MP Board Class 12 Chemistry Previous Year Paper 2027 — Solved Questions with Answers

MP Board Class 12 Chemistry Previous Year Paper 2027 — Chemistry is one of the most scoring subjects in MP Board Class 12 exams, and solving previous year papers is the most effective way to boost your score. This blog post provides the complete MP Board Class 12 Chemistry solved paper with answers, marking scheme, chapter-wise weightage, and exam tips. Practicing with actual board exam papers helps you understand question patterns, time management, and important topics. Whether you are aiming for 80/80 or just want to pass with good marks, this guide has everything you need.

📋 Paper Overview & Marking Scheme

The MP Board Class 12 Chemistry paper (Code: 403) is conducted for a total of 70 marks with a duration of 3 hours and 15 minutes. The paper follows the latest NCERT-based syllabus for the 2026–27 academic session. Below is the complete marking scheme:

Section Type of Questions No. of Questions Marks per Q Total Marks
A Multiple Choice / Objective 10 1 10
B Very Short Answer 5 2 10
C Short Answer 10 3 30
D Long Answer 4 5 20
Total 70

✅ Section A: Objective Questions (1 Mark Each)

These MCQs test your basic understanding of key concepts from all chapters. Here are 10 important objective questions from the MP Board 2027 Chemistry paper:

Q. No. Question Answer
1 The unit of rate constant for a zero-order reaction is: mol L⁻¹ s⁻¹
2 Which of the following is an example of lyophilic colloid? Starch solution
3 The formula of gypsum is: CaSO₄·2H₂O
4 Which element has the highest electronegativity? Fluorine (F)
5 In DNA, the linkage between nucleotides is: Phosphodiester bond
6 Which of the following is used as a refrigerant? CFC (Chlorofluorocarbon)
7 The total number of electrons in 1.7 g of NH₃ is: 6.022 × 10²³
8 Which of the following is not a type of crystal defect? Magnetic defect
9 The IUPAC name of CH₃-CH₂-CHO is: Propanal
10 Which vitamin is water soluble? Vitamin C
🎯 Exam Tip: MP Board 2027 — For Section A, focus on NCERT in-text questions and examples. Most MCQs are directly picked from NCERT textbook exercises. Practice at least 200 MCQs from all 16 chapters before the exam.

✍️ Section B: Very Short Answer (2 Marks Each)

Very short answer questions require precise, point-wise responses. Each question carries 2 marks — typically split into two sub-parts or two key points.

📝 Q11. Define the terms: (a) Molarity (b) Molality

Answer:
(a) Molarity (M): Number of moles of solute present per litre of solution. Unit: mol/L or M.
(b) Molality (m): Number of moles of solute present per kilogram of solvent. Unit: mol/kg or m.

📝 Q12. What is the difference between Schottky and Frenkel defects?

Property Schottky Defect Frenkel Defect
Definition Equal number of cations and anions missing from lattice Cation leaves its site and occupies interstitial position
Density Decreases Remains unchanged
Found in Ionic crystals with high coordination number (e.g., NaCl) Ionic crystals with low coordination number (e.g., AgCl)
🎯 Exam Tip: In Very Short Answer questions, write in bullet points. Don’t write paragraphs — examiners look for concise, accurate points. Each correct point = 1 mark.

📘 Section C: Short Answer (3 Marks Each)

Short answer questions require detailed explanations with examples, formulas, or reactions. Each 3-mark question expects 3 distinct points or a well-structured answer with a diagram where applicable.

🧪 Q21. Explain the mechanism of SN1 and SN2 reactions with suitable examples.

SN1 (Substitution Nucleophilic Unimolecular): Two-step mechanism involving carbocation intermediate. Rate = k[R-X]. Favored by tertiary alkyl halides and polar protic solvents.
Example: (CH₃)₃C-Br + OH⁻ → (CH₃)₃C-OH + Br⁻

SN2 (Substitution Nucleophilic Bimolecular): Single-step concerted mechanism with inversion of configuration (Walden inversion). Rate = k[R-X][Nu⁻]. Favored by primary alkyl halides and polar aprotic solvents.
Example: CH₃-Br + OH⁻ → CH₃-OH + Br⁻

📘 Key Difference Table — SN1 vs SN2

SN1: 2 steps, carbocation intermediate, racemization, 3°>2°>1°, polar protic solvent
SN2: 1 step, no intermediate, inversion, 1°>2°>3°, polar aprotic solvent

📐 Q22. Derive the integrated rate equation for a first-order reaction. What is the half-life of a first-order reaction?

Answer:
For a first-order reaction: A → Products
Rate = -d[A]/dt = k[A]
Integrating: ∫d[A]/[A] = -k∫dt → ln[A] = -kt + ln[A]₀
k = (2.303/t) log([A]₀/[A])

Half-life (t₁/₂): When [A] = [A]₀/2,
t₁/₂ = (2.303/k) log(2) = 0.693/k

Key point: Half-life of a first-order reaction is independent of initial concentration.

📝 Section D: Long Answer (5 Marks Each)

Long answer questions are the most scoring section. They require detailed explanations with step-by-step derivations, chemical equations, diagrams, and numerical calculations. Each 5-mark question typically has 3-4 sub-parts.

🧪 Q31. (a) State Kohlrausch’s law. What are its applications? (b) Calculate the molar conductivity of NH₄OH given: Λ°(NH₄Cl) = 149.7, Λ°(NaCl) = 126.5, Λ°(NaOH) = 247.7 S cm² mol⁻¹

(a) Kohlrausch’s Law: At infinite dilution, the molar conductivity of an electrolyte equals the sum of the contributions of its individual ions.
Λ° = ν₊λ⁰₊ + ν₋λ⁰₋

Applications:
1. Calculation of Λ° for weak electrolytes (like NH₄OH, CH₃COOH)
2. Determination of degree of dissociation (α = Λₘ/Λ°ₘ)
3. Calculation of dissociation constant (Kₐ) for weak acids/bases

(b) Calculation:
According to Kohlrausch’s law:
Λ°(NH₄OH) = Λ°(NH₄Cl) + Λ°(NaOH) − Λ°(NaCl)
= 149.7 + 247.7 − 126.5
= 270.9 S cm² mol⁻¹

📊 Q32. What are transition elements? Explain the following properties of transition elements: (a) Variable oxidation states (b) Formation of coloured ions (c) Catalytic activity

Transition Elements: Elements of d-block (groups 3-12) that have partially filled d-orbitals in their ground state or common oxidation states. Examples: Fe, Cu, Ni, Mn, Cr, Co.

