29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl's method and the evolved ammonia was absorbed in 20 mL of 0.1 M 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 compounds is

Options

A. `59.0`
B. `47.4`
C. `23.7`
D. `29.5`

Explanation

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

### Step 1: Understand the problem
We have an organic compound containing nitrogen that was digested using Kjeldahl's method. The ammonia evolved was absorbed in HCl, and we need to find the percentage of nitrogen in the compound.

### Step 2: Gather the given data
- Mass of the organic compound (w) = 29.5 mg = 29.5 × 10^-3 g
- Volume of HCl solution (V_HCl) = 20 mL
- Molarity of HCl = 0.1 M
- Volume of NaOH used for neutralization (V_NaOH) = 15 mL
- Molarity of NaOH = 0.1 M

### Step 3: Calculate the normality of HCl
Since HCl is a strong acid, its normality (N) is equal to its molarity (M):
- N_HCl = 0.1 N

### Step 4: Calculate the amount of HCl initially present
Using the formula:
\[ \text{moles of HCl} = \text{normality} \times \text{volume (in L)} \]
\[ \text{moles of HCl} = 0.1 \, \text{N} \times 0.020 \, \text{L} = 0.002 \, \text{moles} \]

### Step 5: Calculate the amount of NaOH used
Using the same formula for NaOH:
\[ \text{moles of NaOH} = 0.1 \, \text{N} \times 0.015 \, \text{L} = 0.0015 \, \text{moles} \]

### Step 6: Calculate the moles of HCl that reacted with ammonia
The moles of HCl that reacted with ammonia is the initial moles of HCl minus the moles of NaOH used for neutralization:
\[ \text{moles of HCl that reacted} = 0.002 \, \text{moles} - 0.0015 \, \text{moles} = 0.0005 \, \text{moles} \]

### Step 7: Relate moles of HCl to moles of nitrogen
According to the Kjeldahl method, each mole of ammonia (NH3) reacts with one mole of HCl. Therefore, the moles of nitrogen in the compound is equal to the moles of HCl that reacted:
\[ \text{moles of nitrogen} = 0.0005 \, \text{moles} \]

### Step 8: Calculate the mass of nitrogen
The molar mass of nitrogen (N) is approximately 14 g/mol. Therefore, the mass of nitrogen in the compound is:
\[ \text{mass of nitrogen} = \text{moles of nitrogen} \times \text{molar mass of nitrogen} \]
\[ \text{mass of nitrogen} = 0.0005 \, \text{moles} \times 14 \, \text{g/mol} = 0.007 \, \text{g} = 7 \, \text{mg} \]

### Step 9: Calculate the percentage of nitrogen in the compound
Using the formula for percentage:
\[ \text{Percentage of nitrogen} = \left( \frac{\text{mass of nitrogen}}{\text{mass of organic compound}} \right) \times 100 \]
\[ \text{Percentage of nitrogen} = \left( \frac{7 \, \text{mg}}{29.5 \, \text{mg}} \right) \times 100 \approx 23.73\% \]

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

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