In Duma's method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25 mm, the percentage of nitrogen in the compound is:

Given:
m=0.25g,V1=40ml,T1=300K,P1=725 mm
𕒶5 mm=700 mm.
Using the ideal gas equation, PV = nRT, we can calculate the volume of nitrogen at STP.
V2 = (P1V1T2)/(P2T1)
V2 = (70040273)/(760300)
V2 = 31.7 ml
At STP, 22400 ml of N2 = 28 g of N2
Therefore, 31.7 ml of N2 = (2831.7)/22400 = 0.04 g of N2
Percentage of nitrogen in the compound = (0.04/0.25)*100 = 16%
The closest answer is 16.76%