A 5g sample containing `Fe_3O_4 (FeO+Fe_2O_3)` and an inert impurity is treated with excess of KI solution in the presence of dilute `H_2SO_4`.The entire Iron converted to Ferrous ion along with liberation of Iodine.The resulting solution is diluted to 100 ml. 20 ml of the diluted solution requires 10 ml of 0.5M `Na_2S_2O_3` solution to reduce the Iodine present.Amongs the following select correct statements.

Options

A. % of `Fe_2O_3` in sample is 40%
B. % of FeO in sample is 28%
C. % of inert impurity in sample is 42%
D. % of inert impurity in sample is 32%

Explanation

To solve the problem step by step, we will analyze the information provided and perform the necessary calculations.

### Step 1: Understanding the Reaction
The sample contains `Fe3O4` (which consists of `FeO` and `Fe2O3`) and an inert impurity. When treated with excess KI in the presence of dilute `H2SO4`, all iron is converted to ferrous ions (`Fe^2+`), and iodine is liberated.

### Step 2: Analyzing the Dilution and Titration
The resulting solution is diluted to 100 mL. From this solution, 20 mL requires 10 mL of 0.5 M `Na2S2O3` to reduce the iodine present.

### Step 3: Calculate Moles of `Na2S2O3`
Using the formula:
\[ \text{Moles of } Na2S2O3 = \text{Volume (L)} \times \text{Molarity} \]
\[ \text{Moles of } Na2S2O3 = 0.010 \, \text{L} \times 0.5 \, \text{mol/L} = 0.005 \, \text{mol} \]

### Step 4: Relating Moles of Iodine to `Na2S2O3`
The reaction between iodine and `Na2S2O3` is:
\[ I_2 + 2Na2S2O3 \rightarrow 2NaI + Na2S4O6 \]
From the stoichiometry, 1 mole of `I2` reacts with 2 moles of `Na2S2O3`. Therefore, the moles of iodine can be calculated as:
\[ \text{Moles of } I2 = \frac{0.005}{2} = 0.0025 \, \text{mol} \]

### Step 5: Concentration of Iodine in the 20 mL Sample
Since 20 mL of the diluted solution contains 0.0025 moles of iodine, the concentration in the 100 mL solution is:
\[ \text{Total moles of } I2 = 0.0025 \, \text{mol} \times \frac{100 \, \text{mL}}{20 \, \text{mL}} = 0.0125 \, \text{mol} \]

### Step 6: Finding the Moles of Iron
The moles of iron in the sample can be determined from the moles of `Fe2O3` and `FeO` in `Fe3O4`. Assuming `x` moles of `Fe2O3` and `x` moles of `FeO`, we have:
- For `Fe2O3`: 1 mole contains 2 moles of Fe
- For `FeO`: 1 mole contains 1 mole of Fe

Thus, the total moles of iron from `Fe2O3` and `FeO` are:
\[ \text{Total Fe moles} = 2x + x = 3x \]

### Step 7: Relating Moles of Iron to Moles of Iodine
Since all iron is converted to ferrous ions, we have:
\[ 3x = 0.0125 \, \text{mol} \implies x = \frac{0.0125}{3} = 0.00417 \, \text{mol} \]

### Step 8: Calculate Mass of `Fe2O3` and `FeO`
- Molar mass of `Fe2O3` = 159.69 g/mol
- Molar mass of `FeO` = 71.84 g/mol

Mass of `Fe2O3`:
\[ \text{Mass of } Fe2O3 = x \times \text{Molar mass} = 0.00417 \times 159.69 = 0.666 \, \text{g} \]

Mass of `FeO`:
\[ \text{Mass of } FeO = x \times \text{Molar mass} = 0.00417 \times 71.84 = 0.299 \, \text{g} \]

### Step 9: Calculate Percentages
- Total mass of the sample = 5 g
- Percentage of `Fe2O3`:
\[ \text{Percentage of } Fe2O3 = \left(\frac{0.666}{5}\right) \times 100 = 13.32\% \]
- Percentage of `FeO`:
\[ \text{Percentage of } FeO = \left(\frac{0.299}{5}\right) \times 100 = 5.98\% \]
- Percentage of inert impurity:
\[ \text{Percentage of inert impurity} = 100 - (13.32 + 5.98) = 80.7\% \]

### Conclusion
Based on the calculations:
1. The percentage of `Fe2O3` is not 40%.
2. The percentage of `FeO` is not 28%.
3. The percentage of inert impurity is not 42%.