The Molarity of a dibasic acid is M. The equivalent weight of the acid is E. The amount of the acid present in 500 ml of the solution is:
A. \[2\times M\times E\]
B. \[M\times E\]
C. \[\dfrac{M\times E}{2}\]
D. \[\dfrac{M\times E}{4}\]

Hint: To solve this question we should know about dibasic acid. If we want to find an amount of acid in solution, we can use the concept of molarity. Molarity equation will lead us to the answer.

Complete step by step answer:

At first, we will know about the dibasic acid.
From the term dibasic acid, we can commonly refer to any substance that can donate two protons or hydrogen (\[{{H}^{+}}\]) ions per molecule in an acid base reaction. Due to this property, the dibasic acid is also known as diprotic acids. A dibasic acid yields two free hydrogen ions in solution for each molecule of acid ionized or in other terms, has two replaceable hydrogen atoms. A simple example of a dibasic acid is sulphuric acid (\[{{H}_{2}}S{{O}_{4}}\text{ }\]).
\[{{H}_{2}}S{{O}_{4}}\text{ }\to \text{ }2{{H}^{+}}+\text{ }SO_{4}^{2-}\]
Number of replaceable hydrogen or the number of equivalents is 2.
We will use molarity to find this question.
$Molarity(M)=\dfrac{number\,of\,moles\,of\,solute(n)}{volume\,of\,solution\,in\,litres}$\[\]
$M=\dfrac{n}{\dfrac{500}{1000}}=2n$ $\to Molarity=2\times moles\,of\,solute$
$\to Number\,of\,moles=\dfrac{Molarity}{2}$
And we know that $\to $$Number\,of\,moles=\dfrac{weight}{molar\,mass}$
\[\to Number\,of\,moles=\dfrac{weight}{molar\,mass}=\dfrac{Molarity}{2}\]
\[\to weight\,\,of\,solute\,present\,in\,solution=\dfrac{Molarity}{2}\times molar\,mass\] (Equation 1st)
Weight of solute present in solution= amount of acid present.
And we know about equivalent weight that,
\[Equivalent\,weight=\dfrac{molar\,mass}{number\,of\,equivalents}\] (Equation 2nd)
Here, molar mass will be of dibasic acid and the number of equivalents will be 2. Because it can donate two protons or hydrogen (H+) ions per molecule in an acid base reaction.
So, our equation 2nd will look like this:
\[Equivalent\,weight=\dfrac{molar\,mass\,of\,dibasic\,acid}{2}\]
So, by this our 1st equation will be like this:
\[\to weight\,\,of\,acid\,present\,in\,solution=Molarity\times \dfrac{molar\,mass\,of\,dibasic\,acid}{2}\]
\[Amount\,of\,acid\,present\,in\,solution=Molarity\times Equivalent\,weight\]
So, our correct option will be B.

Note: In this question we use the concept of both molarity and equivalent weight. There is one important difference we used in this question that is weight and mass. Mass is the amount of matter in something, while weight is the gravitational pull on a mass. We never use molar mass to find actual weight of solute or solvent present in solution.