NCERT Solutions Class 12 Chemistry Solutions

Access comprehensive NCERT Solutions Class 12 Chemistry for Chapter 1: Solutions with detailed step-by-step derivations, numerical solutions, and clear concept explanations. In the updated CBSE syllabus, Solutions is designated as Chapter 1 and carries a substantial weightage of 7 marks in the theory examination. Whether you are preparing for your CBSE board exams or national entrance tests like NEET and JEE, mastering these NCERT Chemistry Class 12 solutions is vital for scoring top grades.

This guide provides complete, curriculum-aligned NCERT Solutions step by step for all in-text and exercise problems. With whiteboard-style step-by-step breakdowns, structured problem-solving templates (Given, To Find, Formula, Substitution, and Answer), and expert exam tips, these CBSE NCERT solutions will help you eliminate common calculation errors and grasp every fundamental concept with clarity.

Chapter Overview

A solution is a homogeneous mixture of two or more chemically non-reacting substances whose composition can be varied within certain limits. In NCERT Solutions Class 12 Chemistry Chapter 1, the syllabus primarily focuses on binary solutions consisting of a single solute and a single solvent. Below are the key topics covered in this chapter:

1. Methods of Expressing Concentration of Solutions

• Mass Percentage (% w/w): Mass of solute per 100 g of solution.
  Mass % of solute = (Mass of solute / Total mass of solution) × 100
• Volume Percentage (% v/v): Volume of solute per 100 mL of solution.
  Volume % of solute = (Volume of solute / Total volume of solution) × 100
• Mass by Volume Percentage (% w/v): Mass of solute dissolved in 100 mL of solution.
  % (w/v) = (Mass of solute in g / Volume of solution in mL) × 100
• Parts Per Million (ppm): Useful for expressing trace concentrations (e.g., water pollutants).
  ppm = (Number of parts of component / Total number of parts of all components) × 106
• Mole Fraction (x): Ratio of the number of moles of a particular component to the total number of moles present in the solution.
  xA = nA / (nA + nB), where xA + xB = 1
• Molarity (M): Number of moles of solute dissolved per litre (or dm³) of solution.
  M = (Moles of solute) / (Volume of solution in Litres) = (w2 × 1000) / (M2 × V in mL)
• Molality (m): Number of moles of solute dissolved per kilogram of solvent.
  m = (Moles of solute) / (Mass of solvent in kg) = (w2 × 1000) / (M2 × w1 in g)

2. Solubility and Henry's Law

The solubility of a solid in a liquid generally increases with a rise in temperature if the dissolution process is endothermic (ΔH_sol > 0), and decreases if exothermic (ΔH_sol < 0). Pressure has a negligible effect on solids and liquids because they are virtually incompressible.

For gases dissolving in liquids, Henry's Law states: 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.

Mathematically: p = K_H × x, where p is the partial pressure of the gas, x is its mole fraction in solution, and KH is the Henry's law constant. Important observations:

• Higher the value of KH at a given pressure, the lower is the solubility of the gas in the liquid.
• KH increases with an increase in temperature, which is why aquatic species are more comfortable in cold water than in warm water.

3. Vapour Pressure and Raoult's Law

For a solution of volatile liquids, Raoult's Law states that the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction in the solution.

For a binary solution of components 1 and 2:
p₁ = p₁° × x₁ and p₂ = p₂° × x₂
Total vapour pressure: p_total = p₁ + p₂ = p₁° × x₁ + p₂° × x₂ = p₁° + (p₂° - p₁°) × x₂

4. Ideal vs Non-Ideal Solutions

• Ideal Solutions: Obey Raoult's law over the entire range of concentration. For ideal solutions, ΔH_mixing = 0, ΔV_mixing = 0, and intermolecular attractive forces between solute-solvent (A-B) are equal to A-A and B-B interactions (e.g., n-hexane + n-heptane, benzene + toluene, bromoethane + chloroethane).
• Non-Ideal Solutions with Positive Deviation: A-B interactions are weaker than A-A and B-B interactions. As a result, vapour pressure is higher than expected from Raoult's law. ΔH_mixing > 0, ΔV_mixing > 0. Example: Ethanol + Acetone, CS₂ + Acetone. They form minimum boiling azeotropes.
• Non-Ideal Solutions with Negative Deviation: A-B interactions are stronger than A-A and B-B interactions (e.g., through hydrogen bonding). Vapour pressure is lower than expected. ΔH_mixing < 0, ΔV_mixing < 0. Example: Chloroform + Acetone, Phenol + Aniline, HNO₃ + Water. They form maximum boiling azeotropes.

