabsorption of water
4. Plant water uptake: Root cells absorb water via osmosis
5. Blood cell integrity: Isotonic solutions prevent cell lysis or crenation
Example: RBCs in hypertonic solution shrink (crenation); in hypotonic solution swell and burst (hemolysis)
Q7. What are colligative properties? Name and explain any two.
Colligative properties are physical properties of solutions that depend only on the number of solute particles, not on their identity.
Two important colligative properties:
1. Boiling Point Elevation (ΔT_b):
Boiling point of solution is higher than pure solvent.
Formula: ΔT_b = K_b × m
where K_b = boiling point elevation constant, m = molality
Cause: Solute particles reduce vapor pressure of solvent, requiring higher temperature to reach atmospheric pressure.
Example: Salt water boils above 100°C
2. Freezing Point Depression (ΔT_f):
Freezing point of solution is lower than pure solvent.
Formula: ΔT_f = K_f × m
where K_f = freezing point depression constant
Cause: Solute disrupts crystal formation of solvent.
Example: Antifreeze (ethylene glycol) lowers freezing point of water below 0°C
Other colligative properties:
3. Osmotic pressure: π = iMRT
4. Vapor pressure lowering: Relative lowering = x₂ (Raoult's Law)
Q8. A solution is prepared by dissolving 8.5 g of NH₄Cl in 200 g of water. Calculate the molality of the solution.
Given: Mass of NH₄Cl = 8.5 g, Mass of water = 200 g = 0.2 kg
Step 1: Calculate molar mass of NH₄Cl
Molar mass = 14 + (4 × 1) + 35.5 = 14 + 4 + 35.5 = 53.5 g/mol
Step 2: Calculate number of moles of NH₄Cl
Moles = mass / molar mass = 8.5 / 53.5 = 0.159 mol
Step 3: Calculate molality
Molality = moles of solute / mass of solvent (kg)
m = 0.159 / 0.2 = 0.795 m ≈ 0.8 m
Answer: Molality = 0.8 m
Solutions — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. What is the formula for calculating mass percentage of a component in a solution?
- A. (Mass of component / Total mass of solution) × 100
- B. (Mass of component / Volume of solution) × 100
- C. (Moles of component / Total moles of solution) × 100
- D. (Volume of component / Total volume of solution) × 100
Answer: A. (Mass of component / Total mass of solution) × 100 — Mass percentage is defined as the mass of the component divided by the total mass of the solution, multiplied by 100.
MCQ 2. Which of the following factors increases the solubility of a gas in a liquid?
- A. Increase in temperature
- B. Decrease in pressure
- C. Increase in pressure
- D. Increase in molecular size of the gas
Answer: C. Increase in pressure — According to Henry's Law, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. A decrease in temperature also increases solubility.
MCQ 3. Calculate the molarity of a solution containing 5g of NaOH in 450 mL of solution. (Molar mass of NaOH = 40 g/mol)
- A. 0.278 M
- B. 0.222 M
- C. 0.111 M
- D. 0.556 M
Answer: A. 0.278 M — Molarity is calculated as moles of solute divided by the volume of solution in liters. For 5g NaOH (0.125 mol) in 450 mL (0.450 L), the molarity is 0.278 M.
MCQ 4. For an ideal solution, which of the following is true?
- A. ΔH_mix > 0
- B. ΔV_mix < 0
- C. ΔH_mix = 0
- D. P_total < P_A°x_A + P_B°x_B
Answer: C. ΔH_mix = 0 — An ideal solution is characterized by zero enthalpy of mixing (ΔH_mix = 0) and zero volume change on mixing (ΔV_mix = 0), and it obeys Raoult's law.
