How to Use This Page
This page provides complete exercise solutions for Chapter 9 — Solutions of the Samacheer Kalvi Class 11 Chemistry textbook. Follow this approach:
1. Key Concepts — revise all concentration formulas and colligative property equations. 2. MCQs through Match the Following — test recall and understanding. 3. Brief and Detailed Answers — practice writing structured exam responses. 4. Worked Numericals — follow step by step; practise the method. 5. Common Mistakes & Exam Checklist — review before your exam.
Key Concepts (Quick Recap)
| Concept | Formula |
|---|
| Molarity (M) | M = moles of solute / volume of solution (L) |
| Molality (m) | m = moles of solute / mass of solvent (kg) |
| Mole fraction (X) | X_A = n_A / (n_A + n_B); X_A + X_B = 1 |
| Mass percentage | (mass of solute / mass of solution) × 100 |
| Volume percentage | (volume of solute / volume of solution) × 100 |
| Parts per million (ppm) | (mass of solute / mass of solution) × 10⁶ |
| Raoult's Law (volatile) | P_total = X_A × P⁰_A + X_B × P⁰_B |
| Raoult's Law (non-volatile solute) | P = X_solvent × P⁰; ΔP = X_solute × P⁰ |
| Relative lowering of VP | ΔP/P⁰ = X_solute = n_solute/(n_solute + n_solvent) ≈ n_solute/n_solvent |
| Boiling point elevation | ΔTb = Kb × m; Kb = ebullioscopic constant |
| Freezing point depression | ΔTf = Kf × m; Kf = cryoscopic constant |
| Osmotic pressure | π = CRT = nRT/V (van't Hoff equation) |
| Van't Hoff factor (i) | i = observed colligative property / calculated for non-electrolyte |
| Modified equations with i | ΔTb = i × Kb × m; ΔTf = i × Kf × m; π = iCRT |
Types of Solutions:
| Solute | Solvent | Example |
|---|
| Solid | Liquid | Salt in water |
| Liquid | Liquid | Alcohol in water |
| Gas | Liquid | Oxygen in water |
| Solid | Solid | Alloys (brass) |
| Gas | Gas | Air (N₂ + O₂) |
I. Choose the Best Answer (MCQs)
1. The molality of a solution containing 29.25 g of NaCl in 500 g of water is:
Answer
(a) 0.5 m (b) 1.0 m (c) 2.0 m (d) 0.25 m
Answer(b)
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Moles of NaCl = 29.25 / 58.5 = 0.5 mol
Mass of solvent = 500 g = 0.5 kg
Molality = moles of solute / mass of solvent (kg)
m = 0.5 / 0.5 = 1.0 m
2. Raoult's law is applicable to:
Answer
(a) Ideal solutions only (b) Dilute solutions of non-volatile solutes only (c) Both ideal solutions and dilute solutions of non-volatile solutes (d) Concentrated solutions only
Answer(c) Raoult's law applies to ideal solutions (where all components follow Raoult's law over the entire composition range) and also to dilute solutions of non-volatile solutes (where the solvent approximately follows Raoult's law).
3. Which of the following is a colligative property?
Answer
(a) Boiling point (b) Freezing point (c) Osmotic pressure (d) Surface tension
Answer(c) Colligative properties depend on the number of solute particles, not their nature. Osmotic pressure, relative lowering of vapour pressure, boiling point elevation, and freezing point depression are the four colligative properties. Boiling point and freezing point themselves are not colligative — only their changes (ΔTb, ΔTf) are.
4. The van't Hoff factor for NaCl in dilute aqueous solution is approximately:
Answer
(a) 0.5 (b) 1.0 (c) 2.0 (d) 3.0
Answer(c) NaCl dissociates almost completely in dilute solution:
NaCl → Na⁺ + Cl⁻
1 mole produces 2 moles of ions
i ≈ 2.0 (for complete dissociation)
5. When a non-volatile solute is added to a solvent, the vapour pressure:
Answer
(a) Increases (b) Decreases (c) Remains the same (d) First increases then decreases
Answer(b) Adding a non-volatile solute reduces the mole fraction of the solvent, which decreases the vapour pressure according to Raoult's law: P = X_solvent × P⁰. Since X_solvent < 1 after adding solute, P < P⁰.
6. The molarity of a solution containing 4 g of NaOH in 500 mL of solution is:
Answer
(a) 0.2 M (b) 0.4 M (c) 0.1 M (d) 2.0 M
Answer(a)
Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol
Moles of NaOH = 4 / 40 = 0.1 mol
Volume of solution = 500 mL = 0.5 L
Molarity = 0.1 / 0.5 = 0.2 M
7. Osmotic pressure of a solution increases with:
Answer
(a) Decreasing temperature (b) Decreasing concentration (c) Increasing concentration (d) Adding a non-volatile solute and then removing it
Answer(c) Osmotic pressure π = CRT, where C is molar concentration. π increases with increasing concentration and temperature. Adding more solute increases concentration and hence osmotic pressure.
