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

Mass of organic compound = 0.25 gExperimental values,               At STP V1=40 mL                              V2=? T1=300 K                              T2=273 K P1=725-25=700 mm      P2=760 mm                                                   P1V1T1=P2V2T2 V2=P1V1T2T1P2=700×40×273300×760=33.52 mL 22400 mL of N2 at STP weighs =28 g ∴ 33.52 mL of N2 at STP weighs =28×33.5222400                                                    =0.0419 g % of N= Mass of nitrogen at STP  Mass of organic compound taken ×100 =0.04190.25×100=16.76%