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Some Basic Concepts Of Chemistry

Nature of Matter

Physical and Chemical Properties


Base or Fundamental Quantities

Base or Fundamental Units

Derived Units

Systems of Units

The SI system

Prefixes used for Multiples and Fractions of base units

Dimensional Analysis  - Unit Factor Method

Laws of Chemical Combination

Law of Conservation of Mass

Law of constant proportions (or constant composition)

Law of Multiple Proportions

Gay Lussac’s Law of Gaseous Volumes

Dalton’s Atomic Theory

Avogadro Law

Some Important Terms


Atomic Mass Unit (amu or u)

Atomic mass of an element



Molecules of Compounds


Gram atomic mass

Molecular mass

Gram molecular mass

Formula unit mass

Mole Concept


Avogadro’s constant or Avogadro’s number

Formulae for Mole Concept

Measurement and Range


Measurement of Mass

Range of Mass





Accuracy and Precision in Measurement

Significant figures

Addition or subtraction with significant figures

Multiplication and division with significant figures

Rounding Off

Result of Multistep Calculation

Scientific Notation and Order of Magnitude

Stoichiometric Calculations

Limiting Reagent

Average Atomic Mass

Percentage Composition

Empirical Formula and Molecular Formula 

Concentration of Solutions

Mass percent or weight per cent (w/w %)


Mole fraction




Equivalent weight or gram equivalent

Basicity of Acids

Acidity of Bases

Laws of Dilution

Parts per million (PPM)


Some Basic Concepts Of Chemistry

Nature of Matter

Anything which has mass and occupies space is called matter.

Matter can exist in three physical states viz. solid, liquid and gas.

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Differences between Compounds and Mixtures

S. No.



Compounds are pure substances.

Mixtures are impure substances.

They are made up of two or more elements combined chemically.

They are made up of two or more substances mixed physically.

The constituents of a compound are present in a fixed ratio.

The constituents of a mixture are present in varying ratios.

Compounds sharp melting and boiling points.

Mixtures do not have sharp melting and boiling points

A compound has properties different from its constituents, as a new compound is formed.

In mixtures, no new substance is formed. It shows the properties of its constituents.

The constituents of a compound can be separated only by chemical methods.

The constituents of a mixture can be separated easily by physical methods.

Physical Properties

Those properties, which can be measured or observed without changing the identity or the composition of the substance are called physical properties. For example colour, odour, melting point, boiling point, density etc.

Chemical Properties

Those properties, which cannot be measured or observed without changing the identity or the composition of the substance, are called chemical properties. The measurement or observation of chemical properties requires a chemical change to occur. For example acidity or basicity, combustibility etc.

Quantitative Properties are those observations or measurements, which can be represented by a number followed by units in which it is measured. For example length, area, volume, etc.


Measurement of any physical quantity involves comparison with certain basic arbitrarily chosen and internationally accepted reference standards called UNIT.

Base or Fundamental quantities
There are certain physical quantities, called basic quantities, on which other physical quantities, called derived quantities are dependent.
Base or Fundamental Units

The units for the fundamental or base quantities are called fundamental or base units. There are also two supplementary units, namely radian (unit of plane angle) and steradian (unit of solid angle).

Derived Units

The units of other physical quantities, namely derived quantities are called derived units and they can be expressed as a combination of the base units.

Systems of Units

A complete set of units, both the base units and derived units, is known as the system of units.

Earlier different systems, the CGS, the FPS (or British) system and the MKS system were in use.

  • In CGS system they were centimetre, gram and second respectively.

  • In FPS system they were foot, pound and second respectively.

  • In MKS system they were metre, kilogram and second respectively

The SI system

To avoid confusion, standard scheme of symbols, units and abbreviations, was developed and recommended by General Conference on Weights and Measures in 1971. This internationally accepted system is called Système Internationale d’ Unites (French for International System of Units), abbreviated as SI.

In SI, there are seven base units. In SI System, there are seven base units.

Fundamental Units

S. No.

Fundamental Quantity

Fundamental Unit




Length, l



The metre is the length of the path travelled by light in vacuum during a time interval of 1299,792,458 of a second. (1983)


Mass, m



The kilogram is equal to the mass of the international prototype of the kilogram (a platinum-iridium alloy cylinder) kept at international Bureau of Weights and Measures, at Sevres, near Paris, France. (1889)


Time, t



The second is the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium-133 atom. (1967)


Electric current, I



The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to 2×10–7 newton per metre of length. (1948)


Temperature, T



The kelvin, is the fraction 1273.16 of the thermo-dynamic temperature of the triple point of water. (1967)


Amount of substance, n



The mole is the amount of substance of a system, which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon - 12. (1971)


Luminous intensity, Iv



The candela is the luminous intensity, in a given intensity direction, of a source that emits monochromatic radiation of frequency 540×1012 hertz and that has a radiant intensity in that direction of 1683 watt per steradian. (1979)

There are prefixes used for multiples and fractions of base units.




