0.2046 g of an organic compound gave `30.4 cm^(3)` of moist nitrogen measured at 288 K and 732.7 mm pressure. Calculate the percentage of nitrogen in the compound (Aqueous tension at 288 K is 12.7 mm)

Answer

`16.68 %`

Explanation

Step I. Volume of `N_(2)` at N.T.P.
`{:("Experimental Conditions","N.T.P. Conditions"),(V_(1)=30.4 cm^(3),V_(2)=?),(P_(1)=732.7-12.7=720 mm,P_(2)=760 mm),(T_(1)=288 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))=((720mm)xx(30.4 cm^(3))xx(273 K))/((288 K)xx(760 mm))=27.3 cm^(3)`
Step II. Percentage of nitrogen
`=28/22400xx("Volume of "N_(2)" at N.T.P.")/("Mass of compound")xx100=28/22400xx27.3/0.2046xx100=16.68 %`