Why does the equivalent mass of KMnO4 change with medium?

Because the electrons Mn gains depend on final oxidation state. In acidic medium Mn(VII) becomes Mn(II) gaining 5 electrons, so E is 158/5. In neutral medium Mn(VII) becomes Mn(IV) gaining 3 electrons.

How do I find moles of a gas not at STP?

Use the ideal gas equation PV equals nRT so n equals PV over RT. The shortcut n equals V over 22.4 only applies at STP.

Mole Concept And Equivalent Concept

Q: Why does 1 mole of any gas occupy 22.4 L at STP?

By Avogadro's hypothesis, equal volumes of all gases at same T and P contain equal molecules. So 1 mole of any ideal gas occupies the same volume: 22.4 L at 273.15 K and 1 atm.

Q: What is the difference between mole and gram atom?

Mole is the general SI term for 6.022 times 10 to 23 entities. Gram atom specifically means 1 mole of atoms with mass equal to atomic mass in grams. Modern usage prefers mole.

Q: Why is H3PO3 dibasic and not tribasic?

In H3PO3 only 2 of the 3 H atoms are bonded to oxygen. The third is bonded directly to phosphorus and cannot ionise. Only H atoms on O can be released as H+, so basicity is 2.

Q: How do I find equivalents from a titration?

At the equivalence point, equivalents of acid equal equivalents of base. Since equivalents equal N times V in litres, N1 V1 equals N2 V2. This is the normality equation.

Q: Why is the equivalent mass of a hydrated salt not the sum of ionic equivalents?

Water of crystallisation adds to formula mass but not to total ionic charge. Directly summing cation and anion equivalents ignores water and undercounts mass. Use formula mass over total positive charge.

Q: What is the difference between equivalent mass and equivalent weight?

Nothing. They are two names for the same quantity. Equivalent weight is older terminology; equivalent mass is modern SI-consistent. Both mean grams per equivalent.

ChemistrySome Basic Concepts of ChemistryFor JEE aspirants

The mole is the chemist's counting unit: $1$ mole of anything contains Avogadro's number, $6.022 \times 1 0^{23}$ , of that thing. It links three ways of measuring stuff - mass, number of particles, and (for gases) volume. The equivalent mass extends this to reactions: it is the mass that combines with or displaces a fixed amount of another substance. In this concept you will master mole calculations for elements and compounds, Avogadro's hypothesis, molar volume at STP, and equivalent masses of elements, ionic compounds, acids, bases, and salts - all of which power titration and redox stoichiometry problems in JEE and NEET.

Key Formulas - Quick Reference

$1$ mole $= 6.022 \times 1 0^{23}$ particles $= N_{A}$ (Avogadro's number)
Moles: $n = \frac{mass}{molar mass} = \frac{particles}{N _{A}} = \frac{V ( gas at STP )}{22.4 L}$
Molar volume of any ideal gas at STP $= 22.4 L mol^{- 1}$ (at $273.15 K$ , $1 atm$ )
Avogadro's hypothesis: equal volumes of gases at the same T and P contain equal numbers of molecules
Equivalent mass (redox): $E = \frac{molar (or atomic) mass}{electrons gained or lost per formula unit}$
Equivalent mass of an element (non-redox): $E = \frac{atomic mass}{valency}$
Equivalent mass of an acid: $E = \frac{molar mass}{basicity}$ ; of a base: $E = \frac{molar mass}{acidity}$
Equivalent mass of an ionic compound: $E = \frac{formula mass}{total positive charge (or total negative)}$

1. The Mole Concept

A mole is the amount of a substance that contains exactly

6.022 \times 1 0^{23}

elementary entities (atoms, molecules, ions, electrons, or formula units). This number is Avogadro's number,

N_{A}

. Since 2019, the SI defines the mole by fixing

N_{A}

at exactly

6.02214076 \times 1 0^{23} mol^{- 1}

Mole of an Element

$1$ mole of atoms of an element $=$ $N_{A}$ atoms $=$ gram atomic mass (atomic mass in grams). It is also called $1$ gram atom.

