In Dumas' method for estimation of nitrogen 0.4 g 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 is (Given: Aqueous tension at 300 K = 15 mm Hg)

In Dumas’ method the nitrogen gas collected over water is always moist, so we must first obtain the volume of dry nitrogen using Dalton’s law of partial pressures.

Total pressure of the gas mixture, $$P_{\text{total}} = 715 \text{ mm Hg}$$
Aqueous tension of water at $$300 \text{ K},\, P_{\text{H}_2\text{O}} = 15 \text{ mm Hg}$$

Hence, pressure of dry nitrogen
$$P_{\text{N}_2} = P_{\text{total}} - P_{\text{H}_2\text{O}} = 715 - 15 = 700 \text{ mm Hg}$$

Convert this pressure into atmospheres because the gas constant $$R$$ in SI units involves atmospheres:
$$1 \text{ atm} = 760 \text{ mm Hg}$$, therefore
$$P = \frac{700}{760} \text{ atm} = 0.921 \text{ atm}$$ (rounded to three significant figures)

The measured volume of nitrogen is
$$V = 60 \text{ mL} = 0.060 \text{ L}$$

Temperature is
$$T = 300 \text{ K}$$

Use the ideal-gas equation $$PV = nRT$$ to find the number of moles of nitrogen.
Gas constant, $$R = 0.0821 \text{ L atm mol}^{-1}\text{K}^{-1}$$

Substituting the values:
$$n = \frac{PV}{RT} = \frac{0.921 \times 0.060}{0.0821 \times 300}$$

Calculate the numerator:
$$0.921 \times 0.060 = 0.0553$$

Calculate the denominator:
$$0.0821 \times 300 = 24.63$$

Hence,
$$n = \frac{0.0553}{24.63} = 0.00224 \text{ mol}$$

Mass of nitrogen present:
Molar mass of $$\text{N}_2 = 28 \text{ g mol}^{-1}$$, so
$$m_{\text{N}_2} = n \times 28 = 0.00224 \times 28 = 0.0628 \text{ g}$$

The original mass of the organic compound taken was $$0.400 \text{ g}$$. Therefore, percentage of nitrogen in the compound is
$$\%\text{N} = \frac{0.0628}{0.400} \times 100 = 15.7\%$$ (to three significant figures)

Thus the percentage composition of nitrogen in the compound is $$\mathbf{15.71\%}$$.

Option A is correct.