In Duma's method for quantitative estimation of nitrogen, 0.5 g of an organic compound gave 100 ml of nitrogen collected at 27°C temperature and 680 mm of Hg pressure. What is the percentage composition of nitrogen in the sample? [Given aqueous tension at 27°C = 20mm Hg]

Options

A. 0.2525
B. 0.1525
C. 0.2875
D. 0.1975

Answer

To solve the problem of determining the percentage composition of nitrogen in the sample using Duma's method, we can follow these steps:

### Step 1: Calculate the Effective Pressure
The effective pressure of nitrogen collected can be calculated by subtracting the aqueous tension from the total pressure.

\[ \text{Effective Pressure} (P) = \text{Total Pressure} - \text{Aqueous Tension} \]
Given:
- Total Pressure = 680 mm Hg
- Aqueous Tension = 20 mm Hg

\[ P = 680 \, \text{mm Hg} - 20 \, \text{mm Hg} = 660 \, \text{mm Hg} \]

### Step 2: Convert Temperature to Kelvin
Convert the temperature from Celsius to Kelvin.

\[ T = 27°C + 273 = 300 \, \text{K} \]

### Step 3: Use the Ideal Gas Law to Find Volume at STP
Using the formula \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \), we can find the volume of nitrogen at standard temperature and pressure (STP).

At STP:
- \( P_2 = 760 \, \text{mm Hg} \)
- \( T_2 = 273 \, \text{K} \)
- \( V_1 = 100 \, \text{ml} \)
- \( P_1 = 660 \, \text{mm Hg} \)
- \( T_1 = 300 \, \text{K} \)

Rearranging the equation gives:

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

Substituting the values:

\[ V_2 = \frac{660 \, \text{mm Hg} \times 100 \, \text{ml} \times 273 \, \text{K}}{760 \, \text{mm Hg} \times 300 \, \text{K}} \]

Calculating \( V_2 \):

\[ V_2 = \frac{660 \times 100 \times 273}{760 \times 300} \]

\[ V_2 = \frac{18007800}{228000} \approx 79.0 \, \text{ml} \]

### Step 4: Calculate the Mass of Nitrogen
Using the molar volume of nitrogen at STP (22.4 L or 22400 ml), we can find the mass of nitrogen corresponding to the volume calculated.

The molar mass of nitrogen (N2) is 28 g/mol.

\[ \text{Mass of N}_2 = \left(\frac{28 \, \text{g}}{22400 \, \text{ml}}\right) \times V_2 \]

Substituting \( V_2 \):

\[ \text{Mass of N}_2 = \left(\frac{28}{22400}\right) \times 79.0 \]

Calculating:

\[ \text{Mass of N}_2 \approx 0.1 \, \text{g} \]

### Step 5: Calculate the Percentage Composition of Nitrogen
Now, we can find the percentage composition of nitrogen in the organic compound.

\[ \text{Percentage of N} = \left(\frac{\text{Mass of N}_2}{\text{Mass of organic compound}}\right) \times 100 \]

Given that the mass of the organic compound is 0.5 g:

\[ \text{Percentage of N} = \left(\frac{0.1}{0.5}\right) \times 100 = 20\% \]

### Final Answer
The percentage composition of nitrogen in the sample is **20%**.

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