(a) Variable Oxidation States: Due to small energy difference between ns and (n-1)d electrons, transition elements show multiple oxidation states. E.g., Mn shows +2 to +7, Fe shows +2 and +3.

(b) Coloured Ions: Due to d-d transitions. When visible light falls on a transition metal ion, electrons absorb energy and jump from lower to higher d-orbitals, emitting complementary colour. E.g., Cu²⁺ (blue), Fe³⁺ (yellow-brown), Cr³⁺ (green).

(c) Catalytic Activity: Transition metals provide large surface area for adsorption of reactants and can adopt multiple oxidation states. E.g., Fe in Haber process (NH₃ synthesis), V₂O₅ in Contact process (H₂SO₄), Ni in hydrogenation.

📊 Chapter-Wise Weightage (MP Board Chemistry 2027)

Knowing the chapter-wise weightage helps you prioritize your study time. Here is the expected weightage for MP Board Class 12 Chemistry 2027 exam based on previous years’ trends:

Chapter Marks Importance
1. Solutions 5–7 ⭐ High
2. Electrochemistry 5–8 ⭐ High
3. Chemical Kinetics 5–7 ⭐ High
4. d & f Block Elements 4–5 ⭐ Medium
5. Coordination Compounds 4–6 ⭐ High
6. Haloalkanes & Haloarenes 3–5 ⭐ Medium
7. Alcohols, Phenols & Ethers 4–6 ⭐ High
8. Aldehydes, Ketones & Acids 4–6 ⭐ High
9. Amines 3–4 ⭐ Medium
10. Biomolecules 3–4 ⭐ Medium
Other Chapters 8–12 ⭐ Mixed

🎯 Exam Preparation Tips

  • Start with NCERT: 90% of MP Board Chemistry questions come directly from NCERT textbook. Read each line carefully, especially in-text questions and examples.
  • Solve 10+ Previous Year Papers: Solve at least 10 previous year papers under timed conditions (3 hours). This builds speed and accuracy.
  • Master Numerical Problems: Practice numerical from Solutions, Electrochemistry, and Chemical Kinetics chapters — these carry 5 marks each in long answer section.
  • Learn Named Reactions: Make a separate list of all named reactions (Sandmeyer, Wurtz, Aldol, Cannizzaro, etc.) with mechanisms and memorize them.
  • Draw Diagrams Clearly: In Organic Chemistry, always draw clear structures. In Physical Chemistry, draw labelled graphs. Diagrams fetch extra marks.
  • Revise NCERT Examples: 80% of numerical problems in the board exam are NCERT examples with changed values. Practice all solved examples from NCERT.
  • Time Management: Section A (10 min), Section B (15 min), Section C (60 min), Section D (45 min), Revision (20 min).
🎯 Exam Tip: In the MP Board 2027 Chemistry exam, write all numerical steps clearly. Even if the final answer is wrong, correct steps fetch 60-70% of the marks. Never leave a question blank — write the relevant formula and known values.

❓ Frequently Asked Questions

Q: How many chapters are there in MP Board Class 12 Chemistry?

Class 12 Chemistry NCERT syllabus has 16 chapters — 9 in Part I (Physical & Inorganic) and 7 in Part II (Organic). The MP Board syllabus covers all 16 chapters for the 2027 board exam.

Q: Is MP Board Chemistry paper tough?

No, the MP Board Chemistry paper is considered of moderate difficulty. Most questions are directly from NCERT. Students who study NCERT thoroughly and solve previous year papers find the exam easy to moderate. The pass percentage in Chemistry is usually 85%+.

Q: What is the passing marks for MP Board Class 12 Chemistry?

The passing marks are 23 out of 70 (theory) and 13 out of 30 (practical). Combined passing requires 33% aggregate in theory + practical. There is no separate passing in theory — aggregate of both components is counted.

Q: Can I get 80/80 in Chemistry?

Yes! Every year thousands of MP Board students score 80/80 (70 theory + 10 internal) or 75+ in Chemistry alone. The key is: (1) Complete NCERT 3 times, (2) Solve 10 previous year papers, (3) Practice all numerical formulas, (4) Write neat diagrams, (5) Revise named reactions daily.

Q: Which chapters have the most numerical questions?

The three chapters with maximum numerical questions are: Solutions (colligative properties, van’t Hoff factor), Electrochemistry (Nernst equation, conductivity, Faraday’s laws), and Chemical Kinetics (rate equations, half-life, activation energy). Master these for the 5-mark numerical section.

Q: How to download MP Board Class 12 Chemistry previous year papers PDF?

You can find MP Board Class 12 Chemistry previous year paper PDFs on mpboard.ai in the Papers section. All papers from 2016 to 2026 are available for free download with complete solutions and marking schemes.

Q: Are there any changes in the MP Board Chemistry syllabus for 2027?

The MP Board follows the NCERT syllabus. There is a 15% reduction in syllabus as per NCERT’s rationalized curriculum. Chapters like “Surface Chemistry” and “Polymers” have reduced content. Always check the latest MPBSE syllabus on the official website mpbse.nic.in before starting your preparation.

Q: What is the best way to revise Chemistry in the last 15 days?

Last 15 days Revision Plan:
Days 1-5: Revise Physical Chemistry (Solutions, Electrochemistry, Kinetics) — formulas + numerical
Days 6-10: Revise Organic Chemistry (name reactions, mechanisms, distinction tests)
Days 11-13: Revise Inorganic (d&f block, coordination, metallurgy)
Days 14-15: Solve 2 full previous year papers under timed conditions

Download MP Board Class 12 Chemistry Previous Year Papers (PDF) — Free
Start your preparation today! Solve at least 10 previous year papers and practice NCERT thoroughly for a score of 75+ in Chemistry.

