Equivalent weight of `KmNO_(4)` when it is convert into `MnSO_(4)` is Where M = molar mass of `KMnO_(4)`.

Options

A. `M//5`
B. `M//6`
C. `M//3`
D. `M//2`

Answer

To find the equivalent weight of \( KMnO_4 \) when it is converted into \( MnSO_4 \), we can follow these steps:

### Step 1: Determine the oxidation states
In \( KMnO_4 \), manganese (Mn) has an oxidation state of +7. When it is converted to \( MnSO_4 \), manganese has an oxidation state of +2.

### Step 2: Calculate the change in oxidation state
The change in oxidation state for manganese is:
\[ \text{Change} = +7 - (+2) = 5 \]
This means that manganese undergoes a reduction of 5 units.

### Step 3: Identify the valency factor (n)
The valency factor (n) is equal to the change in oxidation state, which we calculated as 5. This indicates that 5 moles of electrons are involved in the reduction of 1 mole of \( KMnO_4 \) to \( MnSO_4 \).

### Step 4: Calculate the equivalent weight
The equivalent weight is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{\text{Valency Factor}} \]
Let \( M \) be the molar mass of \( KMnO_4 \). Therefore, the equivalent weight can be expressed as:
\[ \text{Equivalent Weight} = \frac{M}{5} \]

### Final Answer
The equivalent weight of \( KMnO_4 \) when converted into \( MnSO_4 \) is \( \frac{M}{5} \).

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