$ ext{With reference to the reaction } KMnO_4 + H_2C_2O_4 + .... ightarrow MnO_2 + CO_2 + ...., ext{ the equivalent weight of } KMnO_4 ext{ would be :}
ext{[Molar mass of } KMnO_4 = 158]$

Q: $\text{With reference to the reaction } KMnO_4 + H_2C_2O_4 + .... \rightarrow MnO_2 + CO_2 + ...., \text{ the equivalent weight of } KMnO_4 \text{ would be :} \text{[Molar mass of } KMnO_4 = 158]$

$\text{With reference to the reaction } KMnO_4 + H_2C_2O_4 + .... \rightarrow MnO_2 + CO_2 + ...., \text{ the equivalent weight of } KMnO_4 \text{ would be :} \text{[Molar mass of } KMnO_4 = 158]$

Question

$	ext{With reference to the reaction } KMnO_4 + H_2C_2O_4 + .... ightarrow MnO_2 + CO_2 + ...., 	ext{ the equivalent weight of } KMnO_4 	ext{ would be :}
	ext{[Molar mass of } KMnO_4 = 158]$

Accepted Answer

To determine the equivalent weight of $ KMnO_4 $ in the given reaction, we need to understand the role of $ KMnO_4 $ in the reaction and how it changes its oxidation state.

### Explanation:

1. **Reaction Context:**
   - The reaction involves $ KMnO_4 $ reacting with $ H_2C_2O_4 $ (oxalic acid) to produce $ MnO_2 $ and $ CO_2 $.
   - In this reaction, $ KMnO_4 $ acts as an oxidizing agent.

2. **Oxidation State Changes:**
   - In $ KMnO_4 $, manganese ($ Mn $) is in the +7 oxidation state.
   - In $ MnO_2 $, manganese is in the +4 oxidation state.
   - Therefore, the change in oxidation state for manganese is from +7 to +4.

3. **Calculation of Equivalent Weight:**
   - The equivalent weight of a substance in a redox reaction is calculated as the molar mass divided by the change in oxidation number per molecule.
   - For $ KMnO_4 $, the change in oxidation state is $ 7 - 4 = 3 $.
   - The molar mass of $ KMnO_4 $ is given as 158.
   - Therefore, the equivalent weight of $ KMnO_4 $ is calculated as:
     $$
     \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{Change in oxidation number}} = \frac{158}{3} \approx 52.66
     $$

4. **Justification for Option 2:**
   - The calculation above shows that the equivalent weight of $ KMnO_4 $ is approximately 52.66, which matches Option 2.

### Explanation of Incorrect Options:

- **Option 1 (158):**
  - This option suggests that the equivalent weight is equal to the molar mass, which would only be true if there was no change in oxidation state. However, since the oxidation state changes from +7 to +4, this option is incorrect.

- **Option 3 (22.57):**
  - This option implies a much larger change in oxidation state than what occurs in the reaction. The calculation does not support this value, making it incorrect.

- **Option 4 (39.50):**
  - This option suggests a different change in oxidation state than what is observed. The calculation of equivalent weight based on the actual change in oxidation state does not yield this value, so it is incorrect.

In conclusion, the correct equivalent weight of $ KMnO_4 $ in the given reaction is 52.66, which is why Option 2 is the correct answer.