NEET Chemistry · NCERT Class 11

Some Basic Concepts of Chemistry notes

Nature of Matter, Classification, Properties and Changes, Measurement, SI Units, Scientific Notation and Significant Figures, Laws of Chemical Combination, Atomic, Molecular and Formula Masses, Mole Concept, Molar Mass and Avogadro Constant, Percentage Composition, Empirical Formula and Molecular Formula, Balanced Chemical Equations and Stoichiometry, Limiting Reagent, Excess Reagent and Percentage Yield, Concentration Terms and Solution Calculations and Integrated NEET Revision, Mind Map and Numerical Strategy — explained topic by topic with key points, formulas, memory tricks and common mistakes.

  • 10 topics
  • 77 min read
  • 32 formulas
  • 40 PYQs
1

Nature of Matter, Classification, Properties and Changes

6 min

Chemistry begins with matter, which is anything that has mass and occupies space. NCERT classifies matter mainly in two ways: by physical state and by chemical composition. By physical state, matter exists as solid, liquid and gas, while by composition it is divided into pure substances and mixtures. Pure substances include elements and compounds with fixed composition, whereas mixtures contain two or more substances in variable proportions. Physical properties such as colour, density and melting point can be observed without changing composition, while chemical properties describe how a substance reacts. Physical changes alter state or appearance, but chemical changes form new substances. NEET questions often test classification, examples and the difference between physical and chemical changes.

Key points7

  • All compounds are pure substances, but all pure substances are not compounds because elements are also pure substances.
  • A mixture can be separated by physical methods such as filtration, distillation, crystallisation or chromatography.
  • A compound can be separated into elements only by chemical methods.
  • The composition of a compound is fixed, while the composition of a mixture is variable.
  • Melting of ice is a physical change; rusting of iron is a chemical change.
  • States of matter depend on intermolecular forces and kinetic energy of particles.
  • NEET often asks examples: air is a homogeneous mixture, milk is a colloidal mixture, brass is an alloy.

Memory tricks2

Pure vs Mixture Shortcut: Pure has a fixed recipe; mixture has a flexible recipe. If composition can change, think mixture.

Physical Change Clue: If only size, shape or state changes, it is usually physical. If smell, gas, heat, precipitate or colour due to reaction appears, suspect chemical change.

Examples2

Kitchen Chemistry

Dissolving salt in water forms a homogeneous mixture, while cooking food involves chemical changes because new substances and flavours form.

Daily-Life Classification

Gold jewellery is usually a mixture or alloy, pure copper is an element, water is a compound and muddy water is a heterogeneous mixture.

Common mistakes3

Calling Air a Compound

Air is not a compound because its gases are not chemically combined in fixed ratio. It is a homogeneous mixture.

Confusing Molecule and Compound

O2 is a molecule but not a compound. A compound must contain different elements chemically combined.

Assuming All Irreversible Changes Are Chemical

Some physical changes like breaking glass may be irreversible but do not produce a new substance.

Reference tables2

Quick revision7

Formulas2

Density
Density = mass / volume\text{Density }=\text{ mass }/\text{ volume}

Density is a physical property used to identify and compare substances.

Density
Mass per unit volume of a substance
mass
Amount of matter present in the sample
volume
Space occupied by the sample
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2

Measurement, SI Units, Scientific Notation and Significant Figures

8 min

Chemical calculations depend on accurate measurement of mass, volume, temperature, amount of substance and time. NCERT emphasises SI units, prefixes, scientific notation and significant figures because they prevent numerical and conceptual errors. Accuracy means closeness to the true value, while precision means closeness among repeated measurements. A value like 2.00 g is more precise than 2 g because it contains more significant figures. Scientific notation expresses very large or small values as powers of ten, such as 6.022 × 10^23. Dimensional analysis uses unit conversion factors to change one unit into another without changing the physical quantity. NEET often tests significant figure rules, unit conversions and precision versus accuracy.

Key points6

  • Accuracy and precision are different; a measurement can be precise but inaccurate.
  • Exact numbers such as counted atoms or defined conversion factors do not limit significant figures.
  • Prefixes simplify large and small units: kilo = 10^3, milli = 10^-3, micro = 10^-6, nano = 10^-9.
  • Always convert units before substituting in formulas.
  • Scientific notation avoids ambiguity in zeros, especially for numbers like 1000.
  • Dimensional analysis is a powerful NEET shortcut for unit-based numericals.

