Q. If molecular weight of ${KMnO}_{4}\ is\ M,$ then its equivalent weight in acidic medium would be
A) $M$
B) $\frac{M}{2}$
C) $\frac{M}{5}$
D) $\frac{M}{3}$
Explanation
(c)
$MnO_{4}^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2 +} + 4H_{2}O$
Gain electrons=5
Molecular weight=$M$
Equivalent weight=$\frac{molecular\ weight}{gain\ electron} = \frac{M}{5}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.