Example: $1$ mole of oxygen atoms $= 6.022 \times 1 0^{23}$ atoms $= 16 g$ . To verify:

Mass of one O atom $= 16 amu = 16 \times 1.66 \times 1 0^{- 24} g$ .

Mass of $N_{A}$ atoms $= 16 \times 1.66 \times 1 0^{- 24} \times 6.022 \times 1 0^{23} = 16 g$ . Checks out.

Mole of a Compound

$1$ mole of molecules of a compound $=$ $N_{A}$ molecules $=$ gram molecular mass (molecular mass in grams). Also called $1$ gram molecule.

Example: $1$ mole of $O_{2} = 6.022 \times 1 0^{23}$ molecules $= 32 g$ .

Key Interconversions

Figure 1: The mole is the central hub. Convert to and from mass, particles, or volume (for a gas at STP) by using the appropriate factor.

Old vs new STP: older textbooks (and most JEE/NEET problems) use

273.15 K

and

1 atm

, giving molar volume

= 22.4 L

. Since 1982, IUPAC uses

273.15 K

and

1 bar

, giving

22.7 L

. Unless the problem specifies otherwise, use

22.4 L

Solved Example 1

Calculate the number of moles, molecules, and volume (at STP) present in

44 g

CO_{2}

Solution:

Molar mass of $CO_{2} = 44 g mol^{- 1}$ .

Moles $= \frac{44}{44} = 1 mol$ .

Molecules $= 1 \times 6.022 \times 1 0^{23} = 6.022 \times 1 0^{23}$ .

Volume at STP $= 1 \times 22.4 = 22.4 L$ .

Also: number of C atoms $= 6.022 \times 1 0^{23}$ ; number of O atoms $= 2 \times 6.022 \times 1 0^{23} = 1.204 \times 1 0^{24}$ .

Solved Example 2

How many moles of oxygen atoms are present in

180 g

of glucose (

C_{6} H_{12} O_{6}

Solution:

Molar mass of glucose $= 6 (12) + 12 (1) + 6 (16) = 180 g mol^{- 1}$ .

Moles of glucose $= \frac{180}{180} = 1$ .

Each glucose molecule has $6$ O atoms, so moles of O atoms $= 6 \times 1 = 6 mol$ .

2. Avogadro's Hypothesis

Avogadro's hypothesis (1811): Equal volumes of all gases, at the same temperature and pressure, contain equal numbers of molecules.

This hypothesis has two important consequences:

If two gases are at the same T, P, and V, then $n_{1} = n_{2}$ (equal moles).
$1$ mole of any gas occupies the same volume at STP - namely $22.4 L$ - regardless of the gas's identity.

Avogadro's hypothesis was the missing link that reconciled Dalton's atomic theory with Gay-Lussac's law of combining volumes: whole-number volume ratios reflect whole-number molecule ratios in the balanced equation.

Solved Example 3

3.2 g

of oxygen (

O_{2}

) and

0.2 g

of hydrogen (

H_{2}

) are placed in a

1.12 L

flask at

0^{\circ} C

. Find the gas pressure in atm.

Solution:

Moles of $O_{2} = \frac{3.2}{32} = 0.10$ . Moles of $H_{2} = \frac{0.2}{2} = 0.10$ . Total moles $n = 0.20$ .

Using $P V = n R T$ : $P = \frac{n R T}{V} = \frac{0.20 \times 0.0821 \times 273.15}{1.12} = 4.00 atm$ .

3. Equivalent Mass (Equivalent Concept)

The equivalent mass of a substance is the mass that combines with or displaces

1 g

of hydrogen,

8 g

of oxygen, or

35.5 g

of chlorine - or, more generally, that supplies or accepts one mole of

H^{+}

OH^{-}

, or electrons.

Equivalent mass depends on the reaction the substance is taking part in. The same compound can have different equivalent masses in different reactions - always check the reaction before writing the formula.

