Molecular weight of `KMnO_(4)` in acidic medium and neutral medium will be respecitvely -

Options

A. 7 `xx` equivalent weight and 2 `xx` equivalent weight
B. 5 `xx` equivalent weight and 4 `xx` equivalent weight
C. 4 `xx` equivalent weight and 25`xx` equivalent weight
D. 2 `xx` equivalent weight and 4 `xx` equivalent weight

Answer

To find the molecular weight of `KMnO4` in acidic and neutral mediums, we will follow these steps:

### Step 1: Determine the balanced redox reaction in acidic medium.
In acidic medium, the reduction of `KMnO4` can be represented as:
\[ \text{MnO}_4^- + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]

### Step 2: Calculate the change in oxidation state of manganese.
In `KMnO4`, manganese (Mn) has an oxidation state of +7. In the product `Mn^{2+}`, the oxidation state is +2. The change in oxidation state is:
\[ +7 \text{ to } +2 = -5 \]
This indicates that 5 electrons are gained.

### Step 3: Determine the n-factor in acidic medium.
The n-factor is defined as the number of electrons transferred in the redox reaction. From the balanced equation, we see that:
\[ n = 5 \]

### Step 4: Calculate the molar mass of `KMnO4`.
The molar mass of `KMnO4` can be calculated as follows:
- Potassium (K) = 39 g/mol
- Manganese (Mn) = 55 g/mol
- Oxygen (O) = 16 g/mol × 4 = 64 g/mol

So, the total molar mass is:
\[ 39 + 55 + 64 = 158 \text{ g/mol} \]

### Step 5: Calculate the equivalent weight in acidic medium.
The equivalent weight is given by:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{n} = \frac{158}{5} = 31.6 \text{ g/equiv} \]

### Step 6: Determine the balanced redox reaction in neutral medium.
In neutral medium, the reaction can be represented as:
\[ \text{MnO}_4^- + 2 \text{H}_2\text{O} + 3 \text{e}^- \rightarrow \text{MnO}_2 + 4 \text{OH}^- \]

### Step 7: Calculate the change in oxidation state of manganese in neutral medium.
Again, manganese starts at +7 in `KMnO4` and goes to +4 in `MnO2`. The change in oxidation state is:
\[ +7 \text{ to } +4 = -3 \]
This indicates that 3 electrons are gained.

### Step 8: Determine the n-factor in neutral medium.
From the balanced equation, we see that:
\[ n = 3 \]

### Step 9: Calculate the equivalent weight in neutral medium.
Using the same molar mass:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{n} = \frac{158}{3} \approx 52.67 \text{ g/equiv} \]

### Step 10: Summarize the molecular weights in both mediums.
- In acidic medium: Molar mass = 158 g/mol, n-factor = 5
- In neutral medium: Molar mass = 158 g/mol, n-factor = 3

### Conclusion:
The molecular weight of `KMnO4` in acidic medium is 158 g/mol, and in neutral medium, it remains 158 g/mol.