MP Board Class 12 Chemistry Chapter 2: Electrochemistry — Complete Notes for 2027 Exam

Chapter 2: Electrochemistry is a high-weightage chapter in the MP Board Class 12 Chemistry syllabus, typically contributing 12–16 marks in the board exam. This chapter bridges physical chemistry with real-world applications — from how batteries power your phone to how metals are purified. These comprehensive notes cover electrochemical cells, Nernst equation, conductance, electrolysis, and batteries with clear derivations and MP Board–focused exam tips.

⚡ Electrochemical Cells

An electrochemical cell is a device that converts chemical energy into electrical energy (or vice versa). There are two types:

  • Galvanic Cell (Voltaic Cell): Spontaneous chemical reaction produces electricity. Example: Daniel cell.
  • Electrolytic Cell: Electrical energy drives a non-spontaneous chemical reaction. Example: Electrolysis of water.

🧪 Daniel Cell — The Classic Example

The Daniel cell consists of a zinc electrode dipped in ZnSO₄ solution (anode) and a copper electrode dipped in CuSO₄ solution (cathode), connected by a salt bridge (KCl or KNO₃).

Component Details
Anode (Oxidation) Zn(s) → Zn²⁺(aq) + 2e⁻
Cathode (Reduction) Cu²⁺(aq) + 2e⁻ → Cu(s)
Net Cell Reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Cell Notation Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)

🔑 Key Terms

  • Salt bridge: Maintains electrical neutrality and completes the circuit. Contains inert electrolyte like KCl, KNO₃, or NH₄NO₃.
  • Electrode potential: The potential difference between the electrode and its electrolyte. Measured in volts (V).
  • Standard electrode potential (E⁰): Measured under standard conditions — 1 M concentration, 1 atm pressure, 298 K.
  • Standard Hydrogen Electrode (SHE): Reference electrode with E⁰ = 0.00 V. Used to measure all other electrode potentials.
🎯 Exam Tip: In MP Board exams, you are often asked to draw a labelled diagram of the Daniel cell and write the cell notation. Practise the cell representation format — it carries 2–3 marks in long-answer questions.

📐 Nernst Equation & EMF of Cell

The Nernst equation relates the electrode potential to the concentration of ions (or pressure of gases) at a given temperature. For a general cell reaction:

Ecell = E⁰cell − (RT / nF) × ln Q

At 298 K, using log₁₀:

Ecell = E⁰cell − (0.0591 / n) × log Q

Symbol Meaning Unit
Ecell Cell potential under non-standard conditions Volts (V)
E⁰cell Standard cell potential Volts (V)
R Gas constant (8.314 J mol⁻¹ K⁻¹) J mol⁻¹ K⁻¹
T Temperature Kelvin (K)
n Number of electrons transferred
F Faraday constant (96485 C mol⁻¹) C mol⁻¹
Q Reaction quotient

🔋 EMF of a Cell

The standard EMF (E⁰cell) of a cell is the difference between the standard reduction potentials of the cathode and anode:

E⁰cell = E⁰cathode − E⁰anode

📘 Gibbs Free Energy Relation

ΔG = −nFEcell and ΔG⁰ = −nFE⁰cell
When ΔG = 0, the cell reaches equilibrium: E⁰cell = (0.0591/n) log Keq

🎯 Exam Tip: MP Board 2027 — A numerical question on EMF calculation using the Nernst equation is almost certain (3–4 marks). Memorise the formula E = E⁰ − (0.0591/n) log Q and practise with different concentrations.

📊 Conductance of Electrolytic Solutions

Conductance (G) is the measure of how easily electricity flows through an electrolytic solution. It is the reciprocal of resistance (R). The key parameters are:

Term Symbol Unit Formula
Resistance R Ω (ohm) R = ρ × ℓ / A
Conductance G S (siemens) G = 1/R
Resistivity ρ Ω m ρ = R × A / ℓ
Conductivity κ (kappa) S m⁻¹ κ = 1/ρ
Molar Conductivity Λm S m² mol⁻¹ Λm = κ / C

📏 Cell Constant

Cell constant (G*) = ℓ / A, where ℓ is the distance between electrodes and A is the cross-sectional area. Experimentally: G* = κ × R. It has units of m⁻¹.

📘 Key Trend: As concentration decreases, molar conductivity (Λm) increases because interionic attractions weaken, allowing ions to move more freely. For strong electrolytes, Λm approaches Λ⁰m (limiting molar conductivity) at infinite dilution.

📜 Kohlrausch’s Law

Kohlrausch’s Law of Independent Migration of Ions: At infinite dilution, the molar conductivity of an electrolyte is equal to the sum of the contributions of its individual ions.

Λ⁰m = ν+λ⁰+ + νλ⁰

Where ν+ and ν are the number of cations and anions per formula, and λ⁰+, λ⁰ are their limiting molar conductivities.

✅ Applications of Kohlrausch’s Law

  • Calculate Λ⁰m for weak electrolytes (e.g., CH₃COOH)
  • Determine degree of dissociation: α = Λm / Λ⁰m
  • Calculate dissociation constant Ka = Cα² / (1 − α)
  • Calculate solubility of sparingly soluble salts: s = κ × 1000 / Λ⁰m
🎯 Exam Tip: Kohlrausch’s Law is a favourite in MP Board 2027 — expect a 3-mark question on its application to calculate Λ⁰m of a weak acid or the degree of dissociation.

💡 Electrolysis & Faraday’s Laws

Electrolysis is the process of decomposing an electrolyte by passing an electric current through its solution (or molten state). Products are obtained at the electrodes through oxidation and reduction.

🧪 Faraday’s First Law

The mass of a substance deposited (or liberated) at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.

m = Z × Q = Z × I × t

Where: m = mass (g), Z = electrochemical equivalent (g C⁻¹), Q = charge (Coulomb), I = current (A), t = time (s).