Memory tricks3

Accuracy vs Precision: Accuracy is hitting the bullseye; precision is grouping the arrows together.

Zeros Rule: Leading zeros are just placeholders; trapped zeros are important; decimal-ending zeros are deliberate and significant.

SI Prefix Ladder: King Henry Died By Drinking Chocolate Milk helps remember kilo, hecto, deca, base, deci, centi, milli.

Examples2

Significant Figure Example

2.5 × 3.42 = 8.55, but final answer is 8.6 because 2.5 has only two significant figures.

Unit Conversion Example

25 mL = 25 × 10^-3 L = 0.025 L. This conversion is essential before calculating molarity.

Common mistakes3

Rounding Too Early

Do not round intermediate steps in NEET numericals. Round only the final answer according to significant figures.

Mixing mL and L in Molarity

Molarity always uses litre of solution, not millilitre, unless converted.

Treating 0.0500 as Two Significant Figures

0.0500 has three significant figures because the two zeros after 5 are after the decimal and are significant.

Reference tables3

Quick revision8

Formulas3

Scientific Notation
Number =N×10n\text{Number }= N × 10^n

Used to express very large or very small chemical quantities compactly.

N
Coefficient between 1 and 10
n
Integer power of ten
Percentage Error
Percentage error =∣experimental value − true value∣/ true value ×100\text{Percentage error }= |\text{experimental value }-\text{ true value}| /\text{ true value }× 100

Measures how far an experimental result is from the accepted value.

experimental value
Measured value obtained in experiment
true value
Accepted or standard value
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3

Laws of Chemical Combination

7 min

The laws of chemical combination explain why substances combine in fixed, simple and predictable ratios. The law of conservation of mass says mass is neither created nor destroyed in a chemical reaction. The law of definite proportions says a pure compound always contains the same elements in the same mass ratio. The law of multiple proportions states that when two elements form more than one compound, the masses of one element combining with a fixed mass of the other are in simple whole-number ratios. Gay-Lussac’s law deals with simple volume ratios of reacting gases, while Avogadro’s law says equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. These laws directly support atomic theory, formula writing and stoichiometry.

Key points6

  • Law of conservation of mass is obeyed in ordinary chemical reactions, not nuclear reactions.
  • Law of definite proportions applies only to pure compounds, not mixtures.
  • For multiple proportions, always keep the mass of one element fixed before comparing the other.
  • Gas volume laws are valid only when gases are measured under same temperature and pressure.
  • Avogadro’s law explains why gaseous volume ratios correspond to molecular ratios.
  • Balanced chemical equations obey conservation of atoms and mass.

Memory tricks2

Laws Order Mnemonic: Conserve Definite Multiple Gas Avogadro = CDMGA: Count Definite Masses, Gases Agree.

Multiple Proportions Trick: Fix one element first, then compare the other. Without fixing one element, the ratio is meaningless.

Examples2

Water Composition

Pure water from any source contains hydrogen and oxygen in the mass ratio 1:8, illustrating definite proportions.

CO and CO2

In CO and CO2, fixed carbon mass combines with oxygen masses in the ratio 1:2, showing multiple proportions.

Common mistakes3

Applying Definite Proportions to Mixtures

Salt solution can have variable salt content, so the law of definite proportions does not apply to it.

Forgetting Same Temperature and Pressure

Gay-Lussac and Avogadro gas volume comparisons are valid only at the same temperature and pressure.

Using Grams Instead of Volumes in Gay-Lussac Law

Gay-Lussac’s law is about gaseous volumes, not masses.

Reference tables2

Quick revision7

Formulas3

Law of Conservation of Mass
Total mass of reactants = Total mass of products\text{Total mass of reactants }=\text{ Total mass of products}

Mass remains constant during a chemical reaction in a closed system.

reactants
Substances consumed in a chemical reaction
products
Substances formed in a chemical reaction
Avogadro's Law
V∝n at constant TandPV ∝ n\text{ at constant }T and P

Volume of a gas is directly proportional to the number of moles at constant temperature and pressure.

