0.2325 g of an organic compound was analysed for nitrogen by Duma's method. 31.7 mL of moist nitrogen gas was collected at `25^(@)C` and 755.8 mm Hg pressure. Determine the percentage of nitrogen in the compound. The aqueous tension of water is 23.8 mm Hg at `25^(@)C`.

Answer

`15.1 %`

Explanation

Step I. Volume of `N_(2)` at N.T.P.
`{:("Experimental Conditions","N.T.P. Conditions"),(V_(1)=31.7 mL,V_(2)=?),(P_(1)=755.8-23.8=732 mm,P_(2)=760 mm),(T_(1)=25+273=298 K,T_(2)=273 K):}`
`(P_(1)V_(1))/T_(1)=(P_(2)V_(2))/T_(2)` or `V_(2)=(P_(1)V_(1)T_(2))/(T_(1)P_(2))=((732mm)xx(31.7 mL)xx(273 K))/((298 K)xx(760 mm))=27.97 mL`
Step II. Percentage of nitrogen
`=28/22400xx("Volume of "N_(2)" at N.T.P.")/("Mass of compound")xx100=28/22400xx27.97/0.2325xx100=15.1 %`