The total volume of `0.1M KMnO_(4)` solution that are needed to oxidize 100mg each of ferrius oxalate and ferrous sulphate in a mixture in acidic medium is:

Options

A. 1.096mL
B. 1.32mL
C. 5.48mL
D. none of these

Answer

To solve the problem of determining the total volume of 0.1M KMnO4 solution needed to oxidize 100 mg each of ferrous oxalate (FeC2O4) and ferrous sulfate (FeSO4) in an acidic medium, we will follow these steps:

### Step 1: Calculate the number of moles of ferrous oxalate (FeC2O4)

1. **Given weight of FeC2O4** = 100 mg = 0.1 g
2. **Molecular weight of FeC2O4** = 144 g/mol
3. **Number of moles of FeC2O4** = Given weight / Molecular weight
\[ \text{Number of moles of FeC2O4} = \frac{0.1 \text{ g}}{144 \text{ g/mol}} = 0.000694 \text{ mol} \]

### Step 2: Calculate the gram equivalents of ferrous oxalate (FeC2O4)

1. **n-factor for FeC2O4**: The oxidation state change is 3 (1 for Fe and 2 for C2O4).
2. **Gram equivalents of FeC2O4** = Number of moles × n-factor
\[ \text{Gram equivalents of FeC2O4} = 0.000694 \text{ mol} \times 3 = 0.00208 \text{ eq} \]

### Step 3: Calculate the number of moles of ferrous sulfate (FeSO4)

1. **Given weight of FeSO4** = 100 mg = 0.1 g
2. **Molecular weight of FeSO4** = 152 g/mol
3. **Number of moles of FeSO4** = Given weight / Molecular weight
\[ \text{Number of moles of FeSO4} = \frac{0.1 \text{ g}}{152 \text{ g/mol}} = 0.000658 \text{ mol} \]

### Step 4: Calculate the gram equivalents of ferrous sulfate (FeSO4)

1. **n-factor for FeSO4**: The oxidation state change is 1 (from Fe2+ to Fe3+).
2. **Gram equivalents of FeSO4** = Number of moles × n-factor
\[ \text{Gram equivalents of FeSO4} = 0.000658 \text{ mol} \times 1 = 0.000658 \text{ eq} \]

### Step 5: Calculate the total gram equivalents

1. **Total gram equivalents** = Gram equivalents of FeC2O4 + Gram equivalents of FeSO4
\[ \text{Total gram equivalents} = 0.00208 + 0.000658 = 0.002738 \text{ eq} \]

### Step 6: Calculate the gram equivalents of KMnO4

1. **n-factor for KMnO4**: The change in oxidation state is 5 (from Mn7+ to Mn2+).
2. **Gram equivalents of KMnO4** = Concentration × Volume × n-factor
\[ \text{Gram equivalents of KMnO4} = 0.1 \text{ mol/L} \times V \text{ (L)} \times 5 \]

### Step 7: Set the equation for equivalence

1. **Setting the total gram equivalents of KMnO4 equal to the total gram equivalents needed**:
\[ 0.1 \times V \times 5 = 0.002738 \]

### Step 8: Solve for V (volume of KMnO4)

1. Rearranging gives:
\[ V = \frac{0.002738}{0.5} = 0.005476 \text{ L} = 5.476 \text{ mL} \]

### Final Result

The total volume of 0.1M KMnO4 solution needed is approximately **5.48 mL**.

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