Power of 10



or deca-











































Power of 10










































Mass and Weight

Mass of a substance is the amount of matter present in it while weight is the force exerted by gravity on an object. The mass of a substance is constant whereas its weight may vary from one place to another due to change in gravity.

Measurement of Mass

Normal masses are measured in Kilogram (kg) – Measured using weighing scales.

Atomic and subatomic masses are measured in Unified Atomic Mass Units (u) – Using mass spectrograph, in which the radius of the trajectory is proportional to the mass of a charged particle moving in uniform electric and magnetic field.

1 u = 112th of the mass of an atom of carbon-12 isotope C612, including the mass of electrons

=1.66 × 10–27 kg

Mass of electron = 9.10938356 × 10-31 kg

Mass of proton = 1.6726219 × 10-27 kg

Mass of neutron = 1.6749 x 10-27 kg

Large masses in the universe like planets, stars, etc., based on Newton’s law of gravitation can be measured by using gravitational method.

Range of Masses

Mass of the electron order of ≈ 10-30 kg

Mass of known universe ≈ 1055 kg

Measurement of Time

Atomic standard of time is based on the periodic vibrations produced in a cesium atom. This is the basis of the cesium clock, called atomic clock, used in the national standards.

In the cesium atomic clock, the second is taken as the time needed for 9,192,631,770 vibrations of the radiation corresponding to the transition between the two hyperfine levels of the ground state of cesium-133 atom.

The cesium atomic clocks are very accurate. They impart the uncertainty in time realisation as ±1 × 10–13, i.e. 1 part in 1013. They lose or gain no more than 3 μs in one year.

Range of time

From life of most unstable particle (10-24 s) to Age of universe (1017 s)


Volume has the units of (length)3. SI unit of volume is m3. Other units are litre (L) and cm3

1 m3 = 1000 L,

1 L = 1000 cm3 = dm3


Density of a substance is its amount of mass per unit volume. SI unit of density is kg m–3

1 kg /m3 = 1000 g/1000000 cm3 = 10-3 g cm-3


There are three common scales to measure temperature — °C (degree celsius), °F (degree fahrenheit) and K (kelvin). K is the SI unit.

C100=F-32180 =K-273100


Accuracy & Precision in measurement

Accuracy refers to the closeness of a measurement to the true value of the physical quantity. That is, if the average value of different measurements is close to the correct value, the measurement is said to be accurate.

Precision refers to the resolution or the limit to which the quantity is measured. That is if values of different measurements are close to each other and the average value, the measurement is said to be accurate.

Significant figures

Significant figures are meaningful digits which are known with certainty. The uncertainty is indicated by writing the certain digits and the last uncertain digit. In 11.2 mL, 11 is certain and 2 is uncertain and the uncertainty would be ±1 in the last digit.

Hence, significant figures are defined as the total number of digits in a number including the last digit that represents the uncertainty of the result.

Rules for determining significant figures

  1. All non-zero digits are significant. For example in 285 cm, there are three significant figures and in 0.25 mL, there are two significant figures.

  2. Zeros preceding to first non-zero digit are not significant. Such zero indicates the position of decimal point. Thus, 0.03 has one significant figure and 0.0052 has two significant figures.

  3. Zeros between two non-zero digits are significant. Thus, 2.005 has four significant figures.

  4. Zeros at the end or right of a number are significant provided they are on the right side of the decimal point. For example, 0.200 g has three significant figures.

    But, if otherwise, the zeros are not significant. For example, 100 has only one significant figure.

  5. Exact numbers have an infinite number of significant figures. For example, in 2 balls or 20 eggs, there are infinite significant figures as these are exact numbers and can be represented by writing infinite number of zeros after placing a decimal i.e., 2 = 2.000000 or 20 = 20.000000

  6. When numbers are written in scientific notation, the number of digits between 1 and 10 gives the number of significant figures.

    Thus, 4.01×102 has three significant figures, and 8.256 × 10–3 has four significant figures.

Addition or subtraction with significant figures

In addition or subtraction, the final result should retain as many decimal places as are there in the number with the least decimal places.

