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Some Basic Concepts of Chemistry and Redox Reactions

NEET > Chemistry > Some Basic Concepts In Chemistry

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Chapter Snapshot - Some Basic Concepts of Chemistry and Redox Reactions

A foundational chapter that anchors the entire NEET Chemistry syllabus. Mole concept calculations, stoichiometry (gravimetric and volumetric), equivalent mass and n-factor determination, oxidation number assignment, balancing redox equations (oxidation-number and ion-electron methods), limiting reagent problems, and concentration expressions (normality, molarity) form the core toolkit. The chapter also covers laws of chemical combination (conservation of mass, definite proportions, multiple proportions, equivalence), Dalton's atomic hypothesis, volume strength of H2O2, and percentage labelling of oleum. Mastery here directly accelerates problem-solving speed across physical, inorganic, and organic chemistry.

āœ“ Use This To Plan Your First 2–3 Hours
Expected Questions (Typical)
Q
2-3
NEET draws 2 to 3 questions from this chapter every year. Typical formats: mole concept numerical (calculate moles, number of atoms, or molecules from given mass), stoichiometry via limiting reagent, oxidation state determination in a complex ion, and n-factor calculation for a redox titration.
Time Required (Practical)
ā±
12-15 hrs
Classification of matter and Dalton's hypothesis 1 hr; atomic and molecular masses with mole concept 2 hrs; equivalent mass and n-factor 2.5 hrs; laws of chemical combination 1 hr; stoichiometry (gravimetric + volumetric) 3 hrs; oxidation number rules and redox balancing 2.5 hrs; H2O2 volume strength and oleum labelling 1 hr; MCQ practice 2 hrs.
Difficulty Level
⚔
Moderate
Concepts are straightforward but the sheer range of calculation types demands systematic practice. The n-factor for partial redox and disproportionation reactions is where most students falter. Oxidation number assignment in species like HN3 or HCN requires careful application of rules.
Most Asked Style: Numerical MCQ: calculate number of moles or atoms from given mass; find oxidation state of an element in a complex ion; determine n-factor of a substance in a given redox reaction; limiting reagent identification; normality and molarity interconversion for titration problems.Biggest Trap: Confusing n-factor calculation in partial redox reactions versus complete redox reactions. When only part of a reactant undergoes oxidation state change (e.g., HCl in the KMnO4 reaction where only 10 out of 16 Cl atoms are oxidised), the n-factor is a fraction (5/8), not a whole number. NEET distractors use the whole-number n-factor.Fast Win: Memorise five key results: (1) Moles = mass / molar mass; (2) Number of particles = moles times 6.022 x 10^23; (3) At STP, 1 mole gas = 22.4 L; (4) Normality = molarity times n-factor; (5) Volume strength of H2O2 = 5.6 times normality. These five formulas solve over 70% of the numericals from this chapter.Revision-Friendly: Yes. Core formulas fit on two cards: mole concept relationships (mass, moles, particles, gas volume) on one, and n-factor rules (acids, bases, salts, redox) on the other. A 30-minute revision sweep of these plus one worked example per stoichiometry type covers the full scoring range.

Subtopics - Some Basic Concepts of Chemistry and Redox Reactions (NEET)

Eight topic blocks: matter classification (physical and chemical), atomic and molecular masses with mole concept, Dalton's atomic hypothesis, laws of chemical combination, stoichiometry (gravimetric and volumetric analysis with n-factor), oxidation state and types of redox reactions, balancing methods (oxidation number and ion-electron), and SI units with measurement.

Revision tip: Before attempting any stoichiometry or redox numerical: (1) write the balanced equation, (2) identify the limiting reagent if multiple reactants are given, (3) determine n-factor based on the specific reaction (not a generic value), (4) confirm units match (grams vs moles vs equivalents). This four-step protocol eliminates the three most common errors in NEET numericals from this chapter.
NCERT LinesMCQsQuick Test

1) Matter and Its Classification

Covers two classification schemes for matter. Physical classification divides matter into solids (fixed shape and volume, particles close and orderly), liquids (fixed volume but no fixed shape, particles mobile), and gases (neither fixed shape nor volume, particles far apart and fast-moving). Chemical classification divides matter into elements, compounds, and mixtures. The three states are interconvertible under different conditions of temperature and pressure. Plasma and Bose-Einstein condensate are additional states existing only under extreme conditions.

Solid: fixed shape + volumeGas vs Vapour distinctionThree states interconvertiblePlasma and BEC: extreme conditions
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Physical Classification of MatterMatter is classified into solids, liquids, and gases based on physical characteristics. Solids have particles held close in an orderly fashion with no freedom of movement. Liquids have particles close but mobile. Gases have particles far apart with easy, fast movement. The distinction between gas and vapour: a substance in the gaseous state at room temperature is a gas (e.g., ammonia), while the gaseous form of a substance that is liquid at room temperature is called vapour (e.g., water vapour). Two additional states, plasma and Bose-Einstein condensate, exist under extreme temperature and pressure.
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Chemical Classification of MatterAt the macroscopic level, matter is classified by chemical composition into elements (single type of atom), compounds (two or more elements chemically combined in fixed ratio), and mixtures (physical combination of two or more substances in variable ratio). Mixtures are further divided into homogeneous (uniform composition, e.g., solutions) and heterogeneous (non-uniform composition, e.g., suspensions). This classification is the foundation for understanding pure substances versus mixtures in analytical chemistry.

