Normality of 0.3 M phosphorus acid (${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}}$) is:
(A) 0.5
(B) 0.6
(C) 0.9
(D) 0.1

Hint: Normality is a term used for indicating the concentration of a solution. Here, you can use the formula ${{Normality = Molarity \times n-factor}}$ for finding the normality from the given data.

Complete step by step answer: Here, in the question, the molarity of the solution is given. So, we can use a simple formula for normality. For using the formula:${{Normality = Molarity \times n-factor}}$
We need to know what n-factor is. N-factor is the acidity of a base or the basicity of an acid.
The acid given is Phosphorus acid whose n-factor is 2. Phosphorus acid has a basicity of 2. This is because ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}}$ can donate two ${\text{O}}{{\text{H}}^{\text{ - }}}$ ions or it contains 2 replaceable ${{\text{H}}^{\text{ + }}}$ions.
So, by substituting the above values in the formula, we get:
${{Normality = Molarity \times n - factor = 0}}{{.3 \times 2 = 0}}{\text{.6N}}$${{Normality = Molarity \times n - factor = 0}}{{.3 \times 2 = 0}}{\text{.6N}}$

So, the normality of the given phosphorus acid is 0.6 N.

Additional Information:
Normality is mainly used as a measure of reactive species in a solution and during titration reactions or particularly in situations involving acid-base chemistry.
As per the standard definition of normality, it is described as the number of gram or mole equivalents of solute present in one litre of a solution. When we say equivalent, it is the number of moles of reactive units in a compound.

Note: When you are taking the values for n-factor, it is the basicity or the number of replaceable ${{\text{H}}^{\text{ + }}}$ ions in the acid. In case of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}}$, you may think that there is 3 replaceable ${{\text{H}}^{\text{ + }}}$. But when you see the structure of phosphorus acid, there are only 2 replaceable ${{\text{H}}^{\text{ + }}}$.