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NCERT Class 12 · Chemistry Based on NCERT Class 12 Chemistry textbook · Free CBSE study kit

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Chapter Notes

Overview
This chapter explores the nature and properties of liquid solutions, which are homogeneous mixtures of two or more components. It covers how to express concentration in various units, the laws governing vapour pressure (Henry's law and Raoult's law), and the classification of solutions as ideal or non-ideal. The central idea is that colligative properties—which depend only on the number of solute particles—allow us to determine the molar mass of solutes and explain abnormal behaviour in solutions.
Key Concepts (26)
Solution
A homogeneous mixture of two or more than two components.
e.g. Sugar dissolved in water is a solution.
Solvent
The component of a solution that is present in the largest quantity and determines the physical state in which the solution exists.
e.g. In a salt solution, water is the solvent.
Solute
The component(s) of a solution other than the solvent.
e.g. In a salt solution, salt is the solute.
Binary solution
A solution consisting of two components.
e.g. A mixture of ethanol and water is a binary solution.
Mass percentage (w/w)
The mass of a component divided by the total mass of the solution, multiplied by 100.
e.g. 10% glucose in water means 10 g of glucose dissolved in 90 g of water to give 100 g of solution.
Volume percentage (V/V)
The volume of a component divided by the total volume of the solution, multiplied by 100.
e.g. 10% ethanol solution means 10 mL of ethanol dissolved in water to make 100 mL of solution.
Mass by volume percentage (w/V)
The mass of solute dissolved in 100 mL of the solution.
e.g. A 5% (w/V) solution of NaCl means 5 g of NaCl in 100 mL of solution.
Parts per million (ppm)
Number of parts of a component per 10^6 total parts of all components of the solution.
e.g. 5.8 ppm of oxygen in sea water means 5.8 g of O₂ per 10⁶ g of sea water.
Mole fraction
The ratio of the number of moles of a component to the total number of moles of all components in the solution.
e.g. In a binary mixture of A and B, x_A = n_A / (n_A + n_B).
Molarity (M)
Number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.
e.g. 0.25 M NaOH means 0.25 mol of NaOH in 1 L of solution.
Molality (m)
Number of moles of solute per kilogram of solvent.
e.g. 1.00 m KCl means 1 mol of KCl dissolved in 1 kg of water.
Saturated solution
A solution in which no more solute can be dissolved at the same temperature and pressure, and which is in dynamic equilibrium with undissolved solute.
e.g. A solution of salt in water that has undissolved salt at the bottom is saturated.
Henry's law
At a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of the liquid or solution.
e.g. Carbonated drinks are sealed under high pressure to increase CO₂ solubility.
Henry's law constant (K_H)
The proportionality constant in Henry's law; p = K_H x, where p is partial pressure and x is mole fraction of gas in solution.
e.g. K_H for N₂ in water at 293 K is 76.48 kbar.
Raoult's law
For a solution of volatile liquids, the partial vapour pressure of each component is directly proportional to its mole fraction in the solution.
e.g. For a binary solution, p₁ = x₁ p₁⁰ and p₂ = x₂ p₂⁰.
Ideal solution
A solution that obeys Raoult's law over the entire range of concentration, with ΔmixH = 0 and ΔmixV = 0.
e.g. A mixture of benzene and toluene is nearly ideal.
Non-ideal solution
A solution that does not obey Raoult's law over the entire range of concentration, showing either positive or negative deviation.
e.g. A mixture of ethanol and acetone shows positive deviation.
Positive deviation from Raoult's law
When the vapour pressure of a solution is higher than that predicted by Raoult's law, due to weaker A-B interactions compared to A-A and B-B interactions.
e.g. Ethanol and acetone mixture shows positive deviation.
Negative deviation from Raoult's law
When the vapour pressure of a solution is lower than that predicted by Raoult's law, due to stronger A-B interactions compared to A-A and B-B interactions.
e.g. Phenol and aniline mixture shows negative deviation.
Azeotrope
A binary mixture that has the same composition in liquid and vapour phase and boils at a constant temperature.
e.g. Ethanol-water mixture forms a minimum boiling azeotrope at about 95% ethanol by volume.
Colligative properties
Properties of solutions that depend on the number of solute particles irrespective of their nature relative to the total number of particles present.
