In the estimation of nitrogen present in an organic compound by Dumas method 0.35 g yielded 20.7 mL of nitrogen at `15^(@)C` and 760 mm Hg pressure. Calculate the percentage of nitrogen in the compound.

Explanation

Weight of organic compound= 0.35 g
`"Volume of moist nitrogen" (V_(1)) = 20.7 ml = 20.7 xx 10^(-3) L`
`"Temperature"= T_(1) =15^(@)C = 273 + 15^(@)C = 288K`
`"Pressure of moist nitrogen" P_(1)=760 "mmHg"`
`(P_(1)V_(1))/(T_(1))=(P_(0)V_(0))/(T_(0))`
`V_(0)=(P_(1)V_(1))/(T_(1)) xx (T_(0))/(P_(0))=(760 xx 20.7 xx 10^(-3))/(288) xx (273K)/(760)`
`V_(0)=19.62 xx 10^(-3) L`
`"Percentage of nitrogen"=(28)/(22.4) xx (V_(0))/(W) xx 100=(28)/(22.4) xx (19.62 xx 10^(-3))/(0.35) xx 100`
`=(28)/(22.4) xx (19.62)/(0.35) xx 10^(-1)=56.05 xx 10^(-3) xx 100= 7.007%`
`implies` Percentage of nitrogen= 7.007%