A) M
B) M/3
C) M/5
D) M/7
Correct Answer: A
Solution :
[a]: \[KMn{{O}_{4}}\to {{K}_{2}}Mn{{O}_{4}}\] Equivalent weight of \[KMn{{O}_{4}}\] \[=\frac{\text{Molecular weight}}{\text{Change in oxidation number}}=\text{M}\]You need to login to perform this action.
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