e.g. Relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure.
Relative lowering of vapour pressure
The ratio of the decrease in vapour pressure of the solvent to the vapour pressure of the pure solvent, equal to the mole fraction of the solute.
e.g. Δp₁ / p₁⁰ = x₂.
Elevation of boiling point (ΔTb)
The increase in boiling point of a solvent when a non-volatile solute is dissolved in it, directly proportional to molality.
e.g. ΔTb = Kb m.
Boiling Point Elevation Constant (Kb)
The molal elevation constant; the elevation in boiling point for a 1 molal solution. Unit: K kg mol⁻¹.
e.g. Kb for water is 0.52 K kg mol⁻¹.
Depression of freezing point (ΔTf)
The decrease in freezing point of a solvent when a non-volatile solute is dissolved in it, directly proportional to molality.
e.g. ΔTf = Kf m.
Freezing Point Depression Constant (Kf)
The molal depression constant; the depression in freezing point for a 1 molal solution. Unit: K kg mol⁻¹.
e.g. Kf for water is 1.86 K kg mol⁻¹.
Formulas & Equations (17)
Mass % of a component = (Mass of the component in the solution / Total mass of the solution) × 100
Reads as: Mass percentage equals mass of component divided by total mass of solution, multiplied by 100.
Variables: Mass of component (g), Total mass of solution (g).
Use when: Calculating concentration by mass in industrial or laboratory contexts.
⚠ Using volume instead of mass, or forgetting to multiply by 100.
Volume % of a component = (Volume of the component / Total volume of solution) × 100
Reads as: Volume percentage equals volume of component divided by total volume of solution, multiplied by 100.
Variables: Volume of component (mL), Total volume of solution (mL).
Use when: Expressing concentration of liquid-liquid solutions.
⚠ Adding volumes of solute and solvent instead of using final solution volume.
Parts per million (ppm) = (Number of parts of the component / Total number of parts of all components of the solution) × 10⁶
Reads as: Parts per million equals number of parts of component divided by total parts, multiplied by 10 to the power 6.
Variables: Number of parts of component, Total number of parts.
Use when: Expressing very small concentrations, e.g., pollutants in water or air.
⚠ Using 10⁵ or 10⁷ instead of 10⁶.
Mole fraction of a component = Number of moles of the component / Total number of moles of all the components
Reads as: Mole fraction equals moles of component divided by total moles.
Variables: n_i = moles of component i; total moles = sum of all n_i.
Use when: Relating vapour pressure to concentration, or in gas mixtures.
⚠ Using mass instead of moles, or forgetting that sum of mole fractions = 1.
Molarity (M) = Moles of solute / Volume of solution in litre
Reads as: Molarity equals moles of solute divided by volume of solution in litres.
Variables: Moles of solute (mol), Volume of solution (L).
Use when: Calculating concentration for reactions in solution, especially volumetric analysis.
⚠ Using volume of solvent instead of volume of solution, or not converting mL to L.
Molality (m) = Moles of solute / Mass of solvent in kg
Reads as: Molality equals moles of solute divided by mass of solvent in kilograms.
Variables: Moles of solute (mol), Mass of solvent (kg).
Use when: Calculating colligative properties, as molality is temperature-independent.
⚠ Using mass of solution instead of mass of solvent, or not converting g to kg.
Henry's law: p = K_H x
Reads as: Partial pressure of gas equals Henry's law constant times mole fraction of gas in solution.
Variables: p = partial pressure of gas (bar or Pa), K_H = Henry's law constant (bar or Pa), x = mole fraction of gas in solution.
Use when: Calculating solubility of a gas in a liquid at a given partial pressure.
⚠ Confusing K_H with solubility; higher K_H means lower solubility.
Raoult's law for volatile component: p₁ = x₁ p₁⁰
Reads as: Partial vapour pressure of component 1 equals mole fraction of component 1 times vapour pressure of pure component 1.
Variables: p₁ = partial vapour pressure of component 1, x₁ = mole fraction of component 1 in liquid, p₁⁰ = vapour pressure of pure component 1.
Use when: Calculating vapour pressure of a component in a liquid-liquid solution.
⚠ Using mole fraction in vapour phase instead of liquid phase.
Total vapour pressure: p_total = p₁ + p₂ = x₁ p₁⁰ + x₂ p₂⁰
Reads as: Total vapour pressure equals sum of partial pressures, which equals mole fraction times pure vapour pressure for each component.