2) Atomic and Molecular Masses

Defines atomic mass as the average relative mass of atoms compared to 1/12 the mass of carbon-12. The fundamental unit is amu (1.6604 x 10^-24 g). Gram atomic mass (GAM) is atomic mass expressed in grams. Molecular mass indicates how many times heavier a molecule is compared to 1/12 of carbon-12. Gram molecular mass (GMM) is molecular mass in grams. Key relationships: number of gram atoms = mass/GAM; mass of one atom = GAM/Avogadro's number; atomic mass of gaseous element = molecular mass/atomicity.

1 amu = 1.6604 x 10^-24 gGAM = atomic mass in gramsGMM = molecular mass in gramsAtomic mass = mol mass / atomicity
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Atomic Mass and Gram Atomic MassAtomic mass of an element is defined as the average relative mass of its atoms compared to the mass of a carbon atom taken as 12. The unit is amu, approximately 1.6604 x 10^-24 g. Gram atomic mass (GAM) is the atomic mass expressed in grams and equals one mole of atoms. Number of gram atoms = mass of element / GAM. Number of atoms in 1 g of element = 6.02 x 10^23 / atomic mass. Mass of one atom in grams = GAM / 6.02 x 10^23. For gaseous elements: atomic mass = molecular mass / atomicity.
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Molecular Mass and Gram Molecular MassMolecular mass equals the number indicating how many times heavier a molecule is compared to 1/12 the mass of one carbon-12 atom. Expressed in amu. Gram molecular mass (GMM) is the molecular mass expressed in grams and represents one mole of molecules. Number of gram molecules = mass of substance / GMM. Number of atoms in a substance = number of GMM times 6.02 x 10^23 times atomicity. Number of electrons = number of GMM times 6.02 x 10^23 times electron count per molecule.

3) Dalton's Atomic Hypothesis

John Dalton proposed this hypothesis to provide theoretical justification for the laws of chemical combination. Six postulates: (1) elements are composed of extremely small particles called atoms, (2) all atoms of a given element are identical, (3) atoms of different elements have different properties and masses, (4) atoms are indestructible and cannot be created or destroyed in chemical reactions, (5) atoms combine to form molecules in compounds, (6) relative number and kinds of atoms in a compound are constant.

6 postulatesAtoms are indestructibleSame element = identical atomsFixed atom ratio in compounds
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Postulates of Dalton's Atomic HypothesisDalton's six postulates: each element is composed of extremely small particles called atoms that participate in chemical combination; all atoms of a given element are identical but differ from atoms of other elements; atoms of different elements possess different properties including different masses; atoms are neither created nor destroyed in chemical reactions; atoms of elements combine to form molecules in compounds; in a given compound the relative number and kinds of atoms are constant. These postulates explain the law of conservation of mass and the law of definite proportions directly.

4) Laws of Chemical Combination

Four fundamental laws governing chemical reactions. Law of conservation of mass: total mass of reactants equals total mass of products. Law of definite proportions: a given compound always contains the same elements in the same proportion by weight regardless of source. Law of multiple proportions: when two elements form more than one compound, the masses of one element combining with a fixed mass of the other are in a ratio of small whole numbers. Law of equivalence: at the endpoint of titration, equivalents of all reactants and products are equal.

Mass conserved in reactionsFixed composition by massSmall whole number ratiosEquivalents equal at endpoint
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Law of Conservation of MassIn any chemical reaction, mass is neither created nor destroyed. The total mass of reactants equals the total mass of products. This law was established by Lavoisier and forms the basis for balancing chemical equations. Every balanced equation must account for all atoms on both sides.
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Law of Definite ProportionsA given compound always contains exactly the same proportion of elements by weight, regardless of its source or method of preparation. For example, water always contains hydrogen and oxygen in the mass ratio 1:8 whether synthesised in the laboratory or obtained from a natural source.
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Law of Multiple ProportionsWhen two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers. For example, carbon forms CO and CO2 with oxygen: for a fixed mass of carbon, the oxygen masses are in the ratio 1:2.
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Law of EquivalenceThe fundamental law governing all titrations. At the endpoint of a titration, the number of equivalents (or milliequivalents) of one reactant equals the number of equivalents of every other reactant and every product. For A + B + C giving D + E + F, the equivalents of A, B, C, D, E, and F are all equal to each other, irrespective of the molar ratio. The advantage: a balanced chemical equation is not needed when working with equivalents.

5) Mole Concept

The mole is the SI base unit for amount of a chemical species, defined as Avogadro's number (6.022 x 10^23) of particles. Mole = mass in grams / molar mass = number of particles / Avogadro's number. At STP (273 K, 1 atm), one mole of an ideal gas occupies 22.4 L. Equivalent mass is defined as the mass that combines with or displaces 1.008 parts of hydrogen, 8.0 parts of oxygen, or 35.5 parts of chlorine. Number of equivalents = number of moles times n-factor.

1 mol = 6.022 x 10^23 particlesMoles = mass / molar mass1 mol gas = 22.4 L at STPEquivalents = moles x n-factor
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Mole and Avogadro's NumberThe mole (mol) is the SI base unit for amount of substance. One mole equals 6.022 x 10^23 particles (atoms, molecules, ions, or formula units) represented by the chemical formula. Molar mass is the mass in grams of one mole. Moles of a substance = mass in grams / molar mass = number of particles / Avogadro's number. At STP (273 K, 1 atm), the volume of one mole of an ideal gas is 22.4 L. These three interconversion routes (mass, particle count, gas volume) form the central calculation framework for all stoichiometry.
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Equivalent Mass and Gram EquivalentEquivalent mass (EM) is the number of parts by mass that combines with or displaces 1.008 parts of hydrogen, 8.0 parts of oxygen, or 35.5 parts of chlorine. Gram equivalent mass (GEM) is the equivalent mass expressed in grams. Number of GEM = mass in grams / GEM. Number of equivalents = number of moles times n-factor. The equivalent concept simplifies titration calculations because at any reaction endpoint, equivalents of all species are equal, eliminating the need for a balanced equation.

