Question

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)

Answer 1

Here, mass of organic substance = 0.36 gmVolume of nitrogen collected V1= 55 mlRoom temperature T1 = 300 K and vapor pressure of water at 300K = 15 mmSo, actual pressure of dry nitrogen gas P​1= 715 - 15 = 700mm​At STP conditions we know that , P2​= 760 mm and T​2= 273 K, then V1= ?Now, we have P1V1/ T1= P2V2/ T2or, 700 x 55 / 300 = 760 x V2/ 273Hence, ​ V2 = 46.09 mlHence, % age of N = 28 x 46.09 x 100 / 22400= 16.45 %

Answer 2

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 P​1= 715 - 15 = 700mm

​At STP conditions we know that , P2​= 760 mm and T​2= 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%