In dumas method of estimation of nitrogen , 0.35g of an organic compound gave 55ml of nitrogen collected at 300K temperature and 715mm presure. The % compostion of nitrogen in the compund would be ( aqueous tension at 300K = 15mm)
Here, mass of organic substance = 0.36 gm
Volume of nitrogen collected V1 = 55 ml
Room temperature T1 = 300 K and vapor pressure of water at 300K = 15 mm
So, actual pressure of dry nitrogen gas P1 = 715 - 15 = 700mm
At STP conditions we know that , P2 = 760 mm and T2 = 273 K, then V1= ?
Now, we have P1V1/ T1 = P2V2/ T2
or, 700 x 55 / 300 = 760 x V2 / 273
Hence, V2 = 46.09 ml
Hence, % age of N = 28 x 46.09 x 100 / 22400 = 16.45 %
Volume of nitrogen collected V1 = 55 ml
Room temperature T1 = 300 K and vapor pressure of water at 300K = 15 mm
So, actual pressure of dry nitrogen gas P1 = 715 - 15 = 700mm
At STP conditions we know that , P2 = 760 mm and T2 = 273 K, then V1= ?
Now, we have P1V1/ T1 = P2V2/ T2
or, 700 x 55 / 300 = 760 x V2 / 273
Hence, V2 = 46.09 ml
Hence, % age of N = 28 x 46.09 x 100 / 22400 = 16.45 %