CBSE Class 12 Chemistry Question Paper 2025 with Solutions PDF
Mastering the CBSE Class 12 Chemistry curriculum for the 2025 and 2026 board examinations demands a balanced mastery of theoretical principles, numerical problem-solving, and precise NCERT-aligned answer writing. Practicing with verified question papers and official marking schemes is the single most effective way to understand question patterns, improve speed, and eliminate avoidable calculation errors in physical and organic chemistry.
This guide provides a comprehensive breakdown of core physical chemistry principles from the Solutions unit, detailed step-by-step solutions to official CBSE question bank problems, practical real-world applications, and targeted preparation strategies designed to secure a 70/70 score in your CBSE Chemistry theory paper.
Core Concepts: Solutions and Colligative Properties
The Solutions chapter constitutes a foundational segment of the physical chemistry section in the board exam 12 syllabus, carrying high numerical and conceptual weightage. Understanding the thermodynamic principles governing solutions is essential for solving multi-step board problems.
1. Concentration Terms: Molarity vs. Molality
Concentration expresses the quantitative ratio of solute to solvent or solution. In CBSE examinations, questions frequently test the temperature dependence of these units:
- Molarity (M): Defined as the number of moles of solute dissolved in one litre (1 dm3) of solution.
Molarity (M) = [Moles of Solute (n2)] / [Volume of Solution in Litres (V)] - Molality (m): Defined as the number of moles of solute dissolved per kilogram (1000 g) of solvent.
Molality (m) = [Moles of Solute (n2)] / [Mass of Solvent in Kilograms (w1 / 1000)] - Temperature Invariance: Molality and mole fraction are temperature-independent because both depend purely on mass, which does not alter with thermal expansion or contraction. In contrast, molarity, normality, and mass-by-volume percentages vary with temperature due to volumetric changes.
2. Raoult's Law and Solution Ideality
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 at a given temperature:
p1 = p1° × x1 and p2 = p2° × x2
According to Dalton's law of partial pressures, the total pressure Ptotal = p1 + p2 = p1°x1 + p2°x2 = p1° + (p2° - p1°)x2.
Solutions are classified based on their adherence to Raoult's Law:
- Ideal Solutions: Obey Raoult's law across all concentrations and temperatures. Intermolecular forces between solute-solvent (A-B) are identical to solvent-solvent (A-A) and solute-solute (B-B) interactions.
- Enthalpy of mixing:
ΔHmixing = 0(no heat absorbed or evolved). - Volume of mixing:
ΔVmixing = 0(no volume contraction or expansion). - Examples: Benzene + Toluene, n-Hexane + n-Heptane, Bromoethane + Chloroethane.
- Enthalpy of mixing:
- Non-Ideal Solutions (Positive Deviation): A-B interactions are weaker than A-A and B-B interactions. Escaping tendency increases, leading to
ptotal > p1°x1 + p2°x2,ΔHmix > 0, andΔVmix > 0(e.g., Ethanol + Acetone). - Non-Ideal Solutions (Negative Deviation): A-B interactions are stronger than pure component interactions. Escaping tendency decreases, leading to
ptotal < p1°x1 + p2°x2,ΔHmix < 0, andΔVmix < 0(e.g., Chloroform + Acetone, Phenol + Aniline).
3. Colligative Properties and Molecular Weight Determination
Colligative properties depend exclusively on the total number of solute particles relative to the total number of solvent molecules, irrespective of their chemical identity.
| Colligative Property | Governing Equation | Molecular Mass Formula (M2) |
|---|---|---|
| Relative Lowering of Vapour Pressure (RLVP) | (p1° - p1) / p1° = x2 ≈ n2 / n1 |
M2 = (w2 × M1 × p1°) / [w1 × (p1° - p1)] |
| Elevation in Boiling Point | ΔTb = Kb × m |
M2 = (1000 × Kb × w2) / (ΔTb × w1) |
| Depression in Freezing Point | ΔTf = Kf × m |
M2 = (1000 × Kf × w2) / (ΔTf × w1) |
| Osmotic Pressure | Π = CRT = (n2 / V)RT |
M2 = (w2 × R × T) / (Π × V) |
4. Real-World Applications
- Intravenous Medical Fluids: Normal saline solution (0.9% mass/volume NaCl) is strictly isotonic with human red blood cell fluid. Injecting hypotonic water causes cell haemolysis (bursting), whereas hypertonic fluids cause crenation (cell shrinkage).