Variables: p_total = total vapour pressure, p₁, p₂ = partial pressures, x₁, x₂ = mole fractions in liquid, p₁⁰, p₂⁰ = pure vapour pressures.
Use when: Calculating total vapour pressure of a binary liquid mixture.
⚠ Forgetting that x₁ + x₂ = 1, or using mole fractions from vapour phase.
Relative lowering of vapour pressure: (p₁⁰ - p₁) / p₁⁰ = x₂
Reads as: Relative lowering of vapour pressure equals mole fraction of solute.
Variables: p₁⁰ = vapour pressure of pure solvent, p₁ = vapour pressure of solvent in solution, x₂ = mole fraction of solute.
Use when: Calculating molar mass of a non-volatile solute from vapour pressure data.
⚠ Using p₁ instead of (p₁⁰ - p₁) in numerator, or forgetting that x₂ = n₂/(n₁+n₂).
For dilute solutions: (p₁⁰ - p₁) / p₁⁰ = n₂ / n₁ = (w₂/M₂) / (w₁/M₁)
Reads as: Relative lowering equals moles of solute divided by moles of solvent, which equals mass of solute over molar mass of solute divided by mass of solvent over molar mass of solvent.
Variables: n₂ = moles of solute, n₁ = moles of solvent, w₂ = mass of solute, M₂ = molar mass of solute, w₁ = mass of solvent, M₁ = molar mass of solvent.
Use when: Calculating molar mass of solute from vapour pressure lowering data.
⚠ Neglecting n₂ in denominator when n₂ is not negligible compared to n₁.
Elevation of boiling point: ΔTb = Kb m
Reads as: Delta T b equals K b times molality.
Variables: ΔTb = elevation of boiling point (K), Kb = molal elevation constant (K kg mol⁻¹), m = molality (mol kg⁻¹).
Use when: Calculating boiling point elevation or molar mass from boiling point data.
⚠ Using molarity instead of molality, or forgetting to convert mass of solvent to kg.
Molar mass from boiling point elevation: M₂ = (Kb × w₂ × 1000) / (ΔTb × w₁)
Reads as: Molar mass of solute equals K b times mass of solute times 1000 divided by delta T b times mass of solvent.
Variables: M₂ = molar mass of solute (g mol⁻¹), Kb = molal elevation constant (K kg mol⁻¹), w₂ = mass of solute (g), ΔTb = elevation (K), w₁ = mass of solvent (g).
Use when: Determining molar mass of a non-volatile solute experimentally.
⚠ Forgetting the factor 1000 to convert g to kg, or using wrong units.
Depression of freezing point: ΔTf = Kf m
Reads as: Delta T f equals K f times molality.
Variables: ΔTf = depression of freezing point (K), Kf = molal depression constant (K kg mol⁻¹), m = molality (mol kg⁻¹).
Use when: Calculating freezing point depression or molar mass from freezing point data.
⚠ Using molarity instead of molality, or confusing Kf with Kb.
Molar mass from freezing point depression: M₂ = (Kf × w₂ × 1000) / (ΔTf × w₁)
Reads as: Molar mass of solute equals K f times mass of solute times 1000 divided by delta T f times mass of solvent.
Variables: M₂ = molar mass of solute (g mol⁻¹), Kf = molal depression constant (K kg mol⁻¹), w₂ = mass of solute (g), ΔTf = depression (K), w₁ = mass of solvent (g).
Use when: Determining molar mass of a non-volatile solute using cryoscopy.
⚠ Forgetting to subtract the freezing point of solution from that of pure solvent to get ΔTf.
Kf = (R × M₁ × Tf²) / (1000 × ΔfusH)
Reads as: K f equals R times molar mass of solvent times freezing point squared divided by 1000 times enthalpy of fusion.
Variables: R = gas constant (8.314 J mol⁻¹ K⁻¹), M₁ = molar mass of solvent (g mol⁻¹), Tf = freezing point of pure solvent (K), ΔfusH = enthalpy of fusion (J mol⁻¹).
Use when: Calculating Kf from thermodynamic data.
⚠ Using Tb instead of Tf, or forgetting to convert units.
Kb = (R × M₁ × Tb²) / (1000 × ΔvapH)
Reads as: K b equals R times molar mass of solvent times boiling point squared divided by 1000 times enthalpy of vaporisation.
Variables: R = gas constant (8.314 J mol⁻¹ K⁻¹), M₁ = molar mass of solvent (g mol⁻¹), Tb = boiling point of pure solvent (K), ΔvapH = enthalpy of vaporisation (J mol⁻¹).
Use when: Calculating Kb from thermodynamic data.
⚠ Using Tf instead of Tb, or forgetting to convert units.
How It Works
Formation of a saturated solution
  1. Step 1: Solid solute is added to solvent; dissolution begins as solute particles enter the solution.
  2. Step 2: Simultaneously, some solute particles in solution collide with undissolved solid and crystallise out.
  3. Step 3: A dynamic equilibrium is reached when the rate of dissolution equals the rate of crystallisation.
  4. Step 4: At equilibrium, the concentration of solute remains constant, and the solution is saturated.
💡 The equilibrium is dynamic — both processes continue at equal rates, not that nothing happens.
Effect of pressure on solubility of a gas (Henry's law)
