What is the ‘n’ factor for acid and how is it calculated for oxalic acid dihydrate?

Hint: The n-factor is defined as the number of $ {H^ + } $ ions replaced by $ 1 $ mole of acid in a reaction and the number of $ O{H^ - } $ replaced by $ 1 $ mole of base in a reaction.The molecular formula of oxalic acid dihydrate is $ {C_2}{H_2}{O_4}.2{H_2}O $ .

Complete answer:
For an acid ‘n’ factor is the number of $ {H^ + } $ ions replaced by $ 1 $ mole of an acid
The factors related to oxalic acid are:
Oxalic acid is a dibasic acid.
Its basicity is 2 because it has two protons that can be replaced.
For an oxalic acid let us consider the following reaction as example:
 $ COOH - COOH \to 2C{O_2} + 2{e^ - } + 2{H^ + } $
From the above reaction we can say that the “n” factor for oxalic acid dihydrate is $ 2 $ as $ 2 $ hydrogen ions are replaced in the reaction.
Equivalent weight of acid = $ \dfrac{\text{Molecular weight of acid}} {\text{Number of Replaceable hydrogen ions}} $
In Oxalic acid, the number of replaceable hydrogen ions is $ 2 $ .
Molecular weight of oxalic acid is $ 126 $ .
Equivalent weight of acid $ = \dfrac{{126}}{2} $
Equivalent weight of oxalic acid $ = 63 $
Oxalic acid is acidic in nature. As oxalic acid has $ 2 $ moles of acidic $ {H^ + } $ ions, which means one mole of oxalic acid can neutralize two moles of $ O{H^ - } $ .
Hence, the n-factor of oxalic acid is $ 2 $ .

Note:
Oxalic acid is a dicarboxylic acid. It is also known as Ethanedioic acid. It is called a dibasic acid because it has the ability to donate two $ {H^ + } $ ions. One molecule of oxalic acid releases two hydrogen ions. Hence, n-factor is two.