The ammonia evolved from 2 g of a compound in Kjeldahl's estimation of nitrogen neutralizes 10 mL of 2 M `H_(2)SO_(4)` solution. The weight percentage of nitrogen in the compound is :

Options

A. 28
B. 14
C. 56
D. 7

Explanation

To find the weight percentage of nitrogen in the compound, we will follow these steps:

### Step 1: Calculate the moles of `H2SO4` used
Given:
- Volume of `H2SO4` = 10 mL = 10 × 10^(-3) L
- Molarity of `H2SO4` = 2 M

Using the formula:
\[ \text{Moles of } H2SO4 = \text{Molarity} \times \text{Volume (in L)} \]
\[ \text{Moles of } H2SO4 = 2 \, \text{mol/L} \times 10 \times 10^{-3} \, \text{L} = 0.02 \, \text{mol} \]

### Step 2: Determine the moles of `NH3` neutralized
The reaction between `NH3` and `H2SO4` is:
\[ 2 \, NH3 + H2SO4 \rightarrow (NH4)2SO4 \]
From the stoichiometry of the reaction, 1 mole of `H2SO4` reacts with 2 moles of `NH3`. Therefore, the moles of `NH3` can be calculated as:
\[ \text{Moles of } NH3 = 2 \times \text{Moles of } H2SO4 = 2 \times 0.02 \, \text{mol} = 0.04 \, \text{mol} \]

### Step 3: Calculate the mass of `NH3`
The molar mass of `NH3` (ammonia) is approximately 17 g/mol. Thus, the mass of `NH3` can be calculated as:
\[ \text{Mass of } NH3 = \text{Moles} \times \text{Molar Mass} = 0.04 \, \text{mol} \times 17 \, \text{g/mol} = 0.68 \, \text{g} \]

### Step 4: Calculate the weight percentage of nitrogen in the compound
The compound weighs 2 g, and the nitrogen in `NH3` contributes to the total nitrogen in the compound. The mass of nitrogen in `NH3` can be calculated as:
\[ \text{Mass of Nitrogen} = \text{Moles of } NH3 \times \text{Molar Mass of Nitrogen} = 0.04 \, \text{mol} \times 14 \, \text{g/mol} = 0.56 \, \text{g} \]

Now, we can calculate the weight percentage of nitrogen in the compound:
\[ \text{Weight percentage of Nitrogen} = \left( \frac{\text{Mass of Nitrogen}}{\text{Mass of Compound}} \right) \times 100 = \left( \frac{0.56 \, \text{g}}{2 \, \text{g}} \right) \times 100 = 28\% \]

### Final Answer:
The weight percentage of nitrogen in the compound is **28%**.

---