4 mole of a mixture of Mohr's salt and `Fe_(2)(SO_(4))_(3)` requires 500mL of `1 M K_(2)Cr_(2)O_(7)` for complete oxidation in acidic medium. The mole % of the Mohr's salt in the mixture is:

Options

A. 25
B. 50
C. 60
D. 75

Answer

To solve the problem of finding the mole percent of Mohr's salt in a mixture with Fe₂(SO₄)₃, we can follow these steps:

### Step 1: Understand the components of the mixture
Mohr's salt is represented as FeSO₄·(NH₄)₂SO₄·6H₂O, which contains Fe²⁺ ions. Fe₂(SO₄)₃ contains Fe³⁺ ions. In the presence of potassium dichromate (K₂Cr₂O₇) in acidic medium, only the Fe²⁺ from Mohr's salt will be oxidized to Fe³⁺.

### Step 2: Calculate the equivalents of K₂Cr₂O₇ used
The problem states that 500 mL of 1 M K₂Cr₂O₇ is used. We can calculate the number of equivalents of K₂Cr₂O₇ as follows:

\[ \text{Number of equivalents} = \text{Molarity} \times \text{Volume (L)} \times \text{n factor} \]

The n factor for K₂Cr₂O₇ is 6 (as it reduces from Cr⁶⁺ to Cr³⁺, changing by 6 electrons).

\[ \text{Volume in liters} = \frac{500 \text{ mL}}{1000} = 0.5 \text{ L} \]

Now substituting the values:

\[ \text{Number of equivalents} = 1 \, \text{mol/L} \times 0.5 \, \text{L} \times 6 = 3 \, \text{equivalents} \]

### Step 3: Relate the equivalents of K₂Cr₂O₇ to Mohr's salt
Since the equivalents of K₂Cr₂O₇ will equal the equivalents of Mohr's salt, we can write:

\[ \text{Equivalents of Mohr's salt} = \text{Number of equivalents of K₂Cr₂O₇} = 3 \]

The n factor for Mohr's salt (Fe²⁺ to Fe³⁺) is 1. Therefore, the number of moles of Mohr's salt can be calculated as:

\[ \text{Number of moles of Mohr's salt} = \frac{\text{Equivalents}}{\text{n factor}} = \frac{3}{1} = 3 \, \text{moles} \]

### Step 4: Calculate the number of moles of the mixture
The total number of moles in the mixture is given as 4 moles. Therefore, the number of moles of Fe₂(SO₄)₃ can be calculated as:

\[ \text{Number of moles of Fe₂(SO₄)₃} = \text{Total moles} - \text{Moles of Mohr's salt} = 4 - 3 = 1 \, \text{mole} \]

### Step 5: Calculate the mole percent of Mohr's salt
The mole percent of Mohr's salt in the mixture can be calculated using the formula:

\[ \text{Mole percent of Mohr's salt} = \left( \frac{\text{Moles of Mohr's salt}}{\text{Total moles}} \right) \times 100 \]

Substituting the values:

\[ \text{Mole percent of Mohr's salt} = \left( \frac{3}{4} \right) \times 100 = 75\% \]

### Final Answer
The mole percent of Mohr's salt in the mixture is **75%**.

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