What volume of `HNO_(3) ("sp. gravity" 1.05mL^(-1) "containing "12.6(w//W)of HNO_(3))` that reduce into NO is required to oxidise iron 1g `1g. FeSO_(4).7H_(2)O` in acid medium is:

Options

A. 70mL
B. 0.57mL
C. 80mL
D. 0.65mL

Answer

To solve the problem, we need to determine the volume of HNO₃ required to oxidize 1 g of FeSO₄·7H₂O in an acidic medium. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the number of equivalents of FeSO₄·7H₂O
1. **Calculate the molar mass of FeSO₄·7H₂O:**
- Fe = 55.85 g/mol
- S = 32.07 g/mol
- O₄ = 4 × 16.00 g/mol = 64.00 g/mol
- 7H₂O = 7 × (2 × 1.01 + 16.00) g/mol = 7 × 18.02 g/mol = 126.14 g/mol
- Total molar mass = 55.85 + 32.07 + 64.00 + 126.14 = 278.06 g/mol (approximately 278 g/mol)

2. **Calculate the number of moles of FeSO₄·7H₂O:**
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{1 \text{ g}}{278 \text{ g/mol}} \approx 0.0036 \text{ mol} \]

3. **Determine the number of equivalents for FeSO₄·7H₂O:**
- The n-factor for Fe²⁺ to Fe³⁺ is 1 (since it loses 1 electron).
\[ \text{Equivalents of FeSO₄·7H₂O} = \text{Number of moles} \times \text{n-factor} = 0.0036 \text{ mol} \times 1 = 0.0036 \text{ equivalents} \]

### Step 2: Determine the number of equivalents of HNO₃
1. **Determine the n-factor for HNO₃ reducing to NO:**
- The n-factor for HNO₃ reducing to NO is 3 (it gains 3 electrons).

2. **Calculate the mass of HNO₃ in the solution:**
- Given that the specific gravity of the solution is 1.05 g/mL and it contains 12.6% w/w HNO₃:
\[ \text{Mass of solution} = 1.05 \text{ g/mL} \times V \text{ mL} = 1.05V \text{ g} \]
\[ \text{Mass of HNO₃} = \frac{12.6}{100} \times 1.05V = 0.1323V \text{ g} \]

3. **Calculate the number of moles of HNO₃:**
- The molar mass of HNO₃ is approximately 63 g/mol.
\[ \text{Number of moles of HNO₃} = \frac{0.1323V \text{ g}}{63 \text{ g/mol}} \approx 0.0021V \text{ mol} \]

4. **Calculate the number of equivalents of HNO₃:**
\[ \text{Equivalents of HNO₃} = \text{Number of moles} \times \text{n-factor} = 0.0021V \text{ mol} \times 3 = 0.0063V \text{ equivalents} \]

### Step 3: Set up the equation for equivalents
- Since the equivalents of HNO₃ must equal the equivalents of FeSO₄·7H₂O:
\[ 0.0063V = 0.0036 \]

### Step 4: Solve for V
\[ V = \frac{0.0036}{0.0063} \approx 0.5714 \text{ mL} \]

### Final Answer
The volume of HNO₃ required is approximately **0.57 mL**.

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