A) \[M\]
B) \[\frac{M}{3}\]
C) \[\frac{M}{5}\]
D) \[\frac{M}{7}\]
Correct Answer: A
Solution :
In this conversion, the oxidation-state change of Mn is equal to 1. \[\overset{+\,7}{\mathop{KM{{O}_{4}}}}\,\xrightarrow{{}}\overset{+\,6}{\mathop{{{K}_{2}}Mn{{O}_{4}}}}\,\] Hence, equivalent weight of \[KMn{{O}_{4}}=\frac{\text{molecular}\,\text{mass}}{\text{change}\,\text{in}\,\text{oxidation}\,\text{state}}\] \[=\frac{M}{1}=M\]
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