0.4 g of an organic compound was kjeldahlised and the ammonia evolved was absorbed into 50 mL of M/4 sulphuric acid solution. The residual acid solution was diluted with distilled water and the volume was made upto 150 mL. 20 mL of this diluted solution required 31 mL of M/20 NaOH solution for complete neutralisation. Calculate the percentage of nitrogen in the compound.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the normality of the sulfuric acid solution
Given that we have 50 mL of M/4 sulfuric acid (H2SO4), we can convert this to normality:
- Molarity (M) of H2SO4 = 1/4 M = 0.25 M
- Normality (N) = Molarity × Number of H+ ions = 0.25 M × 2 = 0.5 N

### Step 2: Determine the amount of unused sulfuric acid
Let V mL be the volume of H2SO4 that was unused. After the ammonia is absorbed, the residual acid solution is diluted to 150 mL. We know that 20 mL of this diluted solution required 31 mL of M/20 NaOH for complete neutralization.

### Step 3: Calculate the normality of the NaOH solution
Since the NaOH solution is M/20, its normality is equal to its molarity:
- Normality (N) of NaOH = 1/20 N = 0.05 N

### Step 4: Use the neutralization equation
Using the formula for neutralization:
\[ V_1 \times S_1 = V_2 \times S_2 \]
Where:
- \( V_1 = 20 \) mL (volume of diluted H2SO4)
- \( S_1 \) = Normality of diluted H2SO4 (unknown)
- \( V_2 = 31 \) mL (volume of NaOH)
- \( S_2 = 0.05 \) N (normality of NaOH)

Plugging in the values:
\[ 20 \times S_1 = 31 \times 0.05 \]
\[ S_1 = \frac{31 \times 0.05}{20} = \frac{1.55}{20} = 0.0775 \, \text{N} \]

### Step 5: Relate the normality of the diluted H2SO4 to the unused H2SO4
The normality of the diluted solution is related to the volume of unused H2SO4:
\[ V \, \text{mL} \, \text{of} \, 0.5 \, \text{N} \, \text{H2SO4} \, \text{is diluted to} \, 150 \, \text{mL} \]
Thus,
\[ V \times 0.5 = 150 \times 0.0775 \]
\[ V = \frac{150 \times 0.0775}{0.5} = \frac{11.625}{0.5} = 23.25 \, \text{mL} \]

### Step 6: Calculate the volume of H2SO4 that was used
The total volume of H2SO4 was 50 mL, so the volume used is:
\[ \text{Used H2SO4} = 50 - 23.25 = 26.75 \, \text{mL} \]

### Step 7: Calculate the amount of ammonia produced
The amount of ammonia (NH3) produced is equivalent to the amount of H2SO4 used:
\[ 26.75 \, \text{mL of} \, 0.5 \, \text{N} \, \text{H2SO4} = 26.75 \, \text{mL of} \, 0.5 \, \text{N} \, \text{NH3} \]

### Step 8: Calculate the mass of nitrogen in the ammonia
The amount of nitrogen in the ammonia can be calculated as follows:
- The equivalent weight of nitrogen (N) = 14 g/mol
- The mass of nitrogen in 26.75 mL of 0.5 N NH3:
\[ \text{Mass of N} = \frac{14}{2} \times \frac{26.75}{1000} = 0.18725 \, \text{g} \]

### Step 9: Calculate the percentage of nitrogen in the compound
Finally, we can calculate the percentage of nitrogen in the organic compound:
\[ \text{Percentage of N} = \left( \frac{0.18725}{0.4} \right) \times 100 = 46.81\% \]

### Final Answer
The percentage of nitrogen in the compound is approximately **46.8%**.

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