V
Volume of gas
n
Number of moles of gas
T
Temperature
P
Pressure
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4

Atomic, Molecular and Formula Masses

7 min

Atoms are extremely small, so their masses are expressed relative to the atomic mass unit, u. One atomic mass unit is defined as one-twelfth of the mass of one carbon-12 atom. Relative atomic mass is the average mass of atoms of an element compared with 1 u and accounts for natural isotopic abundance. Molecular mass is the sum of atomic masses of all atoms in a molecule, such as H2O or CO2. Formula mass is used for ionic compounds such as NaCl because they do not exist as discrete molecules; it is calculated from the empirical formula unit. These ideas connect microscopic particles with measurable gram quantities and prepare students for mole concept calculations.

Key points6

  • Atomic mass and molar mass have same number but different units: u for one particle, g mol^-1 for one mole.
  • Chlorine atomic mass is about 35.5 u because natural chlorine is a mixture of isotopes.
  • Molecular mass is used for covalent molecules; formula mass is preferred for ionic compounds.
  • For hydrates, include water of crystallisation in formula mass.
  • Always multiply atomic mass by subscript before adding.
  • Brackets in formulae multiply all atoms inside, as in Ca3(PO4)2.

Memory tricks2

Molecular Mass Rule: Subscript means multiply first, then add. Bracket means multiply everything inside.

u vs g mol^-1: One particle talks in u; one mole talks in grams per mole.

Examples2

Calcium Carbonate

Formula mass of CaCO3 = 40 + 12 + 48 = 100 u, making it useful for quick percentage composition calculations.

Carbon Dioxide

Molecular mass of CO2 = 12 + 32 = 44 u; molar mass = 44 g mol^-1.

Common mistakes3

Ignoring Brackets

In Ca3(PO4)2, oxygen atoms are 8, not 4. The outside 2 multiplies both P and O.

Using Molecular Mass for Ionic Compounds

NaCl does not exist as separate NaCl molecules in solid state; use formula mass.

Forgetting Isotopic Abundance

Average atomic mass is not a simple average unless isotopes are equally abundant.

Reference tables2

Quick revision7

Formulas3

Atomic Mass Unit
1u=1/12× mass of one atom of carbon−121 u = 1/12 ×\text{ mass of one atom of carbon}-12

Defines the relative scale used for atomic masses.

u
Atomic mass unit
carbon-12
Isotope of carbon used as standard
Average Atomic Mass
Average atomic mass =Σ(isotopic mass × fractional abundance)\text{Average atomic mass }= Σ(\text{isotopic mass }×\text{ fractional abundance})

Used when an element has naturally occurring isotopes.

isotopic mass
Mass of a particular isotope
fractional abundance
Abundance percentage divided by 100
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5

Mole Concept, Molar Mass and Avogadro Constant

9 min

The mole connects the microscopic world of atoms, molecules and ions with measurable laboratory quantities. One mole contains exactly 6.022 × 10^23 elementary entities, known as the Avogadro constant. The mass of one mole of a substance is called molar mass and is expressed in g mol^-1. For example, one mole of carbon atoms has a mass of 12 g, while one mole of water molecules has a mass of 18 g. At STP in many school-level problems, one mole of an ideal gas occupies 22.4 L. Mole concept is the most important calculation tool in this chapter because it links mass, particles, gas volume and chemical equations. NEET repeatedly tests these interconversions.

Key points6

  • Mole concept is like a dozen, but a mole counts 6.022 × 10^23 entities.
  • One mole of O atoms and one mole of O2 molecules are different: O2 contains two moles of O atoms.
  • Mass depends on molar mass, but particle count for one mole is always Avogadro constant.
  • The number of atoms in molecules requires multiplying molecules by atoms per molecule.
  • Use 22.4 L mol^-1 only when the question specifies STP or compatible ideal gas conditions.
  • Mole map questions are common in NEET and can be solved by converting everything to moles first.

Memory tricks3

Mole Triangle: Mass on top, moles and molar mass at bottom: mass = moles × molar mass.

Mole Conversion Mantra: Mass? Divide by M. Particles? Divide by NA. Gas volume at STP? Divide by 22.4.

