A solution of `10 ml M/10 FeSO_(4)` war titrated with `KMnO_(4)` solution in acidic medium. The amount of `KMNO_(4)` used will be

Options

A. (a)` 5` ml of` 0.1 M`
B. (b)` 10` ml of `1.1 M`
C. (c )` 10 `ml of `0.5 M`
D. (d)`10` ml of `0.02 M `

Explanation

To determine the amount of KMnO4 used in the titration of 10 ml of M/10 FeSO4 solution in acidic medium, we can follow these steps:

### Step 1: Calculate the number of moles of FeSO4
Given:
- Molarity of FeSO4 = M/10 = 0.1 M
- Volume of FeSO4 solution = 10 ml = 10 × 10^(-3) L

Using the formula for moles:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (in L)} \]
\[ \text{Number of moles of FeSO4} = 0.1 \, \text{mol/L} \times 10 \times 10^{-3} \, \text{L} = 1 \times 10^{-3} \, \text{moles} \]

### Step 2: Determine the valence factors
- For KMnO4:
- Mn changes from +7 in MnO4^- to +2 in Mn2+.
- Change in oxidation number = 7 - 2 = 5.
- Therefore, the valence factor of KMnO4 = 5.

- For FeSO4:
- Fe changes from +2 in Fe2+ to +3 in Fe3+.
- Change in oxidation number = 3 - 2 = 1.
- Therefore, the valence factor of FeSO4 = 1.

### Step 3: Establish the mole ratio between KMnO4 and FeSO4
From the valence factors:
- 1 mole of KMnO4 reacts with 5 moles of FeSO4.
- Therefore, 1 mole of FeSO4 reacts with 1/5 moles of KMnO4.

### Step 4: Calculate the number of moles of KMnO4 required
Using the number of moles of FeSO4 calculated:
\[ \text{Number of moles of KMnO4} = \frac{1}{5} \times \text{Number of moles of FeSO4} \]
\[ \text{Number of moles of KMnO4} = \frac{1}{5} \times 1 \times 10^{-3} = 2 \times 10^{-4} \, \text{moles} \]

### Step 5: Convert moles of KMnO4 to volume of KMnO4 solution
Let’s assume we have a KMnO4 solution of concentration C (in mol/L). The volume V (in L) can be calculated using:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume} \]
\[ 2 \times 10^{-4} = C \times V \]

### Step 6: Check the options provided
To find the correct concentration of KMnO4 that gives us \(2 \times 10^{-4}\) moles, we can check the options given in the problem:

1. **5 ml of 0.1 M KMnO4**:
\[ \text{Moles} = 0.1 \times 5 \times 10^{-3} = 5 \times 10^{-4} \, \text{moles} \quad \text{(not matching)} \]

2. **10 ml of 1.1 M KMnO4**:
\[ \text{Moles} = 1.1 \times 10 \times 10^{-3} = 11 \times 10^{-3} \, \text{moles} \quad \text{(not matching)} \]

3. **10 ml of 0.5 M KMnO4**:
\[ \text{Moles} = 0.5 \times 10 \times 10^{-3} = 5 \times 10^{-3} \, \text{moles} \quad \text{(not matching)} \]

4. **10 ml of 0.02 M KMnO4**:
\[ \text{Moles} = 0.02 \times 10 \times 10^{-3} = 2 \times 10^{-4} \, \text{moles} \quad \text{(matching)} \]

### Conclusion
The amount of KMnO4 used in the titration is **10 ml of 0.02 M KMnO4**.

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