CBSE Class 12 Chemistry Question Paper 2025 with Solutions

The CBSE Class 12 Chemistry question paper 2025 with solutions is the most valuable reference for students preparing for their upcoming board examinations. Analyzing the official 2025 Chemistry paper provides clear visibility into the actual exam pattern, chapter-wise weightage, step-wise marking scheme, and the exact difficulty level set by the Central Board of Secondary Education. Practicing previous year question paper CBSE sets allows you to evaluate your conceptual clarity across Physical, Inorganic, and Organic Chemistry while refining your time management for the 3-hour theory exam.

Accessing the verified answer key and solution PDF helps students understand how CBSE evaluators award step marks for numerical derivations, chemical reactions, mechanism pathways, and reasoning questions. Whether you are aiming for a 95+ score or looking to consolidate core passing concepts, mastering the official 2025 question paper is an indispensable milestone in your board preparation roadmap.

Question Paper Structure

The CBSE Class 12 Chemistry theory paper carries a total of 70 marks with an allotted time of 3 hours (180 minutes), while internal assessment and practical examinations account for the remaining 30 marks. The minimum qualifying mark for the theory paper is 23 marks (33%). There is no overall choice in the paper, but internal choices are provided in select questions across different sections.

According to the official CBSE Class 12 Chemistry question paper 2025 blueprint, the exam is divided into five distinct sections comprising 33 compulsory questions:

Students also receive an additional 15 minutes of cool-off reading time prior to the start of the examination to review the paper and plan their answering strategy.

Complete Questions with Detailed Solutions

Below are essential questions extracted from the official CBSE Class 12 Chemistry question repository, complete with step-by-step working, chemical equations, mathematical derivations, and scoring points aligned with the official CBSE marking scheme.

Question 1: Raoult's Law for Volatile Liquids

Question: State Raoult's Law for a solution containing volatile liquids. Write its mathematical expression.

Solution:

1. Statement: Raoult's Law states that 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 liquid phase at a given temperature.
2. Mathematical Expression: For a binary solution containing two volatile components 1 and 2:
   p₁ = p₁° × x₁
   p₂ = p₂° × x₂
   where p₁ and p₂ are the partial vapour pressures of components 1 and 2 in the solution, p₁° and p₂° are the vapour pressures of pure components 1 and 2 at that temperature, and x₁, x₂ are their respective mole fractions in the solution.
3. Total Pressure (Dalton's Law):
   P_total = p₁ + p₂ = (p₁° × x₁) + (p₂° × x₂) = p₁° + (p₂° − p₁°)x₂

Examiner Tip: Always write both the verbal statement and the algebraic definition with defined variables to secure full credit.

Question 2: Molarity vs. Molality and Temperature Dependence

Question: Define molarity and molality. Which one among the two is temperature independent and why?

Solution:

1. Molarity (M): Molarity is defined as the number of moles of solute dissolved in one litre (1 dm³) of the solution.
   Formula: Molarity (M) = (Moles of solute) / (Volume of solution in litres)
   Unit: mol L⁻¹ or M.
2. Molality (m): Molality is defined as the number of moles of solute dissolved per kilogram (1000 g) of the solvent.
   Formula: Molality (m) = (Moles of solute) / (Mass of solvent in kg)
   Unit: mol kg⁻¹ or m.
3. Temperature Independence: Molality is temperature independent.
   Reason: Molality involves only mass quantities (mass of solute and mass of solvent), which do 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: Ideal Solutions and Characteristics

Question: What is an ideal solution? Give two key thermodynamic characteristics of an ideal solution.

Solution:

1. Definition: An ideal solution is a homogeneous mixture in which the solute and solvent molecules follow Raoult's Law over the entire range of concentrations and temperatures. In an ideal binary solution of components A and B, the intermolecular attractive forces between A–B are identical in magnitude to the A–A and B–B intermolecular attractions.
2. Key Characteristics:
   • Enthalpy of mixing is zero: Δ_mixH = 0. No heat is evolved or absorbed when pure components are mixed to form the solution.
   • Volume change on mixing is zero: Δ_mixV = 0. The total volume of the solution equals the exact sum of the individual volumes of the components mixed.
3. Examples: Benzene + Toluene, n-hexane + n-heptane, Bromoethane + Chloroethane.