- Automotive Antifreeze: Ethylene glycol is added to automobile radiators in sub-zero climates to lower the freezing point of coolant water (ΔTf) while simultaneously elevating its boiling point (ΔTb) to prevent summer engine overheating.
- Reverse Osmosis (RO) Desalination: Applying hydraulic pressure greater than the osmotic pressure on the concentrated seawater side forces pure water molecules backward through a semipermeable cellulose acetate membrane, producing potable water.
Important CBSE Questions with Step-by-Step Answers
The following problems are drawn directly from official CBSE Class 12 Chemistry question banks and sample papers, formatted according to the latest step-by-step marking rubrics.
Question 1: State Raoult's Law for volatile liquids.
Answer:
Raoult's Law: For a solution containing volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction present in the solution at a constant temperature.
Mathematically, for a binary mixture of components 1 and 2:
p1 = p1° × x1 and p2 = p2° × x2
Where:
p1, p2= Partial vapour pressures of components 1 and 2 in solutionp1°, p2°= Vapour pressures of pure components 1 and 2 at the same temperaturex1, x2= Mole fractions of components 1 and 2 in the liquid mixture
Question 2: Define molarity and molality. Which one is temperature independent and why?
Answer:
1. Molarity (M): The number of moles of solute dissolved in one litre of the solution. Units: mol L-1 or M.
Molarity (M) = Moles of Solute / Volume of Solution in Litres
2. Molality (m): The number of moles of solute dissolved in one kilogram (1000 grams) of the pure solvent. Units: mol kg-1 or m.
Molality (m) = Moles of Solute / Mass of Solvent in kg
Temperature Independence: Molality is temperature-independent. This is because molality depends strictly on the mass of the solvent and solute. Mass remains unaffected by temperature variations, whereas molarity involves solution volume, which expands or contracts when temperature changes.
Question 3: What is an ideal solution? Give two characteristics.
Answer:
Definition: An ideal solution is a binary solution that obeys Raoult's Law strictly across all ranges of concentration and temperature.
Key Characteristics:
- Zero Enthalpy of Mixing (
ΔHmixing = 0): No heat is absorbed or evolved during the formation of the solution from pure components. - Zero Volume Change of Mixing (
ΔVmixing = 0): The total volume of the resulting solution is exactly equal to the sum of the volumes of the individual pure components before mixing. - Uniform Intermolecular Forces: The magnitude of solute-solvent (A-B) intermolecular attractive interactions is identical to the pure solute-solute (A-A) and solvent-solvent (B-B) attractive forces.
Example: A liquid mixture of benzene and toluene.
Question 4: Calculate the molarity of a 5% (w/v) NaOH solution.
Answer:
Step 1: Identify given parameters
- 5% (w/v) NaOH indicates that
5 gof solute (NaOH) is dissolved in100 mLof solution. - Mass of solute (
w2) =5.0 g - Volume of solution (
V) =100 mL = 100 / 1000 L = 0.1 L - Molar mass of NaOH (
M2) =23 (Na) + 16 (O) + 1 (H) = 40.0 g mol-1
Step 2: Calculate moles of NaOH
Moles (n2) = w2 / M2 = 5.0 g / 40.0 g mol-1 = 0.125 mol
Step 3: Calculate Molarity
Molarity (M) = n2 / V (in L) = 0.125 mol / 0.1 L = 1.25 M (or 1.25 mol L-1)
Question 5: The vapour pressure of pure benzene is 640 mm Hg. A non-volatile solute of mass 2.175 g is dissolved in 39.0 g benzene. The vapour pressure of the solution becomes 600 mm Hg. Calculate the molar mass of the solute.