🧪 Faraday’s Second Law

When the same quantity of electricity is passed through different electrolytes, the masses of substances deposited are proportional to their chemical equivalents.

m₁ / m₂ = E₁ / E₂

Where E is the equivalent weight of the substance. Also: w = (E × I × t) / 96500

📘 1 Faraday (F) = 96485 C = charge of 1 mole of electrons. This can deposit 1 gram-equivalent of any substance.

⚙️ Products of Electrolysis

The products depend on the nature of the electrolyte and the electrode material:

Electrolyte At Cathode At Anode
Molten NaCl Na(s) Cl₂(g)
Aqueous NaCl H₂(g) Cl₂(g)
Aqueous Na₂SO₄ H₂(g) O₂(g)
Aqueous CuSO₄ (Pt electrodes) Cu(s) O₂(g)
Aqueous CuSO₄ (Cu electrodes) Cu(s) Cu²⁺(aq) (anode dissolves)

🔋 Batteries & Fuel Cells

🔋 Primary Batteries

Non-rechargeable. Example: Dry cell (Leclanché cell) — Zn anode, graphite cathode, MnO₂ + NH₄Cl paste as electrolyte. Produces 1.5 V.

🔋 Secondary Batteries

Rechargeable. Example: Lead-acid battery — Pb anode, PbO₂ cathode, H₂SO₄ electrolyte. Produces 2 V per cell (6 cells = 12 V car battery). Nickel-Cadmium (Ni-Cd) battery — NiO(OH) cathode, Cd anode, KOH electrolyte.

⛽ Fuel Cells

Fuel cells convert chemical energy of a fuel (like H₂) directly into electrical energy. The Hydrogen-Oxygen fuel cell uses H₂ at the anode and O₂ at the cathode with KOH as electrolyte. Produces electricity + water as the only by-product — clean and efficient (60–70% efficiency).

Feature Primary Battery Secondary Battery Fuel Cell
Rechargeable? No Yes Continuous fuel supply
Example Dry cell, Mercury cell Lead-acid, Ni-Cd, Li-ion H₂-O₂ fuel cell
Lifespan Single use 500–1000 cycles As long as fuel is supplied

🛡️ Corrosion

Corrosion is the process where metals are oxidized by atmospheric oxygen, moisture, or other chemicals. The most common example is rusting of iron:

  • Anode: Fe → Fe²⁺ + 2e⁻
  • Cathode: O₂ + 4H⁺ + 4e⁻ → 2H₂O (in acidic medium)
  • Overall: 2Fe + O₂ + 2H₂O → 2Fe(OH)₂ → Fe₂O₃·xH₂O (rust)

🛡️ Methods to Prevent Corrosion

  • Galvanization: Coating iron with zinc. Zn acts as a sacrificial anode.
  • Tinning: Coating with tin (used in food cans).
  • Electroplating: Depositing a protective metal layer (Cr, Ni) on the surface.
  • Sacrificial anode method: Attaching a more reactive metal (Mg, Zn) to protect the main structure.
  • Painting or oiling: Creating a barrier against air and moisture.
  • Alloying: Mixing with other metals (e.g., stainless steel = Fe + Cr + Ni).

📝 Important Questions for MP Board 2027

✏️ Very Short Answer (1 Mark)

  1. What is the function of a salt bridge in a galvanic cell?
  2. Define molar conductivity.
  3. What is the unit of cell constant?
  4. Give one example of a primary battery.
  5. State Faraday’s First Law of Electrolysis.

📝 Short Answer (2–3 Marks)

  1. Write the cell reaction and calculate E⁰cell for the Daniel cell. Given: E⁰Cu²⁺/Cu = +0.34 V, E⁰Zn²⁺/Zn = −0.76 V.
  2. Explain why molar conductivity increases with dilution.
  3. State and explain Kohlrausch’s Law with one application.
  4. Distinguish between a primary and secondary battery.
  5. Calculate the mass of copper deposited when 0.5 A current is passed through CuSO₄ solution for 30 minutes. (Atomic mass of Cu = 63.5 g, F = 96500 C)

📚 Long Answer (5–6 Marks)

  1. Explain the working of a Daniel cell with a labelled diagram. Derive the expression for its EMF using the Nernst equation.
  2. What is electrolysis? Explain Faraday’s Laws of Electrolysis with equations and one numerical example.
  3. Describe the construction and working of a hydrogen-oxygen fuel cell. What are its advantages?
  4. Explain the mechanism of corrosion of iron. Discuss any four methods to prevent corrosion.

📘 Quick Revision Tips

  • Memorise the standard reduction potential series — it helps predict cell reactions
  • Nernst equation derivation appears every year — practice with different concentrations
  • For electrolysis, remember: cathode attracts cations (reduction), anode attracts anions (oxidation)
  • Practice at least 3 numerical problems: EMF calculation, Faraday’s laws, Kohlrausch’s Law

📖 More Resources: Download MP Board Class 12 Chemistry previous year papers and practice sets from mpboard.ai. For video explanations and chapter-wise PYQ discussions, visit our Class 12 Chemistry Course.

MP Board Class 12 Chemistry Chapter 1: Solutions — Complete Notes for 2027 Exam

📘 MP Board Class 12 Chemistry Chapter 1: Solutions — Complete Notes for 2027 Exam

Chapter 1: Solutions is one of the most scoring chapters in your MP Board Class 12 Chemistry syllabus. This chapter lays the foundation for physical chemistry and carries significant weight in board exams (typically 10–14 marks). In these comprehensive notes, we cover every key concept — from concentration terms to colligative properties — with clear formulas and practical examples tailored to the MP Board 2027 exam pattern.

🧪 1. Types of Solutions

A solution is a homogeneous mixture of two or more substances. The component present in larger quantity is called the solvent, and the one present in smaller quantity is called the solute.

State of Solute State of Solvent Example
Gas Gas Air (O₂ + N₂)
Gas Liquid Soda water (CO₂ + H₂O)
Liquid Liquid Ethanol + Water
Solid Liquid Salt + Water
Solid Solid Alloys (Brass — Cu + Zn)

⚡ MP Board Tip: Questions on classifying solution types appear frequently in 1-mark and 2-mark sections. Remember that gaseous solutions can be gas-gas (air), gas-liquid (soda water), or gas-solid (hydrogen in palladium).

