3 " mol of "a mixture of `FeSO_(4)` and `Fe_(2)(SO_(4))_(3)` requried 100 " mL of " 2 M `KMnO_(4)` solution in acidic medium. Hence, mole fraction of `FeSO_(4)` in the mixture is

Options

A. `(1)/(3)`
B. `(2)/(3)`
C. `(2)/(5)`
D. `(3)/(5)`

Explanation

To solve the problem, we need to find the mole fraction of \( \text{FeSO}_4 \) in a mixture of \( \text{FeSO}_4 \) and \( \text{Fe}_2(\text{SO}_4)_3 \) that reacts with \( \text{KMnO}_4 \). Here’s a step-by-step solution:

### Step 1: Understand the Reaction
The reaction involves \( \text{KMnO}_4 \) acting as an oxidizing agent in acidic medium. The relevant reactions are:
- \( \text{Fe}^{2+} \) from \( \text{FeSO}_4 \) is oxidized to \( \text{Fe}^{3+} \).
- \( \text{KMnO}_4 \) is reduced from \( \text{Mn}^{7+} \) to \( \text{Mn}^{2+} \).

### Step 2: Write the Balanced Reaction
The balanced reaction for \( \text{KMnO}_4 \) with \( \text{FeSO}_4 \) in acidic medium is:
\[ 2 \text{KMnO}_4 + 8 \text{H}_2\text{SO}_4 + 10 \text{FeSO}_4 \rightarrow K_2\text{SO}_4 + 2 \text{MnSO}_4 + 5 \text{Fe}_2(\text{SO}_4)_3 + 8 \text{H}_2\text{O} \]

### Step 3: Calculate the Equivalents of \( \text{KMnO}_4 \)
Given that 100 mL of 2 M \( \text{KMnO}_4 \) is used:
- Convert volume to liters: \( 100 \, \text{mL} = 0.1 \, \text{L} \)
- Calculate the equivalents of \( \text{KMnO}_4 \):
\[ \text{Normality} = \text{Molarity} \times n \text{-factor} \]
For \( \text{KMnO}_4 \), the \( n \)-factor is 5 (since it can oxidize 5 moles of \( \text{Fe}^{2+} \)):
\[ \text{Normality} = 2 \, \text{M} \times 5 = 10 \, \text{N} \]
Now, calculate the equivalents:
\[ \text{Equivalents of } \text{KMnO}_4 = \text{Normality} \times \text{Volume in L} = 10 \, \text{N} \times 0.1 \, \text{L} = 1 \, \text{equivalent} \]

### Step 4: Equivalents of \( \text{FeSO}_4 \)
From the stoichiometry of the reaction, the equivalents of \( \text{FeSO}_4 \) will equal the equivalents of \( \text{KMnO}_4 \):
\[ \text{Equivalents of } \text{FeSO}_4 = 1 \, \text{equivalent} \]

### Step 5: Calculate Moles of \( \text{FeSO}_4 \)
The \( n \)-factor for \( \text{FeSO}_4 \) is 1 (since it provides 1 mole of \( \text{Fe}^{2+} \)):
\[ \text{Moles of } \text{FeSO}_4 = \frac{\text{Equivalents}}{n \text{-factor}} = \frac{1}{1} = 1 \, \text{mol} \]

### Step 6: Determine Moles of the Mixture
The total moles of the mixture (which consists of \( \text{FeSO}_4 \) and \( \text{Fe}_2(\text{SO}_4)_3 \)) is given as 3 moles.

### Step 7: Calculate the Mole Fraction of \( \text{FeSO}_4 \)
The mole fraction of \( \text{FeSO}_4 \) is calculated as:
\[ \text{Mole Fraction of } \text{FeSO}_4 = \frac{\text{Moles of } \text{FeSO}_4}{\text{Total Moles of Mixture}} = \frac{1}{3} \]

### Final Answer
The mole fraction of \( \text{FeSO}_4 \) in the mixture is \( \frac{1}{3} \).

---