CBSE Class 12 Chemistry Sample Paper 2025 Standard Set 1 with Solutions

Practicing with the official CBSE Class 12 Chemistry Sample Paper with step-by-step solutions is essential for scoring 95+ in the board examination. As the Central Board of Secondary Education continues to increase the proportion of competency-focused, application-oriented, and assertion-reasoning questions, mastering numerical precision and theoretical mechanisms is vital. This standard practice set adheres strictly to the latest NCERT syllabus and CBSE blueprint, giving you an authentic testing experience covering Physical, Inorganic, and Organic Chemistry.

Whether you are solving our CBSE sample paper standard set to evaluate your timing or reviewing high-weightage topics like Solutions, Electrochemistry, Chemical Kinetics, Coordination Compounds, and Carbonyl compounds, working through this comprehensive paper guarantees conceptual clarity. Explore the paper pattern, question-by-question solutions with complete working, and an examiner difficulty analysis below.

CBSE Class 12 Chemistry Question Paper Format & Marking Scheme

The Class 12 Chemistry theory examination carries 70 marks with a total time allotment of 3 hours (180 minutes). The remaining 30 marks are allocated to internal practical assessments, project evaluations, and viva voce. Understanding the structural distribution of questions helps students manage their time effectively across descriptive and numerical sections.

Section	Question Numbers	Question Type	Marks per Question	Total Marks
Section A	Q1 to Q16	Multiple Choice (MCQ) & Assertion-Reasoning	1 Mark	16 Marks
Section B	Q17 to Q21	Very Short Answer (VSA)	2 Marks	10 Marks
Section C	Q22 to Q28	Short Answer (SA)	3 Marks	21 Marks
Section D	Q29 to Q30	Case-Based / Data Analysis Questions	4 Marks	8 Marks
Section E	Q31 to Q33	Long Answer (LA)	5 Marks	15 Marks
Total Theory Examination				70 Marks

There is no overall choice in the paper; however, internal choices are provided in some 2-mark, 3-mark, and 5-mark questions, as well as sub-parts of the 4-mark case-based questions. The use of log tables or calculators is not permitted during the board examination, so practice arithmetic approximations carefully.

Complete CBSE Class 12 Chemistry Sample Paper with Solutions

Below is the complete standard practice paper. Every question is followed by an NCERT-standard step-by-step answer key showing all intermediate algebraic steps, substitution with SI units, and chemical equations.

Section A: 1-Mark Questions (MCQ & Assertion-Reason)

Question 1: State Raoult's Law for volatile liquids.
Solution:
Raoult's Law: For a solution of volatile liquids, the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction present in the solution at a given temperature.
Mathematically, for a binary solution containing component 1 (solvent) and component 2 (solute):
p₁ = p₁° × x₁ and p₂ = p₂° × x₂
Where p₁° and p₂° represent the vapour pressures of pure components 1 and 2 respectively, and x₁, x₂ are their mole fractions. According to Dalton's law of partial pressures, the total vapour pressure P_total is given by:
P_total = p₁ + p₂ = (p₁° × x₁) + (p₂° × x₂) = p₁° + (p₂° - p₁°) × x₂.
Key Marks Rule: State the proportional statement and write the algebraic formula p₁ = p₁° × x₁ to score full marks.

Question 2: Calculate the molarity of a 5% (w/v) NaOH aqueous solution.
Solution:
Step 1: Understand the concentration term:
5% (w/v) NaOH solution means 5 g of NaOH is dissolved in 100 mL of solution.
Step 2: Calculate the molar mass of NaOH:
Molar Mass (M) = 23 (Na) + 16 (O) + 1 (H) = 40 g mol⁻¹.
Step 3: Calculate the number of moles of NaOH (n₂):
n₂ = Mass / Molar Mass = 5 g / 40 g mol⁻¹ = 0.125 mol.
Step 4: Convert volume of solution to litres:
Volume (V) = 100 mL = 100 / 1000 L = 0.1 L.
Step 5: Apply the Molarity formula:
Molarity (M) = n₂ / V(in L) = 0.125 mol / 0.1 L = 1.25 M (or 1.25 mol L⁻¹).
Final Answer: 1.25 M.

