- Normality Definition and Formula
- Difference Between Normality and Molarity
- Relation between Normality and Molarity
- How to calculate Normality of the Solution
- Solved Examples
- Normality equation
- Acid-Base Titration equation using Normality
- FORMALITY (F)
- Normaility Quiz

where n = number of H

Equivalent Weight is new thing in the formula. Lets check how it is defined

Equivalent Weight is defined the as ratio of Molar Mass of substance to the valence of the substance.Valence is also denoted as equivalence factor. The valence is the number of H

- Normality unit is N or eq/L .It depends on the chemical reaction being studied and depends on temperature too. So ,it is meaning with respect to the chemical reaction

- It is the reactive capacity of the solution and it is called equivalent concentration

- This is used in acid-base reactions and redox reactions

Molarity |
Normality |

number of moles of solute per liter of solution. |
number of gram equivalent of solute per liter of solution. |

Molarity is a measurement of the Moles in the total volume of the solution, |
Normality is a measurement of the gram equivalent in relationship to the total volume of the solution. |

Unit is Moles L^{-1} |
Unit is eq L^{-1} |

$\text{Normality} = n \times \text{Molarity}$

where n = number of H

a. Find the equivalent weight of the solute based on the chemical reaction it is going to be used

b. Calculate No. of Gram Equivalent of Solute

c. Calculate the volume in litre

d. Calculate the Normality using the formula given above

Lets Calculate Normality of HCL using this

Suppose we have 36.5 gram of 1 L HCL solution.How can we calculate the Normality of HCL solution

a. Molecular Mass of HCL= 36.5

Number of H+ ion=1

So equivalent Weight = Molecular mass/1 = 36.5

b. No. of Gram Equivalent of Solute = Mass of Solute/Equivalent Weight = 36.5 /36.5 = 1

c. Volume here is given as 1 L

d. Normality will be calculated as

Normality = No. of Gram Equivalent of Solute/ Volume of Solution in L

=1 N

Calculate the equivalent weight of following:-

(i) $\text{H}_2 \text{SO}_4$

(ii)$\text{NaCl}$

(iii) $\text{NaOH}$

(iv) $\text{Na}_2 \text{CO}_3$

(i) Molar Mass of $\text{H}_2 \text{SO}_4$=98

n = 2 as it has 2 H+ ion

$ \text{Equivalent Weight}= \frac {\text{Molar Mass}}{n} = \frac {98}{2} = 49$

(ii)Molar Mass of NaCl =23+35.5= 58.5

n = 1

$ \text{Equivalent Weight}= \frac {\text{Molar Mass}}{n} = \frac {58.5}{1} = 58.5$

(iii)

Molar Mass of $\text{NaOH}$=40

n = 1 as it has 1 OH- ion

$ \text{Equivalent Weight}= \frac {\text{Molar Mass}}{n} = \frac {40}{1} = 40$

(iv)

Molar Mass of $\text{Na}_2 \text{CO}_3$=106

The salt $\text{Na}_2 \text{CO}_3$ ionizes to form $2\text{Na}^+$ and $\text{CO}_3^{-2}$. So, the charge present on both is 2

n = 2

$ \text{Equivalent Weight}= \frac {\text{Molar Mass}}{n} = \frac {106}{2} = 53$

Calculate the normality of $\text{NaOH}$ solution Formed by dissolving 0.2 gm $\text{NaOH}$ to make 250 ml solution

Normality(N) = (no.Gram Equivalent of solute)/(Volume of Solution in litre)

No. of Gram Eq. of Solute = weight/Equivalent weight

Now, Equivalent weight=$ \frac {\text{Molar Mass}}{n} = \frac {23+16+1}{1}=40$

So, N = (No.gram eq.mass)/(Vol (l)

=Weight/(Equivalent weight) × 1000/(V ml)

=(2/40) X (1000/250)

= .2 N

Calculate the number of moles & molarity of N/2 500 ml solution of $\text{H}_2 \text{SO}_4$

N = n ×M

1/2=2 ×M

1/4=M

Now Molarity of solution

= No of Moles/Volume in litre

1/4=n/.5

or n = 1/8

Find the equivalent Weight of Potassium permanganate($\text{KMnO}_4$) in redox reaction with Oxalic Acid($\text{H}_2\text{C}_2\text{O}_4.\text{H}_2\text{0}$) in presence of dilute $\text{H}_2 \text{SO}_4$

$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{0}$

$\text{KMnO}_4$ acts as oxidizing agent in the acidic medium and gain electron. Here it gain 5 electron.So valence or equivalence factor is 5

Now Molar Mass of $\text{KMnO}_4$=158

$ \text{Equivalent Weight}= \frac {\text{Molar Mass}}{n} = \frac {158}{5} = 31.6$

N

where N

N

V

V

Suppose we have mix three solution of same solute of Normality N

Normality of the resulting solution is given by

$N_R= \frac {N_1V_1 +N_2V_2 + N_3V_3}{V_1 + V_2 + V_3}$

Suppose we have mix three solution of different solute(acid) of Molarity M

and no of H

$N_R= \frac {n_1M_1V_1 +n_2M_2V_2 + n_3M_3V_3}{V_1 + V_2 + V_3}$

Find out the volumes of two $\text{HCl}$ solutions P (0.5 N) and Q (0.1 N) to be mixed for preparing 2 L of 0.2 N HCl solution

Let y L of P and (2 -y) L of Q are mixed.

N

$0.5 \times y + 0.1 \times (2 -y) = 0.2 \times 2$

$(0.5 -0.1) y = 0.4 -0.2$

$y = 0.5 L$

Therefore,

0.5 L of A and 1.5 L of B should be mixed.

N

Where

N

V

N

V

Molar mass of ionic compound is known as formula Mass i. e. (FM)

So it is the number of Formula mass in gram of solute in 1 litre solution

unit = ‘F’ or ‘Moles L

**Notes**- Some Basic Concepts of Chemistry
- Physical States of matter
- Chemical Classification Of matter
- The International System of Units (SI units)
- Significant Figures
- Laws of Chemical Combination
- Atomic Mass
- Molecular Mass
- Mole Concept (Avogadro Constant) And Molar mass
- Gram Atomic Mass
- Gram Molecular Mass
- Percentage composition
- StioChiometry
- Mole Fraction
- Molarity Definition
- Molality Definition
- Normality Formula

**Questions**

Class 11 Maths Class 11 Physics Class 11 Chemistry Class 11 Biology

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