When 0.75 gm of a substance was kjeldalised, it produced `NH_(3)`. Which neutralizes of 20ml of 0.25 N sulphuric acid. The percentage of nitrogen in the organic compound is `:`

Options

A. 14
B. 11
C. 1
D. None

Answer

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

### Step 1: Calculate the milli-equivalents of H2SO4 used
The formula for calculating milli-equivalents (mEq) is:
\[ \text{mEq} = \text{Volume (L)} \times \text{Normality (N)} \]
Given:
- Volume of H2SO4 = 20 mL = 0.020 L
- Normality of H2SO4 = 0.25 N

Calculating the milli-equivalents:
\[ \text{mEq} = 0.020 \, \text{L} \times 0.25 \, \text{N} = 0.005 \, \text{equivalents} \]

### Step 2: Convert milli-equivalents to moles
Since 1 equivalent of H2SO4 corresponds to 1 mole, we can say:
\[ \text{moles of H2SO4} = \text{mEq} = 0.005 \, \text{moles} \]

### Step 3: Calculate the amount of ammonia (NH3) produced
According to the Kjeldahl method, each mole of NH3 produced corresponds to 1 mole of H2SO4 consumed. Therefore, the moles of NH3 produced is also 0.005 moles.

### Step 4: Calculate the mass of nitrogen in the ammonia
The molar mass of nitrogen (N) is approximately 14 g/mol. Since each mole of NH3 contains one mole of nitrogen:
\[ \text{Mass of nitrogen} = \text{moles of NH3} \times \text{molar mass of N} = 0.005 \, \text{moles} \times 14 \, \text{g/mol} = 0.07 \, \text{g} \]

### Step 5: Calculate the percentage of nitrogen in the organic compound
The percentage of nitrogen in the organic compound is calculated using the formula:
\[ \text{Percentage of N} = \left( \frac{\text{mass of N}}{\text{mass of organic compound}} \right) \times 100 \]
Given that the mass of the organic compound is 0.75 g:
\[ \text{Percentage of N} = \left( \frac{0.07 \, \text{g}}{0.75 \, \text{g}} \right) \times 100 = 9.33\% \]

### Final Answer
The percentage of nitrogen in the organic compound is **9.33%**.

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