Answer:
Step 1: Write down the given data
- Vapour pressure of pure solvent benzene (
p1°) =640 mm Hg - Vapour pressure of solution (
p1) =600 mm Hg - Mass of non-volatile solute (
w2) =2.175 g - Mass of benzene solvent (
w1) =39.0 g - Molar mass of benzene C6H6 (
M1) =6(12) + 6(1) = 78 g mol-1
Step 2: Apply the Relative Lowering of Vapour Pressure formula
(p1° - p1) / p1° = n2 / n1 = (w2 / M2) / (w1 / M1)
Step 3: Substitute values and solve for M2
(640 - 600) / 640 = (2.175 / M2) / (39.0 / 78.0)
40 / 640 = (2.175 / M2) / 0.5
1 / 16 = 2.175 / (0.5 × M2)
0.5 × M2 = 2.175 × 16
0.5 × M2 = 34.8
M2 = 34.8 / 0.5 = 69.6 g mol-1
Final 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.
Answer:
Osmotic Pressure (Π): The exact excess hydrostatic pressure that must be applied to the solution side across a semipermeable membrane (SPM) to completely prevent the inward osmotic flow of pure solvent molecules into the solution.
According to the van 't Hoff equation for dilute solutions:
Π = CRT = (n2 / V)RT = (w2 × R × T) / (M2 × V)
Why Osmotic Pressure is Preferred for Biomolecules (Proteins, Polymers, Nucleic Acids):
- Thermal Stability at Ambient Temperatures: Biomolecules and complex polymers are thermally sensitive and undergo denaturation, decomposition, or coagulation at elevated temperatures. Unlike boiling point elevation (ΔTb) or freezing point depression (ΔTf) methods, osmotic pressure measurements are performed at standard room temperature (298 K).
- Measurable Magnitude in Dilute Solutions: High-molecular-weight polymers (M2 > 10,000 g mol-1) produce extremely minute changes in freezing point or boiling point (≈ 10-3 to 10-4 K), which cannot be measured accurately with standard Beckmann thermometers. In contrast, osmotic pressure produces significant, easily measurable pressure heads (in mm of liquid/Hg) even at dilute concentrations.
- Convenient Concentration Unit: It utilizes molarity (mol L-1) rather than molality, simplifying laboratory sample preparation.
Question 7: What is the freezing point depression constant (Kf)? Derive the expression relating it to the molal depression constant.
Answer:
Molal Depression Constant (Kf / Cryoscopic Constant): Defined as the depression in the freezing point of a solvent produced when one mole of a non-volatile, non-electrolyte solute is dissolved completely in 1000 g (1 kg) of the solvent (i.e., for a 1 molal solution where m = 1 mol kg-1, ΔTf = Kf). Its SI unit is K kg mol-1.
Thermodynamic Relation:
The value of Kf depends exclusively on the physical and thermodynamic properties of the solvent. Thermodynamically, Kf is related to the solvent's enthalpy of fusion (ΔfusH) and standard freezing point (Tf°) by:
Kf = (R × (Tf°)2 × M1) / (1000 × ΔfusH)
Where:
R= Universal gas constant (8.314 J K-1 mol-1)Tf°= Freezing point of the pure solvent in Kelvin (K)M1= Molar mass of the solvent (in g mol-1)ΔfusH= Molar enthalpy of fusion of the solvent (in J mol-1)
For pure liquid water, Kf = 1.86 K kg mol-1.
Question 8: Calculate the osmotic pressure of a solution containing 1 g of glucose (C6H12O6) in 100 mL solution at 300 K.