3.1 Equivalent Mass in a Redox Change

E = \frac{molar (or atomic) mass}{number of electrons gained or lost per formula unit}

Example: in $KMnO_{4}$ acting as an oxidiser in acidic medium, $Mn^{7 +} \to Mn^{2 +}$ , gaining $5$ electrons. So

$E (KMnO_{4}) = \frac{158}{5} = 31.6 g eq^{- 1}$

In neutral or feebly alkaline medium, $Mn^{7 +} \to Mn^{4 +}$ (gain of $3$ electrons), so $E = \frac{158}{3} = 52.7$ . Same compound, different reaction, different $E$ .

3.2 Equivalent Mass of an Element (Non-Redox)

E = \frac{atomic mass}{valency}

Example: for Ca (valency $2$ ), $E = \frac{40}{2} = 20$ . For Al (valency $3$ ), $E = \frac{27}{3} = 9$ .

3.3 Equivalent Mass of an Ionic Compound

E = \frac{formula mass}{total positive charge (or total negative charge)}

Or equivalently:

$E_{ionic compound} = E_{cation} + E_{anion} (not valid for hydrated salts)$

Examples:

$E (NaCl) = \frac{58.5}{1} = 58.5$
$E (AlCl_{3}) = \frac{133.5}{3} = 44.5$ (also $E_{Al} + E_{Cl} = \frac{27}{3} + 35.5 = 44.5$ )
$E (Al_{2} (SO_{4})_{3}) = \frac{342}{6} = 57$ (six positive charges total from $2 Al^{3 +}$ )
Potash alum $K_{2} SO_{4} \cdot Al_{2} (SO_{4})_{3} \cdot 24 H_{2} O$ : total positive charge from $2 K^{+} + 2 Al^{3 +} = 2 + 6 = 8$ , so $E_{alum} = \frac{M}{8}$

3.4 Equivalent Mass of an Acid

E_{acid} = \frac{molar mass of acid}{basicity}

Basicity is the number of ionisable (replaceable)

H^{+}

per molecule of acid.

Examples: $E (HCl) = 36.5$ (basicity $1$ ); $E (H_{2} SO_{4}) = \frac{98}{2} = 49$ ; $E (H_{3} PO_{4}) = \frac{98}{3} = 32.67$ .

Only H atoms attached to O are replaceable in oxyacids. In

H_{3} PO_{3}

(phosphorous acid), only

2

of the

3

H atoms are on O, so basicity is

2

E = \frac{82}{2} = 41

. In

H_{3} PO_{2}

(hypophosphorous acid), only

1

H is on O: basicity

= 1

E = 66

3.5 Equivalent Mass of a Base

E_{base} = \frac{molar mass of base}{acidity}

Acidity is the number of ionisable

OH^{-}

per formula unit.

Examples: $E (NaOH) = 40$ ; $E (Ca (OH)_{2}) = \frac{74}{2} = 37$ ; $E (Al (OH)_{3}) = \frac{78}{3} = 26$ .

Reaction-dependent: if only some $OH$ groups react, use that number instead.

Example: $Al (OH)_{3} + HCl \to Al (OH)_{2} Cl + H_{2} O$ . Only one OH reacts, so $E (Al (OH)_{3}) = \frac{78}{1} = 78$ for this specific reaction.

3.6 Equivalent Mass of an Ion or Radical

E_{ion} = \frac{formula mass of ion}{∣ charge on ion ∣}

Examples: $E (Na^{+}) = 23$ ; $E (Ca^{2 +}) = 20$ ; $E (SO_{4}^{2 -}) = \frac{96}{2} = 48$ ; $E (PO_{4}^{3 -}) = \frac{95}{3} = 31.67$ .

3.7 Equivalent Mass of an Acid Salt

An acid salt contains replaceable H. Its equivalent mass depends on how many H are replaced in the reaction.

