Board Paper Blueprint and Solutions
This opening chapter sets out the CBSE Class XII Chemistry (043) 2026-27 paper design as published by CBSE, shows how each kind of question is answered for full marks, and then teaches Unit 1, Solutions, which carries 7 of the 70 theory marks. Solutions is the first of the book's numerical units: its concentration conversions, Raoult's law, colligative properties and van't Hoff factor reappear in electrochemistry and in the practical examination, so accuracy here pays off twice.
KarmSakha preparation topics covered
- Theory and practical marks, section-wise pattern and unit-wise marks for 2026-27
- Question types: MCQ, assertion-reason, very short, short, case-based and long answer
- Time plan and no-calculator arithmetic in a 3-hour paper
- Expressing concentration: mass %, ppm, mole fraction, molarity, molality and their interconversion
- Solubility of gases and Henry's law
- Vapour pressure, Raoult's law, ideal and non-ideal solutions, azeotropes
- Colligative properties: relative lowering, elevation of boiling point, depression of freezing point, osmotic pressure
- Determination of molar mass and the abnormal molar mass / van't Hoff factor
The 2026-27 Chemistry paper at a glance
CBSE Class XII Chemistry (subject code 043) is assessed for 100 marks: a 70-mark theory paper written in 3 hours and a 30-mark practical. The practical splits into volumetric analysis (8), salt analysis (8), a content-based experiment (6), project work (4) and the class record with viva (4). Theory and practical are separate components, so a strong practical score is the cheapest way to lift the total, and chapter 7 of this guide covers it. Everything below follows the 2026-27 curriculum and sample question paper (SQP) published by the CBSE Academic Unit; the SQP footnote states that there is no change in the Question Paper Design and Assessment Pattern for Academic Session 2026-27.
Questions are also tagged by competency. Remembering and understanding account for 28 marks (40 per cent), applying for 21 marks (30 per cent), and analysing, evaluating and creating for the remaining 21 marks (30 per cent). Practically, this means only about two-fifths of the paper can be earned by memorising statements. The rest asks you to use a rule on an unfamiliar compound, compare two cases, or interpret data. The curriculum does not prescribe chapter-wise weightage beyond the unit marks shown in the table of the third section, so no unit can safely be skipped. Calculators and log tables are not allowed, so numericals simplify by hand and any logarithm needed is supplied in the question. In the sample paper, Q1 to Q12 are plain MCQs and Q13 to Q16 are assertion-reason items.
| Section | Questions | Marks each | Section marks | Format |
|---|---|---|---|---|
| A | 16 | 1 | 16 | Q1-12 MCQs; Q13-16 assertion-reason with four standard options |
| B | 5 | 2 | 10 | Very short answer |
| C | 7 | 3 | 21 | Short answer |
| D | 2 | 4 | 8 | Case-based, with sub-parts |
| E | 3 | 5 | 15 | Long answer |
| Total | 33 | - | 70 | All compulsory; internal choice; no calculators or log tables |
How the sections test you
Section A (16 marks) rewards precision under speed. A plain MCQ gives one mark for choosing the single correct option, with three distractors that are common mistakes: the wrong sign, a missing factor of i, a swapped definition. Example: among 0.1 m aqueous solutions of glucose, NaCl and CaCl2, which has the lowest freezing point? The answer is CaCl2, because it supplies about three particles per formula unit. Rule: work out the answer before reading the options, then match. An assertion-reason item gives two statements and asks whether each is true and whether the reason explains the assertion. Example: A: 'Molality does not change with temperature.' R: 'Molality is defined using the mass of solvent.' Both are true and R explains A, so the first standard option is correct. Check the assertion first, then the reason, then ask 'does R cause A?'
Sections B and C reward the marking-scheme habit of one mark per distinct point. A 2-mark very short question such as 'Define molality and state why it is independent of temperature' earns one mark for the definition with unit (moles of solute per kilogram of solvent, mol/kg) and one for the reason (mass does not change with temperature, unlike volume). A 3-mark question such as 'Why does a mixture of ethanol and acetone show positive deviation from Raoult's law?' needs the cause (acetone breaks some of the hydrogen bonds of ethanol, so A–B forces are weaker than A–A and B–B forces), the consequence (higher vapour pressure than predicted) and the associated signs (ΔmixH and ΔmixV positive). Rule: in numericals, write the formula, the substitution with units and the final answer with unit.
