In Duma's method 0.52g of an organic compound on combustion gave 68.6 mL N_(2) at 27^(@)C and 756mm pressure. What is the percentage of nitrogen in the compound?

In Duma's method 0.52g of an organic compound on combustion gave 68.6

1.	In Duma's method 0.52g of an organic compound on combustion gave 68.6 mL N_(2) at 27^(@)C and 756mm pressure. What is the percentage of nitrogen in the compound?
Answer» 0.1222 0.1493 0.1584 0.1623 Solution :`V_(1)=68.6mL,P_(1)=756mm,T_(1)=300K` `V_(2)=?,P_(2)=760MM,T_(2)=273K` `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))` At NTP, vol. of `N_(2),V_(2)=(P_(1)V_(1))/(T_(1))*(T_(2))/(P_(2))=(756xx68.6)/(300)xx(273)/(760)` =62.09 mL Percentage of nitrogen in organic compound `=(28)/(22400)xx(V_(2))/(w)xx100=(28)/(22400)xx(62.09)/(0.52)xx100=14.93%`