6) Stoichiometry

Quantitative relationships in chemical reactions. Gravimetric analysis (Stoichiometry-I) uses mass-mass, mass-volume, and volume-volume relationships from balanced equations. Volumetric analysis (Stoichiometry-II) uses standard solutions and titrations. The n-factor determines equivalent weight for different species: for acids it equals basicity, for bases it equals acidity, for salts it depends on the total charge on cation or anion or the electron transfer in redox. Limiting reagent is the reactant consumed first, determining maximum product yield.

Gravimetric: mass ratios from equationVolumetric: N1V1 = N2V2n-factor governs equivalentsLimiting reagent caps product
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Calculation of n FactorThe n-factor determines how many equivalents one mole of a substance provides. For elements: n-factor = valency. For acids: n-factor = number of replaceable H+ ions (basicity), e.g., HCl (1), H2SO4 (2), H3PO4 (3), H3PO3 (2). For bases: n-factor = number of replaceable OH- ions (acidity), e.g., NaOH (1), Ba(OH)2 (2). For salts in non-redox reactions: n-factor = total positive or negative charge. In redox reactions: n-factor = total electrons lost or gained per formula unit. For partial redox (e.g., HCl with KMnO4 where only 10 of 16 Cl atoms are oxidised), n-factor = 5/8 per HCl. For disproportionation (e.g., H2O2 decomposing where one mole is oxidised and one reduced), n-factor equals the electrons exchanged per mole of reacting substance.
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Limiting Reagent and Yield of ProductWhen reactants are not in stoichiometric proportion, one is consumed first and limits the product formed. This is the limiting reagent. To identify it: calculate moles of each reactant, divide by the stoichiometric coefficient, and the smallest quotient identifies the limiting reagent. Percentage yield = (actual yield / theoretical maximum yield) times 100. The theoretical yield is calculated from the limiting reagent.
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Chemical Stoichiometry - Gravimetric Analysis (Stoichiometry-I)Calculations based on balanced chemical equations using three relationship types: mass-mass (convert mass of one substance to mass of another via molar ratios), mass-volume (convert between mass and gas volume at STP), and volume-volume (gas volume ratios at same T and P equal molar ratios). A balanced equation gives the molar ratio of reactants and products, the mass ratio, and the volume ratio for gaseous species.
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Volumetric Analysis (Stoichiometry-II)Quantitative determination through volume measurements using standard solutions. Key formula: N1V1 = N2V2 at endpoint. Types include redox titrations (e.g., KMnO4 vs FeSO4, K2Cr2O7 vs FeSO4), acid-base titrations, and precipitation titrations. Normality = molarity times n-factor. Strength = normality times equivalent weight = molarity times molecular weight. Number of milliequivalents = normality times volume in mL.
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Some Important General Reactions Used in StoichiometryThermal decomposition patterns of salts essential for stoichiometric calculations. Carbonates: Na2CO3 and K2CO3 are thermally stable; other metal carbonates decompose to oxide + CO2; all bicarbonates decompose to carbonate + CO2 + H2O. Nitrates: NH4NO3 gives N2O + H2O; NaNO3/KNO3 give nitrite + O2; heavy metal nitrates give oxide + NO2 + O2. Sulphates: Na2SO4 and K2SO4 are stable; others decompose to oxide + SO3. Hydrated halides on heating give oxide + HX + H2O. These patterns are tested as stoichiometric calculations.

7) Oxidation State and Redox Reactions

Oxidation is loss of electrons; reduction is gain of electrons. Oxidation number is assigned by seven rules (free element = 0, monoatomic ion = its charge, alkali metals +1, alkaline earths +2, hydrogen +1 except hydrides -1, fluorine always -1, oxygen -2 except peroxides -1 and superoxides -1/2). Types of redox reactions: combination, decomposition, displacement (metal and non-metal), and disproportionation. Two balancing methods: oxidation number method and ion-electron (half-equation) method.