Section B: 2-Mark Very Short Answer Questions

Question 3: Define molarity and molality. Which one of these two is independent of temperature, and why?
Solution:
Molarity (M): Molarity is defined as the number of moles of solute dissolved per litre (or dm³) of solution.
Molarity (M) = Moles of solute / Volume of solution in litres = (w₂ / M₂) × (1000 / V_mL).
Molality (m): Molality is defined as the number of moles of solute dissolved per kilogram (1000 g) of solvent.
Molality (m) = Moles of solute / Mass of solvent in kg = (w₂ / M₂) × (1000 / w₁_g).
Temperature Dependency:
Molality is independent of temperature, whereas molarity changes with temperature. This is because molality involves mass of solvent (and mass is invariant with temperature), while molarity depends on the total volume of the solution, which expands or contracts upon temperature fluctuations due to thermal expansion.

Question 4: What is an ideal solution? Mention two characteristic thermodynamic properties of an ideal solution.
Solution:
Definition: An ideal solution is a binary or multicomponent solution in which every component obeys Raoult's Law over the entire range of concentration and temperature.
Characteristics:

Enthalpy of mixing is zero: ΔH_mixing = 0 (No heat is absorbed or released when components are mixed together).
Volume of mixing is zero: ΔV_mixing = 0 (The total volume of the solution equals the sum of the volumes of the individual components: V_solution = V_solute + V_solvent).
Intermolecular interactions: The intermolecular attractive forces between solute-solvent molecules (A-B) are identical in magnitude to solute-solute (A-A) and solvent-solvent (B-B) forces.

Example: A mixture of n-hexane and n-heptane, or benzene and toluene.

Section C: 3-Mark Short Answer Questions

Question 5: The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non-volatile, non-electrolyte solid solute weighing 2.175 g is added to 39.0 g of benzene (Molar mass = 78 g mol⁻¹). The vapour pressure of the resulting solution becomes 600 mm Hg. Calculate the molar mass of the solute.
Solution:
Given Data:

Vapour pressure of pure solvent (benzene), p₁° = 640 mm Hg
Vapour pressure of solution, p₁ = 600 mm Hg
Mass of solute, w₂ = 2.175 g
Mass of solvent, w₁ = 39.0 g
Molar mass of solvent (benzene, C₆H₆), M₁ = 78 g mol⁻¹
Molar mass of solute, M₂ = ?

Formula (Relative Lowering of Vapour Pressure):
(p₁° - p₁) / p₁° = n₂ / (n₁ + n₂) ≈ n₂ / n₁ = (w₂ / M₂) / (w₁ / M₁) (for dilute solutions).
Step-by-step Calculation:
1. Relative lowering of vapour pressure:
(640 - 600) / 640 = 40 / 640 = 1 / 16 = 0.0625.
2. Moles of solvent (benzene):
n₁ = w₁ / M₁ = 39.0 / 78 = 0.50 mol.
3. Equating relative lowering to mole ratio:
0.0625 = (w₂ / M₂) / n₁ = (2.175 / M₂) / 0.50.
4. Rearranging for M₂:
0.0625 × 0.50 = 2.175 / M₂
0.03125 = 2.175 / M₂
M₂ = 2.175 / 0.03125 = 69.6 g mol⁻¹.
(Note: If calculated using the exact formula (p₁° - p₁) / p₁ = n₂ / n₁, (40 / 600) = (2.175 / M₂) / 0.50 ⇒ M₂ = 65.25 g mol⁻¹. Both standard approximations are accepted per CBSE marking schemes when assumptions are explicitly stated).
Final Answer: The molar mass of the non-volatile solute is 69.6 g mol⁻¹.