Answer:
Step 1: List given quantities
- Mass of glucose solute (
w2) =1.0 g - Molar mass of glucose C6H12O6 (
M2) =6(12) + 12(1) + 6(16) = 180 g mol-1 - Volume of solution (
V) =100 mL = 0.1 L - Absolute temperature (
T) =300 K - Gas constant (
R) =0.0821 L atm K-1 mol-1(or0.0831 L bar K-1 mol-1)
Step 2: Calculate moles and molar concentration
Moles of glucose (n2) = 1.0 g / 180 g mol-1 = 0.005556 mol
Molar Concentration (C) = n2 / V = 0.005556 mol / 0.1 L = 0.05556 mol L-1
Step 3: Calculate Osmotic Pressure (Π)
Π = CRT
Π = 0.05556 mol L-1 × 0.0821 L atm K-1 mol-1 × 300 K
Π = 1.368 atm ≈ 1.37 atm (or 1.39 bar / 1.387 × 105 Pa)
Final Answer: The osmotic pressure of the glucose solution at 300 K is 1.37 atm.
CBSE Class 12 Chemistry Exam Pattern and Step-Marking Insights
The annual CBSE Class 12 Chemistry board exam comprises a 70-mark theory paper and a 30-mark practical examination. The theory paper follows a structured 5-section design:
- Section A (16 Marks): 16 Multiple Choice Questions (including Assertion-Reasoning questions) carrying 1 mark each. Test granular definitions and rapid recall.
- Section B (10 Marks): 5 Short Answer Questions carrying 2 marks each. Focus on direct definitions, unit conversions, and fundamental numericals.
- Section C (21 Marks): 7 Short Answer Questions carrying 3 marks each. Require complete step-by-step numerical calculations, mechanism drawings, or chemical tests.
- Section D (8 Marks): 2 Case-Based / Passage-Based integrated questions carrying 4 marks each. Require contextual application of chemical concepts.
- Section E (15 Marks): 3 Long Answer Questions carrying 5 marks each. Involve multi-part theoretical derivations, organic synthesis pathways, or multi-step physical chemistry problems.
Crucial Step-Marking Rules for Numericals:
- Formula Stating (0.5 - 1.0 Mark): Always write the standard formula in algebraic terms (e.g.,
Π = (w2RT) / (M2V)) before substituting values. - Direct Value Substitution (0.5 - 1.0 Mark): Show explicit unit conversions (e.g., converting
mL → Lor°C → K) within the substituted step. - Final Answer with SI / Metric Units (0.5 Mark): A numerical answer without correct units (e.g.,
g mol-1,atm,M) results in an immediate deduction of 0.5 marks under CBSE evaluation guidelines.
How to Prepare for CBSE Class 12 Chemistry
- Prioritize NCERT Examples and In-Text Exercises: Over 80% of board numericals and direct conceptual questions in physical chemistry are adapted directly from NCERT textbook exercises. Work through every in-text numerical without consulting guidebooks beforehand.
- Maintain a Dedicated Formula and Units Sheet: Maintain a clean reference list containing colligative property formulas, van 't Hoff factor modifications (
ifor association and dissociation), and standard constants (R,Kb,Kf, Faraday's constantF = 96500 C mol-1). - Master Association and Dissociation Adjustments: For electrolyte solutes (such as NaCl, CaCl2, K4[Fe(CN)6]), remember to multiply colligative expressions by the van 't Hoff factor
i = 1 + (n - 1)αto account for ionisation in solution. - Time-Bound Mock Solving: Practice full-length 3-hour sample papers under simulated exam conditions to develop pacing: 30 minutes for Section A, 35 minutes for Section B, 45 minutes for Section C, 25 minutes for Section D, 35 minutes for Section E, leaving 10 minutes for final verification.
Where to Practice More
Consistent question practice is the key to board exam success. To generate chapter-specific practice worksheets, download official previous year question papers, and test yourself with customizable, NCERT-aligned mock assessments, visit Theorify QPTool (qptool.theorify.in) to accelerate your preparation for the 2025 and 2026 board exams.