Question

The normality of $0.3M$ $H_{3}P{O}_3$ acid solution is:

• A. 0.1
• B. 0.9
• C. 0.3
• D. 0.6

Answer 1

The correct option is D 0.6

Molarity is the number of moles of solute present in 1 litre of the solution.

$\text{Molarity}=\frac{\text{Number of moles of solute}}{\text{Volume of solution in litre}}$

Normality is the number of gram equivalents of the solute present in 1 litre of the solution.

$\text{Normality}=\frac{\text{Number of equivalents of solute}}{\text{Volume of solution in litre}}$

Given,
$\text{Molarity of phosphorous acid}=0.3M$

We know,
The relation between normality and molarity is,
$\text{Normality}=\text{Molarity}\times \text{n-factor}$
For an acid, the n-factor will be the number of $H^{+}$ released.
$H_{3}P{O}_3$ is a dibasic acid.
Hence, the n-factor is 2
Therefore,
$\text{Normality}=\text{Molarity}\times 2$
$\text{Normality}=0.3\times 2$
Normality = 0.6 N