To solve this problem, we will use the ideal gas equation and the concept of aqueous tension.$\\$Given:$\\$• Mass of the organic compound $= 0.25$ g$\\$• Volume of nitrogen collected $= 40$ mL$= 0.040$ L$\\$• Temperature $= 300$ K$\\$• Total pressure $= 725$ mm Hg$\\$• Aqueous tension at 300 K $= 25$ mm Hg$\\$First, calculate the pressure of dry nitrogen gas:$\\ P_{dry} = P_{total} -$Aqueous tension$\\ P_{dry} = 725 - 25 = 700$ mm Hg$\\$Convert pressure from mm Hg to atm:$\\ P_{dry} = \frac{700}{760}$ atm$\\$Using the ideal gas equation $PV = nRT,$ solve for $n: \\ n = \frac{P_{dry} \cdot V}{R \cdot T} \\$Substitute the known values:$\\ n = \frac{\left(\frac{700}{760}\right) \cdot 0.040}{0.0821 \cdot 300} \\$Calculate $n: \\ n = \frac{0.9211 \cdot 0.040}{24.63} \\ n \approx 0.0015$ moles$\\$Calculate the mass of nitrogen using its molar mass (28 g/mol):$\\$Mass of $N_2 = n \cdot$Molar mass of $N_2 \\$Mass of $N_2 = 0.0015 \cdot 28 \\$Mass of $N_2 \approx 0.042$ g$\\$Calculate the percentage of nitrogen in the compound:$\\$Percentage of $N = (\frac{Mass \ of \ N_2}{Mass \ of \ compound}) \times 100 \\$Percentage of $N = \left(\frac{0.042}{0.25}\right) \times 100 \\$Percentage of $N = 16.8\% \\$Therefore, the percentage of nitrogen in the compound is approximately $16.76\%. \\$This corresponds to Option 2.