`0.5g` of an organic compound on `Kjeldahl's` analysis gave enough ammonia to just neutralize `10 cm^(3)` of `1 M H_(2)SO_(4)`. The percentage of nitrogen in the compound is

Options

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

Explanation

To find the percentage of nitrogen in the organic compound using Kjeldahl's analysis, we can follow these steps:

### Step 1: Understand the Reaction
In Kjeldahl's analysis, ammonia (NH₃) is produced from the nitrogen in the organic compound. This ammonia then reacts with sulfuric acid (H₂SO₄) in a neutralization reaction.

### Step 2: Calculate Moles of H₂SO₄
We are given that `10 cm³` of `1 M H₂SO₄` is used. First, we need to convert the volume from cm³ to liters:
\[ 10 \, \text{cm}^3 = 10 \times 10^{-3} \, \text{L} = 0.01 \, \text{L} \]

Now, calculate the moles of H₂SO₄:
\[ \text{Moles of H₂SO₄} = \text{Molarity} \times \text{Volume} = 1 \, \text{mol/L} \times 0.01 \, \text{L} = 0.01 \, \text{mol} \]

### Step 3: Determine Moles of NH₃
The balanced reaction between NH₃ and H₂SO₄ is:
\[ \text{2 NH₃} + \text{H₂SO₄} \rightarrow \text{(NH₄)₂SO₄} \]
From the equation, we see that 1 mole of H₂SO₄ reacts with 2 moles of NH₃. Therefore, the moles of NH₃ produced are:
\[ \text{Moles of NH₃} = 2 \times \text{Moles of H₂SO₄} = 2 \times 0.01 \, \text{mol} = 0.02 \, \text{mol} \]

### Step 4: Calculate Mass of Nitrogen
The molar mass of NH₃ is approximately `17 g/mol`, and since each molecule of NH₃ contains one nitrogen atom, the moles of nitrogen will be the same as the moles of NH₃:
\[ \text{Mass of nitrogen} = \text{Moles of NH₃} \times \text{Molar mass of N} = 0.02 \, \text{mol} \times 14 \, \text{g/mol} = 0.28 \, \text{g} \]

### Step 5: Calculate Percentage of Nitrogen in the Compound
Now, we can calculate the percentage of nitrogen in the organic compound:
\[ \text{Percentage of nitrogen} = \left( \frac{\text{Mass of nitrogen}}{\text{Mass of organic compound}} \right) \times 100 = \left( \frac{0.28 \, \text{g}}{0.5 \, \text{g}} \right) \times 100 = 56\% \]

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

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