Some Basic Concepts of Chemistry - Result Question 23

23. The equivalent weight of $MnSO _4$ is half of its molecular weight, when it converts to

(1988,1 M)

(a) $Mn _2 O _3$

(b) $MnO _2$

(c) $MnO _4^{-}$

(d) $MnO _4^{2-}$

Show Answer

Answer:

Correct Answer: 23. (b)

Solution:

Equivalent weight in redox system is defined as:

$\hspace{12mm} E= \frac{\text{Molar mass}}{n-factor}$

Here n-factor is the net change in oxidation number per formula unit of oxidising or reducing agent. In the present case, n-factor is 2 because equivalent weight is half of molecular weight. Also,

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