🔢 2. Expressing Concentration of Solutions

MP Board exams frequently ask numerical problems on concentration terms. Memorise the following formulas — they are your key to scoring full marks in the numerical section.

Term Formula Unit / Remarks
Mass % (w/w) (Mass of solute ÷ Mass of solution) × 100 %
Volume % (v/v) (Volume of solute ÷ Volume of solution) × 100 %
Mole Fraction (x) x_A = n_A / (n_A + n_B) No unit (dimensionless)
Molarity (M) Moles of solute ÷ Volume of solution (L) mol/L or M
Molality (m) Moles of solute ÷ Mass of solvent (kg) mol/kg or m
Normality (N) Gram equivalents ÷ Volume of solution (L) eq/L or N

📘 Key Distinction: Molarity depends on temperature (since volume changes with temperature), while molality is temperature-independent (mass is constant). MP Board often asks: “Which concentration term is preferred when studying temperature-dependent properties?” Answer: Molality.

💧 3. Solubility — Factors & Henry’s Law

Factors Affecting Solubility

  • Nature of solute and solvent — “Like dissolves like” (polar solutes in polar solvents, non-polar in non-polar)
  • Temperature — For most solids, solubility increases with temperature (endothermic dissolution). For gases, solubility decreases with increasing temperature.
  • Pressure — Affects solubility of gases only. Governed by Henry’s Law.

Henry’s Law

Statement: At constant temperature, the mass of a gas dissolved in a given volume of liquid is directly proportional to the pressure of the gas above the liquid.

Mathematical form: p = KH × x

Where p = partial pressure of the gas, KH = Henry’s law constant, x = mole fraction of the gas in solution.

⚡ MP Board Exam Pattern: Henry’s Law application questions appear in 2-mark and 3-mark sections. Common applications: (a) Carbonated beverages — CO₂ is bottled under high pressure (follows Henry’s Law), (b) Scuba divers — increased pressure leads to more N₂ dissolving in blood, causing “bends” if ascent is too rapid, (c) Aquatic life — oxygen solubility in water is higher at lower temperatures (cold-water fish thrive).

Important: Higher the value of KH, lower the solubility of the gas at the same partial pressure. For example, KH for N₂ (6.5×10⁴ bar) > KH for O₂ (3.3×10⁴ bar), meaning O₂ is more soluble than N₂ in water.

📊 4. Vapour Pressure & Raoult’s Law

Vapour Pressure of Liquid Solutions

When a non-volatile solute is dissolved in a volatile solvent, the vapour pressure of the solution is lower than that of the pure solvent. This is the basis for all colligative properties.

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.

pA = pA° × xA  and  pB = pB° × xB

Total vapour pressure: ptotal = pA + pB = pA°xA + pB°xB

Type Deviation from Raoult’s Law Example ΔHmix
Ideal Solution No deviation Benzene + Toluene ΔH = 0
Positive Deviation ptotal > calculated Ethanol + Water, Acetone + CS₂ ΔH > 0
Negative Deviation ptotal < calculated Chloroform + Acetone, HNO₃ + H₂O ΔH < 0

📘 Important: Azeotropes are mixtures that boil at constant temperature. Minimum-boiling azeotropes (e.g., ethanol-water: 95.4% ethanol, 78.15°C) show positive deviation. Maximum-boiling azeotropes (e.g., HNO₃-H₂O: 68% HNO₃, 120.5°C) show negative deviation. This is a frequently asked concept in MP Board 5-mark questions.

📐 5. Colligative Properties

Colligative properties depend only on the number of solute particles, not their chemical identity. These are the most numerically important sections for the MP Board Chemistry exam.

Property Formula Symbol Meaning
Relative Lowering of Vapour Pressure (p° − p) / p° = xB p° = pure solvent V.P.
p = solution V.P.
xB = mole fraction of solute
Elevation of Boiling Point (ΔTb) ΔTb = Kb × m Kb = ebullioscopic constant
m = molality
Depression of Freezing Point (ΔTf) ΔTf = Kf × m Kf = cryoscopic constant
m = molality
Osmotic Pressure (π) π = iCRT i = van’t Hoff factor
C = molarity
R = 0.0821 L·atm/mol·K
T = temperature (K)

Key Points for Numerical Problems

  • Use the formula: Molar mass (M) = (Kf × wB × 1000) / (ΔTf × wA), where wB = mass of solute (g), wA = mass of solvent (g)
  • Osmotic pressure method is preferred for determining molar mass of macromolecules (proteins, polymers) because its magnitude is large even at low concentrations
  • Q10: The temperature coefficient — most colligative property problems are solved with standard Kb and Kf values given in the question paper

🧮 6. Abnormal Molar Masses — van’t Hoff Factor

When a solute undergoes association (molecules combine) or dissociation (molecules break into ions) in solution, the observed molar mass differs from the theoretical value. This is corrected using the van’t Hoff factor (i).

i = Normal molar mass / Observed molar mass = Observed colligative property / Normal colligative property

For dissociation: i > 1 (e.g., NaCl → Na⁺ + Cl⁻, so i = 2)
For association: i < 1 (e.g., benzoic acid in benzene dimerises, so i = 0.5)

📘 Modified Formulas with van’t Hoff Factor:
ΔTb = i × Kb × m
ΔTf = i × Kf × m
π = iCRT
(p° − p) / p° = i × (nB / (nA + nB))

Solute Behaviour van’t Hoff Factor (i)
NaCl Dissociation (2 ions) ≈ 2
K₂SO₄ Dissociation (3 ions) ≈ 3
Benzoic Acid (C₆H₅COOH) in benzene Association (dimerisation) ≈ 0.5
Glucose (C₆H₁₂O₆) No dissociation or association 1

❓ 7. Frequently Asked Questions (FAQs)

Q1: What is the difference between molarity and molality?

Answer: Molarity (M) is moles of solute per litre of solution — it depends on temperature because volume changes with temperature. Molality (m) is moles of solute per kg of solvent — it is temperature-independent because mass is constant. For MP Board numericals, use molality when dealing with colligative properties like boiling point elevation or freezing point depression.