  1. Step 1: A gas above a liquid exerts a partial pressure; gas molecules strike the liquid surface and dissolve.
  2. Step 2: At equilibrium, the rate of gas molecules entering the solution equals the rate of those leaving.
  3. Step 3: Increasing the pressure (by compressing the gas) increases the number of gas molecules per unit volume above the solution.
  4. Step 4: More gas molecules strike the surface per unit time, increasing the rate of dissolution.
  5. Step 5: A new equilibrium is reached with a higher concentration of dissolved gas — solubility increases.
💡 Solubility of a gas increases with pressure, but decreases with temperature because dissolution is exothermic.
Vapour pressure of a binary liquid solution (Raoult's law)
  1. Step 1: Two volatile liquids are mixed in a closed container; both evaporate.
  2. Step 2: At equilibrium, the partial vapour pressure of each component is proportional to its mole fraction in the liquid: p₁ = x₁ p₁⁰, p₂ = x₂ p₂⁰.
  3. Step 3: Total vapour pressure is the sum of partial pressures: p_total = p₁ + p₂.
  4. Step 4: The composition of the vapour phase is determined by Dalton's law: y₁ = p₁ / p_total, y₂ = p₂ / p_total.
💡 The vapour phase is always richer in the more volatile component (higher p⁰).
Determination of molar mass using relative lowering of vapour pressure
  1. Step 1: Measure vapour pressure of pure solvent (p₁⁰) and of solution (p₁) at the same temperature.
  2. Step 2: Calculate relative lowering: (p₁⁰ - p₁) / p₁⁰.
  3. Step 3: For dilute solutions, this equals n₂ / n₁ = (w₂/M₂) / (w₁/M₁).
  4. Step 4: Rearrange to solve for M₂: M₂ = (w₂ × M₁ × p₁⁰) / (w₁ × (p₁⁰ - p₁)).
💡 This method works only for non-volatile solutes; the solute must not contribute to vapour pressure.
Determination of molar mass using boiling point elevation
  1. Step 1: Measure the boiling point of pure solvent (Tb⁰) and of the solution (Tb).
  2. Step 2: Calculate ΔTb = Tb - Tb⁰.
  3. Step 3: Use ΔTb = Kb × m, where m = (w₂ × 1000) / (M₂ × w₁).
  4. Step 4: Rearrange to solve for M₂: M₂ = (Kb × w₂ × 1000) / (ΔTb × w₁).
💡 Ensure the solute is non-volatile; otherwise, the boiling point may not elevate as expected.
Determination of molar mass using freezing point depression
  1. Step 1: Measure the freezing point of pure solvent (Tf⁰) and of the solution (Tf).
  2. Step 2: Calculate ΔTf = Tf⁰ - Tf.
  3. Step 3: Use ΔTf = Kf × m, where m = (w₂ × 1000) / (M₂ × w₁).
  4. Step 4: Rearrange to solve for M₂: M₂ = (Kf × w₂ × 1000) / (ΔTf × w₁).
💡 Freezing point depression is more sensitive than boiling point elevation because Kf is usually larger than Kb.
Exam Traps — Don't Make These Mistakes
⚠ Confusing molarity and molality in colligative properties
✗ Using molarity (M) instead of molality (m) in formulas ΔTb = Kb m or ΔTf = Kf m.
→ Colligative properties depend on the number of solute particles per mass of solvent, not per volume of solution. Molarity changes with temperature because volume changes, while molality does not.
✓ Always use molality (m) for boiling point elevation and freezing point depression calculations.
⚠ Forgetting to convert mass of solvent to kg
✗ Plugging mass of solvent in grams directly into the molality formula: m = moles of solute / mass of solvent in g.
→ Molality is defined as moles of solute per kilogram of solvent. Using grams gives a value 1000 times too large.
✓ Convert mass of solvent to kg: m = (moles of solute) / (mass of solvent in g / 1000).
⚠ Using mole fraction in vapour phase instead of liquid phase in Raoult's law
✗ Writing p₁ = y₁ p₁⁰, where y₁ is the mole fraction in vapour phase.
→ Raoult's law relates the partial vapour pressure to the mole fraction of the component in the liquid phase, not the vapour phase. The vapour phase composition is given by Dalton's law: y₁ = p₁ / p_total.
✓ Use x₁ (liquid mole fraction) in Raoult's law: p₁ = x₁ p₁⁰.
⚠ Confusing positive and negative deviations from Raoult's law
✗ Stating that stronger A-B interactions than A-A and B-B cause positive deviation.
→ Positive deviation occurs when A-B interactions are weaker than A-A and B-B interactions, making it easier for molecules to escape, thus increasing vapour pressure. Negative deviation occurs when A-B interactions are stronger.
✓ Positive deviation: weaker A-B interactions → higher vapour pressure. Negative deviation: stronger A-B interactions → lower vapour pressure.
⚠ Misinterpreting Henry's law constant
✗ Stating that a higher K_H value means higher solubility of the gas.
→ From p = K_H x, at a given pressure p, x = p / K_H. So higher K_H means lower mole fraction (lower solubility).
