Question

Equivalent weight of $\text{ CuS}{{\text{O}}_{\text{4}}}\text{ }$ in terms of its molecular weight M in the following reaction is:
A) M
B) $\text{ }\dfrac{\text{M}}{2}\text{ }$
C) $\text{ }\dfrac{\text{M}}{4}\text{ }$
D) $\text{ }\dfrac{\text{M}}{8}\text{ }$

Answer 1

Hint: Equivalent weight of the compound is defined as the molecular weight of the compound per number of equivalent molecules of it. Equivalent moles are also called the valence factor which depends upon the chemical properties of the compound. For example, for ions the number of electrons gain or the loss by the species.
$\text{ Equivalent weight=}\dfrac{\text{Molecular weight}}{\text{No of equivalent moles}}$

Complete Step by step answer:
Equivalent weight is of an element, molecule ,ion or the radical is equal to the atomic weight or the molecular weight divided by its valence number .Valence number are also known as the n factor or the number of equivalent of moles .the value of valence number depend on the type of the reaction in which the it is employed. Let consider a molecular weight of copper sulphate as M .we know that the equivalent weight of a compound is written in terms of molecular weight of copper sulphate as follows,
$\text{ Equivalent weight of CuS}{{\text{O}}_{\text{4}}}\text{ = }\dfrac{\text{Molecular weight of CuS}{{\text{O}}_{\text{4}}}}{\text{Valence Factor of CuS}{{\text{O}}_{\text{4}}}}\text{= }\dfrac{\text{M}}{\text{Valence Factor of CuS}{{\text{O}}_{\text{4}}}}\text{ }$
We are interested to determine the valence factor for the copper sulphate $\text{ CuS}{{\text{O}}_{\text{4}}}\text{ }$ .The valance factor for the compounds which are made of the ions is equal to the valency or the number of electrons loss or gain by the species.
Here, information of copper sulphate copper loses its two electrons an exhibits the $\text{ +2 }$ charge on it.Thus $\text{ +2 }$ is a valence factor for the copper .similarly the sulphate ion takes this two electrons. Thus the total number of electrons lost by copper and gain by sulphate is two. Thus the valence factor for copper sulphate 2 .Thus the equivalent weight of copper sulphate is equal to,$\text{ Equivalent weight of CuS}{{\text{O}}_{\text{4}}}=\dfrac{\text{M}}{2}\text{ }$
Thus, the equivalent weight of $\text{ CuS}{{\text{O}}_{\text{4}}}\text{ }$ in terms of its molecular weight M in the following reaction is $\text{ }\dfrac{\text{M}}{2}\text{ }$ .

Hence, (B) is the correct option.

Note: Note that, the concept of equivalent weight is used to determine the fact that atoms combine in a fixed number ratios to form a molecule, not in the mass ratio. It means that when certain masses of the elements combine even have the difference in the masses they combine in such a way that the number of atoms is used to determine how much the element reacts to give the mass. Here copper sulphate equivalent weight depends on the number of electrons. For acid and bases, it is based on the basicity and acidity respectively.