Q2: What are ideal and non-ideal solutions?

Answer: Ideal solutions obey Raoult’s law at all concentrations — the intermolecular forces between A-A, B-B, and A-B are equal (e.g., benzene + toluene, n-hexane + n-heptane). Non-ideal solutions deviate from Raoult’s law — they can show either positive deviation (A-B interactions are weaker than A-A and B-B) or negative deviation (A-B interactions are stronger).

Q3: Why is sea water not suitable for drinking?

Answer: Sea water has a high osmotic pressure due to dissolved salts. When consumed, the osmotic pressure difference causes dehydration of body cells (exosmosis). This is why drinking sea water increases thirst and can lead to kidney damage. Reverse osmosis is used to desalinate sea water for drinking purposes.

Q4: What is the significance of the van’t Hoff factor?

Answer: The van’t Hoff factor (i) corrects the colligative properties for the actual number of particles in solution. When a solute dissociates (like NaCl splitting into Na⁺ and Cl⁻), i > 1, and the observed depression in freezing point is larger than expected. When a solute associates (like benzoic acid forming dimers in benzene), i < 1, and the observed colligative effect is smaller. This concept appears in almost every MP Board numerical section.

Q5: How is freezing point depression used to determine molar mass?

Answer: By measuring ΔTf experimentally (using a Beckmann thermometer or cryoscope), the molar mass of the solute can be calculated using: MB = (Kf × wB × 1000) / (ΔTf × wA). This cryoscopic method is widely used for determining molar masses of non-volatile, non-electrolytic substances in the laboratory.

Q6: What is an azeotropic mixture?

Answer: An azeotrope is a mixture of two liquids that boils at a constant temperature and has the same composition in the liquid and vapour phases. The ethanol-water azeotrope (95.4% ethanol, 4.6% water, bp 78.15°C) is the most famous example — this is why 100% ethanol cannot be obtained by simple fractional distillation. Azeotropes can be minimum-boiling (positive deviation) or maximum-boiling (negative deviation).

Q7: What is reverse osmosis and where is it used?

Answer: When pressure greater than the osmotic pressure is applied on the solution side, solvent molecules move from the solution to the pure solvent through a semipermeable membrane — this is reverse osmosis. It is used for desalination of sea water, water purification (RO water filters), and wastewater treatment. MP Board often includes this as a 2-mark application question.

🎯 Key Exam Tips for MP Board Class 12 Chemistry 2027:

  • Memorise all concentration formulas — they are easy marks
  • Practice at least 5 numericals on Raoult’s Law and Colligative Properties
  • Learn the van’t Hoff factor values for common electrolytes (NaCl, CaCl₂, K₂SO₄)
  • Revise Henry’s Law applications — a sure-shot 2-3 mark question
  • Focus on differences: molarity vs molality, ideal vs non-ideal, positive vs negative deviation

📚 MP Board Class 12 Chemistry Chapter 1 Solutions Notes — Prepared for MP Board 2027 Exam. Visit mpboard.ai for more chapter-wise notes, PYQs, and study material.

MP Board Class 12 Chemistry Solved Paper 2024 — Previous Year Question Paper with Step-by-Step Solutions

The MP Board Class 12 Chemistry exam is a crucial subject for science stream students. Solving the 2024 previous year question paper is one of the best ways to understand the exam pattern, marking scheme, and the types of questions asked. This post provides a complete question-by-question solved paper with detailed step-by-step solutions to help you score maximum marks in the 2027 board exams.

MP Board Class 12 Chemistry Solved Paper 2024

📋 MP Board Class 12 Chemistry Exam 2024 — Paper Overview

  • Subject: Chemistry (रसायन शास्त्र)
  • Year: 2024
  • Class: 12
  • Total Marks: 100 (Theory: 70 + Practical: 30)
  • Time Allowed: 3 Hours
  • Total Questions: 25 (Section A: 15 MCQs, Section B: 5 Short Answer, Section C: 3 Long Answer, Section D: 2 Numerical)

📌 Important Instructions for Students

  1. Read the entire question paper carefully before starting
  2. All questions are compulsory
  3. Draw neat labeled diagrams for chemical reactions and structures
  4. Use blue/black pen only; diagrams in pencil
  5. Show all calculations stepwise for numerical problems

✅ Section A: Multiple Choice Questions (1 Mark Each)

Q1. The number of atoms in a face-centred cubic unit cell is:
(a) 1
(b) 2
(c) 4
(d) 6
Answer: (c) 4
Solution: In FCC, atoms at corners contribute 8 × ⅛ = 1, atoms at face centres contribute 6 × ½ = 3. Total = 1 + 3 = 4 atoms per unit cell.

Q2. Which of the following is an example of a colligative property?
(a) Surface tension
(b) Vapour pressure lowering
(c) Viscosity
(d) Optical activity
Answer: (b) Vapour pressure lowering
Solution: Colligative properties depend only on the number of solute particles, not their identity. The four colligative properties are: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure.

Q3. The correct order of basic strength in gaseous phase for the following is:
(a) NH₃ > C₂H₅NH₂ > (CH₃)₂NH > (CH₃)₃N
(b) (CH₃)₃N > (CH₃)₂NH > C₂H₅NH₂ > NH₃
(c) (CH₃)₂NH > C₂H₅NH₂ > (CH₃)₃N > NH₃
(d) C₂H₅NH₂ > (CH₃)₂NH > NH₃ > (CH₃)₃N
Answer: (b) (CH₃)₃N > (CH₃)₂NH > C₂H₅NH₂ > NH₃
Solution: In the gaseous phase, basicity depends on the +I effect of alkyl groups. More alkyl groups → more electron density on N → stronger base. Hence, tertiary > secondary > primary > ammonia.

Q4. The IUPAC name of CH₃CH₂CH=CH₂ is:
(a) But-1-ene
(b) But-2-ene
(c) 1-Butene
(d) Butene
Answer: (a) But-1-ene
Solution: The parent chain has 4 carbons (but-), the double bond is at carbon 1, making it but-1-ene. The IUPAC systematic name uses the lowest locant for the functional group.

