question_answer

In alkaline medium, KMn04 reacts as follows\[2KMn{{O}_{4}}+2KOH\xrightarrow[{}]{{}}2{{K}_{2}}Mn{{O}_{4}}\] \[+{{H}_{2}}O+[O]\] Therefore, the equivalent weight of\[KMn{{O}_{4}}\]will be

A)  31.6                                      

B)  52.7

C)  7.0                                        

D)  158.0

Correct Answer: D

Solution :

                Equivalent wt. of\[KMn{{O}_{4}}=\frac{molecular\text{ }weight}{n}\]                                                 \[=\frac{158}{1}=158.0\]