Oxidation = electron loss7 rules for oxidation number4 types of redox reactionsTwo balancing methods
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Oxidation and ReductionOxidation is the loss of electrons by a reactant; reduction is the gain of electrons by another reactant. These always occur simultaneously in redox reactions. The substance that loses electrons is the reducing agent (gets oxidised). The substance that gains electrons is the oxidising agent (gets reduced). The total electrons lost by one substance always equals the total electrons gained by the other.
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Oxidation Number and Oxidation StateOxidation number indicates electrons lost, gained, or shared when an element goes from free state to a compound. Seven rules for assignment: (1) free element = 0, (2) monoatomic ion = its charge, (3) alkali metals = +1 and alkaline earths = +2 in compounds, (4) hydrogen = +1 except -1 in hydrides, (5) fluorine = -1 always, (6) oxygen = -2 except -1 in peroxides and -1/2 in superoxides, (7) algebraic sum of oxidation numbers equals net charge on the species. Oxidation state refers to charge per atom; oxidation number is total charge on all atoms of that kind.
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Types of Redox ReactionsFour categories: (1) Combination reactions where two or more substances form a single product with at least one element in elemental form. (2) Decomposition reactions where a compound breaks into two or more products with at least one in elemental form. (3) Displacement reactions: metal displacement (more reactive metal displaces less reactive from its compound) and non-metal displacement (metals displacing hydrogen from water, steam, or acid). (4) Disproportionation reactions where the same element is simultaneously oxidised and reduced.
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Balancing Redox Reactions Via Oxidation NumbersFive-step method: (1) write the skeletal equation, (2) assign oxidation numbers and identify atoms undergoing oxidation and reduction, (3) find the change in oxidation number and equalise by multiplying with suitable integers, (4) complete balancing by inspection, balancing atoms that changed oxidation state first, then other atoms, then H and O using H2O, (5) for ionic equations, ensure net charges balance using H+ in acidic medium or OH- in basic medium.
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Balancing Redox Reactions Via Ion-Electron (Half-Equation) MethodSix-step method: (1) write two separate half-equations for oxidation and reduction, (2) balance atoms being oxidised or reduced, (3) add H+ (acidic) or OH- (basic) to balance oxygen atoms via water molecules, (4) add electrons to balance charge on both sides, (5) multiply each half-equation by the smallest integer to equalise electrons, (6) add both equations and cancel common species. Example: 3Cu + 8H+ + 2NO3- gives 3Cu2+ + 2NO + 4H2O.
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Volume Strength of H2O2 and Percentage Labelling of OleumH2O2 labelled as x volumes means 1 mL of that solution liberates x mL of O2 at STP on complete decomposition. The relation: volume strength = 5.6 times normality. Since n-factor of H2O2 = 2, normality = 2 times molarity, so volume strength = 11.2 times molarity. For oleum labelled as y%: 100 g of oleum on dilution with water produces y g of H2SO4. Oleum contains H2SO4 and SO3; on adding water, SO3 + H2O gives H2SO4, increasing total H2SO4 mass. Percentage label = mass of H2SO4 initially present + mass of H2SO4 produced on dilution.

8) SI Units and Measurement

Seven base SI units: metre (length), kilogram (mass), second (time), kelvin (temperature), ampere (current), candela (luminous intensity), mole (amount). Standard prefixes from yocto (10^-24) to yotta (10^24). Key conversion factors: 1 atm = 101325 Pa, 1 eV = 1.602 x 10^-19 J, 1 cal = 4.184 J, 1 amu = 931.5016 MeV, 1 Angstrom = 10^-10 m.

7 base SI unitsPrefixes: nano to giga1 atm = 101325 Pa1 eV = 1.602 x 10^-19 J
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Basic SI UnitsSeven fundamental SI units form the measurement backbone of chemistry: metre (m) for length, kilogram (kg) for mass, second (s) for time, kelvin (K) for temperature, ampere (A) for electric current, candela (cd) for luminous intensity, and mole (mol) for amount of substance. Derived units include newton (kg m s^-2), pascal (N m^-2), and joule (kg m^2 s^-2).
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Standard Prefixes and Conversion FactorsPrefixes scale base units from 10^-24 (yocto) to 10^24 (yotta). Most relevant for NEET: nano (10^-9), micro (10^-6), milli (10^-3), kilo (10^3), mega (10^6). Critical conversion factors: 1 litre = 1000 mL; 1 atm = 101325 Pa; 1 bar = 10^5 Pa; 1 cal = 4.184 J; 1 eV = 1.602 x 10^-19 J; 1 eV/atom = 96.5 kJ/mol; 1 amu = 931.5016 MeV; 1 Angstrom = 10^-10 m; 1 nm = 10^-9 m; V.D. times 2 = molecular weight (gases only).

Some Basic Concepts of Chemistry and Redox Reactions Download Notes & Weightage Plan

For each topic in the Some Basic Concepts of Chemistry and Redox Reactions chapter below, you get (2) the exact resources to download and how to use them, and (3) a simple importance & time plan so NEET students know what to do first and what to revise last.

2 Downloads

Matter and Its Classification

Physical and chemical classification of matter. Foundation for understanding pure substances, mixtures, and states of matter.

Conceptual recall onlyGas vs vapour testedPlasma/BEC: rareQuick 1-mark pickup

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Physical classification: solids (close, orderly, no movement freedom), liquids (close, orderly, mobile), gases (far apart, fast movement). Three states are interconvertible. Gas vs vapour: vapour is gaseous form of a substance that is liquid at room temperature. Chemical classification: elements, compounds, mixtures (homogeneous and heterogeneous). Two additional states: plasma and Bose-Einstein condensate under extreme conditions.
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Draw a two-branch tree diagram: Physical (solid, liquid, gas) and Chemical (element, compound, mixture). Annotate each with one defining characteristic. Know the gas vs vapour distinction cold.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions0-1Rarely tested as a standalone NEET question. When it appears, the question tests the distinction between gas and vapour or asks to classify a substance.
Time Required0.5 hrs15 min reading; 15 min memorising classification trees and the gas vs vapour distinction.
DifficultyEasyPure conceptual recall. No calculations involved.
  • Scoring Focus: Gas vs vapour distinction. This is the only point from this topic that appears in NEET options.
  • High-risk Area: Confusing gas and vapour. Ammonia is a gas, not a vapour, because it exists as a gas at room temperature.
  • Best Practice Style: If the substance is liquid at room temperature, its gaseous form is vapour. If it is already gaseous at room temperature, it is a gas.
Priority rule: Low priority. Spend 30 minutes then move to mole concept and stoichiometry.

Atomic and Molecular Masses

Defines amu, atomic mass, molecular mass, GAM, GMM, and their interrelationships. The calculation backbone for all mole concept work.

