What volume at N.T.P. of gaseous `NH_(3)` will be required to be passed into 30 ml of `N-H_(2)SO_(4)` solution to bring down the acid strength of this solution to 0.2 N

Options

A. 357.2 ml
B. 444.4 ml
C. 537.6 ml
D. 495.6 ml

Explanation

To solve the problem of determining the volume of gaseous ammonia (NH₃) required to reduce the normality of a sulfuric acid (H₂SO₄) solution from 1 N to 0.2 N, we can follow these steps:

### Step-by-step Solution:

1. **Determine the Initial Moles of H₂SO₄:**
- Given that the volume of H₂SO₄ solution is 30 mL and its normality is 1 N, we can calculate the equivalents of H₂SO₄ present.
- Formula:
\[ \text{Equivalents of H₂SO₄} = \text{Volume (L)} \times \text{Normality (N)} \]
- Calculation:
\[ \text{Equivalents of H₂SO₄} = 0.030 \, \text{L} \times 1 \, \text{N} = 0.030 \, \text{equivalents} \]

2. **Determine the Final Moles of H₂SO₄:**
- After passing NH₃, the desired normality of the solution is 0.2 N.
- Using the same volume (30 mL or 0.030 L), we calculate the final equivalents.
- Calculation:
\[ \text{Equivalents of H₂SO₄ (final)} = 0.030 \, \text{L} \times 0.2 \, \text{N} = 0.006 \, \text{equivalents} \]

3. **Calculate the Change in Equivalents:**
- The change in equivalents of H₂SO₄ due to the addition of NH₃ is:
- Calculation:
\[ \text{Change in equivalents} = \text{Initial equivalents} - \text{Final equivalents} = 0.030 - 0.006 = 0.024 \, \text{equivalents} \]

4. **Relate NH₃ to H₂SO₄:**
- The reaction between NH₃ and H₂SO₄ shows that 1 equivalent of NH₃ neutralizes 1 equivalent of H₂SO₄.
- Therefore, the equivalents of NH₃ required will be equal to the change in equivalents of H₂SO₄:
- \[ \text{Equivalents of NH₃} = 0.024 \, \text{equivalents} \]

5. **Calculate the Mass of NH₃ Required:**
- Using the formula:
- \[ W = \text{Equivalents} \times \text{Molar mass} \]
- The molar mass of NH₃ is approximately 17 g/mol.
- Calculation:
\[ W = 0.024 \, \text{equivalents} \times 17 \, \text{g/equivalent} = 0.408 \, \text{g} \]

6. **Calculate the Volume of NH₃ at NTP:**
- At NTP (Normal Temperature and Pressure), 1 mole of gas occupies 22.4 L.
- To find the volume of NH₃:
- \[ \text{Volume} = \frac{W}{\text{Molar mass}} \times 22.4 \, \text{L} \]
- Calculation:
\[ \text{Volume} = \frac{0.408 \, \text{g}}{17 \, \text{g/mol}} \times 22.4 \, \text{L} = 0.5376 \, \text{L} \]
- Converting to mL:
\[ \text{Volume} = 0.5376 \, \text{L} \times 1000 = 537.6 \, \text{mL} \]

### Final Answer:
The volume of gaseous NH₃ required is **537.6 mL**.