Step 1: Identify the oxidation state change.
In acidic medium, potassium permanganate acts as a strong oxidising agent, and manganese is reduced from the +7 state in \( \text{MnO}_4^- \) down to the +2 state as \( \text{Mn}^{2+} \).
Step 2: Write the half-reaction and find the n-factor.
\[ \text{MnO}_4^- + 8H^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4H_2O \]
Each \( \text{MnO}_4^- \) ion gains 5 electrons, so the n-factor is 5.
Step 3: Calculate the equivalent weight.
Equivalent weight equals molar mass divided by the n-factor. The molar mass of \( \text{KMnO}_4 \) is 158 g/mol, so
\[ \text{Equivalent weight} = \frac{158}{5} = 31.6 \text{ g/equiv} \]
Step 4: Final Answer.
\[ \boxed{31.6} \]