0.2325 g of an organic compound was analysed for nitrogen by Duma's method. 31.7 mL of moist nitrogen was collected at 25°C and 755.8 mm pressure. Calculate the percentage of nitrogen in the sample (aq. tension of water at 25°C and 23.8 mm).

Explanation

To calculate the percentage of nitrogen in the organic compound using Duma's method, we can follow these steps:

### Step 1: Gather the given data
- Mass of organic compound = 0.2325 g
- Volume of moist nitrogen collected (V1) = 31.7 mL
- Temperature (T1) = 25°C = 298 K (converted to Kelvin)
- Pressure (P1) = 755.8 mm Hg
- Aqueous tension of water at 25°C = 23.8 mm Hg

### Step 2: Calculate the pressure of dry nitrogen
To find the pressure of the dry nitrogen gas, we need to subtract the aqueous tension from the total pressure:
\[ P_{\text{dry}} = P_1 - \text{Aqueous tension} \]
\[ P_{\text{dry}} = 755.8 \, \text{mm Hg} - 23.8 \, \text{mm Hg} = 732 \, \text{mm Hg} \]

### Step 3: Use the Ideal Gas Law to find the volume of nitrogen at STP
We can use the formula derived from the Ideal Gas Law:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Where:
- \( P_2 \) = 760 mm Hg (standard pressure)
- \( T_2 \) = 273 K (standard temperature)
- \( V_2 \) = volume of nitrogen at STP

Rearranging the formula to solve for \( V_2 \):
\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]

Substituting the known values:
\[ V_2 = \frac{732 \, \text{mm Hg} \times 31.7 \, \text{mL} \times 273 \, \text{K}}{760 \, \text{mm Hg} \times 298 \, \text{K}} \]

Calculating \( V_2 \):
\[ V_2 = \frac{732 \times 31.7 \times 273}{760 \times 298} \approx 28 \, \text{mL} \]

### Step 4: Calculate the mass of nitrogen gas
The molar mass of nitrogen gas (N₂) is 28 g/mol. At STP, 22.4 L (or 22400 mL) of nitrogen gas corresponds to 28 g. Thus, we can find the mass of nitrogen in 28 mL:
\[ \text{Mass of N}_2 = \frac{28 \, \text{g}}{22400 \, \text{mL}} \times 28 \, \text{mL} \]
\[ \text{Mass of N}_2 = \frac{28 \times 28}{22400} \approx 0.035 \, \text{g} \]

### Step 5: Calculate the percentage of nitrogen in the organic compound
Now, we can calculate the percentage of nitrogen in the sample:
\[ \text{Percentage of N} = \left( \frac{\text{Mass of N}_2}{\text{Mass of organic compound}} \right) \times 100 \]
\[ \text{Percentage of N} = \left( \frac{0.035 \, \text{g}}{0.2325 \, \text{g}} \right) \times 100 \approx 15.1\% \]

### Final Result
The percentage of nitrogen in the organic compound is **15.1%**.

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