Q5. Which of the following solutions has the highest osmotic pressure?
(a) 0.1 M NaCl
(b) 0.1 M Glucose
(c) 0.1 M CaCl₂
(d) 0.1 M Urea
Answer: (c) 0.1 M CaCl₂
Solution: Osmotic pressure (π) = iCRT. CaCl₂ dissociates into 3 ions (Ca²⁺ + 2Cl⁻), giving i = 3. NaCl gives i = 2 (Na⁺ + Cl⁻), while glucose and urea are non-electrolytes (i = 1). Higher i → higher osmotic pressure.

Q6. Galvanization of iron involves coating with:
(a) Copper
(b) Zinc
(c) Tin
(d) Nickel
Answer: (b) Zinc
Solution: Galvanization is the process of coating iron or steel with a layer of zinc to prevent rusting. Zinc acts as a sacrificial anode — it oxidises preferentially, protecting the iron underneath.

Q7. The activation energy of a reaction can be determined using:
(a) Gibbs equation
(b) Nernst equation
(c) Arrhenius equation
(d) van’t Hoff equation
Answer: (c) Arrhenius equation
Solution: The Arrhenius equation k = Ae^(−Ea/RT) relates the rate constant k to activation energy Ea. Taking log, ln k = ln A − Ea/RT, so a plot of ln k vs 1/T gives slope = −Ea/R, from which Ea is calculated.

Q8. The complex K₃[Fe(CN)₆] is named as:
(a) Potassium ferrocyanide
(b) Potassium ferricyanide
(c) Potassium hexacyanoferrate(II)
(d) Potassium hexacyanoferrate(III)
Answer: (d) Potassium hexacyanoferrate(III)
Solution: Fe in [Fe(CN)₆]³⁻ has oxidation state +3. In IUPAC nomenclature, the complex anion is named as hexacyanoferrate(III). The counterion K⁺ is named first: potassium hexacyanoferrate(III).

Q9. Which of the following is NOT a biodegradable polymer?
(a) Nylon-2-nylon-6
(b) PHBV
(c) Teflon
(d) Dextron
Answer: (c) Teflon
Solution: Teflon (polytetrafluoroethylene) is a synthetic fluoropolymer with strong C-F bonds that are highly resistant to degradation. Nylon-2-nylon-6, PHBV (polyhydroxybutyrate-co-valerate), and Dextron are biodegradable polymers.

Q10. The rate of a chemical reaction doubles for every ___ rise in temperature:
(a) 5°C
(b) 10°C
(c) 15°C
(d) 20°C
Answer: (b) 10°C
Solution: According to the temperature coefficient rule (van’t Hoff rule), the rate of a chemical reaction approximately doubles for every 10°C rise in temperature (temperature coefficient, Q₁₀ ≈ 2).

✅ Section B: Short Answer Questions (2-3 Marks Each)

Q11. Explain Henry’s Law and give one application.

Solution: Henry’s Law states that at constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. Mathematically: p = K_H × x, where p is the partial pressure, K_H is Henry’s constant, and x is the mole fraction.

Application: Carbonated soft drinks are bottled under high CO₂ pressure. When the bottle is opened, pressure decreases, and CO₂ comes out of solution — this is why fizz is observed.

Q12. Distinguish between lyophilic and lyophobic colloids with two examples each.

PropertyLyophilic ColloidsLyophobic Colloids
MeaningSolvent-lovingSolvent-hating
ReversibilityReversible (can be redispersed)Irreversible
ViscosityHigher than solventSame as solvent
ExamplesGelatin, starch in waterGold sol, Fe(OH)₃ sol

Q13. Write the mechanism of SN1 reaction with an example.

Solution: SN1 (Substitution Nucleophilic Unimolecular) proceeds in two steps:

  1. Formation of carbocation (slow, rate-determining step): (CH₃)₃C-Br → (CH₃)₃C⁺ + Br⁻
  2. Attack of nucleophile (fast step): (CH₃)₃C⁺ + OH⁻ → (CH₃)₃C-OH

Example: Hydrolysis of tert-butyl bromide (2-bromo-2-methylpropane) with aqueous KOH gives tert-butyl alcohol. The reaction follows first-order kinetics (rate depends only on [substrate]). Carbocation rearrangement may occur, leading to racemization.

Q14. Explain the structure of chlorine molecule using MOT.

Solution: Cl₂ has 17 + 17 = 34 electrons. The Molecular Orbital configuration of Cl₂ is:
σ1s² σ*1s² σ2s² σ*2s² σ2pz² π2px² = π2py² π*2px² = π*2py² σ*2pz²
Bond Order = (10 − 8) / 2 = 1.
Cl₂ has a single covalent bond. The filled π* and σ* orbitals make it diamagnetic (no unpaired electrons).

Q15. What is the role of depressants in froth flotation? Give an example.

Solution: Depressants are chemicals that prevent certain minerals from forming froth, allowing selective separation. In froth flotation of sulphide ores, NaCN (sodium cyanide) is used as a depressant. When PbS (galena) and ZnS (sphalerite) are present together, NaCN prevents ZnS from floating by forming a complex [Zn(CN)₄]²⁻ on its surface, while PbS remains unaffected and forms froth with collectors.

✅ Section C: Long Answer Questions (5 Marks Each)

Q16. Explain the chemistry of H₂O₂ as an oxidising and reducing agent with examples. How is H₂O₂ prepared by the auto-oxidation method?

Solution:

H₂O₂ as an oxidising agent: H₂O₂ → 2H⁺ + O₂ + 2e⁻ (E° = 1.77 V in acidic medium). It readily accepts electrons and gets reduced to H₂O (acidic) or OH⁻ (basic).
Example: 2Fe²⁺ + H₂O₂ + 2H⁺ → 2Fe³⁺ + 2H₂O
(Fe²⁺ is oxidized to Fe³⁺, H₂O₂ is reduced to H₂O)

H₂O₂ as a reducing agent: H₂O₂ can also act as a reducing agent by donating electrons. It gets oxidized to O₂.
Example: 2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O
(KMnO₄ is reduced to Mn²⁺, H₂O₂ is oxidized to O₂)

Preparation by auto-oxidation: BaO₂ + H₂SO₄ → BaSO₄ + H₂O₂. Barium peroxide reacts with dilute sulphuric acid at 0°C. BaSO₄ precipitates and H₂O₂ is obtained in the filtrate. This is called the auto-oxidation method.