E = \frac{molar mass of acid salt}{replaceable H atoms}

For $NaH_{2} PO_{4}$ : if both remaining H are replaced (in $Na_{3} PO_{4}$ formation), $E = \frac{M}{2} = \frac{120}{2} = 60$ . If only one is replaced ( $Na_{2} HPO_{4}$ formed), $E = \frac{120}{1} = 120$ .

Note: $Na_{2} HPO_{3}$ is not an acid salt, because $H_{3} PO_{3}$ is dibasic (only $2$ H are replaceable), and both are already used in this salt.

3.8 Equivalent Mass of a Basic Salt

A basic salt contains replaceable OH.

E = \frac{molar mass of basic salt}{replaceable OH groups}

Example: $E (Al (OH)_{2} Cl) = \frac{M}{2}$ if both OH are replaced; $E = M$ if only one is replaced.

Summary Table

Substance	Equivalent mass formula	Example
Element (non-redox)	Atomic mass / valency	$E (Al) = 27/3 = 9$
Element/compound (redox)	Molar mass / electrons transferred	$E (KMnO_{4}) = 158/5 = 31.6$
Acid	Molar mass / basicity	$E (H_{2} SO_{4}) = 98/2 = 49$
Base	Molar mass / acidity	$E (Ca (OH)_{2}) = 74/2 = 37$
Salt/ionic compound	Formula mass / total positive charge	$E (Al_{2} (SO_{4})_{3}) = 342/6 = 57$
Ion	Formula mass / \|charge\|	$E (CO_{3}^{2 -}) = 60/2 = 30$

Solved Example 4

0.20 g

of a metal completely reacts with dilute HCl to give

112 mL

of hydrogen at STP. Calculate the equivalent mass of the metal.

Solution:

Volume of $H_{2}$ at STP $= 112 mL = 0.112 L$ .

Since $11.2 L$ of $H_{2}$ at STP $= 1$ equivalent, moles of H displaced $= \frac{0.112}{11.2} = 0.010 eq$ .

Equivalent mass of metal $= \frac{mass of metal}{equivalents of H displaced} = \frac{0.20}{0.010} = 20 g eq^{- 1}$ .

(This is consistent with calcium, $E (Ca) = 40/2 = 20$ .)

Solved Example 5

Write the equivalent mass of

H_{3} PO_{4}

in each of these reactions:
(a)

H_{3} PO_{4} + NaOH \to NaH_{2} PO_{4} + H_{2} O

(b)

H_{3} PO_{4} + 2 NaOH \to Na_{2} HPO_{4} + 2 H_{2} O

(c)

H_{3} PO_{4} + 3 NaOH \to Na_{3} PO_{4} + 3 H_{2} O

Solution:

Molar mass of $H_{3} PO_{4} = 98 g mol^{- 1}$ .

(a) Only $1$ H is replaced, so $E = \frac{98}{1} = 98$ .

(b) $2$ H are replaced, so $E = \frac{98}{2} = 49$ .

Same acid, three different equivalent masses depending on the reaction. This is why the equation matters.

4. Connecting Mole and Equivalent

The two ideas link through valency:

$equivalents = moles \times n where n = \frac{molar mass}{equivalent mass}$

$n$ is the number of $H^{+}$ , $OH^{-}$ , or electrons per formula unit. So molarity ( $M$ ) and normality ( $N$ ) are related by:

$N = M \times n$

Examples: $1 M H_{2} SO_{4}$ is $2 N$ ; $1 M Ca (OH)_{2}$ is $2 N$ ; $1 M KMnO_{4}$ (acid medium) is $5 N$ .

Dulong and Petit's law (atomic mass

\times

specific heat

\approx 6.4

) links atomic mass to a measurable physical property. It was covered in Concept 2 (Atomic Mass); see there for a worked example.

