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What is Normality?

We define normality as the number of grams equivalent to solute that is present in a one-liter solution. So, the unit of normality is gram/liter.

We denote normality with the letter ‘N’.

We can write the Normality Formula as: 


N  =  No. of gram equivalent of the solute/volume of the solution in Liters

         No. of gram equivalent of the solute per volume of the solution in Liters

Now, you might be wondering what gram equivalent is and how it is related to the number of moles.

Let’s understand what is equivalent and its significance as well.

Let’s say we have an equation:

       NaCl              +     H2SO4          →     Na2SO4                +      HCl

(Sodium Chloride) (Sulphuric Acid)   (Sodium Sulfate)    (Hydrochloric Acid)


You can see that this equation is not balanced. Now, let’s balance this equation:

                                 2NaCl  + H2SO4 → Na2SO4 + 2HCl

No of moles (n):   (2 moles)   (1 mole)     (1 mole)  (2 moles)

So, 2 moles of NaCl react with 1 mole of H2SO4 to give 1 mole of Na2SO4 and 2 moles of HCl.

This means, without balancing the equation, we can’t determine the quantity of reactant (or moles) that undergoes a reaction to form a product.

Number of Gram Equivalent

So, let’s get forward with understanding the concept of gram equivalence.

We know that no of moles = Mass/Molecular weight

Number of gram equivalent = Mass/Equivalent weight, 

and Equivalent weight = Molecular weight X

(X = valence factor, where valence factor for acids and bases is the number of H+ and OH- ions they release in the solution, respectively).

We’ll understand these two formulas with an example.

Let’s find out Gram Equivalent

For example, Find the number of gram equivalents present in 0.5 g of HCl.

HCl releases one H+ ion in the solution, so its valence factor = 1.

The molecular weight of HCl = 36.46 g.

So, equivalent weight =

Molecular weight X = 36.46/1 = 36.46 g, and

Number of gram equivalent

= Mass/Equivalent weight 

= 0.5/36.46 = 0.0137

So, we get the number of gram equivalent  = 0.0137

Let us take another example of 1.06 g of Na2CO3 to understand this concept clearly

We are given the mass of Na2CO3 = 1.06 g.

Firstly, Find the equivalent weight of Na2CO3

Since Na2CO3 is a salt, so the number of positive charges on the cation gives X = 2

Molecular weight = 106 g

So, Equivalent Weight = 

Molecular weightX

= 106/2 = 53 g, 

and number of gram equivalent is:

Mass/Equivalent weight

= 1.06/53 = 0.02

In a chemical equation, the number of grams equivalent of both reactions always remains the same.

Normality Chemistry

There are three types of Normality

  1. Seminormal - The solutions whose normality is ½ or N/2.

  2. Binormal - The solutions having normality as 2 or 2 N.

  3. Decinormal - Normality is 1/10 or N/10.

  4. CentiNormal - Normality is 1/100 or N/100.

Normality Calculation Formula

Let’s take an example of how to calculate normality:

If 13 g of N2O4 is present in 500 ml of solution. Find normality.

We are given a mass of N2O4 = 0.65 g, and volume = 500 ml = 0.5 l.

We know that normality, N = no of gram equivalent/volume of solution in liters

Let’s find out equivalent weight to find out the number of gram equivalent for N2O4

Molecular weight of N2O4 = (2 x 14) + (4 x 16) = 28 + 64 = 92 g.

Since the number of negative charges on oxygen = 4. So, X = 4

So, equivalent weight = Molecular weight X = 92/4 = 13 g, 

and Number of gram equivalent =

Mass/Equivalent weight

= 13/13 = 1

Now, let’s calculate the normality, by the formula given below:

N =  No of gram equivalent / Volume of solution in liters = 1/0.5   


N = 2 gram / liter.

Here, the Normality is N = 2, which means the solution of N2O4 is BiNormal.

Normality Equation

For deriving the normality equation, let’s understand normality in mixtures 

Let’s consider two ideal solutions having their normalities as Na and Nb, and the volume as Va and Vb respectively as shown below:

Where Na = The normality of the acidic solution,

             Va = Volume of the acidic solution, 

             Nb = Normality of the basic solution, and

             Vb = Volume of the basic solution.

