Question

Normality of 2M sulphuric acid is:
$A.2N \\\ B.4N \\\ C.\dfrac{N}{2} \\\ D.\dfrac{N}{4} \\\$

Answer 1

Hint – In this question, it is important to know the formula of Normality in terms of Morality that is Normality $=$ ${e^ - }$ transfer $\times$ Morality.

Complete answer:
Normality is defined 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.
As we know that the chemical name of the chemical formula, ${H_2}S{O_4}$ that is Sulphuric acid.
And here we have to find the Normality (N) can be defined as the number of gram or mole equivalents of solute present in 1 litre of solution.
And in terms of Molarity it is given by
Normality (N) $=$ Molarity $\times$ No. of Hydrogen (or hydroxide) ions.
Here number of replaceable ${H^ + }$ions in ${H_2}S{O_4}$that is
$n = 2$
Equivalent weight (E) of sulphuric acid $=$ molecular weight of sulphuric acid
$\dfrac{M}{{1 \times 2}} = \dfrac{M}{2}$
So $M = 2E$
$2M$ sulphuric acid means,
$1L$solution contains $2M$ sulphuric acid $= 2 \times \left( {2E} \right)$sulphuric acid
$= 4E$ sulphuric acid
And, Molarity (M) is also $2M$
Therefore, Normality $\left( n \right)$will be equal to $2 \times 2 = 4n$
Hence, B is the correct option.

Note – Whenever we come up with this type of problem, one must know that the morality of an acid or base solution, one can easily convert it to Normality by multiplying Morality by the number of hydrogen ions in the acid.