A sample of `1.0` gm of organic compound was treated with `NaOH` and `NH_(3)` evolved was absorbed in `70ml 0.5M H_(2)SO_(4)`. The residual acid required `65 ml` of `0.5MNaOH` solution of neutraization in Kjeldahl's method. What is the percentage of nitrogen in organic compound?

Explanation

To find the percentage of nitrogen in the organic compound, we can follow these steps:

### Step 1: Calculate the initial moles of H₂SO₄
Given:
- Volume of H₂SO₄ = 70 mL = 0.070 L
- Molarity of H₂SO₄ = 0.5 M

Using the formula:
\[ \text{Moles of H₂SO₄} = \text{Volume (L)} \times \text{Molarity (mol/L)} \]
\[ \text{Moles of H₂SO₄} = 0.070 \, \text{L} \times 0.5 \, \text{mol/L} = 0.035 \, \text{mol} \]

Since H₂SO₄ is a dibasic acid, it can absorb two moles of ammonia (NH₃) for each mole of H₂SO₄. Therefore, the total moles of ammonia that can be absorbed is:
\[ \text{Moles of NH₃ absorbed} = 2 \times 0.035 \, \text{mol} = 0.070 \, \text{mol} \]

### Step 2: Calculate the moles of NaOH used for neutralization
Given:
- Volume of NaOH = 65 mL = 0.065 L
- Molarity of NaOH = 0.5 M

Using the formula:
\[ \text{Moles of NaOH} = \text{Volume (L)} \times \text{Molarity (mol/L)} \]
\[ \text{Moles of NaOH} = 0.065 \, \text{L} \times 0.5 \, \text{mol/L} = 0.0325 \, \text{mol} \]

### Step 3: Calculate the moles of H₂SO₄ that remained after ammonia absorption
The moles of H₂SO₄ that reacted with NaOH is equal to the moles of NaOH used since NaOH is a monoacidic base:
\[ \text{Moles of H₂SO₄ remaining} = 0.0325 \, \text{mol} \]

### Step 4: Calculate the moles of H₂SO₄ that reacted with ammonia
The moles of H₂SO₄ that reacted with ammonia is:
\[ \text{Moles of H₂SO₄ reacted with NH₃} = \text{Initial moles} - \text{Remaining moles} \]
\[ \text{Moles of H₂SO₄ reacted with NH₃} = 0.035 \, \text{mol} - 0.0325 \, \text{mol} = 0.0025 \, \text{mol} \]

### Step 5: Calculate the moles of NH₃ generated
Since each mole of H₂SO₄ reacts with 2 moles of NH₃:
\[ \text{Moles of NH₃ generated} = 2 \times 0.0025 \, \text{mol} = 0.005 \, \text{mol} \]

### Step 6: Calculate the mass of NH₃ generated
Using the molar mass of NH₃ (17 g/mol):
\[ \text{Mass of NH₃} = \text{Moles} \times \text{Molar mass} = 0.005 \, \text{mol} \times 17 \, \text{g/mol} = 0.085 \, \text{g} \]

### Step 7: Calculate the percentage of nitrogen in the organic compound
The mass of nitrogen in NH₃ can be calculated as follows (since nitrogen is 14 g in NH₃):
\[ \text{Mass of N} = \text{Mass of NH₃} \times \frac{14}{17} = 0.085 \, \text{g} \times \frac{14}{17} \approx 0.070 \, \text{g} \]

Now, to find the percentage of nitrogen in the organic compound:
\[ \text{Percentage of N} = \left( \frac{\text{Mass of N}}{\text{Mass of organic compound}} \right) \times 100 = \left( \frac{0.070 \, \text{g}}{1.0 \, \text{g}} \right) \times 100 = 7.0\% \]

### Final Answer:
The percentage of nitrogen in the organic compound is approximately **7.0%**.

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