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.

In Duma's method 0.206 g of an organic compound gave 18.8 cm^(3) of mo

1.	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.
Answer» Solution :Mass of ORGANIC compound =0.206 G Calculation of the volume of `N_(2)` at N.T.P. `{:("Experimental Conditions","N.T.P. conditions"),(V_(1)=18.8 cm^(3),V_(2)=?),(P_(1)=(760-14.5)mm Hg,P_(2)=760 mm Hg),(=745.5 mm Hg,),(T_(1)=(273+17)K=290 K,T_(2)=273 K):}` By applying gas EQUATION, `(P_(1)V_(1))/T_(1)=(P_(2)V_(2))/T_(2)` `V_(2)=(P_(1)V_(1)T_(2))/(T_(1)P_(2))=((745.5mmHg)xx(18.8 cm^(3))xx(273 K))/((290 K)xx(760 mm Hg))=17.37 cm^(3)` Step II. Calculation of percentage of nitrogen `22400 cm^(3)` of `N_(2)` at N.T.P. has a mass =28 g `17.37 cm^(3)` of `N_(2)` at N.T.P. has a mass `=((28g)xx(17.37 cm^(3)))/((22400 cm^(3)))=0.0217 g` Percentage of nitrogen `=("Mass of "N_(2)" produced "xx100)/("Mass of organic compound")=((0.0217g)xx100)/((0.206 g))=10.54`