`KMnO_(4)` (m.w.=158) oxidises oxalic acid in acid medium to `CO_(2)` and water as follows :
`5C_(2)O_(4)^(2-)+2MnO_(4)^(-)+16H^(+) to 10 CO_(2)+2Mn^(2+)+8H_(2)O`
What is the equivalent weigth of `KMnO_(4)` ?

Options

A. 158
B. 31.6
C. 39.5
D. 79

Answer

To find the equivalent weight of `KMnO4`, we can follow these steps:

### Step 1: Understand the Reaction
The reaction given is:
\[ 5C_2O_4^{2-} + 2MnO_4^{-} + 16H^{+} \rightarrow 10CO_2 + 2Mn^{2+} + 8H_2O \]

### Step 2: Determine the Change in Oxidation State
In the reaction, manganese in `KMnO4` (where Mn is in +7 oxidation state) is reduced to `Mn^{2+}` (where Mn is in +2 oxidation state).

- Change in oxidation state for Mn:
\[ \text{Change} = +7 - (+2) = 5 \]

### Step 3: Calculate the Equivalent Weight
The equivalent weight of a substance is calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n}} \]
where \( n \) is the number of electrons transferred per molecule of the substance in the reaction.

For `KMnO4`, the molecular weight is given as 158 g/mol, and since 2 moles of `KMnO4` are involved in the reaction, the total change in oxidation state for manganese is:
\[ n = 2 \times 5 = 10 \]

Now, substituting the values into the formula:
\[ \text{Equivalent weight of KMnO4} = \frac{158}{5} = 31.6 \text{ g/equiv} \]

### Final Calculation
Thus, the equivalent weight of `KMnO4` is:
\[ \text{Equivalent weight} = 31.6 \text{ g/equiv} \]

### Conclusion
The equivalent weight of `KMnO4` is 31.6 g/equiv.

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