5. Colligative Properties and Molar Mass Determination

Colligative properties depend solely on the number of solute particles (molecules or ions) present in a given amount of solvent and not upon their chemical nature. The four colligative properties are:

1. Relative Lowering of Vapour Pressure: (p1° - p1) / p1° = x2 ≈ (w2 × M1) / (M2 × w1)
2. Elevation of Boiling Point: ΔTb = Tb - Tb° = Kb × m = (1000 × Kb × w2) / (M2 × w1)
3. Depression of Freezing Point: ΔTf = Tf° - Tf = Kf × m = (1000 × Kf × w2) / (M2 × w1)
4. Osmotic Pressure (π): π = C R T = (n2 / V) R T = (w2 R T) / (M2 V)

6. Abnormal Molecular Masses & van 't Hoff Factor (i)

When a solute undergoes dissociation or association in solution, the observed colligative property deviates from the theoretical value. The van 't Hoff factor (i) is defined as:

i = (Normal molar mass) / (Abnormal molar mass) = (Observed colligative property) / (Calculated colligative property) = (Total moles of particles after association/dissociation) / (Number of moles of particles before association/dissociation)

• For dissociation (e.g., NaCl, CaCl₂): i > 1. Degree of dissociation: α = (i - 1) / (n - 1), where n is the number of ions formed per formula unit.
• For association (e.g., benzoic acid in benzene): i < 1. Degree of association: α = (1 - i) / (1 - 1/n), where n is the association number (e.g., n = 2 for dimerization).
• For non-electrolytes (e.g., urea, glucose, sucrose): i = 1.

Exercise Solutions

Here are detailed, step-by-step whiteboard solutions for the most crucial CBSE board exam and NCERT questions from this chapter. Follow the structured method shown below to secure full marks in descriptive answers.

Question 1: State Raoult's Law for volatile liquids.

Solution:

• Statement: For a solution of volatile liquids, Raoult's Law states that the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in the solution at a given temperature.
• Mathematical Formulation: Let a binary solution contain two volatile components 1 and 2 with mole fractions x₁ and x₂.
  p1 ∝ x1 ⇒ p1 = p1° × x1
p2 ∝ x2 ⇒ p2 = p2° × x2
  Where p1° and p2° represent the vapour pressures of pure components 1 and 2 respectively.
• Total Pressure: According to Dalton's Law of partial pressures:
  ptotal = p1 + p2 = (p1° × x1) + (p2° × x2)
  Since x1 = 1 - x2:
  p_total = p₁° + (p₂° - p₁°) × x₂

Question 2: Define molarity and molality. Which one is temperature independent and why?

Solution:

1. Molarity (M): It is defined as the number of moles of solute dissolved in one litre (1 L or 1 dm³) of the solution.
   Molarity (M) = (Moles of solute) / (Volume of solution in Litres) [Unit: mol L^-1 or M]
2. Molality (m): It is defined as the number of moles of solute dissolved in one kilogram (1 kg or 1000 g) of the solvent.
   Molality (m) = (Moles of solute) / (Mass of solvent in kg) [Unit: mol kg^-1 or m]
3. Temperature Dependence:
   Molality is independent of temperature.
   Reason: Molality involves only mass measurements (mass of solute and mass of solvent), and mass does not change with temperature. In contrast, molarity depends on the volume of the solution, which expands or contracts with changes in temperature according to thermal expansion principles.

Question 3: What is an ideal solution? Give two characteristics and one example.

Solution:

• Definition: An ideal solution is a solution that obeys Raoult's Law strictly over the entire range of concentration and temperature.
• Characteristics:
  1. Enthalpy of mixing is zero (ΔH_mixing = 0): No heat is absorbed or evolved when the components are mixed to form the solution.
  2. Volume of mixing is zero (ΔV_mixing = 0): The total volume of the solution is exactly equal to the sum of the volumes of the unmixed components.
  3. Intermolecular interactions: The intermolecular attractive forces between solute and solvent (A-B) are identical in magnitude to the pure solute-solute (A-A) and pure solvent-solvent (B-B) interactions.
• Example: A mixture of Benzene and Toluene, or n-Hexane and n-Heptane, or Bromoethane and Chloroethane.

Question 4: Calculate the molarity of 5% (w/v) NaOH solution.

Solution:

• Given:
  • Concentration of NaOH = 5% (w/v)
  • Mass of solute (NaOH), w₂ = 5 g
  • Volume of solution, V = 100 mL = 0.1 L
  • Molar mass of NaOH (M₂) = 23 (Na) + 16 (O) + 1 (H) = 40 g mol^-1
• To Find: Molarity (M) of the solution.
• Formula:
  Molarity (M) = (Number of moles of solute) / (Volume of solution in L) = (w2 × 1000) / (M2 × V in mL)
• Substitution & Calculation:
  Number of moles of NaOH (n2) = 5 / 40 = 0.125 mol
Molarity (M) = 0.125 mol / 0.1 L = 1.25 mol L-1
• Answer: The molarity of 5% (w/v) NaOH solution is 1.25 M.

