For the estimation of nitrogen, 1.4 g of an organic compound was digested by Kjeldahl method and the evolved ammonia acid absorbed in 60 mL of `M/10` sulphuric acid. The unreacted acid required 20 mL of `M/10` sodium hydroxide for complete neutralization. the percentage of nitrogen in the compound is

Options

A. `5 %`
B. `6 %`
C. `10 %`
D. `3 %`

Explanation

To calculate the percentage of nitrogen in the organic compound using the Kjeldahl method, follow these steps:

### Step 1: Calculate the moles of sulfuric acid used
Given that 60 mL of M/10 sulfuric acid was used to absorb the ammonia, we first convert this volume to liters:
\[ \text{Volume of H}_2\text{SO}_4 = 60 \, \text{mL} = 0.060 \, \text{L} \]

Now, calculate the moles of sulfuric acid:
\[ \text{Moles of H}_2\text{SO}_4 = \text{Molarity} \times \text{Volume (in L)} = \frac{1}{10} \times 0.060 = 0.006 \, \text{mol} \]

### Step 2: Calculate the moles of unreacted sulfuric acid
Next, we need to calculate the moles of sulfuric acid that reacted with the sodium hydroxide. Given that 20 mL of M/10 sodium hydroxide was used:
\[ \text{Volume of NaOH} = 20 \, \text{mL} = 0.020 \, \text{L} \]

Calculate the moles of sodium hydroxide:
\[ \text{Moles of NaOH} = \text{Molarity} \times \text{Volume (in L)} = \frac{1}{10} \times 0.020 = 0.002 \, \text{mol} \]

### Step 3: Calculate the moles of sulfuric acid that reacted with ammonia
The reaction between sulfuric acid and sodium hydroxide is a 1:1 reaction. Therefore, the moles of sulfuric acid that reacted with ammonia can be calculated as:
\[ \text{Moles of H}_2\text{SO}_4 \text{ that reacted with NH}_3 = \text{Moles of H}_2\text{SO}_4 \text{ initially} - \text{Moles of H}_2\text{SO}_4 \text{ that reacted with NaOH \]
\[ = 0.006 - 0.002 = 0.004 \, \text{mol} \]

### Step 4: Calculate the mass of nitrogen in the compound
From the stoichiometry of the reaction, 1 mole of \( \text{H}_2\text{SO}_4 \) reacts with 2 moles of ammonia (NH₃), which means:
\[ \text{Moles of NH}_3 = 2 \times \text{Moles of H}_2\text{SO}_4 \text{ that reacted with ammonia} = 2 \times 0.004 = 0.008 \, \text{mol} \]

Now, calculate the mass of nitrogen in the ammonia:
\[ \text{Moles of N} = \text{Moles of NH}_3 = 0.008 \, \text{mol} \]
\[ \text{Mass of N} = \text{Moles of N} \times \text{Molar mass of N} = 0.008 \times 14 = 0.112 \, \text{g} \]

### Step 5: Calculate the percentage of nitrogen in the compound
Finally, we can calculate the percentage of nitrogen in the organic compound:
\[ \text{Percentage of N} = \left( \frac{\text{Mass of N}}{\text{Mass of compound}} \right) \times 100 = \left( \frac{0.112}{1.4} \right) \times 100 \approx 8.00\% \]

### Final Answer:
The percentage of nitrogen in the compound is approximately **8.00%**.