MCQ 5. An aqueous solution of a protein contains 0.6 g of protein per 100 mL of the solution at 27°C. If the osmotic pressure of the solution is 0.25 atm, what is the molar mass of the protein? (R = 0.0821 L atm mol⁻¹ K⁻¹)
- A. 591 g/mol
- B. 580 g/mol
- C. 620 g/mol
- D. 720 g/mol
Answer: A. 591 g/mol — Using the osmotic pressure formula π = (w/M) * (RT/V), where w=0.6g, R=0.0821, T=300K, π=0.25 atm, V=0.1L, the molar mass M calculates to approximately 591 g/mol.
MCQ 6. When two liquids are completely miscible with each other, they form a:
- A. Suspension
- B. Colloidal solution
- C. True solution
- D. Emulsion
Answer: C. True solution — A true solution is a homogeneous mixture where components are uniformly distributed at a molecular level, as seen when two completely miscible liquids mix.
MCQ 7. The solubility of oxygen gas in water at 293 K is given by Henry's law constant k_H = 2.3 x 10⁻⁴ mol L⁻¹ atm⁻¹. Calculate the concentration of oxygen in mg/L in water exposed to air at 1 atm and 293 K. (Partial pressure of O2 in air = 0.2 atm, Molar mass of O2 = 32 g/mol)
- A. 1.47 mg/L
- B. 2.94 mg/L
- C. 5.88 mg/L
- D. 0.73 mg/L
Answer: A. 1.47 mg/L — Using Henry's Law, C = k_H * P_O2, where P_O2 is 0.2 atm and k_H is 2.3 x 10⁻⁴ mol L⁻¹ atm⁻¹. The concentration in mol/L is 4.6 x 10⁻⁵ mol/L, which converts to 1.47 mg/L.
MCQ 8. Why does the boiling point of water increase upon adding a non-volatile solute?
- A. The vapor pressure of the solution decreases, requiring higher temperature to reach atmospheric pressure.
- B. The surface tension of water increases, making it harder to boil.
- C. The solute particles physically block the escape of solvent molecules.
- D. The density of the solution increases, requiring more energy to boil.
Answer: A. The vapor pressure of the solution decreases, requiring higher temperature to reach atmospheric pressure. — A non-volatile solute lowers the vapor pressure of the solvent. A higher temperature is then required for the solution's vapor pressure to equal the atmospheric pressure, hence the elevation in boiling point.
MCQ 9. Which of the following 0.1 M aqueous solutions will have the highest freezing point? (Assume complete dissociation for electrolytes)
- A. Glucose
- B. NaCl
- C. CaCl2
- D. Al2(SO4)3
Answer: A. Glucose — The freezing point depression (ΔTf) is directly proportional to the van't Hoff factor (i). A non-electrolyte like glucose has i=1, leading to the least freezing point depression and thus the highest freezing point.
MCQ 10. The sum of the mole fractions of all components in a solution is always equal to:
- A. 0
- B. 1
- C. depends on temperature
- D. depends on pressure
Answer: B. 1 — Mole fraction is defined as the moles of a component divided by the total moles of all components in the solution. Therefore, the sum of all mole fractions must be 1.
MCQ 11. Which of the following represents a solution showing negative deviation from Raoult's law?
- A. Ethanol + Acetone
- B. Benzene + Toluene
- C. Chloroform + Acetone
- D. Carbon tetrachloride + Benzene
Answer: C. Chloroform + Acetone — In a solution of chloroform and acetone, hydrogen bonding forms between chloroform and acetone, leading to lower vapor pressure than expected from Raoult's law, indicating negative deviation.
MCQ 12. The relative lowering of vapour pressure is equal to the mole fraction of the solute. This statement is known as:
- A. Henry's Law
- B. Dalton's Law of Partial Pressures
- C. Raoult's Law
- D. Osmosis
Answer: C. Raoult's Law — Raoult's Law states that for a solution of a non-volatile solute, the relative lowering of vapor pressure is equal to the mole fraction of the solute.