1 amu = 1.6604 x 10^-24 gGAM and GMM definitionsMass of 1 atom formulaAtomicity link

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Atomic mass: average relative mass of atoms vs 1/12 of carbon-12. Unit: amu = 1.6604 x 10^-24 g. GAM = atomic mass in grams = 1 mole of atoms. Number of gram atoms = mass/GAM. Number of atoms in 1 g = 6.02 x 10^23 / atomic mass. Mass of one atom = GAM / 6.02 x 10^23. Molecular mass: how many times heavier a molecule is vs 1/12 of C-12. GMM = molecular mass in grams = 1 mole of molecules. Atomicity links atomic and molecular mass for gaseous elements.
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Write the four GAM relationships and four GMM relationships on one card. Solve 3 quick problems: mass of one atom of Fe, number of atoms in 10 g of Ca, molecular mass of O2 from atomic mass and atomicity.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions0-1Typically combined with mole concept questions rather than standalone. May ask for mass of one atom or molecule.
Time Required1 hr30 min on definitions and formulas; 30 min on practice calculations.
DifficultyEasyStraightforward definitions and arithmetic. The only subtlety is atomicity for gaseous elements.
  • Scoring Focus: Mass of one atom = GAM / Avogadro's number. This calculation is tested directly.
  • High-risk Area: Forgetting atomicity when converting between atomic mass and molecular mass for gaseous elements. O2 has atomicity 2, so molecular mass = 2 times atomic mass.
  • Best Practice Style: Always check: is the question asking about atoms or molecules? Use GAM for atoms and GMM for molecules.
Priority rule: Medium priority. Quick topic that feeds into the more important mole concept calculations.

Dalton's Atomic Hypothesis

Six postulates providing the theoretical basis for laws of chemical combination.

6 postulatesExplains conservation lawExplains definite proportionsConceptual only

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Six postulates: (1) elements composed of atoms, (2) atoms of same element are identical, (3) different elements have different atoms, (4) atoms are indestructible in reactions, (5) atoms combine to form molecules in compounds, (6) fixed relative number and kind of atoms in a compound. Postulate 4 explains law of conservation of mass. Postulate 6 explains law of definite proportions.
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: List the 6 postulates and map each to the law it explains. Postulate 4 maps to conservation of mass, postulate 6 to definite proportions.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions0Almost never tested directly in NEET. Background knowledge only.
Time Required0.5 hrsQuick read-through. No calculations.
DifficultyEasyPure recall of postulates.
  • Scoring Focus: Know which postulate explains which law. This is the only examinable angle.
  • High-risk Area: None significant. This topic is almost never tested.
  • Best Practice Style: Read once, connect postulates to laws, move on.
Priority rule: Lowest priority. Spend minimal time here.

Laws of Chemical Combination

Four laws that govern how substances combine: conservation of mass, definite proportions, multiple proportions, and equivalence.

Conservation: mass in = mass outDefinite: fixed % by weightMultiple: small whole ratiosEquivalence: the titration law

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Law of conservation of mass: total mass of reactants = total mass of products. Law of definite proportions: a compound has a fixed composition by weight regardless of source. Law of multiple proportions: masses of one element combining with fixed mass of another are in small whole number ratios. Law of equivalence: at endpoint of titration, equivalents of all species are equal. The equivalence law eliminates the need for a balanced equation in titration calculations.
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Make a four-row table: Law | Statement | Example. Practise one numerical on multiple proportions (e.g., CO and CO2 oxygen ratios). Know that law of equivalence is the basis for N1V1 = N2V2.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions0-1Occasionally tested as a conceptual MCQ asking to identify which law a given example illustrates.
Time Required1 hr30 min on law statements with examples; 30 min on MCQ practice.
DifficultyEasyConceptual recall with simple arithmetic for multiple proportions examples.
  • Scoring Focus: Law of equivalence: at endpoint, meq of acid = meq of base. This feeds directly into volumetric analysis.
  • High-risk Area: Confusing law of definite proportions with law of multiple proportions. Definite proportions applies to one compound; multiple proportions applies to two or more compounds of the same elements.
  • Best Practice Style: Definite = one compound, fixed ratio. Multiple = two or more compounds, small whole number ratio between them.
Priority rule: Low priority. Quick conceptual topic.

Mole Concept

Central calculation framework: moles, Avogadro's number, molar mass, equivalent mass, and n-factor. The backbone of all NEET chemistry numericals.

Moles = mass / mol mass1 mol = 6.022 x 10^2322.4 L at STPEquivalents = moles x n

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Mole = SI unit for amount. 1 mole = 6.022 x 10^23 particles. Molar mass = mass of 1 mole in grams. Moles = mass/molar mass = particles/Avogadro's number = volume at STP/22.4 L. Equivalent mass = mass combining with 1.008 H or 8 O or 35.5 Cl. GEM = equivalent mass in grams. Number of equivalents = mass/GEM = moles x n-factor. Normality = number of equivalents per litre = molarity x n-factor.
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Draw the mole concept triangle: mass at top, moles at centre, particles on left, gas volume on right. Write the conversion formula on each arrow. Solve 5 problems covering all three conversions. Memorise: Normality = Molarity x n-factor.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1-2One to two questions per NEET paper. Formats: calculate moles from mass, number of atoms in a given mass, convert between normality and molarity, or find equivalents.
Time Required2 hrs45 min on mole concept with Avogadro's number; 45 min on equivalent mass and n-factor; 30 min on MCQ practice.
DifficultyModerateThe formulas are simple but keeping track of units (grams, moles, equivalents, litres) and choosing the correct n-factor for a given reaction requires careful attention.
  • Scoring Focus: Three key conversions: mass to moles, moles to particles, moles to volume at STP. Plus: Normality = Molarity x n-factor. These four relationships solve most NEET problems from this topic.
  • High-risk Area: Using the wrong n-factor. The n-factor of H3PO3 is 2 (not 3) because only two H atoms are replaceable. H3PO4 has n-factor 3.
  • Best Practice Style: For every calculation, write the formula with units first, substitute, then solve. Circle the unit of the answer to confirm it matches what was asked.
Priority rule: Highest priority. This topic is the foundation for all quantitative chemistry. Master it before moving to stoichiometry.

