The Molarity of a dibasic acid is $M$. The equivalent weight of the acid is$E$. The amount of the acid present in  $500 \mathrm{ml}$  of the solution is

We know that, Molarity $\left(M\right)=\frac{\text{ number of moles of solute }(\text{ n) }}{\text{ volume of solution in litres }}$

where,

$M=$ Molarity (Given)

$\mathrm{n}=$ no. of moles

$\mathrm{volume} =500\mathrm{ml} =\frac{500}{1000} \mathrm{l}$ (Given)

$M=\frac{n}{\frac{500}{1000}}=2n\to$ Molarity $=2\times$ moles of solute

$\to Number of moles$  $=\frac{\text{ Molarity }}{2}$

And we know that$\to$ Number of moles $=\frac{\text{ weight }}{\text{ molar mass }}$

$\to Number of moles$ $=\frac{\text{ weight }}{\text{ molar mass }}=\frac{\text{ Molarity }}{2}$

$\to$ weight of solute present in solution $=\frac{\text{ Molarity }}{2}\times$ molar mass (Equation1st)

Weight of solute present in solution= amount of acid present.

And we know about equivalent weight that,

Equivalent weight $=\frac{\text{ molar mass }}{\mathrm{basicity}}$(Equation 2nd)

Here, the number of equivalents will be 2 of dibasic acid because per molecule, it can donate two protons or hydrogen$\left({\mathrm{H}}^{+}\right)$ ions in the acid base reaction.

Rearranging the equation 2 we have:

Equivalent weight $=\frac{\text{ molar mass of dibasic acid }}{2}$

Again, rearranging equation 1 we have:

$\to$ weight of acid present in solution $=$Molarity $\times \frac{\text{ molar mass of dibasic acid }}{2}$

Amount of acid present in solution$=$$\mathrm{Molarity} \times \mathrm{Equivalent} \mathrm{weight}$ 

Hence, option(b) is correct.