Entity Alert: A mole counts whatever entity is written after it: atoms, molecules, ions, electrons or formula units.

Examples2

Particles in Water

9 g H2O = 9/18 = 0.5 mol molecules = 0.5 × 6.022 × 10^23 molecules.

Atoms in Oxygen Gas

16 g O2 = 0.5 mol O2 molecules. It contains 1 mol O atoms because each O2 molecule has 2 O atoms.

Common mistakes3

Confusing Oxygen Atom and Oxygen Molecule

1 mol O2 molecules contains 2 mol O atoms. Always check whether O or O2 is written.

Using 22.4 L for Liquids and Solids

22.4 L mol^-1 is for gases at STP, not for water liquid, sodium chloride solid or other condensed substances.

Not Converting Mass to Grams

Molar mass in g mol^-1 requires mass in grams unless units are consistently changed.

Reference tables2

Quick revision7

Formulas4

Moles from Mass
n=m/Mn = m / M

Converts a given mass into amount of substance.

n
Number of moles
m
Given mass in grams
M
Molar mass in g mol^-1
Particles from Moles
N=n×NAN = n × NA

Gives number of atoms, molecules, ions or formula units.

N
Number of particles
n
Number of moles
NA
Avogadro constant, 6.022 × 10^23 mol^-1
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6

Percentage Composition, Empirical Formula and Molecular Formula

8 min

Percentage composition tells the mass percentage of each element in a compound. It is calculated from the formula using atomic masses and molar mass, or experimentally from elemental analysis. Empirical formula gives the simplest whole-number ratio of atoms in a compound, while molecular formula gives the actual number of atoms in one molecule. For example, CH2O is the empirical formula of glucose, but its molecular formula is C6H12O6. To find an empirical formula from percentage data, assume 100 g sample, convert masses to moles, divide by the smallest mole value and convert to whole numbers. Molecular formula is obtained by multiplying empirical formula by a whole-number factor derived from molar mass.

Key points6

  • Empirical and molecular formulae may be the same, as in H2O and CO2.
  • Compounds with different molecular formulae can have the same empirical formula.
  • Percentage composition is useful for checking purity and determining formulae.
  • Rounding must be logical; ratios like 1.5 require multiplying all values by 2.
  • Do not multiply only one element while converting fractional ratios.
  • The molecular formula factor must be a whole number.

Memory tricks2

Empirical Formula Steps: Percent to mass, mass to mole, divide by small, multiply to whole, write the formula.

Molecular Formula Factor: Big mass divided by small formula mass tells how many small units make the big molecule.

Examples2

Glucose

Glucose has empirical formula CH2O and molecular mass 180. Empirical formula mass = 30, so n = 6 and molecular formula is C6H12O6.

Water Percentage Composition

In H2O, hydrogen percent = 2/18 × 100 = 11.1 percent and oxygen percent = 16/18 × 100 = 88.9 percent.

Common mistakes3

Using Percentage Directly as Mole Ratio

Percentage gives mass ratio, not atom ratio. Always convert each mass to moles.

Rounding 1.5 to 2

Do not round 1.5 to 2. Multiply all ratios by 2 to get whole numbers.

Forgetting to Calculate Molecular Formula

If molar mass is given, empirical formula alone is usually not the final answer.

Reference tables2

Quick revision8

Formulas3

Percentage Composition
Mass percent of element = mass of element in 1 mol compound / molar mass of compound ×100\text{Mass percent of element }=\text{ mass of element in }1\text{ mol compound }/\text{ molar mass of compound }× 100

Calculates how much of a compound’s mass is due to a particular element.

mass of element
Total mass contributed by that element in one mole of compound
molar mass
Mass of one mole of the compound
Empirical Formula Ratio
Simplest mole ratio = moles of each element / smallest mole value\text{Simplest mole ratio }=\text{ moles of each element }/\text{ smallest mole value}

Converts elemental mole amounts into simplest atomic ratio.

moles of each element
Mass of element divided by atomic mass
smallest mole value
Lowest mole value among all elements
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7

Balanced Chemical Equations and Stoichiometry

9 min

Stoichiometry is the calculation of reactants and products in a chemical reaction using a balanced chemical equation. A balanced equation obeys the law of conservation of mass because the number of atoms of each element is equal on both sides. The coefficients in a balanced equation represent mole ratios, not mass ratios. For example, in 2H2 + O2 → 2H2O, 2 moles of hydrogen react with 1 mole of oxygen to form 2 moles of water. Most NEET stoichiometry problems become easy when every given quantity is first converted into moles, then the balanced equation mole ratio is applied, and finally the answer is converted into the required unit such as grams, particles or gas volume.

