A) molecular weight
B) \[\frac{molecular\text{ }weight}{2}\]
C) \[\frac{molecular\text{ }weight}{3}\]
D) \[\frac{molecular\text{ }weight}{5}\]
Correct Answer: C
Solution :
Equivalent weight of KMn04 in neutral medium is as: \[\overset{+7}{\mathop{2KMn{{O}_{4}}}}\,+{{H}_{2}}O\xrightarrow{{}}2KOH+\overset{+4}{\mathop{2Mn{{O}_{2}}}}\,+3\,[O]\] \[\therefore \] \[Equivalent\text{ }weight=\frac{molecular\text{ }weight}{3}\]You need to login to perform this action.
You will be redirected in
3 sec