Question 5: The vapour pressure of pure benzene is 640 mm Hg. A non-volatile solute 2.175 g is dissolved in 39.0 g benzene. Vapour pressure of the solution becomes 600 mm Hg. Calculate the molar mass of the solute.

Solution:

• Given:
  • Vapour pressure of pure benzene (p₁°) = 640 mm Hg
  • Vapour pressure of solution (p) = 600 mm Hg
  • Mass of non-volatile solute (w₂) = 2.175 g
  • Mass of solvent benzene (w₁) = 39.0 g
  • Molar mass of benzene C₆H₆ (M₁) = (6 × 12) + (6 × 1) = 78 g mol^-1
• To Find: Molar mass of solute (M₂).
• Formula:
  From Raoult's law for relative lowering of vapour pressure:
  (p1° - p) / p1° = n2 / (n1 + n2)
  For a dilute solution, n2 ≪ n1, so:
  (p1° - p) / p1° ≈ n2 / n1 = (w2 / M2) / (w1 / M1) = (w2 × M1) / (M2 × w1)
• Substitution & Step-by-Step Calculation:
  1. Calculate moles of solvent (n₁):
  n1 = 39.0 / 78 = 0.5 mol
  2. Substitute values into the relative lowering formula:
  (640 - 600) / 640 = (2.175 × 78) / (M2 × 39.0)
40 / 640 = (2.175 × 2) / M2
1 / 16 = 4.35 / M2
M2 = 4.35 × 16 = 69.6 g mol-1
• Answer: The molar mass of the non-volatile solute is 69.6 g mol^-1.

Question 6: Explain osmotic pressure and its application in determining the molecular mass of biomolecules.

Solution:

• Definition of Osmotic Pressure: Osmotic pressure (π) is the excess hydrostatic pressure that must be applied to the solution side across a semi-permeable membrane (SPM) to prevent the inward flow of pure solvent into the solution via osmosis.
  According to the van 't Hoff equation for dilute solutions:
  π = C R T = (n₂ / V) R T = (w₂ R T) / (M₂ V)
  Where C = molar concentration, R = universal gas constant (0.0821 L atm K^-1 mol^-1), T = absolute temperature in Kelvin, w2 = mass of solute, and M2 = molar mass of solute.
• Why Osmotic Pressure is Preferred for Biomolecules: Determining the molar mass of polymers, proteins, nucleic acids, and other biomolecules via osmotic pressure offers distinct advantages over other colligative properties (ΔT_b, ΔT_f, Δp):
  1. Measurable Magnitude at Room Temperature: Biomolecules have very high molecular masses (10,000 to 1,000,000 g mol^-1), making their molar concentration extremely small. The changes in boiling point and freezing point are negligibly small (of the order of 10^-3 to 10^-4 K), whereas osmotic pressure is large enough to be measured accurately with water or mercury manometers at room temperature.
  2. Thermal Stability: Biomolecules degrade or denature at elevated temperatures. Boiling point elevation requires heating, which destroys protein structure, whereas osmotic pressure measurements are safely conducted at ambient room temperature (298 K).
  3. Use of Molarity: Osmotic pressure uses molarity (mol L^-1) instead of molality, which allows measurements directly on prepared volumetric liquid solutions.

Question 7: What is the freezing point depression constant (K_f)? Derive the expression relating it to the molal depression constant.

Solution:

• Definition: The freezing point depression constant (also known as the Molal Depression Constant or Cryoscopic Constant, K_f) is defined as the depression in freezing point produced when one mole of a non-volatile solute is dissolved in 1000 g (1 kg) of the solvent (i.e., for a 1 molal solution).
  ΔTf = Kf × m
  When m = 1 mol kg-1, ΔTf = Kf.
  Unit of K_f: K kg mol^-1 or °C kg mol^-1.
• Thermodynamic Relation: From thermodynamics, K_f depends solely on the physical properties of the solvent:
  K_f = (R × M₁ × T_f°²) / (1000 × ΔH_fus)
  Where:
  • R = Universal gas constant (8.314 J K^-1 mol^-1)
  • M1 = Molar mass of the solvent (in g mol^-1)
  • Tf° = Freezing point of the pure solvent in Kelvin
  • ΔHfus = Enthalpy of fusion of the solvent per mole (in J mol^-1)

  If latent heat of fusion per gram (l_fus = ΔH_fus / M₁) is used:
  K_f = (R × T_f°²) / (1000 × l_fus)

  For pure water, with T_f° = 273.15 K and ΔH_fus = 6008 J mol^-1, K_f = 1.86 K kg mol^-1.

