Chemistry > Stoichiometry > 4.0 Volumetric Analysis

  Stoichiometry
    1.0 The Mole
    2.0 The Limiting Reagent
    3.0 Gravimetric Analysis
    4.0 Volumetric Analysis
    5.0 Calculation of n-factor
    6.0 Redox Reactions
    7.0 Titration
    8.0 Iodimetric and Iodometric Titrations
    9.0 Volume strength peroxide solution
    10.0 Percentage Labeling of Oleum
    11.0 Hardness of Water

4.4 Normality $(N)$

No of equivalents of solute present in one litre of the solution is known as Normality $(N).$

Normality $(N)$ = $\frac{{{\text{No}}{\text{. of equivalents of solute}}}}{{{\text{Volume of solution in litre}}}}$

= $\frac{{{\text{ Wt}}{\text{. of solute/equivalent weight of solute}}}}{{{\text{Volume of solution in litre}}}}$

No. of equivalent of solute $=$ Volume of solution (in litre) $ \times $ normality of solution

No. of milli equivalents of solute $=$ Normality of solution $ \times $ Volume of solution in millilitres
Improve your JEE MAINS score
10 Mock Test
Increase JEE score
by 20 marks
Detailed Explanation results in better understanding
Exclusively for
JEE MAINS and ADVANCED
9 out of 10 got
selected in JEE MAINS
Lets start preparing
DIFFICULTY IN UNDERSTANDING CONCEPTS?
TAKE HELP FROM THINKMERIT DETAILED EXPLANATION..!!!
9 OUT OF 10 STUDENTS UNDERSTOOD