Key points6

  • Balanced chemical equations are quantitative recipes of reactions.
  • A coefficient before a formula multiplies the entire formula.
  • For gases at same temperature and pressure, mole ratio equals volume ratio.
  • Mass ratio must be calculated using mole ratio and molar masses.
  • In NEET, mole ratio errors are more common than arithmetic errors.
  • The balanced equation is the bridge between reactant and product quantities.

Memory tricks2

Stoichiometry Five Steps: Balance, convert to moles, compare coefficients, calculate required moles, convert to asked unit.

Coefficient Meaning: Coefficients speak in moles, molecules and gas volumes, but not directly in grams.

Examples2

Water Formation

If 4 g H2 reacts completely with excess O2, moles H2 = 2. From 2H2 → 2H2O, moles H2O = 2, so mass water = 36 g.

CO2 from Calcium Carbonate

CaCO3 → CaO + CO2. 100 g CaCO3 gives 1 mol CO2, which is 44 g or 22.4 L at STP.

Common mistakes3

Changing Subscripts While Balancing

Changing H2O to H2O2 changes the compound. Only coefficients may be changed.

Using Unbalanced Equation

A mole ratio from an unbalanced equation gives wrong answers even if arithmetic is correct.

Using Mass Ratio as Coefficient Ratio

In 2H2 + O2 → 2H2O, the mole ratio H2:O2 is 2:1, but the mass ratio is 4:32.

Reference tables2

Quick revision7

Formulas3

Mole Ratio from Equation
moles required = moles given × coefficient required / coefficient given\text{moles required }=\text{ moles given }×\text{ coefficient required }/\text{ coefficient given}

Core formula for stoichiometric conversion between substances.

moles required
Moles of substance asked in the question
moles given
Moles of substance provided in the question
coefficient required
Balanced equation coefficient of required substance
coefficient given
Balanced equation coefficient of given substance
Mass from Stoichiometric Moles
mass = moles × molar mass\text{mass }=\text{ moles }×\text{ molar mass}

Converts calculated moles into gram mass.

mass
Mass in grams
moles
Amount of substance
molar mass
Mass of one mole in g mol^-1
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8

Limiting Reagent, Excess Reagent and Percentage Yield

8 min

In real stoichiometry problems, reactants are often not present in exact balanced-equation proportions. The reactant that gets completely consumed first is called the limiting reagent, and it determines the maximum amount of product formed. The other reactant is present in excess and remains partly unused. To identify the limiting reagent, convert each reactant into moles and compare available mole ratio with the balanced equation ratio, or calculate product possible from each reactant; the smaller product amount identifies the limiting reagent. The theoretical yield is the maximum calculated product, while actual yield is the amount obtained experimentally. Percentage yield compares actual yield with theoretical yield and is usually less than 100 percent due to losses and side reactions.

Key points6

  • Never choose limiting reagent by smaller mass alone; compare moles and coefficients.
  • If reactants are in exact stoichiometric ratio, no reactant is in excess.
  • Limiting reagent controls product formation even if present in larger mass.
  • Percentage yield cannot normally exceed 100 percent in ideal reporting; if it does, impurity or measurement error is likely.
  • Theoretical yield must be calculated from the limiting reagent.
  • Excess left can be calculated by subtracting amount consumed from amount initially present.

Memory tricks2

Limiter Rule: The smaller product wins: calculate product from each reactant, choose the smaller amount.

Yield Formula: Actual is what you got; theoretical is what you thought. Got/thought × 100.

Examples2

Hydrogen and Oxygen

For 2H2 + O2 → 2H2O, if 3 mol H2 and 1 mol O2 are present, O2 is limiting because 1 mol O2 can react with only 2 mol H2.