For example if A = 334.5 kg; B = 23.43kg then

A + B = 334.5 kg + 23.43 kg = 357.93 kg

The result with significant figures is 357.9 kg

Multiplication and division with significant figures

In multiplication or division, the final result should retain as many significant figures as are there in the original number with the least significant figures.

For example if the mass of an object is 4.237 g and the volume is 2.51 cm3, then, density = 4.3272.51 should be reported with three significant digits, that is 1.69 g/cm3.

Similarly, if the speed of light is given as 3.00 × 108 ms-1 (three significant figures) and one year (1 y = 365.25 d) has 3.1557 × 107s (five significant figures), the light year is 9.47 × 1015 m (three significant figures).

Rounding Off

While rounding off measurements the following rules are applied

Rule I: If the digit to be dropped is less than 5, then the preceding digit should be left unchanged. For example, 9.32 is rounded off to 9.3

Rule II: If the digit to be dropped is greater than 5, then the preceding digit should be raised by 1 For example 8.27 is rounded off to 8.3

Rule III: If the digit to be dropped is 5 then,

  1. If the preceding digit is even, the insignificant digit is simply dropped and,

  2. If the preceding digit is odd, the preceding digit is raised by 1.

For example the number 2.745 rounded off to three significant figures becomes 2.74. On the other hand, the number 2.735 rounded off to three significant figures becomes 2.74.

Result of Multistep Calculations

  1. Decide how many significant digits the answer should have.

  2. Before carrying out the calculation, every number should be rounded off to contain one significant digit more than the answer should have.

  3. The answer should then be rounded off to contain required number of significant digits.

Scientific Notation

Very large and very small numbers are represented in scientific notation, i.e., exponential notation in which any number can be represented in the form N × 10n where n is an exponent having positive or negative values and N is a real number which can vary between 1 to 10.

232.508 = 2.32508 ×102

Order of Magnitude

In the number a × 10b, round off the number a to 1 (for a ≤ 5) and to 10 (for 5 < a ≤ 10). Then the number can be expressed approximately as 10b in which the exponent (or power) b of 10 is called order of magnitude of the physical quantity. Order of magnitude of 6.2 × 103 is 4.

Dimensional Analysis

The method used to convert a quantity from one unit system to another is called factor label method or unit factor method or dimensional analysis.


A piece of metal is 3 inch (represented by in) long. What is its length in cm?

We know that 1 in = 2.54 cm

From this equivalence, we can write

1 in2.54 cm= 1 =2.54 cm1 in

Both of these are called unit factors.

If a number is multiplied by any of these unit factors (i.e. 1) it will not be affected.


3 in=3 in× 2.54 cm1 in=7.62 cm

Example – Convert 2 L to m3.

1 L1000 cm3= 1 =1000 cm31 L  

Also  1 m100 cm= 1 =100 cm1 m  

  1 m100 cm3= 1 =100 cm1 m3 

  1 m31000000 cm3= 1 =1000000 cm31 m3

Now, 2 L=2 L× 1000 cm31 L ×1 m31000000 cm3 

=2 ×10-3 m3

Laws Of Chemical Combinations

1. Law of conservation of mass

This law was stated by Lavoisier in 1789. It states that

“In all physical and chemical changes, the total mass of reactants is equal to the total mass of products.”


“Matter can neither be created nor be destroyed in a chemical reaction.”

Note: Mass and energy are inter-convertible as per Einstein’s equation,

E = mc2

But the total of the mass and enrgy remains constant

2. Law of constant proportions (or constant composition):

This law was first stated by Proust in 1797. According to the law

“In a chemical substance the elements are always present in definite proportions by mass”.

For example the ratio of hydrogen and oxygen in pure water is always 1:8 by mass.

This law is also called law of definite proportions.


  1. This law is not applicable if an element occurs as different isotopes.

  2. In case of isomers, although the proportions of elements are in same proportions, but the compounds formed may be different.

3. Law of Multiple Proportions

This law was proposed by Dalton in 1803.

“If two elements can 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, hydrogen combines with oxygen to form two compounds, namely, water and hydrogen peroxide.

Hydrogen + Oxygen → Water

2g 16g 18g

Hydrogen + Oxygen → Hydrogen Peroxide

2g 32g 34g

Here, the masses of oxygen (i.e. 16 g and 32 g) which combine with a fixed mass of hydrogen (2g) are in a simple ratio, i.e. 16:32 or 1: 2.

