0.22 g of an organic compound on combustion in an atmosphere of `CO_2` gave 34 `cm^3` of moist `N_2` at `17^@ C` and 733.4 mm pressure. If the aqueous tension at `17^@ C` is 13.4 mm, calculate the percentage of nitrogen in the compound.

Explanation

Mass of organic compound= 0.22g Pressure od dry `N_(2)= 733.4-13.4= 720mm Hg`.
Calculation of volume of `N_(2)` at S.T.P. Here, `P_(1)= 720mm Hg, P_(2)= 760mm Hg`
`V_(1)= 34 cm^(3), V_(2)`= ?
`T_(1)= 273 + 17= 290K`
`T_(2)= 273K`
According to the gas equation. `(P_(1)V_(1))/(T_(1))= (P_(2)V_(2))/(T_(2))`
OR
`V_(2)= (P_(1)V_(1))/(T_(1)) xx (T_(2))/(P_(2))= (720 xx 34 xx 273)/(290 xx 760)= 30.32cm^(3)`
Calculation of percentage of N Since 22400 `cm^(3)` of `N_(2)` at S.T.P. weighs 28g, `30.32cm^(3)` of `N_(2)` at S.T.P. will weigh `(28)/(22400) xx 30.32= 0.038g`
Hence, percentage of N in the given compound `=(0.038)/(0.22) xx 100= 17.27%`