Percentage Yield

If theoretical yield is 10 g and actual yield is 8 g, percentage yield = 8/10 × 100 = 80 percent.

Common mistakes3

Choosing Smaller Mass as Limiting Reagent

A smaller mass may have more moles if molar mass is low. Always convert to moles.

Calculating Theoretical Yield from Excess Reagent

Theoretical yield must come from limiting reagent only.

Ignoring Balanced Coefficients

Equal moles do not always mean exact reaction; the equation may require 2:1 or 3:2 ratios.

Reference tables2

Quick revision8

Formulas3

Product Possible from Reactant
moles product = moles reactant × coefficient product / coefficient reactant\text{moles product }=\text{ moles reactant }×\text{ coefficient product }/\text{ coefficient reactant}

Calculate product from each reactant; the smaller value determines limiting reagent.

moles product
Product moles possible from a reactant
moles reactant
Available moles of the reactant
coefficient product
Balanced equation coefficient of product
coefficient reactant
Balanced equation coefficient of reactant
Percentage Yield
Percentage yield = actual yield / theoretical yield ×100\text{Percentage yield }=\text{ actual yield }/\text{ theoretical yield }× 100

Measures efficiency of a chemical preparation.

actual yield
Product obtained experimentally
theoretical yield
Maximum product calculated from limiting reagent
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9

Concentration Terms and Solution Calculations

9 min

A solution contains solute dissolved in solvent, and its concentration tells how much solute is present in a given amount of solution or solvent. NCERT introduces several concentration terms: mass percentage, volume percentage, mass by volume percentage, parts per million, mole fraction, molarity and molality. Molarity is moles of solute per litre of solution and changes with temperature because volume changes. Molality is moles of solute per kilogram of solvent and is temperature independent because mass does not change with temperature. Mole fraction compares moles of one component with total moles of all components. NEET commonly asks direct formula application, dilution, and choosing the temperature-independent concentration term.

Key points7

  • Solution mass = solute mass + solvent mass.
  • Molarity uses volume of solution, not volume of solvent.
  • Molality uses mass of solvent, not mass of solution.
  • Mole fractions of all components add up to 1.
  • ppm is used for very dilute solutions, such as pollutants in water.
  • For reactions in solution, moles = molarity × volume in litres.
  • Always convert mL to L before using molarity.

Memory tricks3

Molarity vs Molality: Molarity uses litres of solution; molality uses kilograms of solvent. Lowercase m is mass-based.

Dilution Mantra: Dilution changes volume and concentration, not moles of solute: M1V1 = M2V2.

Mole Fraction Check: All mole fractions must add to 1. If not, recheck moles.

Examples2

Molarity Calculation

If 0.5 mol NaCl is dissolved to make 2 L solution, molarity = 0.5/2 = 0.25 M.

Dilution Calculation

To prepare 500 mL of 0.1 M solution from 1 M stock: V1 = M2V2/M1 = 0.1 × 500/1 = 50 mL.

Common mistakes3

Using Solvent Volume for Molarity

Molarity uses total solution volume, not only solvent volume.

Using Solution Mass for Molality

Molality uses kg of solvent, not kg of solution.

Forgetting mL to L Conversion

250 mL is 0.250 L. Using 250 L gives an answer off by a factor of 1000.

Reference tables2

Quick revision8

Formulas5

Mass Percentage
Mass percent = mass of solute / mass of solution ×100\text{Mass percent }=\text{ mass of solute }/\text{ mass of solution }× 100

Expresses solute mass as percentage of total solution mass.

mass of solute
Mass of dissolved substance
mass of solution
Mass of solute plus solvent
Mole Fraction
χA=nA/(nA+nB+...)χA = nA / (nA + nB + ...)

Fraction of total moles contributed by component A.

χA
Mole fraction of component A
nA
Moles of component A
nB
Moles of component B
Molarity
M=n/V(L)M = n / V(L)

Moles of solute present per litre of solution.

