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

Options

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

Answer

To solve the problem of estimating the percentage of nitrogen in the organic compound using Kjeldahl's method, we can follow these steps:

### Step 1: Determine the amount of unreacted sulfuric acid
The total volume of sulfuric acid used is 60 mL of \( M/10 \) sulfuric acid. The unreacted acid required 20 mL of \( M/10 \) sodium hydroxide for neutralization.

### Step 2: Calculate the moles of unreacted sulfuric acid
The molarity of the sulfuric acid is \( M/10 \), which means:
\[ \text{Molarity} = \frac{1}{10} \text{ mol/L} \]
The volume of unreacted sulfuric acid is 20 mL of \( M/10 \) sodium hydroxide. We can calculate the moles of sodium hydroxide used for neutralization:
\[ \text{Moles of NaOH} = \text{Molarity} \times \text{Volume (L)} = \frac{1}{10} \times \frac{20}{1000} = 0.002 \text{ moles} \]

### Step 3: Relate moles of NaOH to moles of sulfuric acid
The reaction between sulfuric acid and sodium hydroxide is:
\[ H_2SO_4 + 2 NaOH \rightarrow Na_2SO_4 + 2 H_2O \]
From the balanced equation, 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide. Therefore, the moles of sulfuric acid that reacted can be calculated as:
\[ \text{Moles of } H_2SO_4 = \frac{0.002}{2} = 0.001 \text{ moles} \]

### Step 4: Calculate the moles of sulfuric acid initially present
The total moles of sulfuric acid initially present in 60 mL can be calculated as:
\[ \text{Moles of } H_2SO_4 = \frac{1}{10} \times \frac{60}{1000} = 0.006 \text{ moles} \]

### Step 5: Calculate the moles of sulfuric acid that reacted
The moles of sulfuric acid that reacted to form ammonia can be calculated as:
\[ \text{Moles of } H_2SO_4 \text{ that reacted} = 0.006 - 0.001 = 0.005 \text{ moles} \]

### Step 6: Calculate the equivalent of ammonia produced
Since 1 mole of sulfuric acid produces 1 mole of ammonia, the moles of ammonia produced is also 0.005 moles.

### Step 7: Calculate the mass of ammonia produced
The molar mass of ammonia (\( NH_3 \)) is approximately 17 g/mol. Therefore, the mass of ammonia produced is:
\[ \text{Mass of } NH_3 = \text{Moles} \times \text{Molar mass} = 0.005 \times 17 = 0.085 \text{ g} \]

### Step 8: Calculate the percentage of nitrogen in the organic compound
The formula for calculating the percentage of nitrogen in the compound is given by:
\[ \text{Percentage of nitrogen} = \left( \frac{\text{Mass of nitrogen}}{\text{Mass of organic compound}} \right) \times 100 \]
The mass of nitrogen in ammonia is:
\[ \text{Mass of nitrogen} = 0.085 \times \frac{14}{17} \approx 0.070 \text{ g} \]
Thus, the percentage of nitrogen is:
\[ \text{Percentage of nitrogen} = \left( \frac{0.070}{1.4} \right) \times 100 \approx 5\% \]

### Final Answer
The percentage of nitrogen in the organic compound is approximately **5%**.

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