✓ Higher K_H indicates lower solubility of the gas in the liquid at a given pressure.
⚠ Forgetting that azeotropes cannot be separated by fractional distillation
✗ Stating that components of an azeotropic mixture can be separated by fractional distillation.
→ An azeotrope has the same composition in liquid and vapour phase, so boiling does not change the composition. Hence, fractional distillation cannot separate the components beyond the azeotropic composition.
✓ Azeotropes boil at a constant temperature and have identical composition in liquid and vapour phases, making separation by fractional distillation impossible.
Compare & Contrast
Dependence on temperature
Molarity (M)
Molality (m)
Molarity depends on temperature because volume changes with temperature; molality is independent of temperature because it uses mass of solvent.
Obeying Raoult's law
Ideal solution
Non-ideal solution
An ideal solution obeys Raoult's law over the entire concentration range with ΔmixH = 0 and ΔmixV = 0; a non-ideal solution does not obey Raoult's law and shows either positive or negative deviation.
Intermolecular interactions
Positive deviation from Raoult's law
Negative deviation from Raoult's law
Positive deviation occurs when A-B interactions are weaker than A-A and B-B interactions (vapour pressure higher); negative deviation occurs when A-B interactions are stronger (vapour pressure lower).
Type of deviation from Raoult's law
Minimum boiling azeotrope
Maximum boiling azeotrope
Minimum boiling azeotropes form from solutions showing large positive deviation (e.g., ethanol-water); maximum boiling azeotropes form from solutions showing large negative deviation (e.g., nitric acid-water).
Exam Focus
Marks pattern: Typically 2-3 MCQs (1 mark each), 1-2 short answer questions (2-3 marks), and 1 long answer numerical (5 marks). Total approximately 8-10 marks.
Most asked: Numerical problems on molarity, molality, mole fraction, and mass percentage · Henry's law and its applications · Raoult's law and vapour pressure calculations · Ideal and non-ideal solutions, deviations, and azeotropes · Colligative properties: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point · Determination of molar mass using colligative properties
Don't skip: Numerical problems on molar mass determination using boiling point elevation or freezing point depression; Conceptual questions on deviations from Raoult's law with examples; Application-based questions on Henry's law (e.g., bends, soft drinks)
One-Line Revision (read aloud before the exam)
  1. A solution is a homogeneous mixture of two or more components; the component in largest quantity is the solvent.
  2. Mass percentage = (mass of component / total mass of solution) × 100.
  3. Volume percentage = (volume of component / total volume of solution) × 100.
  4. Parts per million (ppm) = (number of parts of component / total parts) × 10⁶.
  5. Mole fraction of a component = moles of component / total moles; sum of all mole fractions = 1.
  6. Molarity (M) = moles of solute / volume of solution in litres; it is temperature-dependent.
  7. Molality (m) = moles of solute / mass of solvent in kg; it is temperature-independent.
  8. Henry's law: p = K_H x; solubility of a gas increases with pressure and decreases with temperature.
  9. Raoult's law: p₁ = x₁ p₁⁰; total vapour pressure of a binary solution = x₁ p₁⁰ + x₂ p₂⁰.
  10. Ideal solutions obey Raoult's law with ΔmixH = 0 and ΔmixV = 0; e.g., benzene + toluene.
  11. Non-ideal solutions show positive deviation (weaker A-B interactions) or negative deviation (stronger A-B interactions).
  12. Azeotropes are constant-boiling mixtures with same composition in liquid and vapour phase; cannot be separated by fractional distillation.
  13. Colligative properties depend only on the number of solute particles: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure.
  14. Relative lowering of vapour pressure = (p₁⁰ - p₁) / p₁⁰ = x₂ (mole fraction of solute).
  15. Elevation of boiling point: ΔTb = Kb m; depression of freezing point: ΔTf = Kf m.
  16. Molar mass from boiling point elevation: M₂ = (Kb × w₂ × 1000) / (ΔTb × w₁).
  17. Molar mass from freezing point depression: M₂ = (Kf × w₂ × 1000) / (ΔTf × w₁).
  18. Kf and Kb are solvent-specific constants; Kf for water is 1.86 K kg mol⁻¹, Kb is 0.52 K kg mol⁻¹.
  19. The vapour phase is always richer in the more volatile component.
  20. Henry's law explains why scuba divers use helium-oxygen mixtures to avoid bends.

