Q1. In an alkaline medium, the equivalent weight of KMnO4 is:
A \frac{M}{3} (52.66) Correct
B \frac{M}{5} (31.60)
C \frac{M}{6} (26.33)
D M (158.00)
Explanation
In an alkaline medium, permanganate ion is reduced from manganese in the +7 oxidation state to manganese dioxide, where manganese is +4. The change in oxidation number is 3, so the n-factor of KMnO4 is 3. Equivalent weight equals molar mass divided by n-factor; with molar mass about 158, this gives M/3, or 52.66. M/5 applies to acidic medium, where permanganate is reduced to Mn2+. M/6 has no role in this standard reaction. M would mean n-factor 1, which is not the alkaline reduction.