4. Gay Lussac’s Law of Gaseous Volumes

This law was given by Gay Lussac in 1808.

“When gases combine or are produced in a chemical reaction they do so in a simple ratio by volume provided all gases are at same temperature and pressure.”

Hydrogen + Oxygen → Water

100 mL 50 mL 100 mL

Thus, the volumes of hydrogen and oxygen which combine together (i.e. 100 mL and 50 mL) to form water bear a simple ratio of 2:1.

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5. Dalton’s Atomic Theory (1808)

  1. Matter consists of indivisible atoms.

  2. All the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass.
    1. Compounds are formed when atoms of different elements combine in a fixed ratio.

    2. Chemical reactions involve reorganization of atoms. These are neither created nor destroyed in a chemical reaction.
  • Dalton’s theory could explain the laws of chemical combination.

6. Avogadro Law (1811)

“Equal volumes of gases at the same temperature and pressure should contain equal number of molecules.”

Some Important Terms

Atom: It is the smallest particle of an element which can take part in a chemical change. It may or may not be capable of independent existence. It cannot be seen by naked eye.

Atomic Mass Unit (amu or u) - One atomic mass unit is a mass equal to exactly one-twelfth 112 th the mass of one atom of carbon-12. 1 amu or u = 1.66056×10–24 g

Atomic mass of an element: The atomic mass of an element is the number which indicates how many times an atom of an element is heavier than 112 th of mass of an atom of carbon-12.

Molecule: It is the smallest particle of an element or compound that is capable of independent existence and shows all the properties of that substance.

[The molecule of an element is made up of same type of atoms, while the molecule of a compound is made up of different types of atoms]

Atomicity: The number of atoms present in a molecule of an element or a compound is known as its atomicity, e.g. the atomicity of oxygen is 2 while atomicity ozone is 3.

Molecules of Compounds: Atoms of different elements join together in definite proportions to form molecules of compounds.

Ion: It is an electrically charged atom or group of atoms. It is formed by the loss or gain of electrons by an atom. Ions are of two types,

  1. Cation: It is positively charged ion and is formed by the loss of electron from an atom e.g.,

    H+, Na+, Ca2+, Al3+, NH4+ etc.

  2. Anion: It is negatively charged ion and is formed by the gain of electrons by an atom, e.g.,

    Cl-, O2-, F-, CO32-, PO43- etc.

Gram atomic mass: The atomic mass of an element expressed in grams is known as gram atomic mass. (Gram atomic mass is also known as gram atomic weight).

Molecular mass: The number of times a molecule of a compound is heavier than 112th of the mass of C-12 atom, is known as its molecular mass.

The molecular mass is equal to the sum of the atomic masses of all atoms present in one molecule of the substance. For example H2S contains two atoms of hydrogen and one atom of S, so molecular mass of H2S is 2×1 + 32 = 34.

Gram Molecular mass is the molecular mass expressed in grams.

Formula unit mass: It is equal to the sum of atomic masses of all the atoms in a formula unit of ionic compounds, as they do not contain discrete molecules as their constituent units. Formula unit mass of NaCl is 23 + 35.5 = 58.5.

Mole Concept: One mole of any species (atoms, molecules, ions or particles) is that quantity in number having a mass equal to its atomic or molecular mass in grams. Irrespective of the substance under consideration 1 mole is equal to 6.022 × 1023 a specie.

The mass of 1 mole of particles (or simply is equal to its mass in grams.

1 mole = 6.023 × 1023 particles

1 mole atoms = 6.023 × 1023 atoms

1 mole electrons = 6.023 × 1023 electrons

1 mole protons = 6.023 × 1023 protons

1 mole ions = 6.023 × 1023 ions

1 mole molecules = 6.023 × 1023 molecules

Avogadro’s constant or Avogadro’s number: The number of particles present in one mole (i.e. 6.023 × 1023) is called Avogadro’s number or Avogadro’s constant.


1 mole = 6.023 × 1023 particles.

Mass of 1 mole = mass of 1 mole particles in grams

Volume of 1 mole of any gas = 22.4L of gas at NTP

Mass of 1 mole atoms = gram atomic mass

Mass of 1 mole molecules = gram molecular mass

NTP stands for normal temperature (0°C) and normal pressure (1 atmosphere or 760 mm of Hg)

Formulae for mole concept

Number of moles = Mass of substance in gramsGram molecular mass 

Number of moles = Volume of gas in litre at N.T.P.22.4

Number of moles = Number ofparticlesAvogadro's Number 

The number of atomsmolecules or ions etc.= Given massMolar mass ×Avogadro number

Given mass=Molar mass ×Number of moles

Average atomic mass of an atom with isotopes

Average Atomic Mass=A 1 ×percentage+A2 ×percentage100 

Where A1 and A2 are mass numbers of isotope1 isotope2 etc.

