M is molecular weight of `KMnO_(4)`. The equivalent weight of `KMnO_(4)` when it is converted into `K_(2)MnO_(4)` is

Options

A. Molecular weight /10
B. M/3
C. M/5
D. M/7

Explanation

To find the equivalent weight of `KMnO_(4)` when it is converted into `K_(2)MnO_(4)`, we can follow these steps:

### Step 1: Determine the Molecular Weight of `KMnO_(4)`
The molecular weight (M) of `KMnO_(4)` can be calculated by adding the atomic weights of its constituent elements:
- Potassium (K): approximately 39.1 g/mol
- Manganese (Mn): approximately 54.9 g/mol
- Oxygen (O): approximately 16.0 g/mol (and there are 4 oxygen atoms)

So, the molecular weight is calculated as:
\[ M = 39.1 + 54.9 + (4 \times 16.0) = 39.1 + 54.9 + 64.0 = 158.0 \text{ g/mol} \]

### Step 2: Identify the Change in Oxidation State
Next, we need to determine the change in oxidation state of manganese when `KMnO_(4)` is converted to `K_(2)MnO_(4)`:
- In `KMnO_(4)`, the oxidation state of Mn is +7.
- In `K_(2)MnO_(4)`, the oxidation state of Mn is +6.

The change in oxidation state is:
\[ \Delta \text{Oxidation State} = +7 - (+6) = 1 \]

### Step 3: Determine the n Factor
The n factor is defined as the number of moles of electrons exchanged per mole of the substance during the reaction. In this case, since the change in oxidation state is 1, the n factor is 1.

### Step 4: Calculate the Equivalent Weight
The equivalent weight is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{n \text{ factor}} \]
Substituting the values we have:
\[ \text{Equivalent Weight} = \frac{M}{1} = M \]
Thus, the equivalent weight of `KMnO_(4)` when converted to `K_(2)MnO_(4)` is equal to its molecular weight.

### Final Answer
The equivalent weight of `KMnO_(4)` when it is converted into `K_(2)MnO_(4)` is **M** (the molecular weight of `KMnO_(4)`).

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