Stoichiometry

Gravimetric and volumetric calculations from balanced equations. Includes n-factor for different reaction types, limiting reagent, and titration formulas.

Gravimetric: 3 relationship typesN1V1 = N2V2n-factor: partial redox fractionLimiting reagent: smallest quotient

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

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Topic Notes (Condensed)Gravimetric analysis uses mass-mass, mass-volume, and volume-volume relationships from balanced equations. Volumetric analysis: N1V1 = N2V2 at endpoint. Key formulas: Normality = strength/eq wt = molarity x n-factor. Milliequivalents = normality x volume in mL. n-factor for acids = basicity, for bases = acidity, for salts in non-redox = total cation/anion charge. In redox: n-factor = electrons transferred per formula unit. Partial redox example: HCl with KMnO4, n-factor of HCl = 5/8 because only 10 of 16 Cl atoms are oxidised. Disproportionation: n-factor of H2O2 = 1 (one mole oxidised, one reduced, each exchanging 1 electron). Limiting reagent: divide moles by stoichiometric coefficient; smallest value determines the limiting reagent.
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Make a decision tree for n-factor: Is it acid/base? Use basicity/acidity. Is it a non-redox salt? Use total charge. Is it redox? Count electrons per mole. Is it partial redox? Need balanced equation first. Solve one problem of each type.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1-2One to two questions per paper. Formats: find the volume of titrant needed, calculate mass of product from limiting reagent, or determine n-factor in a specific reaction.
Time Required3 hrs1 hr on gravimetric problems; 1 hr on volumetric problems with normality; 1 hr on n-factor calculation (including partial redox and disproportionation).
DifficultyModerate-HardThe main challenge is n-factor for partial redox and disproportionation reactions. These require a balanced equation and careful counting of electrons exchanged.
  • Scoring Focus: N1V1 = N2V2 for titration problems. Limiting reagent identification for product yield problems. n-factor for KMnO4 in acidic medium (5), K2Cr2O7 (6), and HCl in the KMnO4 reaction (5/8).
  • High-risk Area: Using a generic n-factor without checking the specific reaction. The n-factor of the same substance can be different in different reactions. H2O2 has n-factor 2 as an oxidising agent (reduces to H2O) but n-factor 1 in disproportionation.
  • Best Practice Style: Always write the specific reaction first, then determine n-factor from that reaction. Never memorise a single n-factor for a substance across all reactions.
Priority rule: Highest priority alongside mole concept. These two topics together account for the majority of marks from this chapter.

Oxidation State and Redox Reactions

Assignment of oxidation numbers, types of redox reactions, and two systematic balancing methods.

7 rules for ox number4 redox reaction typesOx-number balancing: 5 stepsIon-electron: 6 steps

1) Download Packs For This Topic (And How To Use Them)

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Topic Notes (Condensed)Seven rules for oxidation number: free element = 0; monoatomic ion = charge; alkali metals +1; alkaline earths +2; H = +1 (except -1 in hydrides); F always -1; O = -2 (except -1 in peroxides, -1/2 in superoxides); sum of ox numbers = net charge. Four redox types: combination, decomposition, displacement (metal and non-metal), disproportionation. Oxidation number method: 5 steps (skeletal equation, assign ox numbers, equalise changes, balance by inspection, charge balance with H+/OH-). Ion-electron method: 6 steps (separate half-equations, balance atoms, add H+/OH- for O, add electrons for charge, equalise electrons, combine).
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Practise oxidation number assignment on 10 species including tricky ones (HN3, HCN, FeC2O4, KClO3). Then balance one reaction by each method. The ion-electron method is faster for ionic equations in solution.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1One question per paper: find oxidation state of an element in a given ion or molecule, or identify the type of redox reaction.
Time Required2.5 hrs1 hr on oxidation number rules with examples; 45 min on types of redox reactions; 45 min on balancing practice.
DifficultyModerateOxidation number assignment is systematic but requires attention for unusual species. Balancing is procedural but time-consuming if not practised.
  • Scoring Focus: Oxidation number of N in HN3 is -1/3: this type of fractional oxidation state question is a NEET favourite. Also: identifying disproportionation reactions where the same element is both oxidised and reduced.
  • High-risk Area: Forgetting that hydrogen is -1 in metal hydrides (LiH, NaH, LiAlH4). Using -2 for oxygen in peroxides (it should be -1) or superoxides (-1/2).
  • Best Practice Style: Apply the seven rules in order of priority: fluorine first, then oxygen, then hydrogen, then algebraic sum. The highest-priority rule overrides lower ones.
Priority rule: High priority. Oxidation state determination is tested almost every year. Master the seven rules and practise on 15 to 20 different species.

SI Units and Measurement

Seven base SI units, standard prefixes, and conversion factors used across chemistry calculations.

