0.25 g of an organic compound gave `30 cm^(3)` of moist dinitrogen at 288 K and 745 mm pressure. Calculate the percentage of nitrogen . (Aq tension of 288 K =12.7 mm)

Explanation

Mass of substance =0.25 g
Volume of moist dinitrogen =`30cm^(3)`
Temperature =288 K
Pressure= `745-12.7 =732.3 mm`
Step I . To reduce the volume of `N_(2)` at S.T.P
We know, `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))`
`P_(2),T_(2),V_(2)` refer to S.T.P conditions)
`V_(2)=(P_(1)V_(1))/(T_(1))xx (T_(2))/(P_(2))==(732.3xx30xx273)/(288xx760) =27.4 cm^(3)`
Step II: Calculation of percentage of nitrogen
`22400 cm^(3)` of dinitrogen at S.T.P weight =28 g
`27.4 cm^(3) ` of dinitronge at S.T.P weight `(28xx274)/22400 =0.034 g `
Percentage of nitrogen in organic compound `=0.034/0.25 xx100=13.6`