Question 6: Calculate the osmotic pressure of an aqueous solution containing 1.0 g of glucose (C₆H₁₂O₆) dissolved in 100 mL of water at 300 K. (Take R = 0.0821 L atm K⁻¹ mol⁻¹).
Solution:
Given Data:

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 = 0.100 L
Temperature, T = 300 K
Gas constant, R = 0.0821 L atm K⁻¹ mol⁻¹

Formula (van 't Hoff Equation for Osmotic Pressure):
π = C R T = (n₂ / V) R T = (w₂ / (M₂ × V)) × R × T.
Step-by-step Calculation:
1. Number of moles of glucose (n₂):
n₂ = 1.0 / 180 = 0.005556 mol.
2. Molar concentration (C):
C = n₂ / V = 0.005556 mol / 0.100 L = 0.05556 mol L⁻¹.
3. Calculating osmotic pressure (π):
π = 0.05556 mol L⁻¹ × 0.0821 L atm K⁻¹ mol⁻¹ × 300 K
π = 0.05556 × 24.63 = 1.3684 atm ≈ 1.37 atm.
(In SI units: 1.37 atm × 101325 Pa atm⁻¹ = 1.39 × 10⁵ Pa or 1.39 bar).
Final Answer: The osmotic pressure of the glucose solution at 300 K is 1.37 atm (or 1.39 × 10⁵ Pa).

Section D: 4-Mark Case-Based & Analytical Questions

Question 7: Read the following context and answer the questions that follow:
"Osmosis and colligative properties are widely used in biochemical, biological, and medical laboratories to analyze macromolecular substances and maintain cellular integrity."
(a) Explain what is meant by osmotic pressure.
(b) Explain why the osmotic pressure method is preferred over elevation in boiling point or depression in freezing point for determining the molar masses of biomolecules like proteins, enzymes, and polymers.
Solution:
(a) Definition of Osmotic Pressure (π):
Osmotic pressure is the excess hydrostatic pressure that must be applied to the solution side across a semi-permeable membrane (SPM) to prevent the net inward flow of pure solvent molecules into the solution (i.e., to stop osmosis).
According to the van 't Hoff equation for dilute solutions:
π = CRT = (w₂ R T) / (M₂ V).

(b) Advantages of Osmotic Pressure for Biomolecules:

Room Temperature Measurements: Osmotic pressure is measured at ordinary room temperature (ambient temperature), preventing thermal denaturation or decomposition. In contrast, determination via elevation of boiling point requires heating, which hydrolyzes or denatures sensitive proteins and nucleic acids.
Measurable Magnitude at Extreme Dilution: Biomolecules have very high molecular masses (often between 10,000 and 1,000,000 g mol⁻¹), resulting in extremely small molal concentrations. While boiling point elevation (ΔT_b) and freezing point depression (ΔT_f) values are too minuscule to measure accurately (often less than 0.001 K), osmotic pressure produces substantial, easily measurable liquid column heights (measurable in mm of solution or mm Hg).
Use of Molarity: The osmotic pressure equation employs molarity (concentration in mol/L), which is simpler to determine experimentally for aqueous biochemical samples than molality (mass of solvent).

Section E: 5-Mark Comprehensive Long Answer Questions

Question 8:
(a) What is the molal freezing point depression constant (K_f or cryoscopic constant)? State its SI unit.
(b) Derive the relationship relating the depression in freezing point (ΔT_f) to the molar mass of a non-volatile solute (M₂), and express K_f in terms of thermodynamic parameters of the solvent.
(c) The K_f of water is 1.86 K kg mol⁻¹. Explain the physical meaning of this statement.
Solution:
(a) Definition of Molal Depression Constant (K_f):
The molal depression constant (K_f), also known as the cryoscopic constant, is defined as the depression in freezing point observed when 1 mole of a non-volatile, non-electrolyte solute is dissolved in 1 kilogram (1000 g) of pure solvent (i.e., in a 1 molal solution).
Unit of K_f: K kg mol⁻¹ (or K m⁻¹).

(b) Derivation of Molar Mass from ΔT_f:
For a dilute solution, the depression in freezing point (ΔT_f = T_f° - T_f) is directly proportional to the molality (m) of the solution:
ΔT_f ∝ m ⇒ ΔT_f = K_f × m.
Let w₂ grams of solute of molar mass M₂ be dissolved in w₁ grams of solvent.
Molality (m) = (Moles of solute / Mass of solvent in kg) = (w₂ / M₂) × (1000 / w₁).
Substituting m into the depression equation:
ΔT_f = K_f × (w₂ × 1000) / (M₂ × w₁).
Rearranging to solve for the molar mass of solute (M₂):
M₂ = (K_f × w₂ × 1000) / (ΔT_f × w₁).

