The weight of `KMnO_(4)` required to completely oxidise 0.25 moles of `FeSO_(4)` in acid medium is ………….. (molecular weight of `KmnO_(4)=158`)

Options

A. 5.8
B. 1.5
C. 7.9
D. 0.79

Explanation

To find the weight of KMnO₄ required to completely oxidize 0.25 moles of FeSO₄ in acidic medium, we can follow these steps:

### Step 1: Determine the reaction and electron transfer
In acidic medium, KMnO₄ acts as an oxidizing agent and gets reduced from MnO₄⁻ to Mn²⁺. During this process, it transfers 5 electrons.

### Step 2: Determine the oxidation state change of Fe
In FeSO₄, iron (Fe) is in the +2 oxidation state (Fe²⁺) and gets oxidized to +3 oxidation state (Fe³⁺). The change in oxidation state is:
- From +2 to +3: Change = 1

### Step 3: Calculate the number of equivalents of FeSO₄
The number of equivalents of a substance is calculated using the formula:
\[ \text{Number of equivalents} = \text{n-factor} \times \text{number of moles} \]
For FeSO₄:
- n-factor = 1 (since it changes by 1)
- Number of moles = 0.25

Thus, the number of equivalents of FeSO₄ is:
\[ \text{Number of equivalents of FeSO₄} = 1 \times 0.25 = 0.25 \]

### Step 4: Calculate the number of equivalents of KMnO₄ required
Since 1 equivalent of KMnO₄ can oxidize 5 equivalents of FeSO₄, we can set up the relationship:
\[ \text{Number of equivalents of KMnO₄} = \frac{\text{Number of equivalents of FeSO₄}}{5} \]
Substituting the value:
\[ \text{Number of equivalents of KMnO₄} = \frac{0.25}{5} = 0.05 \]

### Step 5: Calculate the number of moles of KMnO₄
Since we have already calculated the number of equivalents of KMnO₄, we can say that:
\[ \text{Number of moles of KMnO₄} = 0.05 \]

### Step 6: Calculate the weight of KMnO₄ required
To find the weight, we use the formula:
\[ \text{Weight} = \text{Number of moles} \times \text{Molecular weight} \]
Given that the molecular weight of KMnO₄ is 158 g/mol:
\[ \text{Weight} = 0.05 \times 158 = 7.9 \text{ grams} \]

### Final Answer:
The weight of KMnO₄ required to completely oxidize 0.25 moles of FeSO₄ in acidic medium is **7.9 grams**.

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