Q: Find Average mass of chlorine if abundance of Cl-35 in nature is 75% and that of Cl-37 is 25%

Percentage Composition

Mass % of an elementin the compound=mass of that element in one mole of compoundmolar mass of the compound

e.g., % of hydrogen in water=218×100=11.8

Empirical Formula and Molecular Formula

An empirical formula represents the simplest whole number ratio of various atoms present in a compound.

The molecular formula shows the exact number of different types of atoms present in a molecule of a compound.

  • If the mass per cent of various elements present in a compound is known, its empirical formula can be determined.

  • Molecular formula can be obtained if the molar mass is also known.

Step by step calculations with example

A compound contains 4.07 % hydrogen, 24.27 % carbon and 71.65 % chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas?

Step 1 - Conversion of mass per cent to grams.

Assume we have 100 g of the compound as the starting material. Thus, in the 100 g sample of the above compound, hence the % will be directly converted to grams,

Step 2 - Convert into number moles of each element

Divide the masses obtained above by respective atomic masses of various elements.

Step 3 - Divide the mole value obtained above by the smallest number

In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient.

Since 2.021 is smallest value, division by it gives a ratio of 2:1:1 for H:C:Cl .

Name of element




Mass of element in 100g of compound (g)




Number of moles of element

4.071.008= 4.04



Ratio of atoms


2.0212.021= 1


Step 4 - Write empirical formula by mentioning the numbers after writing the symbols of respective elements.


Step 5 - Writing molecular formula

  1. Determine empirical formula mass - Add the atomic masses of various atoms present in the empirical formula.

    For CH2Cl, empirical formula mass = 12.01 + 2 × 1.008 + 35.453 = 49.48 g

  2. Divide Molar mass by empirical formula mass to get the multiplication factor

    n=Molar massEmpirical formula mass=98.9649.48= 2

  3. Multiply empirical formula by n obtained above to get the molecular formula.

    In this case, molecular formula is C2H4Cl2.

Stoichiometric Calculations

Stoichiometry deals with the calculation of masses and volumes of the reactants and the products involved in a chemical reaction.

For this the balanced chemical equation of a given reaction is used.

Let us consider the combustion of methane. A balanced equation for this reaction is,

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)

Reactants Products

The coefficients by which the reactants and products are multiplied are called stoichiometric coefficients. They represent the number of moles taking part in the reaction or formed in the reaction.


  • One mole of CH4(g) reacts with two moles of O2(g) to give one mole of CO2(g) and two moles of H2O(g)

  • One molecule of CH4(g) reacts with 2 molecules of O2(g) to give one molecule of CO2(g) and 2 molecules of H2O(g)

  • At STP, 22.4 L of CH4(g) reacts with 44.8 L of O2 (g) to give 22.4 L of CO2 (g) and 44.8 L of H2O(g)

  • 16 g of CH4 (g) reacts with 2×32 g of O2 (g) to give 44 g of CO2 (g) and 2×18 g of H2O (g)

  • mass ⇌ moles ⇌ no.of molecules ⇌ volume at NTP (only for gases)

Limiting Reagent

The reactant which gets consumed before other reactants, limits the amount of product formed and is called the limiting reagent.

The amount of product formed is limited by this reagent, since the reaction cannot continue without it.

  • Calculate number of moles available for each reactant.

  • Calculate the number of moles of one reactant, required for a given moles of another reactant, as per balanced equation.

  • If the number of moles available is less than the moles required, the reactant is the limiting reagent.

  • Now all the calculations must be done as per the moles of limiting reagent available.

  • Use the following formula for yields

    Percent yield =Actual yieldTheoritical yield×100


50.0 kg of N2 (g) and 10.0 kg of H2 (g) are mixed to produce NH3 (g). Calculate the NH3 (g) formed. Identify the limiting reagent in the production of NH3 in this situation.


The balanced equation for the above reaction

N2 (g) + 3H2 (g) ⇌ 2NH3 (g)

Moles of N2 

= 50.0 kg of N2 ×1000 g of N2 1 kg  of N2  ×1 mol of N2 28.0 g  of N2 

=1.786×103  mol  of N2 

Moles of H2 

     = 10.0 kg of N2 ×1000 g of H2 1 kg  of H2  ×1 mol of H2 2.016 g  of H2 

=4.96×103  mol of H2

According to the above equation, 1 mol N2 (g) requires 3 mol H2 (g), for the reaction.