II. Fill in the Blanks
1. Molarity is defined as the number of moles of solute per litre of solution. 2. Molality is independent of temperature because it involves mass, not volume. 3. The mole fraction of all components in a solution adds up to 1. 4. Raoult's law states that the vapour pressure of a solvent above a solution is proportional to its mole fraction in the solution. 5. The relative lowering of vapour pressure is equal to the mole fraction of the solute. 6. The boiling point of a solution is higher than that of the pure solvent. 7. The freezing point of a solution is lower than that of the pure solvent. 8. The van't Hoff factor for a non-electrolyte like glucose is 1. 9. The van't Hoff factor for CaCl₂ (assuming complete dissociation) is 3. 10. The solute that produces the maximum number of ions will cause the greatest freezing point depression. 11. Osmotic pressure is a colligative property that depends on the number of solute particles. 12. An ideal solution obeys Raoult's law over the entire range of composition. 13. The ebullioscopic constant Kb has the unit K kg mol⁻¹. 14. In reverse osmosis, a pressure greater than the osmotic pressure is applied. 15. Henry's law relates the solubility of a gas to the pressure of the gas above the solution.
III. True or False (with Corrections)
1. Molarity changes with temperature but molality does not.
True. Molarity depends on volume of solution, which changes with temperature (due to thermal expansion). Molality depends on mass of solvent, which does not change with temperature.
2. An ideal solution has ΔH_mix = 0 and ΔV_mix = 0.
True. In an ideal solution, there are no intermolecular interactions different from those in the pure components, so enthalpy of mixing is zero and volume of mixing is zero.
3. The boiling point elevation is directly proportional to the molarity of the solution.
False. Boiling point elevation is directly proportional to the molality of the solution, not molarity. ΔTb = Kb × m. This is because molality is temperature-independent while molarity changes with temperature.
4. A hypotonic solution has a higher osmotic pressure than the reference solution.
False. A hypotonic solution has a lower osmotic pressure (lower solute concentration) than the reference solution. When a cell is placed in a hypotonic solution, water enters the cell (endosmosis), causing it to swell.
5. The van't Hoff factor for an associating solute like acetic acid in benzene is less than 1.
True. When acetic acid associates (dimerises) in benzene, two molecules combine to form one dimer, effectively reducing the number of particles. Hence i < 1 for association.
IV. Match the Following
| Column A | Column B |
|---|
| 1. Molarity | (a) Moles of solute per kg of solvent |
| 2. Molality | (b) Moles of solute per litre of solution |
| 3. Mole fraction | (c) ΔTb = Kb × m |
| 4. Boiling point elevation | (d) n_A / (n_A + n_B) |
| 5. Freezing point depression | (e) π = iCRT |
| 6. Osmotic pressure | (f) ΔTf = Kf × m |
Answer
1 → (b) Molarity = moles of solute / volume of solution (L)
2 → (a) Molality = moles of solute / mass of solvent (kg)
3 → (d) Mole fraction = n_A / (n_A + n_B)
4 → (c) Boiling point elevation: ΔTb = Kb × m
5 → (f) Freezing point depression: ΔTf = Kf × m
6 → (e) Osmotic pressure: π = iCRT
V. Answer Briefly (2-mark)
1. Distinguish between molarity and molality.
Molarity (M): Molality (m):
- moles of solute / volume (L) - moles of solute / mass of solvent (kg)
- Depends on temperature - Independent of temperature
- Unit: mol/L (or M) - Unit: mol/kg (or m)
- Changes with expansion/contraction - Does not change with temperature
2. State Raoult's law for a solution containing a non-volatile solute.
Answer
Raoult's law states that the vapour pressure of a solvent above a solution containing a non-volatile solute is equal to the product of the mole fraction of the solvent and the vapour pressure of the pure solvent.
P = X_solvent × P⁰
Where:
P = vapour pressure of the solution
X_solvent = mole fraction of the solvent
P⁰ = vapour pressure of the pure solvent
Also: ΔP = P⁰ − P = X_solute × P⁰
Relative lowering of vapour pressure = ΔP/P⁰ = X_solute
3. Define osmotic pressure. Write its mathematical expression.
Answer
Osmotic pressure is the minimum external pressure that must be applied to a solution to prevent the inward flow of solvent molecules across a semipermeable membrane.
