In the Dumas method for the estimation of nitrogen , `0.0237` grams of an organic compound gave `2.21 mL` of nitrogen at `754.32 mm` of `Hg` pressrue at `18^(@)C`. (Aquesous tension at `18^(@)C` is `15.4mm` of `Hg`). Therefore the percentage of nitrogen in the compound is

Options

A. `20.67%`
B. `10.6%`
C. `11.2%`
D. `13.9%`

Answer

To find the percentage of nitrogen in the organic compound using the Dumas method, we will follow these steps:

### Step-by-Step Solution:

1. **Calculate the Effective Pressure (P1)**:
- The total pressure (P_total) is given as 754.32 mm of Hg.
- The aqueous tension (P_aqueous) at 18°C is given as 15.4 mm of Hg.
- Therefore, the effective pressure (P1) can be calculated as:
\[ P1 = P_{total} - P_{aqueous} = 754.32 \, \text{mm Hg} - 15.4 \, \text{mm Hg} = 738.92 \, \text{mm Hg} \]

2. **Convert Temperature to Kelvin (T1)**:
- The temperature is given as 18°C. To convert this to Kelvin:
\[ T1 = 18 + 273 = 291 \, \text{K} \]

3. **Use the Ideal Gas Law to Find V2**:
- We know that at standard temperature and pressure (STP), the conditions are:
- \( P2 = 760 \, \text{mm Hg} \)
- \( T2 = 273 \, \text{K} \)
- Using the equation \( \frac{P1 \cdot V1}{T1} = \frac{P2 \cdot V2}{T2} \), where \( V1 = 2.21 \, \text{mL} \):
\[ \frac{738.92 \cdot 2.21}{291} = \frac{760 \cdot V2}{273} \]
- Rearranging and solving for \( V2 \):
\[ V2 = \frac{738.92 \cdot 2.21 \cdot 273}{291 \cdot 760} \approx 2.01 \, \text{mL} \]

4. **Calculate the Mass of Nitrogen in the Gas**:
- At STP, 1 mole of nitrogen (N2) occupies 22400 mL and has a mass of 28 g.
- Therefore, the mass of nitrogen in 2.01 mL can be calculated as:
\[ \text{Mass of N} = \left(\frac{28 \, \text{g}}{22400 \, \text{mL}}\right) \cdot 2.01 \, \text{mL} \approx 0.0025 \, \text{g} \]

5. **Calculate the Percentage of Nitrogen in the Organic Compound**:
- The mass of the organic compound is given as 0.0237 g.
- The percentage of nitrogen is calculated as:
\[ \text{Percentage of N} = \left(\frac{\text{Mass of N}}{\text{Mass of organic compound}}\right) \times 100 = \left(\frac{0.0025 \, \text{g}}{0.0237 \, \text{g}}\right) \times 100 \approx 10.55\% \]
- Rounding this gives approximately 10.6%.

### Final Answer:
The percentage of nitrogen in the compound is **10.6%**.

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