Question 4: Molarity Calculation for NaOH Solution

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

Solution:

1. Given Data:
   • 5% (w/v) NaOH means 5 g of NaOH is present in 100 mL of solution.
   • Mass of solute (NaOH), w = 5 g
   • Volume of solution, V = 100 mL = 0.100 L
   • Molar mass of NaOH (M) = 23 (Na) + 16 (O) + 1 (H) = 40 g mol⁻¹
2. Step 1: Calculate moles of NaOH:
   Moles (n) = Mass / Molar mass = 5 g / 40 g mol⁻¹ = 0.125 mol
3. Step 2: Calculate Molarity:
   Molarity (M) = Moles of solute / Volume of solution in L
   Molarity = 0.125 mol / 0.100 L = 1.25 M (or 1.25 mol L⁻¹)

Final Answer: The molarity of the 5% (w/v) NaOH solution is 1.25 M.

Question 5: Molar Mass Determination from Vapour Pressure Lowering

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

Solution:

1. Given Data:
   • Vapour pressure of pure solvent (p₁°) = 640 mm Hg
   • Vapour pressure of solution (p₁) = 600 mm Hg
   • Mass of solute (w₂) = 2.175 g
   • Mass of solvent benzene (w₁) = 39.0 g
   • Molar mass of benzene (M₁) = 78 g mol⁻¹
2. Formula: According to Raoult's Law for relative lowering of vapour pressure in a dilute solution:
   (p₁° − p₁) / p₁° = n₂ / n₁ = (w₂ / M₂) / (w₁ / M₁) = (w₂ × M₁) / (M₂ × w₁)
3. Step-by-Step Substitution:
   (640 − 600) / 640 = (2.175 × 78) / (M₂ × 39.0)
   40 / 640 = (2.175 × 2) / M₂    [since 78 / 39.0 = 2]
   1 / 16 = 4.35 / M₂
   M₂ = 4.35 × 16 = 69.6 g mol⁻¹

Final Answer: The molar mass of the non-volatile solute is 69.6 g mol⁻¹.

Question 6: Osmotic Pressure and Molecular Mass of Biomolecules

Question: Explain osmotic pressure. Why is osmotic pressure measurement preferred over other colligative properties for determining the molar mass of biomolecules like proteins and polymers?

Solution:

1. Definition of Osmotic Pressure (π): Osmotic pressure is the minimum excess hydrostatic pressure that must be applied to the solution side across a semi-permeable membrane (SPM) to prevent the inward osmosis of pure solvent molecules into the solution.
   According to van't Hoff's equation:
   π = CRT = (n₂ / V)RT = (w₂ × R × T) / (M₂ × V)
   where C is molar concentration, R is the gas constant, T is temperature in Kelvin, w₂ is solute mass, V is solution volume, and M₂ is molar mass.
2. Reasons for Preferring Osmotic Pressure for Biomolecules:
   • Room Temperature Measurement: Osmotic pressure is measured at ambient room temperature (approx. 298 K). Biomolecules such as proteins, enzymes, and nucleic acids are thermally unstable and undergo denaturation or decomposition at elevated temperatures required for boiling point elevation (ΔT_b).
   • Significant Measurable Magnitude: Biomolecules have high molecular masses (10,000 to 1,000,000+ g mol⁻¹). At dilute concentrations, the freezing point depression (ΔT_f) and boiling point elevation (ΔT_b) values are imperceptibly small (order of 10⁻³ K to 10⁻⁴ K) and cannot be measured accurately. However, osmotic pressure values remain large enough to be measured accurately in mm of solution height.
   • Use of Molarity: The technique relies on molarity (mol L⁻¹) rather than molality, which is straightforward to prepare accurately for biological solutions.

Question 7: Freezing Point Depression Constant (Kf)

Question: What is the molal freezing point depression constant (K_f)? Write its thermodynamic relationship with the enthalpy of fusion.

