In Duma's method 0.206 g of an organic compound gave `18.8 cm^(3)` of moist `N_(2)` at `17^(@)C` and 760 mm Hg pressure. If aqueous at `17^(@)C` is 14.5 mm Hg, calculate the percentage of nitrogen in the given organic compound.

Explanation

Mass of organic compound = 0.22 g
Pressure of dry `N_2 = 733.4 - 13.4 = 720` mm
Calculation of volume of `N_2` at S.T.P.
In the present case,
` P_1 = 720 mm " "P_2 = 760` mm
`V_1 = 34 cm^3 " " V_2 = ? `
`T_1 = 273 +17 = 290 K" " T_2 = 273 K`
According to the gas equation,
or `(P_1 V_1 )/(T_1 ) = (P_2 V_2)/(T_2)`
` V_2=(P_1 V_1)/(T_1) xx(T_2)/(P_2) =(720 xx 34 xx 273 )/(290 xx 760)`
`= 30.32 cm^3`
Calculation of percentage of N
` :. 22400 cm^3` of `N_2` at S.T. P. weigh = 28 g
` :. ` 30.32 `cm^3` of `N_2` at S.T.P. will weigh
`=(28 )/(22400 ) xx 30.32 = 0.038 g `
`therefore `Percentage of N in the given compound
`=( 0.038)/(0.22 ) xx 100 =17.27 `