MCQs — 10 Questions with Answers

Q1. Which of the following is NOT a type of solution according to the material?

  • A. Gas dissolved in liquid
  • B. Solid dissolved in gas
  • C. Liquid mixed with another immiscible liquid forming uniform phase ✓
  • D. Solid dissolved in solid

Answer: C — Immiscible liquids form two separate phases (heterogeneous), not a homogeneous solution; all other options are valid types.

Q2. A solution contains 10% ethanol by volume. This means:

  • A. 10 g ethanol in 100 g solution
  • B. 10 mL ethanol in 100 mL total solution ✓
  • C. 10 mL ethanol in 90 mL water
  • D. Ethanol and water mixed in 1:10 mass ratio

Answer: B — Volume percentage (v/v) is defined as volume of component in 100 mL of total solution.

Q3. If a solution of ethylene glycol (C₂H₆O₂) contains 20% by mass, what is its mole fraction? (Molar mass: C₂H₆O₂ = 62 g/mol, H₂O = 18 g/mol)

  • A. 0.068 ✓
  • B. 0.20
  • C. 0.322
  • D. 0.50

Answer: A — In 100 g solution: 20 g C₂H₆O₂ = 20/62 = 0.322 mol; 80 g H₂O = 80/18 = 4.444 mol; x = 0.322/(0.322+4.444) = 0.068.

Q4. Which concentration unit is independent of temperature?

  • A. Molarity
  • B. Molality
  • C. Mole fraction ✓
  • D. Normality

Answer: C — Mole fraction depends only on number of moles, not volume; all other units depend on volume which changes with temperature.

Q5. A litre of seawater (density 1030 g/L) contains 6 × 10⁻³ g dissolved O₂. Express this concentration in ppm.

  • A. 6 ppm
  • B. 5.8 ppm ✓
  • C. 0.6 ppm
  • D. 58 ppm

Answer: B — ppm = (6 × 10⁻³ g / 1030 g) × 10⁶ = (6/1030) × 10³ ≈ 5.8 ppm.

Q6. What is the molarity of a solution containing 5 g NaOH in 450 mL solution? (Molar mass NaOH = 40 g/mol)

  • A. 0.111 M
  • B. 0.278 M ✓
  • C. 0.5 M
  • D. 1.11 M

Answer: B — Moles of NaOH = 5/40 = 0.125 mol; Molarity = 0.125 mol / 0.45 L = 0.278 M.

Q7. Which statement about parts per million (ppm) is correct? (A) ppm is used only for mass-to-mass ratios. (B) ppm is expressed as (parts of solute / total parts) × 10⁶.

  • A. Both A and B are correct
  • B. Only B is correct ✓
  • C. Only A is correct
  • D. Neither A nor B is correct

Answer: B — Statement B is correct; ppm can be expressed as mass-to-mass, volume-to-volume, or mass-to-volume, so A is incorrect.

Q8. In a binary solution of A and B, if x_A = 0.4, then x_B must be:

  • A. 0.4
  • B. 0.6 ✓
  • C. 0.2
  • D. Cannot be determined without knowing molar masses

Answer: B — For a binary solution, x_A + x_B = 1; therefore x_B = 1 − 0.4 = 0.6.

Q9. Commercial bleaching solution contains 3.62% sodium hypochlorite by mass. If 500 g of this solution is used, what mass of NaClO is present?

  • A. 18.1 g ✓
  • B. 3.62 g
  • C. 36.2 g
  • D. 181 g

Answer: A — Mass of NaClO = (3.62/100) × 500 g = 18.1 g.

Q10. Which of the following solutions requires the smallest total volume to contain the same molar amount of solute compared to a 1 M solution? (A) A 0.5 M solution. (B) A solution with mole fraction x = 0.5.

  • A. Neither — both have the same volume requirement
  • B. Solution (A) requires twice the volume
  • C. Solution (B) comparison is invalid since mole fraction is dimensionless ✓
  • D. Solution (B) requires half the volume of solution (A)

Answer: C — Mole fraction is dimensionless and independent of volume, so it cannot be directly compared to molarity for volume calculations.

