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

The Correct Option is A

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 $

Answer 2

Equivalent mass = \(\frac {Molecular\  mass}{n- factor}\)
In  this, n- factor = change in the O.N of any element 
The oxidation state KMnO₄ is +7 and  that of K₂MnO₄ is +6. 
The value for n-factor is 1 
Equivalent mass =\(\frac M1\)

So, the correct answer is (A): M