0.1688 g when analysed by the Dumas method yield 31.7 mL of moist nitrogen measured at 14^(@)C and 758 mm mercury pressure. Determine the % of N in the substance (Aqueous tension at 14^(@)C =12 mm of Hg).

0.1688 g when analysed by the Dumas method yield 31.7 mL of moist nitr

1.	0.1688 g when analysed by the Dumas method yield 31.7 mL of moist nitrogen measured at 14^(@)C and 758 mm mercury pressure. Determine the % of N in the substance (Aqueous tension at 14^(@)C =12 mm of Hg).
Answer» SOLUTION :Weight of Organic compound =0.168 g `"Volume of moist nitrogen" (V_(1))=31.7 ml=31.7 xx 10^(-3) L` `"Temperature" (T_(1)) = 14^(@)C = 14 + 273 = 287 K` Pressure of Moist nitrogen (P) =758 mm Hg Aqueous tension at `14^(@)C` = 12 mm of Hg `:.` Pressure of DRY nitrogen=(`P-P^(1)`)=758-12=746 mm of Hg `(P_(1)V_(1))/(T_(1))=(P_(0)V_(0))/(T_(0))` `V_(0)=(746 xx 31.7 xx 10^(-3))/(287) xx (273)/(760)` `V_(0)=29.8 xx 10^(-3) L` `"Percentage of nitrogen"=((28)/(22.4) xx (V_(0))/(W)) xx 100` `=(28)/(22.4) xx (29.58 xx 10^(-3))/(0.1688) xx 100=21.90%`