In Duma's method of estimation of nitrogen 0.35 ~g of an organic compound gave 55 ~mL of nitrogen collected at 300 ~K temperature and 715 ~mm pressure. The percentage composition of nitrogen in the compound would be (Aqueous tension at 300 ~K =15 ~mm )

Step-by-step solution

p₁ V₁ T₁ = p₂ V₂ T₂ Where p₂= pressure of N ₂ at S T P=760 ~mm aligned & T₂= Temperature of N ₂ at STP =273 ~K & V₂= ? aligned Volume of N ₂ at STP (By gas equation) ( - ₁ t+273 ) V₁ 273 760 =V₂ where p₁= - ₁ =7 / 5 ~mm (pressure at which N ₂ collected). ₁= aqueous tension of water =15 ~mm T₁=t+273=300 ~K V₁=55 ~mL = volume of moist nitrogen in nitrometer V₂= (715-15) 55 300 273 760 =46.098 ~mL aligned & % of nitrogen in given compound & = 28 22400 V₂ W 100= 28 22400 46.098 0.35 100 & =16.45 % aligned