Flashcards

What is a solution?

A solution is a homogeneous mixture of two or more components with uniform composition and properties throughout.

Define solvent in a solution.

The solvent is the component present in the largest quantity that determines the physical state of the solution.

What is the formula for mass percentage?

Mass % = (mass of component / total mass of solution) × 100.

How is mole fraction defined mathematically?

Mole fraction of component A = n_A / (n_A + n_B), where n is the number of moles.

What does ppm stand for and when is it used?

ppm (parts per million) is used to express trace quantities of solutes where concentration is very small.

Define molarity and write its formula.

Molarity (M) is moles of solute dissolved per litre of solution: M = moles of solute / volume in litres.

Why is mole fraction useful for solutions?

Mole fraction is dimensionless and useful for relating physical properties like vapour pressure to solution composition.

What is the relationship between all mole fractions in a solution?

The sum of all mole fractions in a solution equals 1: x₁ + x₂ + ... + xᵢ = 1.

Distinguish between 10% ethanol (v/v) and 10% glucose (w/w).

10% ethanol (v/v) means 10 mL ethanol in 100 mL total solution; 10% glucose (w/w) means 10 g glucose in 100 g total solution.

What is the significance of fluoride concentration in water?

1 ppm fluoride prevents tooth decay; 1.5 ppm causes mottling; high concentrations are poisonous.

Important Board Questions

Define mole fraction and explain why it is useful in chemistry. Give one example of its application. [2 marks]

State the formula x_A = n_A / (n_A + n_B), mention it is dimensionless and temperature-independent, and explain its use in relating vapour pressure to solution composition.

A solution contains 20 g of glucose (C₆H₁₂O₆) dissolved in 100 g of water. Calculate: (i) mass percentage of glucose, (ii) mole fraction of glucose, and (iii) molarity if the solution volume is 120 mL. (Molar masses: C₆H₁₂O₆ = 180 g/mol; H₂O = 18 g/mol) [5 marks]

For (i) use mass % = (20/120) × 100; for (ii) calculate moles of glucose and water, then apply x_A = n_A/(n_A + n_B); for (iii) use M = moles/volume in litres (0.120 L).

Why are different concentration units used for different applications? Explain with reference to industrial chemicals, pharmaceuticals, and trace pollutants. State the concentration unit used in each case and justify why that particular unit is most appropriate. [6 marks]

Discuss mass percentage for industrial use (e.g., 3.62% NaClO in bleach), mass by volume % or molarity for pharmaceuticals (IV fluids matching plasma ionic concentration), and ppm for trace quantities (e.g., fluoride 1 ppm prevents decay); justify based on practical convenience, precision requirements, and range of concentrations encountered.

Assertion-Reason Questions for Solutions

Q1. Assertion (A):

Mole fraction is a temperature-independent concentration unit.

Reason (R):

Mole fraction depends only on the number of moles of components, not on temperature or pressure changes.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason correctly explains why mole fraction is temperature-independent. The number of moles does not change with temperature, unlike molarity which depends on solution volume that changes with temperature.

Q2. Assertion (A):

A 10% w/w glucose solution has the same concentration as a 10% v/v glucose solution.

Reason (R):

Both weight percentage and volume percentage measure the amount of solute per 100 units of total solution.

Answer: D — A is false but R is true — The assertion is false. A 10% w/w solution contains 10 g solute per 100 g solution, while 10% v/v contains 10 mL solute per 100 mL solution. These are not equivalent because the density of the solute and solution affect the relationship. The reason is also misleading because it ignores the fundamental difference between mass and volume.

Q3. Assertion (A):

Molarity of a solution changes when the solution is heated.

Reason (R):

Heating causes the volume of solution to increase while the number of moles of solute remains constant, decreasing molarity.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason correctly explains the assertion. Molarity = moles/volume; heating increases volume through thermal expansion, thus decreasing molarity, while moles stay constant.

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Case-Based Questions (CBQ) for Solutions

Case Study:

A pharmaceutical company manufactures intravenous (IV) saline solution for hospitals. The solution must match the ionic concentration of blood plasma to prevent cell damage. A typical IV saline contains 0.9% NaCl by mass (also called normal saline). A nurse prepares 500 mL of this solution by dissolving NaCl in sterile water. The density of the resulting solution is approximately 1.005 g/mL. Medical regulations require that the concentration be between 0.85% and 0.95% w/w for patient safety.

Q1. How many grams of NaCl are needed to prepare 500 mL of 0.9% w/w saline solution (density = 1.005 g/mL)?