Section D presents a short passage or data table followed by three sub-parts (typically two 1-mark questions and one 2-mark question with internal choice). Example: a passage on desalination by reverse osmosis, with sub-parts asking what pressure must be exceeded, which solution is hypertonic, and, for the 2-mark part, what happens to a cell in 0.5 per cent saline and why. Rule: answer only from the passage plus syllabus knowledge, and pull out numbers first. Section E (5 marks each) is a multi-part question that may be pure theory (as with the organic and coordination-compound questions in the sample paper) or may mix theory and a numerical (as in the electrochemistry question), for example 'differentiate ideal and non-ideal solutions in terms of ΔmixH, ΔmixV and Raoult's law, and find the molar mass from a depression of freezing point.' Rule: attempt every sub-part; marks are awarded step-wise even if the final value is wrong.
Plan for 180 minutes, at roughly 2.5 minutes per mark. Start with the questions you are surest of so that early marks bank, and mark the numericals for a second pass. In a no-calculator paper arithmetic is part of the test: turn ratios into simple fractions before dividing, cancel powers of ten first and keep two or three significant figures. Any logarithm needed will be supplied, for example 'log 2 = 0.30', so do not try to recall tables. Round only at the end; rounding 0.083 to 0.08 early can push a clean answer off the key.
| Section | Suggested time | Why |
|---|---|---|
| A (16 questions) | 25 min | Under 1.6 min per item; skip and return to calculation-based items |
| B (5 questions) | 25 min | Definition plus reason or a short numerical, 5 min each |
| C (7 questions) | 45 min | Mechanism, reasoning or numerical with three scoring points |
| D (2 case-based) | 25 min | Reading time plus three sub-parts each |
| E (3 long answers) | 45 min | Theory plus numerical, written step-wise |
| Final check | 15 min | Units, signs, internal choice attempted, nothing left blank |
| Total | 180 min | A suggested plan, not a CBSE rule |
What this guide covers and what it leaves out
The 2026-27 Class XII course has ten units that are examined in the written paper, each with a fixed mark allocation, plus four topics that the syllabus includes but treats only formatively. This guide follows that structure: chapter 1 (this one) teaches Solutions, chapter 2 Electrochemistry and Chemical Kinetics, chapter 3 the d- and f-block elements with Coordination Compounds, chapter 4 Haloalkanes, Haloarenes, Alcohols, Phenols and Ethers, chapter 5 Aldehydes, Ketones and Carboxylic Acids, chapter 6 Amines and Biomolecules, and chapter 7 the four formative topics together with the practical. Because unit marks add up to 70, you can use the table below to decide how much revision time each chapter deserves.
Several topics are deliberately missing. Class XI chemistry is not covered, so atomic structure, thermodynamics, equilibrium and organic basics (inductive effect, resonance, hyperconjugation) are assumed. Topics that are not in the 2026-27 Class XII syllabus, such as solid state and the p-block elements, are not taught here. The older 'metallurgy' content survives only as the formatively assessed topic 'General Principles and Processes of Isolation of Elements'. If you are also preparing for NEET or JEE, treat this guide as a Class XII revision layer: those examinations draw on Class XI chemistry and on topics that this CBSE syllabus omits, so you need your own full syllabus on top.
