When 2.5g of a sample of Mohr's salt reacts completely with 50mL of `(N)/(10) KMnO_(4)` solution. The % purity of the sample of Mohr's salt is:

Options

A. 78.4
B. 70
C. 37
D. 40

Answer

To find the % purity of Mohr's salt from the given data, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**: Mohr's salt (FeSO₄·(NH₄)₂SO₄·6H₂O) reacts with KMnO₄ in an acidic medium. The relevant half-reaction for Fe²⁺ is:
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \]
Thus, the n-factor of Fe²⁺ is 1.

2. **Calculate the Equivalent Weight of KMnO₄**: The normality of the KMnO₄ solution is given as \( \frac{N}{10} \). Therefore, the number of equivalents in 50 mL of this solution can be calculated as:
\[ \text{Equivalents of KMnO₄} = \text{Normality} \times \text{Volume (L)} = \frac{1}{10} \times \frac{50}{1000} = 0.005 \text{ equivalents} \]

3. **Calculate the Mass of Mohr's Salt that Reacts**: The reaction between Mohr's salt and KMnO₄ is a 1:1 mole ratio. Therefore, the equivalents of Mohr's salt will also be 0.005.

4. **Determine the Molecular Weight of Mohr's Salt**: The molar mass of Mohr's salt (FeSO₄·(NH₄)₂SO₄·6H₂O) is approximately 392 g/mol.

5. **Calculate the Mass of Mohr's Salt that Reacts**: Using the number of equivalents and the molar mass:
\[ \text{Mass of Mohr's salt} = \text{Equivalents} \times \text{Molar Mass} = 0.005 \times 392 = 1.96 \text{ g} \]

6. **Calculate the % Purity**: The % purity of the Mohr's salt sample can be calculated using the formula:
\[ \text{% Purity} = \left( \frac{\text{Mass of pure Mohr's salt}}{\text{Total mass of sample}} \right) \times 100 \]
Substituting the values:
\[ \text{% Purity} = \left( \frac{1.96}{2.5} \right) \times 100 = 78.4\% \]

### Final Answer:
The % purity of the sample of Mohr's salt is **78.4%**.

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