To solve this question using Dumas' method for nitrogen estimation, we'll use the ideal gas law and make necessary corrections for aqueous tension.
### Step-by-step solution:
**Given Data:**
- Mass of organic compound = 0.3 g
- Volume of nitrogen collected = 50 mL = 0.05 L
- Temperature (T) = 300 K
- Pressure (P) = 715 mm Hg
- Aqueous tension = 15 mm Hg
We need to calculate the percentage composition of nitrogen in the compound.
### Step 1: Correct the pressure for aqueous tension
The total pressure measured includes the pressure due to the nitrogen gas and the vapor pressure (aqueous tension). To find the pressure due to nitrogen alone, we subtract the aqueous tension from the total pressure:
\[
P_{\text{N}_2} = P_{\text{total}} - P_{\text{aqueous}} = 715 \, \text{mm Hg} - 15 \, \text{mm Hg} = 700 \, \text{mm Hg}
\]
Convert the pressure to atmospheres (since 1 atm = 760 mm Hg):
\[
P_{\text{N}_2} = \frac{700 \, \text{mm Hg}}{760 \, \text{mm Hg}} = 0.921 \, \text{atm}
\]
### Step 2: Use the ideal gas law
The ideal gas law is:
\[
PV = nRT
\]
Where:
- \( P \) = Pressure of nitrogen (0.921 atm)
- \( V \) = Volume of nitrogen (0.05 L)
- \( n \) = Moles of nitrogen
- \( R \) = Gas constant = 0.0821 L·atm/(mol·K)
- \( T \) = Temperature (300 K)
Rearrange the equation to solve for \( n \) (moles of nitrogen):
\[
n = \frac{PV}{RT}
\]
Substitute the values:
\[
n = \frac{(0.921 \, \text{atm}) \times (0.05 \, \text{L})}{(0.0821 \, \text{L·atm/(mol·K}}) \times (300 \, \text{K})}
\]
\[
n = \frac{0.04605}{24.63} = 0.00187 \, \text{mol}
\]
### Step 3: Calculate the mass of nitrogen
The molar mass of nitrogen (\(N_2\)) is 28 g/mol. So, the mass of nitrogen is:
\[
\text{Mass of } N_2 = n \times \text{Molar mass of } N_2 = 0.00187 \, \text{mol} \times 28 \, \text{g/mol}
\]
\[
\text{Mass of } N_2 = 0.05236 \, \text{g}
\]
### Step 4: Calculate the percentage of nitrogen
Now, we can calculate the percentage of nitrogen in the compound using the formula:
\[
\% \text{N} = \frac{\text{Mass of } N_2}{\text{Mass of compound}} \times 100
\]
\[
\% \text{N} = \frac{0.05236 \, \text{g}}{0.3 \, \text{g}} \times 100 = 17.45\%
\]
### Final Answer:
The percentage of nitrogen in the compound is approximately **17.46%**, so the correct answer is:
**(B) 17.46%**.