π = CRT = nRT/V
Where:
π = osmotic pressure (atm)
C = molar concentration (mol/L)
R = 0.0821 L·atm mol⁻¹ K⁻¹
T = temperature (K)
n = number of moles of solute
V = volume of solution (L)
4. What is the van't Hoff factor? What are its values for NaCl and glucose?
Answer
The van't Hoff factor (i) is the ratio of the observed colligative property to the calculated colligative property (assuming no dissociation or association).
i = observed colligative property / calculated colligative property
i = actual number of particles / number of formula units dissolved
For NaCl (complete dissociation): NaCl → Na⁺ + Cl⁻; i = 2
For glucose (non-electrolyte): no dissociation; i = 1
VI. Answer in Detail (5-mark)
1. Explain Raoult's law for volatile and non-volatile solutes. Calculate the vapour pressure of a solution prepared by dissolving 18 g of glucose (C₆H₁₂O₆) in 180 g of water at 25°C. Vapour pressure of pure water at 25°C = 23.8 mmHg.
Answer
Raoult's Law for Volatile Solutes:
For a solution of two volatile liquids A and B, the partial vapour pressure of each component is equal to the product of its mole fraction and its vapour pressure in the pure state.
P_A = X_A × P⁰_A
P_B = X_B × P⁰_B
P_total = P_A + P_B = X_A × P⁰_A + X_B × P⁰_B
Raoult's Law for Non-Volatile Solute:
When a non-volatile solute is dissolved in a solvent, the vapour pressure of the solution is lower than that of the pure solvent. The vapour pressure is proportional to the mole fraction of the solvent.
P = X_solvent × P⁰
ΔP = P⁰ − P = X_solute × P⁰
Relative lowering: ΔP/P⁰ = X_solute = n_solute/(n_solute + n_solvent) ≈ n_solute/n_solvent
Numerical Calculation:
Given: Mass of glucose = 18 g, Mass of water = 180 g
P⁰_water = 23.8 mmHg
Step 1: Calculate moles
Molar mass of glucose (C₆H₁₂O₆) = 6(12) + 12(1) + 6(16) = 180 g/mol
Molar mass of water (H₂O) = 2(1) + 16 = 18 g/mol
n_glucose = 18 / 180 = 0.1 mol
n_water = 180 / 18 = 10 mol
Step 2: Calculate mole fraction of water (solvent)
X_water = n_water / (n_water + n_glucose)
X_water = 10 / (10 + 0.1) = 10 / 10.1 = 0.9901
Step 3: Apply Raoult's law
P = X_water × P⁰_water
P = 0.9901 × 23.8
P = 23.56 mmHg
Step 4: Calculate lowering of vapour pressure
ΔP = P⁰ − P = 23.8 − 23.56 = 0.24 mmHg
Relative lowering = ΔP/P⁰ = 0.24/23.8 = 0.0101 ≈ X_solute
2. Explain colligative properties in detail. Calculate the boiling point of a solution containing 15 g of NaCl (i = 2) in 250 g of water. Kb for water = 0.512 K kg mol⁻¹.
Answer
Colligative properties are properties of dilute solutions that depend on the number of solute particles and not on their nature (identity). There are four colligative properties:
1. Relative Lowering of Vapour Pressure:
ΔP/P⁰ = X_solute = n₂/(n₁ + n₂) ≈ n₂/n₁ (for dilute solutions)
Where n₁ = moles of solvent, n₂ = moles of solute
2. Elevation of Boiling Point:
ΔTb = Kb × m (for non-electrolytes)
ΔTb = i × Kb × m (for electrolytes)
Where:
ΔTb = Tb(solution) − Tb(solvent)
Kb = ebullioscopic constant
m = molality
i = van't Hoff factor
The boiling point of a solution is always higher than that of
the pure solvent because the lowered vapour pressure requires a
higher temperature to reach atmospheric pressure.
3. Depression of Freezing Point:
ΔTf = Kf × m (for non-electrolytes)
ΔTf = i × Kf × m (for electrolytes)
Where:
ΔTf = Tf(solvent) − Tf(solution)
Kf = cryoscopic constant
4. Osmotic Pressure:
π = CRT (for non-electrolytes)
π = iCRT (for electrolytes)
Osmotic pressure is the most sensitive colligative property
and is preferred for determining molecular masses of polymers.
Answer
Numerical Calculation:
Given: Mass of NaCl = 15 g, Mass of water = 250 g = 0.25 kg
Kb = 0.512 K kg mol⁻¹, i = 2
Boiling point of pure water = 100°C
Step 1: Calculate moles of NaCl
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
n_NaCl = 15 / 58.5 = 0.2564 mol
Step 2: Calculate molality
m = n_NaCl / mass of solvent (kg)
m = 0.2564 / 0.25 = 1.0256 mol/kg
Step 3: Calculate boiling point elevation
ΔTb = i × Kb × m
ΔTb = 2 × 0.512 × 1.0256
ΔTb = 1.050 K
Step 4: Calculate boiling point of solution
Tb(solution) = Tb(solvent) + ΔTb
Tb(solution) = 100 + 1.050 = 101.05°C
The boiling point of the NaCl solution is 101.05°C.
