Normality of solution is the number of equivalent weights of solute in a liter of solution. Normality of solution is expressed as,

Normality of solution = $\frac{number \ of \ equivalents \ of \ solute}{one \ litre \ of \ solution}$

number of equivalents of solute = $\frac{grams \ of \ solute}{equivalent \ weight \ of \ solute}$

Plug in the values, according to the question, into the Normality calculator and get the normality of the solution. Here you can see two calculators. If the mass is given then you can use the second calculator.

Normality of solution = $\frac{number \ of \ equivalents \ of \ solute}{one \ litre \ of \ solution}$

number of equivalents of solute = $\frac{grams \ of \ solute}{equivalent \ weight \ of \ solute}$

Plug in the values, according to the question, into the Normality calculator and get the normality of the solution. Here you can see two calculators. If the mass is given then you can use the second calculator.

**Step 1 :**From the question, write down the values given.

**Step 2 :**Apply the formula of normality given above and get the normality of solution by substituting the values into the formula.

Solved problems based on normality are given below.

### Solved Examples

**Question 1:**Calculate the normality of the solution that contains 5.300 g/L of $Na_{2}CO_{3}$, carbonate reacts with two protons.

**Solution:**

Equivalent weight = $\frac{105.99}{2}$ = 53.00

Equivalent of solute = $\frac{5.300}{53.00}$ = 0.1000

Normality = $\frac{0.1000}{1L}$ = 0.1000 N

**Question 2:**Calculate the normality of the solution that conatins 5.267 g/L $K_{2}Cr_{2}O_{7}, if $Cr^{6+}$ is reduced to $Cr^{3+}$.

**Solution:**

Equivalent weight = $\frac{294.19 g/mol}{6 eq/mol}$

Equivalent of solute = $5.267 g/$ $\frac{294.19 g/mol}{6 eq/mol}$

Normality = $\frac{294.19 g/mol}{\frac{6 eq/mol}{1L}}$