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. (a) 78.4
B. (b) 70
C. (c) 37
D. (d) 40

Answer

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

### Step 1: Understand the Reaction
Mohr's salt (FeSO4·(NH4)2SO4·6H2O) reacts with potassium permanganate (KMnO4) in an acidic medium. The relevant half-reaction is:
\[ \text{Fe}^{2+} + \text{MnO}_4^{-} \rightarrow \text{Fe}^{3+} + \text{Mn}^{2+} \]

### Step 2: Calculate the Equivalent Weight of Mohr's Salt
The equivalent weight of a substance is calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{n} \]
where \( n \) is the number of electrons transferred per formula unit in the reaction. For Fe in Mohr's salt, it changes from +2 to +3, so \( n = 1 \).

**Molar Mass of Mohr's Salt:**
- Iron (Fe): 56 g/mol
- Sulfur (S): 32 g/mol (2 Sulfur atoms)
- Oxygen (O): 16 g/mol (4 Oxygen atoms)
- Nitrogen (N): 14 g/mol (2 Nitrogen atoms)
- Hydrogen (H): 1 g/mol (12 Hydrogen atoms)

Calculating the molar mass:
\[ \text{Molar Mass} = 56 + (2 \times 32) + (4 \times 16) + (2 \times 14) + (12 \times 1) = 392 \text{ g/mol} \]

Now, calculate the equivalent weight:
\[ \text{Equivalent Weight} = \frac{392 \text{ g/mol}}{1} = 392 \text{ g/equiv} \]

### Step 3: Calculate the Gram Equivalents of KMnO4
Given:
- Normality of KMnO4 = \( \frac{1}{10} \) N
- Volume of KMnO4 = 50 mL = 0.050 L

Using the formula for gram equivalents:
\[ \text{Gram Equivalents} = \text{Normality} \times \text{Volume} = \frac{1}{10} \times 0.050 = 0.005 \text{ equivalents} \]

### Step 4: Calculate the Weight of Pure Mohr's Salt
Using the relationship:
\[ \text{Gram Equivalents of Mohr's Salt} = \frac{\text{Weight of Mohr's Salt}}{\text{Equivalent Weight}} \]
We know the gram equivalents of Mohr's salt is equal to that of KMnO4:
\[ 0.005 = \frac{\text{Weight of Mohr's Salt}}{392} \]
Solving for the weight of pure Mohr's salt:
\[ \text{Weight of Mohr's Salt} = 0.005 \times 392 = 1.96 \text{ g} \]

### Step 5: Calculate the Percentage Purity
Now, we can find the percentage purity of the Mohr's salt sample:
\[ \text{Percentage Purity} = \left( \frac{\text{Weight of Pure Mohr's Salt}}{\text{Total Weight of Sample}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage Purity} = \left( \frac{1.96 \text{ g}}{2.5 \text{ g}} \right) \times 100 = 78.4\% \]

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

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