Q17. Describe the principle, construction, and working of a fuel cell. Write the reactions involved in a hydrogen-oxygen fuel cell.

Solution:

Principle: A fuel cell is an electrochemical cell that converts chemical energy of a fuel directly into electrical energy through a redox reaction. Unlike a battery, it operates continuously as long as fuel is supplied.

Construction: The hydrogen-oxygen fuel cell consists of two porous carbon electrodes impregnated with catalysts (Pt, Pd). The electrolyte is a concentrated KOH solution (30%). Hydrogen gas is passed through the anode compartment, and oxygen through the cathode compartment.

Electrode Reactions:
At anode (oxidation): 2H₂(g) + 4OH⁻(aq) → 4H₂O(l) + 4e⁻
At cathode (reduction): O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)
Overall reaction: 2H₂(g) + O₂(g) → 2H₂O(l)

Advantages: High efficiency (~70%), zero pollution (only water is produced), continuous operation.

Q18. Outline the product formed when D-glucose is treated with the following reagents:

  • (a) Br₂ water: D-glucose is oxidized to gluconic acid (COOH group at C1). Br₂ water specifically oxidizes the aldehyde group (−CHO) to carboxylic acid (−COOH).
  • (b) HNO₃: Both terminal groups are oxidized — D-glucose gives saccharic acid (both C1 and C6 become COOH).
  • (c) H₂N-OH (hydroxylamine): D-glucose forms glucose oxime (C=N-OH bond at C1).
  • (d) (CH₃CO)₂O (acetic anhydride): D-glucose forms pentaacetate — all five −OH groups are acetylated.
  • (e) NH₂OH + H⁺ (phenylhydrazine): D-glucose forms glucosazone (yellow crystalline osazone).

✅ Section D: Numerical Problems (3+4 Marks)

Q19. A 0.5 g sample containing Na₂CO₃·10H₂O and inert impurities is titrated against 0.1 M HCl. The endpoint is reached at 22.4 mL of HCl. Calculate the percentage purity of the sample.

Solution:
Reaction: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

Moles of HCl used = M × V(L) = 0.1 × (22.4/1000) = 0.00224 mol
Moles of Na₂CO₃ = ½ × moles of HCl = 0.00112 mol
Molar mass of Na₂CO₃·10H₂O = (2×23) + 12 + (3×16) + 10×(2+16) = 46 + 12 + 48 + 180 = 286 g/mol
Mass of pure Na₂CO₃·10H₂O = 0.00112 × 286 = 0.3203 g
Percentage purity = (0.3203 / 0.5) × 100 = 64.06%

Q20. The rate constant of a first-order reaction is 2.5 × 10⁻³ s⁻¹ at 300 K. Calculate the activation energy if the pre-exponential factor is 1.2 × 10¹³ s⁻¹. (R = 8.314 J/mol·K)

Solution:
Using Arrhenius equation: k = Ae^(−Ea/RT)

ln(k) = ln(A) − Ea/(RT)
ln(2.5 × 10⁻³) = ln(1.2 × 10¹³) − Ea/(8.314 × 300)
ln(2.5) + ln(10⁻³) = ln(1.2) + ln(10¹³) − Ea/(2494.2)
0.9163 − 6.9078 = 0.1823 + 29.9336 − Ea/2494.2
−5.9915 = 30.1159 − Ea/2494.2
Ea/2494.2 = 30.1159 + 5.9915 = 36.1074
Ea = 36.1074 × 2494.2 = 90,028 J/mol ≈ 90.03 kJ/mol

📊 Topic-Wise Weightage Analysis

TopicQuestion NumbersTotal Marks
Solid StateQ11
SolutionsQ2, Q5, Q117
Chemical KineticsQ7, Q10, Q208
ElectrochemistryQ6, Q177
Surface ChemistryQ123
p-Block ElementsQ165
Coordination CompoundsQ81
Organic Chemistry (Haloalkanes)Q4, Q135
Alcohols, Phenols, Ethers
BiomoleculesQ185
Chemistry in Everyday LifeQ91
MetallurgyQ153
Stoichiometry & TitrationQ194

🎯 Expert Tips to Score 90%+ in Chemistry Board Exam

  1. Master NCERT thoroughly — 80% of questions come directly from NCERT textbook
  2. Practice numerical problems daily — physical chemistry carries 15-20 marks
  3. Learn named reactions with mechanisms — draw arrow pushing mechanisms for all organic reactions
  4. Create a formula sheet — colligative properties, rate equations, Nernst equation, Ksp
  5. Solve at least 10 previous year papers — identify recurring question patterns
  6. Focus on 3-4 mark questions — they are the easiest to score if concepts are clear
  7. Use diagrams and flowcharts — metallurgy flowcharts, crystal structures, cell diagrams
  8. Attempt MCQs first — 15 marks in Section A can be secured in 20 minutes
  9. Write balanced equations every time — partial equations lose marks
  10. Practice time management — allocate 60 min for Section A+B, 60 min for C, 30 min for D, 30 min for revision

🔍 Download MP Board Previous Year Papers

Regular practice of previous year question papers is the most effective strategy for board exam preparation. The 2024 Chemistry paper shows that:

  • Numerical problems focus on solutions, kinetics, and electrochemistry
  • Organic chemistry questions test IUPAC nomenclature, reaction mechanisms, and conversions
  • Inorganic chemistry mainly covers coordination compounds, p-block, and d-and f-block elements
  • Physical chemistry requires strong numerical aptitude and formula application

Practice this solved paper multiple times, time yourself, and check your answers against the detailed solutions provided. Good luck with your 2027 MP Board Chemistry exam!