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

Last updated date: 22nd Jun 2024
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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.
$\text{ Equivalent weight=}\dfrac{\text{Molecular weight}}{\text{No of equivalent moles}}$

$\text{ Equivalent weight of CuS}{{\text{O}}_{\text{4}}}\text{ = }\dfrac{\text{Molecular weight of CuS}{{\text{O}}_{\text{4}}}}{\text{Valence Factor of CuS}{{\text{O}}_{\text{4}}}}\text{= }\dfrac{\text{M}}{\text{Valence Factor of CuS}{{\text{O}}_{\text{4}}}}\text{ }$
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 $\text{ }\dfrac{\text{M}}{2}\text{ }$ .