In Dumas' method for estimation of nitrogen, 0.5 gram of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound (Aqueous tension at 300 K = 15 mm Hg) is

In Dumas’ method, nitrogen is collected over water, so the measured pressure includes the vapour pressure of water (aqueous tension).
To obtain the pressure of dry nitrogen, subtract the aqueous tension from the total pressure.

Total pressure of the gas mixture at $$300\ \text{K} = 715\ \text{mm Hg}$$.
Aqueous tension of water at $$300\ \text{K} = 15\ \text{mm Hg}$$.

Pressure of dry $$N_2$$:
$$P_{N_2}=715\ \text{mm Hg}-15\ \text{mm Hg}=700\ \text{mm Hg}$$.

Convert this pressure to atmospheres (because the gas constant $$R$$ we shall use is in $$\text{L atm mol}^{-1}\text{K}^{-1}$$):
$$P=\frac{700}{760}\ \text{atm}=0.921\ \text{atm}$$.

Volume of nitrogen collected:
$$V =60\ \text{mL}=0.06\ \text{L}$$.

Temperature:
$$T =300\ \text{K}$$.

Ideal gas equation: $$PV = nRT$$.
Gas constant: $$R =0.0821\ \text{L atm mol}^{-1}\text{K}^{-1}$$.

Calculate moles of nitrogen:
$$n=\frac{PV}{RT}= \frac{0.921 \times 0.06}{0.0821 \times 300}$$.

Numerator: $$0.921 \times 0.06 = 0.05526$$.
Denominator: $$0.0821 \times 300 = 24.63$$.
Thus,
$$n = \frac{0.05526}{24.63}=0.002244\ \text{mol}$$.

Molar mass of $$N_2 =28\ \text{g mol}^{-1}$$.
Mass of nitrogen present:
$$m = n \times 28 = 0.002244 \times 28 = 0.06283\ \text{g}$$.

Mass of the organic sample taken: $$0.5\ \text{g}$$.

Percentage of nitrogen in the compound:
$$\%N = \frac{0.06283}{0.5} \times 100 = 12.57\%$$.

The percentage composition of nitrogen in the compound is $$\mathbf{12.57\%}$$.
Hence, the correct choice is Option D.