In Dumas' method of estimation of nitrogen `0.35 g` of an organic compound gave `55mL` of nitrogen collected at `300K` temperature and `715 mm` pressure. The percentage composition of nitrogen in the compound would be `: (` Aqueous tension at `300K =15 mm)`

Options

A. `16.45`
B. `17.45`
C. `14.45`
D. `15.45`

Explanation

To solve the problem of estimating the percentage composition of nitrogen in the organic compound using Dumas' method, we will follow these steps:

### Step 1: Calculate the Effective Pressure of Nitrogen
The total pressure of the gas collected is given as 715 mmHg, and we need to subtract the aqueous tension (15 mmHg) to find the effective pressure of nitrogen.

\[ P_{gas} = P_{total} - P_{aqueous} = 715 \, \text{mmHg} - 15 \, \text{mmHg} = 700 \, \text{mmHg} \]

### Step 2: Use the Ideal Gas Law to Find Volume at STP
We will use the formula relating the pressures and volumes at two different conditions (initial and STP):

\[ P_1 V_1 / T_1 = P_2 V_2 / T_2 \]

Where:
- \( P_1 = 700 \, \text{mmHg} \)
- \( V_1 = 55 \, \text{mL} \)
- \( T_1 = 300 \, \text{K} \)
- \( P_2 = 760 \, \text{mmHg} \) (standard pressure)
- \( T_2 = 273 \, \text{K} \) (standard temperature)
- \( V_2 \) is what we need to find.

Rearranging the equation to solve for \( V_2 \):

\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]

Substituting the known values:

\[ V_2 = \frac{700 \, \text{mmHg} \times 55 \, \text{mL} \times 273 \, \text{K}}{760 \, \text{mmHg} \times 300 \, \text{K}} \]

Calculating \( V_2 \):

\[ V_2 = \frac{700 \times 55 \times 273}{760 \times 300} \]

\[ V_2 \approx 46.099 \, \text{mL} \]

### Step 3: Calculate the Percentage Composition of Nitrogen
Now we can use the formula for the percentage composition of nitrogen:

\[ \text{Percentage of } N = \frac{V_2}{8 \times \text{mass of organic compound}} \times 100 \]

Substituting the values:

\[ \text{Percentage of } N = \frac{46.099 \, \text{mL}}{8 \times 0.35 \, \text{g}} \times 100 \]

Calculating:

\[ \text{Percentage of } N = \frac{46.099}{2.8} \times 100 \approx 16.45\% \]

### Final Answer
The percentage composition of nitrogen in the compound is **16.45%**.

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