Question 8: Calculate the osmotic pressure of a solution containing 1.0 g of glucose (C₆H₁₂O₆) in 100 mL of solution at 300 K.

Solution:

• Given:
  • Mass of glucose (w₂) = 1.0 g
  • Molecular formula of glucose = C₆H₁₂O₆
  • Molar mass of glucose (M₂) = (6 × 12) + (12 × 1) + (6 × 16) = 72 + 12 + 96 = 180 g mol^-1
  • Volume of solution (V) = 100 mL = 0.1 L = 0.1 × 10^-3 m³
  • Temperature (T) = 300 K
  • Gas constant (R) = 0.0821 L atm K^-1 mol^-1 (or 8.314 kPa L K^-1 mol^-1)
• To Find: Osmotic pressure (π).
• Formula:
  π = (w2 × R × T) / (M2 × V)
• Substitution & Step-by-Step Calculation:
  1. Calculate moles of glucose (n₂):
  n2 = 1.0 / 180 = 0.005556 mol
  2. Calculate molar concentration (C):
  C = n2 / V = 0.005556 mol / 0.1 L = 0.05556 mol L-1
  3. Compute osmotic pressure:
  π = 0.05556 × 0.0821 × 300
π = 0.05556 × 24.63 = 1.368 atm (or 1.37 atm)
  In SI units (Pascals): 1.368 × 101325 Pa = 1.386 × 105 Pa = 1.386 bar.
• Answer: The osmotic pressure of the glucose solution at 300 K is 1.37 atm (1.386 × 10⁵ Pa).

Important Formulas and Theorems

Here is an all-in-one formula reference table for rapid revision before your CBSE Class 12 board examination:

Common Mistakes and Tips

Avoid these frequent traps where students lose critical marks in CBSE Chemistry numericals and theory questions:

• Forgetting the van 't Hoff Factor (i): Whenever the solute is an ionic compound (e.g., NaCl, BaCl₂, K₂SO₄, Al₂(SO₄)₃) or an organic acid in non-polar solvent (e.g., CH₃COOH in benzene), you must include i in all colligative property formulas. Omitting i is the #1 reason students lose 2 to 3 marks in numericals.
• Molarity vs Molality Confusion: Remember that molarity (M) uses volume of solution in litres, whereas molality (m) uses mass of solvent in kilograms. Do not use total solution mass when calculating molality. If the density (ρ) of the solution is given, calculate solution mass = ρ × V, and then subtract the solute mass to obtain solvent mass.
• Unit Mismatches in Osmotic Pressure: If you use R = 0.0821 L atm K-1 mol-1, your volume must be in litres (L) and osmotic pressure in atm. If you use R = 8.314 J K-1 mol-1, volume must be in m³ (where 1 L = 10^-3 m³) and pressure in Pascals (Pa).
• Temperature Units: Always convert Celsius to Kelvin: T(K) = T(°C) + 273.15. However, remember that the temperature change ΔT (such as ΔT_b or ΔT_f) has the same numerical value in Kelvin as in Celsius (ΔT in K = ΔT in °C).
• Azeotropic Mixtures: Remember that azeotropes are constant-boiling binary mixtures that distill without change in composition. Minimum boiling azeotropes arise from solutions showing large positive deviations from Raoult's law (e.g., 95% ethanol-water), while maximum boiling azeotropes arise from solutions showing large negative deviations (e.g., 68% nitric acid-water).

Board Exam Relevance

In the CBSE Class 12 Chemistry theory paper (70 marks total), Chapter 1: Solutions consistently carries a weightage of 7 marks. The questions are distributed predictably across different question typologies:

• Multiple Choice Questions / Assertion-Reason (1-2 Marks): Often test Henry's law constant trends (K_H vs solubility/temperature), ideal vs non-ideal characteristics (ΔH, ΔV), or van 't Hoff factor comparison for equimolar solutions (e.g., highest boiling point among 0.1 M NaCl, 0.1 M BaCl₂, 0.1 M Glucose).
• Short Answer Questions (2 Marks): State and explain Raoult's Law, define ebullioscopic/cryoscopic constants, explain why molality is preferred over molarity, or explain applications of osmotic pressure for biomolecules.
• Numerical / Long Answer Questions (3-5 Marks): High-probability 3-mark numericals on elevation in boiling point or depression in freezing point incorporating the van 't Hoff factor (i) and degree of dissociation (α). Case-based questions frequently center around colligative properties and reverse osmosis (desalination of seawater).

More NCERT Solutions and Practice

Consistent numerical practice is the key to securing a full 70/70 in CBSE Class 12 Chemistry. After going through these NCERT Solutions Class 12 Chemistry, test your retention by attempting chapter-wise mock tests, assertion-reason drills, and previous years' board questions under timed conditions.

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