7 base units: m, kg, s, K, A, cd, molKey prefixes memorisedConversion factors on one cardVD x 2 = mol mass (gases)

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Seven SI base units: metre, kilogram, second, kelvin, ampere, candela, mole. Prefixes: nano (10^-9), micro (10^-6), milli (10^-3), kilo (10^3), mega (10^6), giga (10^9). Key conversions: 1 atm = 101325 Pa; 1 bar = 10^5 N/m^2; 1 cal = 4.184 J; 1 eV = 1.602 x 10^-19 J; 1 eV/atom = 96.5 kJ/mol; 1 amu = 931.5016 MeV; 1 Angstrom = 10^-10 m; 1 nm = 10^-9 m; 1 litre atm = 101.3 J; 1 debye = 10^-18 esu cm; Vapour density x 2 = molecular weight (gases).
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Write all seven base units and the 10 most-used conversion factors on one card. The conversions asked in NEET are primarily 1 atm in Pa, 1 eV in J, and prefix conversions.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions0-1Rarely tested as a standalone question. Conversion factors appear embedded within numericals from other topics.
Time Required0.5 hrsQuick memorisation of units and conversion factors.
DifficultyEasyPure memorisation. No conceptual difficulty.
  • Scoring Focus: Know 1 atm = 101325 Pa, 1 eV = 1.602 x 10^-19 J, and VD x 2 = molecular weight. These appear in gas law and thermodynamics problems.
  • High-risk Area: Confusing bar and atm. 1 atm = 101325 Pa but 1 bar = 100000 Pa. They are close but not equal.
  • Best Practice Style: Keep a one-page conversion table. Refer to it during problem solving until the key values are memorised.
Priority rule: Low priority. Memorise the top 5 conversions and move on.

Some Basic Concepts of Chemistry and Redox Reactions Chapter NEET Traps & Common Mistakes (Topic-Wise)

Each subtopic below is of the Some Basic Concepts of Chemistry and Redox Reactions chapter and shows what NEET students usually do wrong in NEET examination, a short example of the mistake, and how NEET frames the question to trick you with close options are given below.

! Avoid Easy Negatives
n-Factor in Partial Redox and Disproportionation
NEETn-factorPartial redoxDisproportionationEquivalents

Mistake Snapshot (What Students Do Wrong)

  • Using whole-number n-factor for partial redox reactions: In the reaction 2KMnO4 + 16HCl, only 10 out of 16 Cl atoms undergo oxidation. The n-factor of HCl is therefore 10/16 = 5/8, not 1 or 2. NEET provides whole-number answers as distractors.
  • Assuming n-factor is the same for a substance in all reactions: H2O2 has n-factor 2 when acting as an oxidising agent (O goes from -1 to -2) but n-factor 1 in its disproportionation (one mole oxidised, one reduced). Students who memorise a single n-factor get the wrong equivalent weight.
2–3 Line Example (Typical Error)

In 2KMnO4 + 16HCl gives 2KCl + 2MnCl2 + 5Cl2 + 8H2O, the 10 electrons lost come from 16 moles of HCl. n-factor of HCl = 10/16 = 5/8. A student using n-factor = 1 calculates equivalent weight of HCl as 36.5, but the correct value here is 36.5 / (5/8) = 58.4, giving a completely different normality.

How NEET Frames The Trap

NEET identifies the reaction and asks for equivalent weight or normality. The student must write the balanced equation, count electrons actually transferred, and divide by total moles of the substance to get the fractional n-factor.

NEET-Style Trap Question Format

Q. In the reaction 2KMnO4 + 16HCl gives 2KCl + 2MnCl2 + 5Cl2 + 8H2O, the n-factor of HCl is:
A. 5/8   B. 1   C. 2   D. 5/16  
Trick: Total electrons lost = 10 (from 10 Cl- oxidised to Cl2). Total moles of HCl = 16. n-factor = 10/16 = 5/8 (Option A). Option B (1) assumes each HCl loses 1 electron. Option C (2) confuses with H2O2 n-factor. Option D (5/16) divides by wrong number.

Quick rule: For partial redox: n-factor = total electrons exchanged divided by total moles of the substance in the balanced equation. Always write the balanced equation first.
Mole Concept Unit Confusions
NEETMole conceptMoles vs moleculesGAM vs GMM

Mistake Snapshot (What Students Do Wrong)

  • Confusing number of moles with number of molecules: 3 moles of H2O is 3 times 6.022 x 10^23 = 1.8066 x 10^24 molecules, not 3 molecules. Students sometimes report moles as the number of particles.
  • Forgetting to account for atomicity when counting atoms: 1 mole of O2 contains 6.022 x 10^23 molecules but 2 times 6.022 x 10^23 = 1.2044 x 10^24 atoms of O. NEET asks for number of atoms, and students who forget atomicity give the molecule count.
2–3 Line Example (Typical Error)

How many oxygen atoms are in 1 mole of O2? Answer: 2 x 6.022 x 10^23 = 1.2044 x 10^24 atoms. A student who forgets atomicity writes 6.022 x 10^23, which is the number of molecules, not atoms. NEET places this as a distractor.

How NEET Frames The Trap

NEET asks for total number of atoms (not molecules) in a given mass of a molecular substance. Students who divide mass by molecular mass get moles of molecules, then must multiply by atomicity to get atoms.

NEET-Style Trap Question Format

Q. The total number of atoms in 0.5 mole of O3 is:
A. 9.033 x 10^23   B. 3.011 x 10^23   C. 1.806 x 10^24   D. 6.022 x 10^23  
Trick: 0.5 mol O3 = 0.5 x 6.022 x 10^23 = 3.011 x 10^23 molecules. Each O3 has 3 atoms. Total atoms = 3 x 3.011 x 10^23 = 9.033 x 10^23 (Option A). Option B gives molecules, not atoms. Option D forgets the 0.5 factor. Option C doubles instead of tripling.

