29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl's method and the evolved ammonia was absorbed in 20mL of 0.1M HCL solution. The excess of the acid required 15 mL of 0.1 M NaOH solution for complete neutralization. The percentage of nitrogen in the compound is:

Options

A. 59
B. 47.4
C. 23.7
D. 29.5

Answer

To find the percentage of nitrogen in the organic compound, we can follow these steps:

### Step 1: Calculate the moles of HCl used
Given:
- Volume of HCl = 20 mL = 0.020 L
- Concentration of HCl = 0.1 M

Using the formula:
\[ \text{Moles of HCl} = \text{Concentration} \times \text{Volume} \]
\[ \text{Moles of HCl} = 0.1 \, \text{mol/L} \times 0.020 \, \text{L} = 0.002 \, \text{mol} = 2 \, \text{mmol} \]

### Step 2: Calculate the moles of NaOH used
Given:
- Volume of NaOH = 15 mL = 0.015 L
- Concentration of NaOH = 0.1 M

Using the same formula:
\[ \text{Moles of NaOH} = 0.1 \, \text{mol/L} \times 0.015 \, \text{L} = 0.0015 \, \text{mol} = 1.5 \, \text{mmol} \]

### Step 3: Calculate the moles of HCl that reacted with ammonia
The moles of HCl that reacted with ammonia can be calculated by subtracting the moles of NaOH from the total moles of HCl:
\[ \text{Moles of HCl reacted with NH}_3 = \text{Moles of HCl} - \text{Moles of NaOH} \]
\[ \text{Moles of HCl reacted with NH}_3 = 2 \, \text{mmol} - 1.5 \, \text{mmol} = 0.5 \, \text{mmol} \]

### Step 4: Calculate the mass of nitrogen in ammonia
The moles of ammonia (NH₃) that reacted are equal to the moles of HCl that reacted, which is 0.5 mmol. Since each mole of NH₃ contains one mole of nitrogen (N), we can find the mass of nitrogen:
\[ \text{Mass of nitrogen} = \text{Moles of N} \times \text{Molar mass of N} \]
\[ \text{Mass of nitrogen} = 0.5 \, \text{mmol} \times 14 \, \text{mg/mmol} = 7 \, \text{mg} \]

### Step 5: Calculate the percentage of nitrogen in the compound
Now, we can find the percentage of nitrogen in the organic compound:
\[ \text{Percentage of nitrogen} = \left( \frac{\text{Mass of nitrogen}}{\text{Mass of organic compound}} \right) \times 100 \]
\[ \text{Percentage of nitrogen} = \left( \frac{7 \, \text{mg}}{29.5 \, \text{mg}} \right) \times 100 \approx 23.73\% \]

### Final Answer
The percentage of nitrogen in the compound is approximately **23.73%**.

---