In Duma's method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25 mm, the percentage of nitrogen in the compound is

Options

A. `16.76`
B. `15.76`
C. `17.36`
D. `18.20`

Answer

A

Explanation

Let us calculate volume of `N_(2)` evolved at N.T.P. Experimental conditions N.T.P. conditions
`{:(V_(1)=40 mL,V_(2)=?),(P_(1)=(725-25)=700mm,P_(2)=760 mm),(T_(1)=300 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))/(P_(2)T_(1))`
`V_(2)=((700mm)xx(40 mL)xx(273 K))/((760mm)xx(300 K))`
`=33.52 mL`
`%` of `N=28/22400xx("Volume of "N_(2)" at N.T.P.")/("Mass of compound")xx100`
`=28/22400xx((33.52 mL))/((0.25 g))xx100`
`=16.76 %`