Quick rule: NEET question says atoms? Multiply moles by Avogadro's number AND by atomicity. Molecules? Just moles times Avogadro's number.
Oxidation Number Assignment in Unusual Species
NEETOxidation numberFractional oxidation statePeroxides

Mistake Snapshot (What Students Do Wrong)

  • Using oxygen as -2 in peroxides and superoxides: In H2O2 and Na2O2 (peroxides), oxygen has oxidation state -1, not -2. In KO2 (superoxide), oxygen is -1/2. Using -2 gives wrong oxidation numbers for other elements in these compounds.
  • Not recognising fractional oxidation states: In HN3 (hydrazoic acid), the average oxidation state of nitrogen is -1/3. Many students assume oxidation states must be integers and reject fractional values, leading to wrong answers.
2–3 Line Example (Typical Error)

Oxidation state of N in HN3: H is +1, so 3x + 1 = 0, giving x = -1/3 per nitrogen atom. Students who expect an integer answer may choose -1 (wrong) or +1 (wrong). The oxidation number of all three N atoms together is -1, but the oxidation state per atom is -1/3.

How NEET Frames The Trap

NEET gives a compound with a fractional oxidation state (HN3, Fe3O4, Pb3O4) and provides only integer options plus the correct fractional option. Students who reject fractions pick the wrong integer.

NEET-Style Trap Question Format

Q. The oxidation state of nitrogen per atom in HN3 is:
A. -1/3   B. -1   C. +1/3   D. 0  
Trick: H = +1, so 1 + 3(x) = 0, x = -1/3 (Option A). Option B (-1) is the total oxidation number of all three N atoms, not per atom. Option C reverses the sign. Option D ignores hydrogen.

Quick rule: Oxidation state CAN be fractional. Apply the sum rule: sum of all oxidation numbers = net charge on the species. Divide by the number of atoms of that element to get the per-atom value.
Equivalent Mass and Normality Interconversion
NEETNormalityMolarityn-factorTitration

Mistake Snapshot (What Students Do Wrong)

  • Using the wrong n-factor for H3PO3 (phosphorous acid): H3PO3 has only 2 replaceable hydrogen atoms (basicity = 2), not 3. Its n-factor is 2. Students who count all three H atoms assign n-factor = 3, giving wrong normality.
  • Confusing normality and molarity in dilution calculations: M1V1 = M2V2 works only for molarity. For normality: N1V1 = N2V2. Mixing these up when the solution involves an acid with n-factor greater than 1 gives wrong concentration after dilution.
2–3 Line Example (Typical Error)

0.1 M H3PO3 solution: Normality = 0.1 x 2 = 0.2 N (n-factor = 2 because only 2 H atoms are ionisable). A student using n-factor = 3 calculates 0.3 N, which is the distractor. H3PO3 is a dibasic acid, not tribasic, because one H is directly bonded to P and is non-ionisable.

How NEET Frames The Trap

NEET gives molarity of phosphorous acid and asks for normality or equivalent weight. The trap relies on students assuming all 3 H atoms are acidic.

NEET-Style Trap Question Format

Q. The normality of 0.2 M H3PO3 solution is:
A. 0.4 N   B. 0.6 N   C. 0.2 N   D. 0.8 N  
Trick: H3PO3 is dibasic (n-factor = 2, not 3). N = M x n = 0.2 x 2 = 0.4 N (Option A). Option B (0.6) uses n = 3. Option C assumes n = 1. Option D uses n = 4.

Quick rule: H3PO3 is dibasic (n = 2). H3PO4 is tribasic (n = 3). The difference is that H3PO3 has one P-H bond (non-ionisable H), while all three H atoms in H3PO4 are on oxygen (all ionisable).
Volume Strength of H2O2 and Oleum Labelling
NEETH2O2Volume strengthOleumPercentage labelling

Mistake Snapshot (What Students Do Wrong)

  • Confusing volume strength formula with concentration: Volume strength = 5.6 x Normality = 11.2 x Molarity. Students who use 5.6 x Molarity get half the correct answer. The n-factor of H2O2 is 2, so normality = 2 x molarity.
  • Not understanding what oleum percentage means: 109% oleum means 100 g of oleum produces 109 g of H2SO4 on dilution. Students assume 109% means 109 g of H2SO4 is already present in 100 g of sample, ignoring that the extra 9 g comes from SO3 reacting with water.
2–3 Line Example (Typical Error)

10V H2O2 solution: Volume strength = 5.6 x N, so N = 10/5.6 = 1.786 N. Molarity = N/2 = 0.893 M. A student using 5.6 x M = 10 gets M = 1.786, which is actually the normality, not molarity. This doubles the true molar concentration.

How NEET Frames The Trap

NEET gives volume strength and asks for molarity. If the student divides by 5.6, they get normality, not molarity, and must divide by 2 again.

NEET-Style Trap Question Format

Q. The molarity of a 11.2V H2O2 solution is:
A. 1 M   B. 2 M   C. 0.5 M   D. 5.6 M  
Trick: Volume strength = 5.6 x N, so N = 11.2/5.6 = 2. Molarity = N/n-factor = 2/2 = 1 M (Option A). Option B (2 M) gives normality, not molarity. Option C halves incorrectly. Option D confuses 5.6 with molarity.

Quick rule: Volume strength to Normality: divide by 5.6. Normality to Molarity: divide by 2 (since n-factor of H2O2 = 2). Shortcut: Molarity = volume strength / 11.2.

Topics

Matter and Its Classification

Atomic and Molecular Masses

Dalton's Atomic Hypothesis

Laws of Chemical Combination

Mole Concept

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Matter and Its Classification

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