0.144 g of pure `FeC_(2)O_(4)` was dissolvedin dilute `H_(2)SO_(4)` and the solution was diluted to 100 ml . What volume in ml of 0.1 M `KMnO_(4)` will be needed to oxidise `FeC_(2)O_(4)` solution

Answer

To solve the problem, we need to determine the volume of 0.1 M KMnO4 required to oxidize 0.144 g of pure FeC2O4 dissolved in dilute H2SO4. Here’s a step-by-step solution:

### Step 1: Calculate the moles of FeC2O4
First, we need to find the molar mass of FeC2O4:
- Molar mass of Fe = 55.85 g/mol
- Molar mass of C = 12.01 g/mol (2 carbon atoms)
- Molar mass of O = 16.00 g/mol (4 oxygen atoms)

Calculating the molar mass:
\[ \text{Molar mass of FeC}_2\text{O}_4 = 55.85 + (2 \times 12.01) + (4 \times 16.00) = 55.85 + 24.02 + 64.00 = 143.87 \text{ g/mol} \]

Now, calculate the moles of FeC2O4:
\[ \text{Moles of FeC}_2\text{O}_4 = \frac{0.144 \text{ g}}{143.87 \text{ g/mol}} \approx 0.001 \text{ moles} \]

### Step 2: Determine the stoichiometry of the reaction
The balanced chemical reaction between FeC2O4 and KMnO4 in acidic medium is:
\[ 10 \text{ FeC}_2\text{O}_4 + 6 \text{ KMnO}_4 \rightarrow 5 \text{ Fe}_2(SO_4)_3 + 20 \text{ CO}_2 + 6 \text{ MnSO}_4 + 3 \text{ K}_2\text{SO}_4 + 24 \text{ H}_2\text{O} \]

From the balanced equation, we see that:
- 10 moles of FeC2O4 react with 6 moles of KMnO4.
- Therefore, 1 mole of FeC2O4 requires \(\frac{6}{10} = 0.6\) moles of KMnO4.

### Step 3: Calculate the moles of KMnO4 required
Using the moles of FeC2O4 calculated:
\[ \text{Moles of KMnO}_4 = 0.001 \text{ moles of FeC}_2\text{O}_4 \times \frac{6 \text{ moles of KMnO}_4}{10 \text{ moles of FeC}_2\text{O}_4} = 0.0006 \text{ moles of KMnO}_4 \]

### Step 4: Calculate the volume of KMnO4 solution needed
Using the formula for molarity:
\[ \text{Molarity} = \frac{\text{Moles}}{\text{Volume (L)}} \]
Rearranging gives:
\[ \text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}} \]
Substituting the values:
\[ \text{Volume (L)} = \frac{0.0006 \text{ moles}}{0.1 \text{ M}} = 0.006 \text{ L} \]

Convert the volume from liters to milliliters:
\[ \text{Volume (mL)} = 0.006 \text{ L} \times 1000 \text{ mL/L} = 6 \text{ mL} \]

### Final Answer
The volume of 0.1 M KMnO4 required to oxidize the FeC2O4 solution is **6 mL**.

---