1. Correct the pressure of dry nitrogen

In Dumas' method, the gas collected over water contains water vapour, so pressure of dry $N_2$ is:

$P_{N_2} = 715 - 15 = 700\ \text{mm Hg}$

Convert into atm:

$P_{N_2} = \frac{700}{760}\ \text{atm}$

2. Given data

• Mass of organic compound $= 0.5\,\text{g}$
• Volume of nitrogen $= 60\,\text{mL} = 0.060\,\text{L}$
• Temperature $= 300\,\text{K}$
• Pressure of dry nitrogen $= \dfrac{700}{760}\,\text{atm}$

3. Use ideal gas equation to calculate moles of nitrogen gas

$n = \frac{PV}{RT}$

Substitute:

$n = \frac{\left(\frac{700}{760}\right)(0.060)}{(0.0821)(300)}$

$n \approx \frac{0.05526}{24.63} \approx 2.243\times 10^{-3}\ \text{mol}$

4. Mass of nitrogen present

The collected gas is molecular nitrogen, $N_2$.

So,

$\text{Mass of }N_2 = n \times 28$

$= 2.243\times 10^{-3} \times 28 \approx 0.0628\,\text{g}$

5. Percentage of nitrogen in the compound

$\%N = \frac{0.0628}{0.5}\times 100$

$\%N \approx 12.56\% \approx 12.57\%$

6. Check options

• A: $20.87$ ❌
• B: $12.57$ ✅
• C: $1.257$ ❌
• D: $18.67$ ❌

Therefore, the correct answer is Option B.