In Dumas' method 292 mg of an organic compound released 50 mL of nitrogen gas ($$N_2$$) at 300 K temperature and 715 mm Hg pressure. The percentage composition of 'N' in the organic compound is _____ %.(Nearest integer)
(Aqueous tension at 300 K = 15 mm Hg)

Mass of organic compound taken: $$m_{\text{compound}} = 292 \text{ mg} = 0.292 \text{ g}$$.

Volume of $$N_2$$ collected: $$V = 50 \text{ mL} = 0.050 \text{ L}$$.
Temperature: $$T = 300 \text{ K}$$.
Total pressure: $$P_{\text{total}} = 715 \text{ mm Hg}$$.
Aqueous tension (vapour pressure of water): $$P_{\text{H}_2O} = 15 \text{ mm Hg}$$.

First remove the vapour-pressure contribution to get the pressure of dry nitrogen.
$$P_{N_2} = P_{\text{total}} - P_{\text{H}_2O} = 715 - 15 = 700 \text{ mm Hg}$$.

Convert this pressure into atmospheres (1 atm = 760 mm Hg):
$$P_{N_2} = \frac{700}{760} \text{ atm} = 0.9211 \text{ atm}$$.

Use the ideal-gas equation $$PV = nRT$$. Take $$R = 0.0821 \text{ L atm mol}^{-1}\text{K}^{-1}$$.

$$n = \frac{P V}{R T} = \frac{0.9211 \times 0.050}{0.0821 \times 300} \text{ mol}$$.

$$n = 0.00187 \text{ mol of } N_2$$.

Molar mass of $$N_2$$ is $$28 \text{ g mol}^{-1}$$, so the mass of nitrogen obtained is
$$m_N = n \times 28 = 0.00187 \times 28 = 0.0524 \text{ g} = 52.4 \text{ mg}$$.

Percentage of nitrogen in the organic compound:
$$\%N = \frac{m_N}{m_{\text{compound}}} \times 100 = \frac{0.0524}{0.292} \times 100 = 17.94 \% \approx 18 \%$$.

Hence, the percentage composition of nitrogen in the organic compound is 18 %.