M
Molarity in mol L^-1
n
Moles of solute
V
Volume of solution in litres
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10

Integrated NEET Revision, Mind Map and Numerical Strategy

6 min

Some Basic Concepts of Chemistry is a calculation-heavy chapter, but all questions follow a connected logic. First, classify the substance and understand whether it is an atom, molecule, ion, formula unit, compound or mixture. Next, measure correctly using SI units and significant figures. Then use atomic and molecular masses to convert given quantities into moles. Once moles are known, apply Avogadro constant, molar volume, concentration formulae or balanced equation coefficients depending on the question. For empirical formula questions, convert composition to mole ratio. For reaction problems, use stoichiometry and check limiting reagent before calculating yield. NEET rewards students who build one mole-centred map rather than memorising isolated formulas.

Key points7

  • The mole is the central hub connecting mass, particles, volume and concentration.
  • Chemical equations give mole ratios, while molar masses convert between moles and grams.
  • Avogadro constant converts between moles and number of particles.
  • Empirical formula problems are composition-to-ratio problems.
  • Stoichiometry problems are given-quantity to required-quantity problems through mole ratio.
  • Limiting reagent problems require checking all reactants, not just the one mentioned first.
  • NCERT examples and in-text questions are highly relevant for NEET basics.

Memory tricks3

Central Mole Mantra: When confused, go to moles. From moles, you can go to mass, particles, gas volume, solution concentration or stoichiometric product.

NEET Calculation Order: Read entity, balance equation, convert units, convert to moles, apply ratio, answer with units.

Final Check: Ask: Is my answer in the unit asked? Did I use the correct entity? Did I round only at the end?

Examples2

Integrated Example

For 50 mL of 0.2 M HCl reacting with NaOH, moles HCl = 0.2 × 0.050 = 0.010 mol. Use balanced equation to find required NaOH moles.

Entity Check Example

If asked for atoms in 0.5 mol CO2, first find molecules = 0.5NA, then multiply by 3 atoms per molecule to get 1.5NA atoms.

Common mistakes3

Formula Hunting Without Concept

Many questions combine two or three formulas. Build the mole map instead of memorising isolated equations.

Ignoring NCERT Language

NEET conceptual questions often use NCERT wording such as formula mass, amount of substance and significant figures.

Skipping Limiting Reagent Check

Whenever two reactants are given, check limiting reagent before calculating product.

Reference tables2

Quick revision8

Formulas3

Master Mole Formula Set
n=m/M;N=nNA; VSTP =22.4n;n=MV(L)n = m/M; N = nNA;\text{ VSTP }= 22.4n; n = MV(L)

A compact formula set covering mass, particles, gas volume and solution moles.

n
Number of moles
m
Mass in grams
M
Molar mass or molarity depending on context
N
Number of particles
NA
Avogadro constant
VSTP
Gas volume at STP
V(L)
Solution volume in litres
Stoichiometry Master Relation
given quantity → moles given → coefficient ratio → moles required → required quantity\text{given quantity }→\text{ moles given }→\text{ coefficient ratio }→\text{ moles required }→\text{ required quantity}

Universal strategy for reaction-based numerical problems.

given quantity
Mass, particles, gas volume or solution data provided
coefficient ratio
Ratio from balanced chemical equation
required quantity
Final answer requested in the question
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NEET PYQs from Some Basic Concepts of Chemistry

The latest 3 of 40 previous-year questions from this chapter.

  1. NEET 2026Set 50EasyQ1

    The numbers $17.0145$ and $21.0235$ were rounded to three figures after the decimal point. The resulting numbers, respectively, are

    • A$17.014$ and $21.023$
    • B$17.015$ and $21.023$
    • C$17.014$ and $21.024$
    • D$17.015$ and $21.024$
  2. NEET 2026Set 50EasyQ2

    The amount of carbon dioxide evolved upon complete combustion of 116 g of n-butane is (Given: atomic mass in amu H = 1, C = 12 and O = 16)

    • A$352\ \mathrm{g}$
    • B$322\ \mathrm{g}$
    • C$176\ \mathrm{g}$
    • D$362\ \mathrm{g}$
  3. NEET 2026Set 11EasyQ3

    The number of hydrogen atoms present in 5.4 g of urea is: (Given: Molar mass of urea = 60 g mol⁻¹; Nₐ = 6.022 × 10²³ particles mol⁻¹)

    • A1.084 × 10²³
    • B1.084 × 10²²
    • C2.168 × 10²²
    • D2.168 × 10²³

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