Equivalent weight of Mohr salt in the titration with `KMnO_(4)` is (M-Molecular weight)

Options

A. `M/1`
B. `M/4`
C. `M/3`
D. `M/2`

Answer

To find the equivalent weight of Mohr's salt in the titration with KMnO4, we can follow these steps:

### Step 1: Understand the Reaction
Mohr's salt, which is represented as \( \text{FeSO}_4 \cdot \text{(NH}_4)_2\text{SO}_4 \cdot 6\text{H}_2\text{O} \), reacts with potassium permanganate (\( \text{KMnO}_4 \)) in an acidic medium. The balanced reaction can be represented as:
\[ 2 \text{KMnO}_4 + 10 \text{FeSO}_4 + 8 \text{H}_2\text{SO}_4 \rightarrow 2 \text{MnSO}_4 + 10 \text{Fe}_2(\text{SO}_4)_3 + K_2\text{SO}_4 + 8 \text{H}_2\text{O} \]

### Step 2: Determine the Oxidation States
In this reaction:
- Manganese in \( \text{KMnO}_4 \) is in the +7 oxidation state and is reduced to +2 in \( \text{MnSO}_4 \).
- Iron in \( \text{FeSO}_4 \) is oxidized from +2 to +3 in \( \text{Fe}_2(\text{SO}_4)_3 \).

### Step 3: Calculate the Change in Oxidation State
The change in oxidation state for iron is:
- From +2 to +3, which is a change of +1.

### Step 4: Determine the Valency Factor
The valency factor (n_f) for Mohr's salt is determined by the total change in oxidation state. Since each iron atom changes from +2 to +3, the total change is +1 for each mole of iron. In the balanced equation, 10 moles of \( \text{FeSO}_4 \) are involved, but since we are looking for the equivalent weight of Mohr's salt, we consider the change for one mole of iron.

Thus, the valency factor \( n_f = 1 \).

### Step 5: Calculate the Equivalent Weight
The equivalent weight (E) is given by the formula:
\[ E = \frac{M}{n_f} \]
where \( M \) is the molecular weight of Mohr's salt.

Since \( n_f = 1 \):
\[ E = \frac{M}{1} = M \]

### Conclusion
The equivalent weight of Mohr's salt in the titration with KMnO4 is equal to its molecular weight \( M \).