Question

The molecular mass of $KMn{{O}_{4}}$ is M. The equivalent weight of $KMn{{O}_{4}}$ , when it is converted into $KMn{{O}_{4}}$ is:

• A. $M$
• B. $\frac{M}{3}$
• C. $\frac{M}{5}$
• D. $\frac{M}{7}$

Answer 1

Answer: (A)

$M$

Answer 2

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 $KMn{{O}_{4}}=\frac{\text{molecular}\,\text{mass}}{\text{change}\,\text{in}\,\text{oxidation}\,\text{state}}$ $=\frac{M}{1}=M$