In Duma's method, 0.3 g of an organic compound gave `50 cm^(3)` of nitrogen collected at 300 K and 715 mm pressure. Calculate the percentage of nitrogen in the compound. Aqueous tension of water at 300 K is 15 mm.

Explanation

To solve the problem using Duma's method, we will follow these steps:

### Step 1: Convert the given volume of nitrogen to liters
Given:
- Volume of nitrogen (V1) = 50 cm³ = 50 mL = 0.050 L

### Step 2: Adjust the pressure for aqueous tension
Given:
- Total pressure (P1) = 715 mm Hg
- Aqueous tension at 300 K = 15 mm Hg

To find the pressure of the dry nitrogen gas (P_gas):
\[ P_{gas} = P_{total} - P_{aqueous} = 715 \, \text{mm Hg} - 15 \, \text{mm Hg} = 700 \, \text{mm Hg} \]

### Step 3: Use the ideal gas law to find the volume at NTP
At Normal Temperature and Pressure (NTP):
- Pressure (P2) = 760 mm Hg
- Temperature (T2) = 273 K

Using the formula:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

We rearrange to find \( V_2 \):
\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]

Substituting the known values:
- \( P_1 = 700 \, \text{mm Hg} \)
- \( V_1 = 0.050 \, \text{L} \)
- \( T_1 = 300 \, \text{K} \)
- \( T_2 = 273 \, \text{K} \)
- \( P_2 = 760 \, \text{mm Hg} \)

Calculating \( V_2 \):
\[ V_2 = \frac{700 \times 0.050 \times 273}{760 \times 300} \]

Calculating the value:
\[ V_2 = \frac{700 \times 0.050 \times 273}{228000} \]
\[ V_2 = \frac{9555}{228000} \approx 0.0419 \, \text{L} \text{ or } 41.9 \, \text{mL} \]

### Step 4: Calculate the moles of nitrogen
Using the molar volume of nitrogen at NTP (22.4 L/mol):
\[ \text{Moles of nitrogen} = \frac{V_2}{22.4} = \frac{0.0419}{22.4} \approx 0.00187 \, \text{mol} \]

### Step 5: Calculate the mass of nitrogen
Using the molar mass of nitrogen (N2 = 28 g/mol):
\[ \text{Mass of nitrogen} = \text{moles} \times \text{molar mass} = 0.00187 \times 28 \approx 0.05236 \, \text{g} \]

### Step 6: Calculate the percentage of nitrogen in the organic compound
Given the mass of the organic compound is 0.3 g:
\[ \text{Percentage of nitrogen} = \left( \frac{\text{mass of nitrogen}}{\text{mass of organic compound}} \right) \times 100 \]
\[ \text{Percentage of nitrogen} = \left( \frac{0.05236}{0.3} \right) \times 100 \approx 17.45\% \]

### Final Answer
The percentage of nitrogen in the organic compound is approximately **17.45%**.

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