MCQ 13. An aqueous solution of glucose is 10% w/w. The density of the solution is 1.2 g/mL. What is the molality of the solution? (Molar mass of glucose = 180 g/mol)
- A. 0.617 mol/kg
- B. 0.556 mol/kg
- C. 0.741 mol/kg
- D. 0.667 mol/kg
Answer: A. 0.617 mol/kg — For a 10% w/w glucose solution, 10g glucose (0.0555 mol) is present in 90g (0.09 kg) of water. Molality is moles of solute per kg of solvent, which calculates to 0.617 mol/kg.
MCQ 14. The phenomenon of solvent molecules passing through a semipermeable membrane from a region of lower solute concentration to a region of higher solute concentration is called:
- A. Diffusion
- B. Reverse osmosis
- C. Osmosis
- D. Effusion
Answer: C. Osmosis — Osmosis is the net movement of solvent molecules across a semipermeable membrane from a region of higher solvent concentration (lower solute concentration) to a region of lower solvent concentration (higher solute concentration).
MCQ 15. A solution of two liquids which boils at a constant temperature without change in composition is called:
- A. Ideal solution
- B. Non-ideal solution
- C. Azeotrope
- D. Colligative solution
Answer: C. Azeotrope — An azeotrope is a mixture of two or more liquids that boils at a constant temperature and whose vapor has the same composition as the liquid.
MCQ 16. 0.1 M solution of K₄[Fe(CN)₆] has an osmotic pressure of 9.84 atm at 27°C. Calculate the van't Hoff factor (i) for this solution. (R = 0.082 L atm mol⁻¹ K⁻¹)
- A. 4.0
- B. 4.5
- C. 5.0
- D. 5.5
Answer: A. 4.0 — Using the formula π = iCRT, we can calculate the van't Hoff factor (i) as π / (CRT). Plugging in the values: i = 9.84 atm / (0.1 M * 0.082 L atm mol⁻¹ K⁻¹ * 300 K) = 4.0.
MCQ 17. Which of the following concentration terms is independent of temperature?
- A. Molarity
- B. Molality
- C. Volume percentage
- D. Parts per million (volume/volume)
Answer: B. Molality — Molality is defined as moles of solute per kilogram of solvent. Both moles and mass are independent of temperature, unlike volume, which changes with temperature.
MCQ 18. At 298 K, the vapor pressure of water is 23.75 mm Hg. Calculate the vapor pressure of a solution containing 5 g of urea (MW=60) dissolved in 100 g of water (MW=18).
- A. 22.95 mm Hg
- B. 23.32 mm Hg
- C. 23.75 mm Hg
- D. 24.12 mm Hg
Answer: B. 23.32 mm Hg — According to Raoult's Law, P_solution = P°_solvent * x_solvent. Calculate moles of urea and water, then the mole fraction of water. Multiplying it by the vapor pressure of pure water gives the solution's vapor pressure, approximately 23.32 mm Hg.
MCQ 19. For an ideal solution, which of the following is true regarding enthalpy and volume change of mixing?
- A. ΔH_mix = 0, ΔV_mix = 0
- B. ΔH_mix > 0, ΔV_mix > 0
- C. ΔH_mix < 0, ΔV_mix < 0
- D. ΔH_mix = 0, ΔV_mix > 0
Answer: A. ΔH_mix = 0, ΔV_mix = 0 — Ideal solutions are characterized by no heat evolution or absorption upon mixing (ΔH_mix = 0) and no change in total volume (ΔV_mix = 0), indicating negligible interaction changes between components.
MCQ 20. Among the following, which 0.1 M aqueous solution will exhibit the largest elevation in boiling point?
- A. Glucose
- B. NaCl
- C. Na₂SO₄
- D. Al₂(SO₄)₃
Answer: D. Al₂(SO₄)₃ — Boiling point elevation is a colligative property directly proportional to the van't Hoff factor (i). Al₂(SO₄)₃ dissociates into 5 ions (2Al³⁺ + 3SO₄²⁻), giving it the largest 'i' (5) among the options, hence the largest elevation.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.