The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was passed in 100 mL of 0.1 M sulphuric acid. The excess of acid required 20 mL of 0.5 M sodium hydroxide solution for complete neutralisation. The organic compound is:
acetamide
benzamide
urea
thiourea

Options

A. acetamide
B. benzamide
C. urea
D. thiourea

Explanation

To determine the organic compound from which ammonia was evolved, we can follow these steps:

### Step 1: Calculate the moles of excess sulfuric acid
Given that the volume of sulfuric acid used is 100 mL (0.1 M) and the excess acid required 20 mL of 0.5 M sodium hydroxide for neutralization, we first calculate the moles of sodium hydroxide used.

\[ \text{Moles of NaOH} = \text{Molarity} \times \text{Volume (in L)} = 0.5 \, \text{mol/L} \times 0.020 \, \text{L} = 0.01 \, \text{mol} \]

### Step 2: Calculate the moles of sulfuric acid neutralized
Since sulfuric acid (H₂SO₄) reacts with sodium hydroxide (NaOH) in a 1:2 ratio, the moles of sulfuric acid neutralized can be calculated as follows:

\[ \text{Moles of H₂SO₄} = \frac{\text{Moles of NaOH}}{2} = \frac{0.01 \, \text{mol}}{2} = 0.005 \, \text{mol} \]

### Step 3: Calculate the initial moles of sulfuric acid
The total moles of sulfuric acid initially present in 100 mL can be calculated as:

\[ \text{Moles of H₂SO₄ (initial)} = \text{Molarity} \times \text{Volume (in L)} = 0.1 \, \text{mol/L} \times 0.1 \, \text{L} = 0.01 \, \text{mol} \]

### Step 4: Calculate the moles of sulfuric acid that reacted with ammonia
The moles of sulfuric acid that reacted with ammonia can be found by subtracting the moles of sulfuric acid that were neutralized from the initial moles:

\[ \text{Moles of H₂SO₄ (reacted with NH₃)} = \text{Moles of H₂SO₄ (initial)} - \text{Moles of H₂SO₄ (excess)} = 0.01 \, \text{mol} - 0.005 \, \text{mol} = 0.005 \, \text{mol} \]

### Step 5: Calculate the moles of ammonia evolved
Since sulfuric acid reacts with ammonia in a 1:1 ratio, the moles of ammonia evolved are equal to the moles of sulfuric acid that reacted:

\[ \text{Moles of NH₃} = 0.005 \, \text{mol} \]

### Step 6: Calculate the mass of nitrogen in ammonia
The molar mass of ammonia (NH₃) is approximately 17 g/mol. The mass of ammonia can be calculated as:

\[ \text{Mass of NH₃} = \text{Moles} \times \text{Molar Mass} = 0.005 \, \text{mol} \times 17 \, \text{g/mol} = 0.085 \, \text{g} \]

### Step 7: Calculate the mass of nitrogen in ammonia
The mass of nitrogen in ammonia can be calculated using the molar mass of nitrogen (N), which is approximately 14 g/mol:

\[ \text{Mass of N} = \text{Moles of NH₃} \times \text{Molar Mass of N} = 0.005 \, \text{mol} \times 14 \, \text{g/mol} = 0.07 \, \text{g} \]

### Step 8: Calculate the percentage of nitrogen in the organic compound
The percentage of nitrogen in the organic compound can be calculated using the formula:

\[ \text{Percentage of N} = \left( \frac{\text{Mass of N}}{\text{Mass of organic compound}} \right) \times 100 = \left( \frac{0.07 \, \text{g}}{0.30 \, \text{g}} \right) \times 100 \approx 23.33\% \]

### Step 9: Identify the organic compound
Now, we compare the calculated percentage of nitrogen with the known percentages in the given options:
- Acetamide: ~17.5%
- Benzamide: ~17.5%
- Urea: ~46.7%
- Thiourea: ~40%

The closest match to our calculated percentage of nitrogen (23.33%) indicates that the organic compound is likely **Urea**, which has a nitrogen content of approximately 46.6%.

### Final Answer:
The organic compound is **Urea**.