Common Mistakes to Avoid

Watch out

Using $22.4 L$ for anything other than a gas at STP. Liquids and solids do not follow the molar-volume rule.
Applying Avogadro's hypothesis to solutions or solids. It is for gases at the same T and P only.
Writing a single "equivalent mass" for $KMnO_{4}$ or $H_{3} PO_{4}$ without checking the reaction. The value depends on how many electrons or H are transferred.
Forgetting that in $H_{3} PO_{3}$ only $2$ H are replaceable (the third is bonded directly to P, not to O). Same for $H_{3} PO_{2}$ (only $1$ replaceable H).
Adding equivalent masses of a cation and anion for a hydrated salt. The formula $E = E_{cation} + E_{anion}$ works for anhydrous ionic compounds only.
Confusing atoms per molecule with molecules per mole. Water has $2$ H and $1$ O per molecule; $1$ mole of $H_{2} O$ has $2 N_{A}$ H atoms and $N_{A}$ O atoms.
Using $22.7 L$ for STP without checking. Legacy JEE/NEET problems and NCERT still use $22.4 L$ (based on $1 atm$ ).

Frequently Asked Questions

Why does 1 mole of any gas occupy 22.4 L at STP?

By Avogadro's hypothesis, equal volumes of all gases at the same T and P contain equal numbers of molecules. So $1$ mole ( $N_{A}$ molecules) of any ideal gas occupies the same volume - $22.4 L$ at $273.15 K$ and $1 atm$ .

What is the difference between mole and gram atom?

"Mole" is the general SI term for $6.022 \times 1 0^{23}$ entities of any type. "Gram atom" specifically means $1$ mole of atoms (mass $=$ atomic mass in grams). Modern usage prefers "mole" everywhere; "gram atom" is largely historical.

Why does the equivalent mass of KMnO4 change with the medium?

Because the number of electrons Mn gains depends on the final oxidation state. In acidic medium $Mn^{7 +} \to Mn^{2 +}$ ( $5 e^{-}$ ), so $E = 158/5$ . In neutral medium $Mn^{7 +} \to Mn^{4 +}$ ( $3 e^{-}$ ), so $E = 158/3$ . In strongly alkaline medium $Mn^{7 +} \to Mn^{6 +}$ ( $1 e^{-}$ ), so $E = 158$ .

Why is H3PO3 dibasic and not tribasic?

In $H_{3} PO_{3}$ , only $2$ of the $3$ H atoms are bonded to oxygen; the third is bonded directly to phosphorus and cannot ionise. Only H atoms on O can be released as $H^{+}$ , so basicity $= 2$ .

How do I find equivalents from a titration?

At the equivalence point, equivalents of acid $=$ equivalents of base. Since equivalents $= N \times V$ (in litres), we get $N_{1} V_{1} = N_{2} V_{2}$ . This is the normality equation, the workhorse of acid-base and redox titrations.

Is the mole a fundamental SI unit?

Yes. Since 2019, the SI defines the mole by fixing Avogadro's number at exactly $6.02214076 \times 1 0^{23} mol^{- 1}$ . It is one of the seven SI base units, alongside metre, kilogram, second, ampere, kelvin, and candela.

Why is the equivalent mass of a hydrated salt not the sum of ionic equivalents?

Water of crystallisation contributes to the formula mass but not to the total ionic charge. So directly summing $E_{cation} + E_{anion}$ ignores the water and undercounts the mass. Always use $E = \frac{formula mass}{total positive charge}$ for hydrated salts.

If a gas is not at STP, how do I find its moles?

Use the ideal gas equation $P V = n R T$ , so $n = \frac{P V}{R T}$ . Only at STP does the shortcut $n = \frac{V}{22.4}$ apply (in litres, with $R = 0.0821 L atm mol^{- 1} K^{- 1}$ ).

What is the difference between equivalent mass and equivalent weight?

Nothing. They are two names for the same quantity. "Equivalent weight" is older terminology; "equivalent mass" is modern SI-consistent. Both mean grams per equivalent.

Previous year questions on Mole Concept And Equivalent Concept

24 questions from past papers, each with a step-by-step solution.

Show all 24 questions

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