Solution:

1. Definition: The molal freezing point depression constant (K_f), also called the cryoscopic constant, is defined as the depression in the freezing point of a solvent produced when 1 mole of a non-volatile, non-electrolyte solute is dissolved in 1 kilogram (1000 g) of the solvent (i.e., for a 1 molal solution).
2. Relation to Depression in Freezing Point:
   ΔT_f = K_f × m
   where m is molality: m = (w₂ × 1000) / (M₂ × w₁).
   Therefore, M₂ = (1000 × K_f × w₂) / (ΔT_f × w₁).
   Unit of K_f: K kg mol⁻¹ (or °C kg mol⁻¹). For water, K_f = 1.86 K kg mol⁻¹.
3. Thermodynamic Relation:
   K_f = (R × T_f² × M₁) / (1000 × Δ_fusH)
   where:
   • R = Universal gas constant (8.314 J K⁻¹ mol⁻¹)
   • T_f = Freezing point of pure solvent (in K)
   • M₁ = Molar mass of solvent (in g mol⁻¹)
   • Δ_fusH = Enthalpy of fusion of the solvent (in J mol⁻¹)

Question 8: Calculation of Osmotic Pressure of Glucose Solution

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

Solution:

1. Given Data:
   • Mass of glucose solute (w₂) = 1.0 g
   • Molar mass of glucose (C₆H₁₂O₆), M₂ = (6 × 12) + (12 × 1) + (6 × 16) = 72 + 12 + 96 = 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
2. Step 1: Calculate moles of glucose (n₂):
   n₂ = w₂ / M₂ = 1.0 / 180 = 0.00556 mol
3. Step 2: Calculate molar concentration (C):
   C = n₂ / V = 0.00556 mol / 0.100 L = 0.0556 mol L⁻¹
4. Step 3: Calculate Osmotic Pressure (π):
   π = CRT
   π = 0.0556 mol L⁻¹ × 0.0821 L atm K⁻¹ mol⁻¹ × 300 K
   π = 0.0556 × 24.63 = 1.369 atm ≈ 1.37 atm

Final Answer: The osmotic pressure of the glucose solution at 300 K is 1.37 atm (or 1.39 bar).

Difficulty Analysis

The CBSE Class 12 Chemistry question paper 2025 maintained a balanced distribution between direct conceptual queries, numerical applications, and analytical reasoning. Here is the section-by-section difficulty evaluation based on student feedback and teacher reviews:

Overall, 30% of the paper was Easy (direct NCERT questions), 50% was Moderate (formula application and mechanism-based), and 20% was HOTS (Higher Order Thinking Skills) requiring multi-concept integration.

Marking Scheme

CBSE follows strict step-wise evaluation. Understanding the official marking scheme ensures you never lose easy marks during board evaluation:

• Physical Chemistry Numericals:
  • Formula statement: ½ mark
  • Correct data substitution: ½ to 1 mark
  • Calculations and intermediate steps: 1 mark
  • Final numerical answer with correct SI units: ½ mark (Missing units lead to a direct ½ mark deduction).
• Organic Chemistry Reactions:
  • Correct chemical equation with reagents and conditions: 1 mark
  • Accurate IUPAC names of primary reactants and major products: ½ mark
  • Mechanism steps (curly arrows, intermediate carbocations/radicals): 1 to 2 marks as per question weightage.
• Inorganic Chemistry & Reasoning:
  • Correct scientific keywords (e.g., lanthanoid contraction, high hydration enthalpy, unhybridized d-orbitals): 1 mark
  • Electronic configurations and oxidation state explanations: ½ to 1 mark.

How to Use This Question Paper for Exam Prep

To maximize your score using previous year question papers, follow this structured 4-step preparation framework:

1. Simulate Actual Exam Conditions: Attempt the entire 2025 question paper in a quiet room within a strict 3-hour window. Do not look at the answer key while writing.
2. Conduct Strict Self-Assessment: Cross-check your responses with the step-by-step solutions provided above. Mark deductions for omitted units, incorrect significant figures, or incomplete chemical equations.
3. Maintain an Error Log: Classify your mistakes into three categories: Conceptual gaps, Calculation slips, or Time pressure omissions. Revise the relevant NCERT textbook sections immediately.
4. Master Organic Conversions & Name Reactions: Compile a separate conversion roadmap notebook for high-frequency reactions from Aldehydes, Ketones, Amines, and Haloalkanes.

More Practice Papers and Solutions

Consistent practice with full-length papers and topic-wise questions is the single most reliable strategy to secure a 70/70 in Class 12 Chemistry. Generate customized mock tests, practice NCERT exemplar problems, and access authentic CBSE sample papers with step-by-step solutions on Theorify QPTool. Build confidence and master every chapter before your board examinations with our curated question paper generator.