| Unit (theory) | Marks | Where in this guide |
|---|---|---|
| Solutions | 7 | Chapter 1 |
| Electrochemistry | 9 | Chapter 2 |
| Chemical Kinetics | 7 | Chapter 2 |
| d- and f-Block Elements | 7 | Chapter 3 |
| Coordination Compounds | 7 | Chapter 3 |
| Haloalkanes and Haloarenes | 6 | Chapter 4 |
| Alcohols, Phenols and Ethers | 6 | Chapter 4 |
| Aldehydes, Ketones and Carboxylic Acids | 8 | Chapter 5 |
| Amines | 6 | Chapter 6 |
| Biomolecules | 7 | Chapter 6 |
| Total theory | 70 | Surface Chemistry, Isolation of Elements, Polymers and Chemistry in Everyday Life are formative: see chapter 7 |
Concentration terms and solubility of gases
A solution is a homogeneous mixture; the component present in larger amount is the solvent and the other is the solute. Concentration can be expressed in several ways, and the board paper tests whether you can move between them. Mass percentage, volume percentage and parts per million (ppm) describe a ratio of amounts. Mole fraction, molarity and molality tie the amount of substance to something measurable. The one conceptual point worth owning is temperature: anything defined by mass (mass %, ppm, mole fraction, molality) is independent of temperature, while anything defined by volume (molarity, volume %) changes because a solution expands when heated. This is why molality, not molarity, is used in freezing-point and boiling-point work.
To convert between molarity and molality, use density d (g/mL) and solute molar mass M2 (g/mol). For a solution of mass percentage w, molarity M = 10 × d × w / M2, because 1000 mL has mass 1000 d g and contains 1000 d w/100 g of solute. Molality from mass percentage is m = 1000 w / [M2 (100 − w)], since 100 g of solution holds w g solute and (100 − w) g solvent. In general, m = 1000 M / (1000 d − M M2) and M = 1000 d m / (1000 + m M2). Mole fraction of a solute is x2 = n2/(n1 + n2) and the mole fractions of all components sum to one. For very dilute aqueous solutions, molarity and molality are nearly equal numerically, because 1 L of solution weighs about 1 kg.
Gases dissolve in liquids to an extent that depends on pressure, temperature and the nature of gas and solvent. Henry's law states that the partial pressure of a gas in the vapour phase is proportional to its mole fraction in solution: p = KH x. KH is the Henry's law constant, in the same pressure unit as p, and it depends on the gas, the solvent and the temperature. A larger KH means lower solubility at the same pressure. The NCERT table at 293 K lists about 34.9 kbar for oxygen and about 76.5 kbar for nitrogen in water, so oxygen is more soluble than nitrogen, as expected. KH rises with temperature, so gases become less soluble in warm water.
Henry's law explains several everyday and exam-favourite situations. Soft drinks are sealed under high pressure of carbon dioxide so that more gas dissolves, and the fizz appears when the cap is opened and the pressure falls. Scuba divers breathe air at high pressure, so more nitrogen dissolves in blood; if they rise too fast, dissolved nitrogen escapes as bubbles (the bends), which is why tanks use helium-diluted air for deep dives, since helium is much less soluble. At high altitude the partial pressure of oxygen is lower, so less dissolves in blood and tissues, causing weakness and unclear thinking (anoxia). Aquatic life is also stressed in warm water because dissolved oxygen falls as temperature rises.
| Term | Definition | Formula | Depends on temperature? |
|---|---|---|---|
| Mass percentage | Mass of component / total mass × 100 | w = (w2/(w1 + w2)) × 100 | No |
| Volume percentage | Volume of component / total volume × 100 | V% = (V2/(V1 + V2)) × 100 | Yes |
| Parts per million | Mass of component per 10^6 parts of solution | ppm = (mass of component/total mass) × 10^6 | No |
| Mole fraction | Moles of component / total moles | x2 = n2/(n1 + n2) | No |
| Molarity (M) | Moles of solute per litre of solution | M = n2 / V(L) | Yes |
| Molality (m) | Moles of solute per kg of solvent | m = n2 / w1(kg) | No |
Vapour pressure, Raoult's law and ideal and non-ideal solutions
The vapour pressure of a pure liquid is the pressure of its vapour in equilibrium with the liquid at a given temperature; it increases with temperature and falls when intermolecular forces are stronger. For a solution of volatile liquids A and B, Raoult's law states that the partial vapour pressure of each component is the product of its mole fraction in the liquid and the vapour pressure of the pure component: pA = xA pA° and pB = xB pB°. By Dalton's law, total pressure is ptotal = pA + pB = pA° + (pB° − pA°) xB. The vapour is richer in the more volatile component, and its composition follows yA = pA / ptotal. For a non-volatile solute, only the solvent contributes, so p1 = x1 p1° and the relative lowering is (p1° − p1)/p1° = x2.
