0.5 g of an organic compound containing N on Kjeldahlising required 29 ml of N/5 `H_(2)SO_(4)` for complete neutrilization of `NH_(3)`. The percentage of N in the compound is

Options

A. 34.3
B. 16. 2
C. 21. 6
D. 14.8

Explanation

To find the percentage of nitrogen (N) in the organic compound, we can use the formula derived from the Kjeldahl method:

\[ \text{Percentage of Nitrogen} = \frac{1.4 \times \text{Normality of Acid} \times \text{Volume of Acid (ml)}}{\text{Weight of Organic Compound (g)}} \]

### Step-by-step Solution:

1. **Identify the Given Values**:
- Weight of the organic compound = 0.5 g
- Volume of N/5 H₂SO₄ = 29 ml
- Normality of H₂SO₄ = N/5 = 0.2 N (since N/5 means 1/5 of 1 N, which is 0.2 N)

2. **Substitute the Values into the Formula**:
\[ \text{Percentage of Nitrogen} = \frac{1.4 \times 0.2 \times 29}{0.5} \]

3. **Calculate the Numerator**:
- First, calculate \(1.4 \times 0.2\):
\[ 1.4 \times 0.2 = 0.28 \]
- Next, multiply by the volume of acid:
\[ 0.28 \times 29 = 8.12 \]

4. **Divide by the Weight of the Organic Compound**:
\[ \text{Percentage of Nitrogen} = \frac{8.12}{0.5} = 16.24 \]

5. **Final Result**:
- The percentage of nitrogen in the organic compound is approximately 16.2%.

### Conclusion:
The percentage of nitrogen in the compound is **16.2%**.