These two concepts are among the most important topics in Mole Concept, Redox Reactions, and Acid-Base Titrations.

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1. What is Normality (N)?

Definition:

Normality is the number of gram equivalents of solute present in 1 litre of solution.$$\boxed{\text{Normality}=\frac{\text{Gram Equivalents}}{\text{Volume (L)}}}$$

Unit

eq/L or simply N

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2. Gram Equivalent

A gram equivalent is the amount of substance equal to its equivalent weight.$$\boxed{\text{Gram Equivalent}=\frac{\text{Mass}}{\text{Equivalent Weight}}}$$

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3. Equivalent Weight

Equivalent weight is the mass of a substance that supplies or reacts with one equivalent.

General formula:$$\boxed{\text{Equivalent Weight}=\frac{\text{Molar Mass}}{n\text{-factor}}}$$

where n-factor depends on the reaction.

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4. Relation between Molarity and Normality

$$\boxed{N=M\times n\text{-factor}}$$

or$$\boxed{M=\frac{N}{n\text{-factor}}}$$​​

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Example

0.5 M H₂S$$N=0.5\times2=1N$$

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5. n-factor (Most Important for NEET/JEE)

The value of n-factor changes according to the reaction.

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A. Acids

n-factor = Number of ionisable H⁺ released

Acid	n-factor
HCl	1
HNO₃	1
CH₃COOH	1
H₂SO₄	2
H₂CO₃	2
H₃PO₄	3

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Example

Equivalent weight of H₂SO₄

Molar mass = 98$$Eq.wt=\frac{98}{2}=49$$

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B. Bases

n-factor = Number of OH⁻ ions furnished

Base	n-factor
NaOH	1
KOH	1
Ba(OH)₂	2
Ca(OH)₂	2
Al(OH)₃	3

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Example

Equivalent weight of Ba(OH)₂

Molar mass =171$$Eq.wt=\frac{171}{2}=85.5$$

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C. Salts

n-factor = Total positive or negative charge involved

Salt	n-factor
NaCl	1
Na₂CO₃	2
Na₂SO₄	2
K₂Cr₂O₇	2
Na₃PO₄	3
Al₂(SO₄)₃	6

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Example

Na₂CO₃

Molar mass =106$$Eq.wt=\frac{106}{2}=53$$

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D. Oxidising Agents (Redox)

n-factor = Electrons gained per molecule

Examples

KMnO₄

Acidic medium

$$Mn^{+7}\rightarrow Mn^{+2}$$

Electrons gained =5

n-factor =5

Equivalent weight$$\frac{158}{5}=31.6$$

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Neutral/Basic medium

$$Mn^{+7}\rightarrow MnO_2$$

Electrons gained =3

n-factor =3

Equivalent weight$$\frac{158}{3}=52.7$$

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K₂Cr₂O₇

Acidic medium

Cr(+6)→Cr(+3)

Each Cr gains 3 electrons.

Two Cr atoms

Total electrons =6

n-factor =6

Equivalent weight$$\frac{294}{6}=49$$

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E. Reducing Agents

n-factor = Electrons lost

Example

Fe²⁺→Fe³⁺

Electron lost =1

n-factor =1

Equivalent weight$$\frac{56}{1}=56$$

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Example

Sn²⁺→Sn⁴⁺

Electron loss =2

n-factor =2

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6. Equivalent Weight of Elements

Equivalent weight$$=\frac{\text{Atomic mass}}{\text{Valency}}$$

Example

Element	Eq. wt
Na	23
Mg	12
Al	9
Fe²⁺	28
Fe³⁺	18.67

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7. Equivalent Weight of Compounds

$$Eq.wt=\frac{\text{Molar mass}}{n\text{-factor}}$$

Examples

Compound	Eq.wt
HCl	36.5
H₂SO₄	49
NaOH	40
Ca(OH)₂	37
Na₂CO₃	53
KMnO₄ (acidic)	31.6
KMnO₄ (basic)	52.7
K₂Cr₂O₇	49

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8. Normality Formulae

From mass

$$\boxed{N=\frac{W\times1000}{Eq.wt\times V(mL)}}$$

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From molarity

$$\boxed{N=M\times n}$$

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Dilution

$$\boxed{N_1V_1=N_2V_2}$$

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9. Titration Formula

At equivalence point$$\boxed{N_1V_1=N_2V_2}$$

Example

20 mL HCl requires 10 mL NaOH (2 N)$$N(HCl)\times20=2\times10$$$$N(HCl)=1N$$

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10. Important NEET/JEE Points

• Equivalent weight = Molar mass ÷ n-factor
• Normality = Molarity × n-factor
• n-factor depends on the reaction
• For acids: ionisable H⁺
• For bases: replaceable OH⁻
• For salts: total ionic charge involved
• For redox: electrons exchanged
• Normality is widely used in titration calculations
• Molarity is fixed for a given solution, but normality can change if the n-factor changes with the reaction.

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Quick Revision Table

Substance	n-factor	Equivalent Weight
HCl	1	36.5
H₂SO₄	2	49
H₃PO₄	3	32.7
NaOH	1	40
Ca(OH)₂	2	37
Na₂CO₃	2	53
Ba(OH)₂	2	85.5
KMnO₄ (acidic)	5	31.6
KMnO₄ (basic/neutral)	3	52.7
K₂Cr₂O₇	6	49
Fe²⁺	1	56
Fe³⁺	3	18.67

NEET/JEE Shortcut

Remember the sequence:

1. Find the reaction.
2. Determine the n-factor from the reaction.
3. Calculate Equivalent Weight = Molar Mass ÷ n-factor.
4. Use Normality = Molarity × n-factor or N₁V₁ = N₂V₂ for titration problems.