Question

In an Alkaline medium ${\mathrm{KMnO}}_4$ reacts as follows: $2{\mathrm{KMnO}}_4+2\mathrm{KOH}\to 2{\mathrm{K}}_2{\mathrm{MnO}}_4+{\mathrm{H}}_2\mathrm{O}+\mathrm{O}$. Find the Equivalent weight of ${\mathrm{KMnO}}_4$

Answer 1

Step 1: Finding molecular weight of ${\mathrm{KMnO}}_4$

Molecular weight$=19+55+4\left(16\right)$

Step 2: Finding the number of electrons gained by ${\mathrm{KMnO}}_4$

${\mathrm{MnO}}_4^{-}+{\mathrm{e}}^{-}\to {\mathrm{MnO}}_4^{2-}$

So, the number of electrons gained per molecule of ${\mathrm{KMnO}}_4$ = 1

Step 3: Finding the Equivalent weight of ${\mathrm{KMnO}}_4$

$\mathrm{Equivalent}\mathrm{weight}\mathrm{of}{\mathrm{KMnO}}_4=\frac{\mathrm{Molecular}\mathrm{weight}}{\mathrm{No}.\mathrm{of}\mathrm{electrons}\mathrm{gained}\mathrm{by}{\mathrm{MnO}}_4}$

The mass of substance in grams that combines with or is equivalent to eight grams of oxygen or one gram of hydrogen.

Therefore, equivalent weight of ${\mathrm{KMnO}}_4$is = 158.