3. Define osmotic pressure. Calculate the osmotic pressure of a 0.5 M glucose solution at 300 K. Also calculate the osmotic pressure of a 0.5 M NaCl solution (i = 2) at 300 K and compare.
Answer
Definition: Osmotic pressure (π) is the minimum hydrostatic pressure that must be applied to the solution side to prevent osmosis (the movement of solvent molecules from the solvent side to the solution side through a semipermeable membrane).
Characteristics of osmotic pressure:
- It is a colligative property (depends on number of particles)
- It increases with concentration and temperature
- It is used to determine molecular masses of polymers and proteins
- π = CRT (van't Hoff equation, analogous to PV = nRT)
Numerical Calculation — Glucose (non-electrolyte, i = 1):
Given: C = 0.5 M, T = 300 K, i = 1 (glucose is a non-electrolyte)
R = 0.0821 L·atm mol⁻¹ K⁻¹
π = iCRT
π = 1 × 0.5 × 0.0821 × 300
π = 12.315 atm
Osmotic pressure of 0.5 M glucose = 12.315 atm
Numerical Calculation — NaCl (electrolyte, i = 2):
Given: C = 0.5 M, T = 300 K, i = 2 (NaCl → Na⁺ + Cl⁻)
R = 0.0821 L·atm mol⁻¹ K⁻¹
π = iCRT
π = 2 × 0.5 × 0.0821 × 300
π = 24.63 atm
Osmotic pressure of 0.5 M NaCl = 24.63 atm
Comparison:
Glucose (i = 1): π = 12.315 atm
NaCl (i = 2): π = 24.63 atm
NaCl produces TWICE the osmotic pressure of glucose at the
same molar concentration because NaCl dissociates into 2 ions
(Na⁺ and Cl⁻), effectively doubling the number of particles.
This demonstrates the importance of the van't Hoff factor:
- For non-electrolytes (glucose, urea): i = 1
- For NaCl type (AB → A⁺ + B⁻): i ≈ 2
- For CaCl₂ type (AB₂ → A²⁺ + 2B⁻): i ≈ 3
- For associating solutes (acetic acid dimer): i < 1
Common Mistakes to Avoid
1. Confusing molarity and molality — molarity uses volume of solution (L), molality uses mass of solvent (kg). They give different values and molality is temperature-independent.
2. Forgetting the van't Hoff factor for electrolytes — for NaCl, i ≈ 2; for CaCl₂, i ≈ 3; for AlCl₃, i ≈ 4. Always multiply the colligative property by i for electrolytes.
3. Using mass of solution instead of mass of solvent for molality — molality = moles of solute / mass of SOLVENT (kg), not total mass of solution.
4. Not converting units properly — always express mass in kg for molality, volume in litres for molarity, and temperature in Kelvin for osmotic pressure calculations.
5. Confusing hypotonic, hypertonic, and isotonic — hypotonic (lower concentration, water enters cell), hypertonic (higher concentration, water leaves cell), isotonic (equal concentration, no net water movement).
6. Adding boiling point elevation instead of subtracting for freezing point — ΔTb is ADDED to the boiling point; ΔTf is SUBTRACTED from the freezing point. Both make the solution's properties deviate from the pure solvent.
7. Forgetting that Raoult's law applies to the solvent, not the solute — the lowering of vapour pressure equals X_solute × P⁰, but the vapour pressure of the solution equals X_solvent × P⁰.
8. Confusing mole fraction with mass fraction — mole fraction uses moles (n = mass/molar mass), not mass directly. Always convert mass to moles before calculating mole fraction.
Exam Checklist
- [ ] Can you calculate molarity, molality, and mole fraction from given data?
- [ ] Can you state and apply Raoult's law for both volatile and non-volatile solutes?
- [ ] Can you calculate the relative lowering of vapour pressure?
- [ ] Can you calculate boiling point elevation (ΔTb = i × Kb × m)?
- [ ] Can you calculate freezing point depression (ΔTf = i × Kf × m)?
- [ ] Can you calculate osmotic pressure (π = iCRT)?
- [ ] Do you know when to use the van't Hoff factor (i) and its values for common electrolytes?
- [ ] Can you determine the molecular mass of a solute from colligative property data?
- [ ] Can you explain the difference between ideal and non-ideal solutions?
- [ ] Can you describe osmosis, reverse osmosis, and their applications?
