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) $MnO _4^{2-}$
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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,
BETA
