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 KMnO₄ when it is converted to MnSO₄, we can follow these steps:

### Step 1: Understand Equivalent Weight
The equivalent weight of a substance is defined as the molar mass divided by the number of equivalents (n factor). The formula is:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{n \text{ factor}} \]

### Step 2: Determine the Molar Mass of KMnO₄
The molar mass of KMnO₄ can be calculated by adding the atomic masses of its constituent elements:
- K (Potassium) = 39 g/mol
- Mn (Manganese) = 55 g/mol
- O (Oxygen) = 16 g/mol × 4 = 64 g/mol

Thus, the molar mass of KMnO₄ is:
\[ \text{Molar Mass of KMnO₄} = 39 + 55 + 64 = 158 \text{ g/mol} \]

### Step 3: Determine the n Factor
The n factor is the number of electrons exchanged in the reaction. To find this, we need to determine the oxidation states of manganese in both KMnO₄ and MnSO₄.

1. **In KMnO₄**:
- Let the oxidation state of Mn be \( x \).
- The equation based on the oxidation states is:
\[ 1 + x + 4(-2) = 0 \quad \Rightarrow \quad 1 + x - 8 = 0 \quad \Rightarrow \quad x = +7 \]

2. **In MnSO₄**:
- The oxidation state of Mn in MnSO₄ is +2 (since SO₄ has a charge of -2).

### Step 4: Calculate the Change in Oxidation State
The change in oxidation state of Mn from KMnO₄ to MnSO₄ is:
\[ \text{Change} = +7 - (+2) = 5 \]
This indicates that 5 electrons are exchanged.

Thus, the n factor is 5.

### Step 5: Calculate the Equivalent Weight
Now that we have the molar mass and the n factor, we can calculate the equivalent weight:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass of KMnO₄}}{n \text{ factor}} = \frac{158 \text{ g/mol}}{5} = 31.6 \text{ g/equiv} \]

### Final Answer
The equivalent weight of KMnO₄ when it is converted into MnSO₄ is \( \frac{M}{5} \), where \( M \) is the molar mass of KMnO₄.

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