(Image Will Be Updated Soon)

(Image Will Be Updated Soon)

On combining these two solutions, we get a mixture whose volume is Va + Vb and the normality as N.

So, we got the Normality Formula for the mixture as:  

N = (Na Va + Nb Vb)/ (Va + Vb)

Let’s consider three Cases:

Case 1: The concentration of the acidic solution > concentration of the basic solution (the release of H+ ions > OH- ions)

So,  Na Va >  Nb Vb

Case 2: The concentration of the basic solution < acidic solution, then

Na Va <  Nb Vb (release of OH- ions > H+ ions)

Case 3: When concentration is equal, then

 Na Va = Nb Vb is the normality equation.

In this case, there is no release of both OH- ions and H+ ions.

This means the number of gram equivalent of H+ ions = the number of gram equivalent of OH- ions. Such a type of solution is neutral and this process is called neutralization.

JEE Main Chemistry Chapter-wise Solutions 2022-23 


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FAQs on Normality

1. What is the simple definition of the Gram Equivalent?

The gram equivalent is the mass or the amount of substance (ion or molecule) that will combine with or displace a fixed quantity of another substance. It is a quantity that is numerically equal to the equivalent. In a reaction, it is the amount of one substance that reacts with one mole of another substance. This is often but not always in the mole ratio of 1:1. To calculate the Gram equivalent weight, the formula used is Eq= MW/n, i.e., the formula is Normality = (Wt. of a solute taken in a solvent in gram * 1000) / (Equivalent Wt. of that solute * Volume of the solution in ml) where equivalent weight is the ratio of Molecular weight of solute to the valency of the compound. 

2. Write an example of Normality.

We know that the normality of a solution is the gram equivalent of a solute in per liter solution. Normality is used specifically for acids and bases. The normality of a solution is never less than its molarity. There are two individual formulas for both an acid solution and a basic solution. For acid solution, n is the number of H+ ions given by a formula unit of acid and for a basic solution, n is the number of OH- ions given by a formula unit of a base. . The valency is generally measured from the no of H+ or OH- ions transferred in a solution. For HCl (Hydrochloric acid) valency is one because when you dissolve it in the water you get one H+ ion. For Sulphuric acid (H2SO4) the valency is 2 as you get 2 H+ ions when you dissolve it in water. For NaOH valency is one as it gives one OH- ion when it dissolves in water. Let’s say a molecule, HNO3 is having a normality of 0.5. So, we can express its concentration as 0.5 N HNO3.

3. What is the Formula for Molarity?

Molarity is defined as the number of moles of solute per liter of solution and is one of the most common units used to measure the concentration of a solution. It can also be used to calculate the volume of a solvent or the amount of a solute. The molarity equation can be rearranged to solve for moles which can then be converted into grams. The Formula for Molarity is:

Molarity (or M) = The number of moles of solute/volume of solution in L.

4. Derive the relationship between Normality and Molarity.

We already know that:

Normality = No of gram equivalent in a solute / Volume of solution (in L), and 

Molarity (or M) = The number of moles of solute / Volume of solution in L

Now, N/M = No of gram equivalent / Number of moles of solute…(1)

Since number of Gram equivalent = Mass / Equivalent mass, and

Number of Moles = Mass / Molecular mass

From (1), N/  M= Molecular mass / Equivalent mass…(2), and

Equivalent mass = Molecular mass / X

From (2), we get,

                N / M = X

This means

     N = M * X

     Normality = Molarity * Valency factor

5. How do we study Normality?

Chemistry is all about formulas, chemical reactions, and equations. The chapter on             Normality is So students must first try and understand the concepts before trying to     memorize the reactions which are not very easy to do. Practice the formulas through   numerical as well as study the chemical reactions through question banks. Also, follow   Previous Years’ question papers, sample papers, and other study materials which are   available on Vedantu’s official website in downloadable PDF format for free so that   students can have easy access to the resources.