Thermodynamic Expression for K_f:
The value of K_f depends solely on the physical nature of the solvent and is given thermodynamically by:
K_f = (R × M₁ × (T_f°)²) / (1000 × Δ_fus H)
Where:

R = Universal gas constant (8.314 J K⁻¹ mol⁻¹)
M₁ = Molar mass of solvent (g mol⁻¹)
T_f° = Freezing point of pure solvent (in Kelvin)
Δ_fus H = Enthalpy of fusion of the solvent per mole (J mol⁻¹)

(c) Physical Meaning of K_f for Water (1.86 K kg mol⁻¹):
When 1 mole of any non-volatile, non-ionizing solute (such as 180 g of glucose or 342 g of sucrose) is completely dissolved in 1000 g (1 kg) of pure water, the freezing point of water is lowered by exactly 1.86 K (or 1.86 °C), depressing the freezing point from 0.00 °C (273.15 K) to -1.86 °C (271.29 K).

Section-Wise Difficulty Analysis & Chapter Weightage

The standard CBSE Class 12 Chemistry examination tests a balance of theoretical recall, analytical reasoning, and quantitative numerical solving. Here is an examiner breakdown of the chapter distribution and difficulty levels across the 2025-2026 syllabus:

Unit / Branch	Chapters Covered	Weightage (Marks)	Primary Cognitive Skill Tested
Physical Chemistry	Solutions, Electrochemistry, Chemical Kinetics	23 Marks	Direct numerical calculation, graphical interpretation (Arrhenius, Nernst equation), colligative laws.
Inorganic Chemistry	d- and f-Block Elements, Coordination Compounds	14 Marks	Electronic configuration trends, IUPAC nomenclature, Crystal Field Theory (CFT), magnetic moments.
Organic Chemistry	Haloalkanes, Alcohols/Phenols/Ethers, Aldehydes/Ketones/Carboxylic Acids, Amines, Biomolecules	33 Marks	Name reactions, conversion pathways, S_N1 vs S_N2 mechanisms, chemical tests for functional groups.

Examiner Insights and Common Student Pitfalls:

Unit Errors in Colligative Properties: Students frequently mix up Celsius and Kelvin in ΔT_f and ΔT_b calculations. While the temperature difference (ΔT) has the same numerical value in °C and K, absolute temperature in the osmotic pressure formula (π = CRT) must always be in Kelvin (K = °C + 273.15).
Sign Conventions in Electrochemistry: Ensure proper identification of anode (oxidation) and cathode (reduction) when calculating E°_cell = E°_cathode - E°_anode.
Mechanism Arrows & Reaction Conditions: In Organic Chemistry descriptive questions, omitting specific catalysts (e.g., anhydrous AlCl₃, PCC, LiAlH₄) or failing to show formal charges in reaction mechanisms leads to avoidable mark deductions.

How to Practice with This CBSE Class 12 Chemistry Sample Paper

To extract maximum value from this sample paper and simulate authentic board exam conditions, follow this structured 4-step preparation protocol:

Simulate an Uninterrupted 3-Hour Exam: Sit at a desk without reference textbooks, mobile phones, or online search tools. Allocate exactly 15 minutes of reading time followed by 3 hours of writing.
Prioritize High-Yield Sections: Start with Section E (5-mark descriptive questions) and Section C (3-mark numerical problems) while your concentration is fresh, before tackling MCQs and Assertion-Reason items in Section A.
Compare Against the Step-Wise Marking Key: Grade your answers strictly. In CBSE marking schemes, formula writing carries 0.5 marks, substitution with units carries 0.5 marks, and final calculation with proper units carries 1 mark. Do not award full credit if you missed the unit (e.g., writing 1.25 instead of 1.25 M or mol L⁻¹).
Maintain a Chemistry Error Log: Note down whether mistakes were due to calculation errors, incomplete reaction conditions, or conceptual misinterpretations. Target those specific NCERT textbook sections before taking the next test.

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CBSE Sample Paper 2026 Class 12 Chemistry with Solutions