0.50 g of an organic compound was Kjeldahlised and the `NH_(3)` evolved was absorbed in50 ml of 0.5 M `H_(2)SO_(4)`. The residual acid required `60cm^(3)` of 0.5 M NaOH . The percentage of nitrogen in the organic compound is

Options

A. 14
B. 28
C. 56
D. 42

Explanation

To solve the problem step by step, we will determine the percentage of nitrogen in the organic compound based on the data provided.

### Step 1: Determine the amount of residual sulfuric acid
We know that 50 mL of 0.5 M H₂SO₄ was used initially. The residual acid required 60 mL of 0.5 M NaOH for neutralization.

### Step 2: Calculate the moles of NaOH used
First, we calculate the number of moles of NaOH used:
\[ \text{Moles of NaOH} = \text{Volume (L)} \times \text{Molarity (mol/L)} = 0.060 \, \text{L} \times 0.5 \, \text{mol/L} = 0.030 \, \text{mol} \]

### Step 3: Calculate the moles of H₂SO₄ neutralized
Since NaOH neutralizes H₂SO₄ in a 2:1 ratio (1 mole of H₂SO₄ reacts with 2 moles of NaOH), the moles of H₂SO₄ neutralized can be calculated as:
\[ \text{Moles of H₂SO₄} = \frac{\text{Moles of NaOH}}{2} = \frac{0.030 \, \text{mol}}{2} = 0.015 \, \text{mol} \]

### Step 4: Calculate the initial moles of H₂SO₄
The initial moles of H₂SO₄ can be calculated from the initial volume and molarity:
\[ \text{Initial moles of H₂SO₄} = \text{Volume (L)} \times \text{Molarity (mol/L)} = 0.050 \, \text{L} \times 0.5 \, \text{mol/L} = 0.025 \, \text{mol} \]

### Step 5: Calculate the moles of H₂SO₄ that reacted with NH₃
The moles of H₂SO₄ that reacted with NH₃ can be found by subtracting the moles of residual H₂SO₄ from the initial moles:
\[ \text{Moles of H₂SO₄ reacted} = \text{Initial moles} - \text{Moles of H₂SO₄ neutralized} = 0.025 \, \text{mol} - 0.015 \, \text{mol} = 0.010 \, \text{mol} \]

### Step 6: Calculate the moles of NH₃ produced
Since 1 mole of H₂SO₄ reacts with 2 moles of NH₃, the moles of NH₃ produced is:
\[ \text{Moles of NH₃} = 2 \times \text{Moles of H₂SO₄ reacted} = 2 \times 0.010 \, \text{mol} = 0.020 \, \text{mol} \]

### Step 7: Calculate the mass of nitrogen in NH₃
The molar mass of NH₃ is approximately 17 g/mol, and the mass of nitrogen (N) in NH₃ can be calculated as follows:
\[ \text{Mass of NH₃} = \text{Moles of NH₃} \times \text{Molar mass of NH₃} = 0.020 \, \text{mol} \times 17 \, \text{g/mol} = 0.34 \, \text{g} \]
Since NH₃ contains one nitrogen atom, the mass of nitrogen is:
\[ \text{Mass of N} = 0.34 \, \text{g} \times \frac{14 \, \text{g/mol}}{17 \, \text{g/mol}} = 0.28 \, \text{g} \]

### Step 8: Calculate the percentage of nitrogen in the organic compound
Finally, we can calculate the percentage of nitrogen in the organic compound:
\[ \text{Percentage of N} = \left(\frac{\text{Mass of N}}{\text{Mass of organic compound}}\right) \times 100 = \left(\frac{0.28 \, \text{g}}{0.50 \, \text{g}}\right) \times 100 = 56\% \]

### Final Answer
The percentage of nitrogen in the organic compound is **56%**.

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