In alkaline medium , `KMnO_(4)` reacts as follows
`2KMnO_(4)+2KOH rarr 2K_(2)MnO_(4)+H_(2)O+O`
Therefore, the equivalent mass of `KMnO_(4)` will be

Options

A. 31.6
B. 52.7
C. `79.0`
D. `158.0`

Answer

To find the equivalent mass of KMnO₄ in the given reaction, we will follow these steps:

### Step 1: Write down the reaction
The reaction in alkaline medium is given as:
\[ 2 \text{KMnO}_4 + 2 \text{KOH} \rightarrow 2 \text{K}_2\text{MnO}_4 + \text{H}_2\text{O} + \text{O} \]

### Step 2: Determine the oxidation states of manganese in KMnO₄ and K₂MnO₄
1. In KMnO₄:
- Let the oxidation state of manganese (Mn) be \( x \).
- The oxidation state of potassium (K) is +1 and that of oxygen (O) is -2.
- The overall charge of KMnO₄ is 0.
- Therefore, the equation is:
\[ +1 + x + 4(-2) = 0 \implies x - 8 + 1 = 0 \implies x = +7 \]

2. In K₂MnO₄:
- Let the oxidation state of manganese (Mn) be \( y \).
- The oxidation state of potassium (K) is +1 and that of oxygen (O) is -2.
- The overall charge of K₂MnO₄ is 0.
- Therefore, the equation is:
\[ 2(+1) + y + 4(-2) = 0 \implies 2 + y - 8 = 0 \implies y = +6 \]

### Step 3: Calculate the change in oxidation state
- The change in oxidation state for manganese from KMnO₄ to K₂MnO₄ is:
\[ \text{Change} = +7 - (+6) = 1 \]
This indicates that one electron is gained in the reaction.

### Step 4: Determine the n-factor
- The n-factor is defined as the number of electrons lost or gained per molecule of the substance. Here, since one electron is gained:
\[ \text{n-factor} = 1 \]

### Step 5: Calculate the molar mass of KMnO₄
- The molar mass of KMnO₄ can be calculated as follows:
- K: 39 g/mol
- Mn: 55 g/mol
- O: 16 g/mol × 4 = 64 g/mol
\[ \text{Molar mass of KMnO}_4 = 39 + 55 + 64 = 158 \text{ g/mol} \]

### Step 6: Calculate the equivalent mass
- The equivalent mass is given by the formula:
\[ \text{Equivalent mass} = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{158 \text{ g/mol}}{1} = 158 \text{ g} \]

### Conclusion
The equivalent mass of KMnO₄ is **158 g**.