In Duma's method 0.52g of an organic compound on combustion gave 68.6 mL `N_(2)` at `27^(@)C and 756mm` pressure. What is the percentage of nitrogen in the compound?

Options

A. 0.1222
B. 0.1493
C. 0.1584
D. 0.1623

Explanation

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

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Mass of the organic compound (m) = 0.52 g
- Volume of nitrogen gas (V) = 68.6 mL
- Temperature (T) = 27°C = 27 + 273 = 300 K
- Pressure (P) = 756 mmHg

2. **Convert Volume of Nitrogen to Liters**:
- Since the volume is given in mL, convert it to liters:
\[ V = 68.6 \, \text{mL} = 68.6 \times 10^{-3} \, \text{L} = 0.0686 \, \text{L} \]

3. **Use the Ideal Gas Law to Calculate Moles of Nitrogen (N₂)**:
- The ideal gas equation is given by:
\[ PV = nRT \]
- Rearranging for n (number of moles):
\[ n = \frac{PV}{RT} \]
- Where:
- P = 756 mmHg (convert to atm: \( \frac{756}{760} \approx 0.9947 \, \text{atm} \))
- R = 0.0821 L·atm/(K·mol)
- T = 300 K
- Substituting the values:
\[ n = \frac{(0.9947 \, \text{atm})(0.0686 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)})(300 \, \text{K})} \]
- Calculate:
\[ n \approx \frac{0.0682}{24.63} \approx 0.00277 \, \text{mol} \]

4. **Calculate the Mass of Nitrogen**:
- The molar mass of nitrogen (N₂) is approximately 28 g/mol.
- Therefore, the mass of nitrogen (m_N) is:
\[ m_N = n \times \text{molar mass of N}_2 = 0.00277 \, \text{mol} \times 28 \, \text{g/mol} \approx 0.07756 \, \text{g} \]

5. **Calculate the Percentage of Nitrogen in the Compound**:
- The percentage of nitrogen in the organic compound is given by:
\[ \text{Percentage of N} = \left(\frac{m_N}{m} \times 100\right) = \left(\frac{0.07756 \, \text{g}}{0.52 \, \text{g}} \times 100\right) \approx 14.9\% \]

### Final Answer:
The percentage of nitrogen in the compound is approximately **14.9%**.