Hence, for 17.86×102 mol of N2, the moles of H2 (g) required would be

= 1.786×103  mol  of N2 × 3 mol  of H21 mol of N2 =5.36×103  mol of H2  

But we have only 4.96 × 103 mol of H2. Hence, dihydrogen is the limiting reagent in this case.

So NH3 (g) would be formed only from that amount of available dihydrogen i.e., 4.96 × 103 mol

Now, since 3 mol of H2 (g) gives 2 mol of NH3 (g)

= 4.96 ×103  mol  of H2 × 2 mol  of NH33 mol of H2 

=3.30×103  mol of NH3

Mass of NH3 produced

1 mol of NH3 (g) = 17.0 g NH3 (g)

So, 3.30×103 mol of NH3 g

= 3.30×103 mol of NH3 ×17 g of NH3 g1 mol of NH3

=56.1 ×103 g of NH3 g= 56.1 kg of NH3 g

Concentration of Solutions

The concentration of a solution or the amount of substance present in its given volume can be expressed in any of the following ways.

No of Moles =Mass of the substanceMolar mass of substance

  1. Mass percent or weight per cent (w/w %)

    Mass percent= Mass of soluteMass of solution ×100

  2. Strength: It is the amount of solute in grams present in 1 L of solution.

    Strength= Mass of solute in gVolume of solution in litre or dm3 

  3. Mole fraction

    Mole fraction of A= No.of Moles of ATotal no.  of moles of the solution =nAnA+nB 

    Mole fraction of B= No.of Moles of BTotal no.  of moles of the solution =nBnA+nB 

  4. Molarity: It is defined as the number of moles of the solute in 1 litre of the solution.

    Molarity M=No. of moles of soluteVolume of solution in litres

  5. Molality: It is defined as the number of moles of solute present in 1 kg of solvent. It is denoted by m.

    Molality m = No. of moles of soluteMass of solvent in kg

  6. Noramality

    Normality N= Gram Equivalents of SoluteVolume of Solution in litres

        = Molarity (M) ×Number of equivalents 

Equivalent weight (also known as gram equivalent) is the mass of one equivalent, that is, the mass of a given substance which will

  • combine or displace directly or indirectly with 1.008 parts by mass of hydrogen or 8 parts by mass of oxygen or 35.5 parts by mass of chlorine – these values correspond to the atomic weight divided by the usual valency; or

  • supply or react with one mole of hydrogen cations (H+) in an acid–base reaction; or

  • supply or react with one mole of electrons (e) in a redox reaction.

  • Like molarity, normality relates the amount of solute to the total volume of solution; however, normality is specifically used for acids and bases.

    The mole equivalents of an acid or base are calculated by determining the number of H+ or OH- ions per molecule:

    Eq.mass of an acid= Molar mass of the acidBasicity

    Eq.mass of a base= Molar mass of the baseAcidity

    Eq.mass of a salt= Molar mass of the saltTotal positive valency of metallic atoms

Laws of Dilution:

  1. If a solution with volume V1 and molarity M1 is diluted to volume V2 with molarity M2, then,

    M1 × V1 = M2 × V2

  2. For a balanced chemical equation, if n1 moles of reactant reacts with n2 moles of reactant 2, then,

    M1 × V1n1 = M2 × V2n2

  3. For a chemical reaction involving an acid and a base,

    na × Ma × Vb = nb × Mb × Vb


na = Basicity of acid

nb = Acidity of base

Va = Volume of acid

Vb = Volume of base

Ma = Molarity of acid

Mb = Molarity of base

  1. If a solution with volume V1 and normality N1 is diluted to volume V2 with normality N2, then,

    N1 × V1 = N2 × V2

Relationship between Molarity and Normality

N = n × M (where n is an integer)

For an acid solution, n is the number of H+ ions provided by a formula unit of acid (i.e. basicity).

Example: A 3M H2SO4 solution is the same as a 6N H2SO4 solution.

For a basic solution, n is the number of OH+ ions provided by a formula unit of base (i.e. acidity).

Example: A 1M Ca(OH)2 solution is the same as a 2N Ca(OH)2 solution

  1. Parts per million (PPM): Parts per million is defined as parts of the solute by mass present in 1 million parts by mass of solution.

    ppm =Mass of soluteMass of solution×106