  • A. 4.5 g
  • B. 4.52 g
  • C. 4.54 g ✓
  • D. 5.0 g

Answer: C

Q2. If a batch of IV saline is found to contain 4.2 g NaCl in 500 mL of solution (density 1.005 g/mL), what is the w/w concentration, and is it within the safe range?

  • A. 0.84% w/w; below safe range ✓
  • B. 0.85% w/w; within safe range
  • C. 0.92% w/w; within safe range
  • D. 0.94% w/w; within safe range

Answer: A

Q3. What is the mole fraction of NaCl (M = 58.5 g/mol) in the standard 0.9% w/w saline solution? (Assume 100 g solution; water M = 18 g/mol)

  • A. 0.0027 ✓
  • B. 0.0054
  • C. 0.018
  • D. 0.27

Answer: A

Q4. If the IV saline solution is accidentally heated from 25°C to 37°C (body temperature), which of the following will change?

  • A. Only mole fraction will change; w/w and molarity remain constant
  • B. All three (molarity, w/w concentration, and mole fraction) will change
  • C. Only molarity will change; w/w concentration and mole fraction remain constant ✓
  • D. Only w/w concentration will change; molarity and mole fraction remain constant

Answer: C

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Previous Year Questions (PYQ) for Solutions

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 56(B) · 1 mark

Which of the following colligative properties can preferably determine the molar mass of proteins ?

  • Relative lowering of vapour pressure
  • Osmotic pressure
  • Elevation of boiling point
  • Depression of freezing point

CBSE 2026 · Set 56(B) · 1 mark

For an electrolyte consisting of a monovalent cation and a monovalent anion undergoing dissociation in water, the van't Hoff factor (i) is :

  • less than 1
  • equal to 1
  • greater than 1
  • greater than 2

CBSE 2026 · Set 56(B) · 2 marks

State Raoult's law for a solution containing volatile components. Write any two characteristics of a solution which obeys Raoult's law at all concentrations.

CBSE 2026 · Set 56(B) · 3 marks

Calculate the molarity and molality of a KI solution if the density of 20% (mass/mass) aqueous solution of KI is 1·2 g mL−1. [Given : Molar mass of KI = 166 g mol−1]

CBSE 2026 · Set 56/1/1 · 1 mark

Which of the following solutions will have the lowest freezing point in water?

  • 0.1 M Glucose
  • 0.1 M CaCl2
  • 0.1 M KCl
  • 0.1 M Urea

CBSE 2026 · Set 56/1/1 · 1 mark

Assertion (A): It is not possible to separate the components of an azeotrope by fractional distillation. Reason (R): Components of an azeotrope have the same composition in liquid and vapour phase and boil at a constant temperature.

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Assertion (A) is false, but Reason (R) is true.

CBSE 2026 · Set 56/1/1 · 2 marks

What type of deviation from Raoult's law is shown by a mixture of phenol and aniline? Give reason. What will happen to the boiling point of the solution on mixing phenol and aniline?

CBSE 2026 · Set 56/1/1 · 3 marks

Calculate the boiling point of a solution containing 0.61 g of benzoic acid (Molar mass = 122 g mol–1) in 5 g of CS2 in which it dimerises to the extent of 88%. The boiling point and Kb of CS2 are 46.2°C and 2.3 K kg mol–1 respectively.

CBSE 2026 · Set 56/1/2 · 1 mark

Which of the following solutions will have the highest osmotic pressure?

  • 0.1 M KCl
  • 0.1 M CaCl2
  • 0.1 M Glucose
  • 0.1 M Urea

CBSE 2026 · Set 56/1/2 · 1 mark

Assertion (A): Components of azeotropes are easily separated by fractional distillation. Reason (R): Components of an azeotrope have same composition in liquid and vapour phase.

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Assertion (A) is false, but Reason (R) is true.

CBSE 2026 · Set 56/1/2 · 2 marks

What type of deviation from Raoult's law is shown by mixture of ethanol and acetone? Give reason. What will happen to the boiling point of the solution on mixing ethanol and acetone?

CBSE 2026 · Set 56/1/2 · 3 marks

Calculate the boiling point of a solution containing 0.61 g of benzoic acid (Molar mass = 122 g mol⁻¹) in 5 g of CS2 in which it dimerises to the extent of 88%. The boiling point and Kb of CS2 are 46.2°C and 2.3 K kg mol⁻¹ respectively.

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Class 12 Chemistry Formula Sheet

Every formula and named reaction from all 10 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

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Class 12 Chemistry Previous Year Questions, Chapter-wise

All 1241 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

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