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

`V_(1) = 40 mL, V_(2)` at NTP = ?
`P_(1) = 725 - 25 = 700 mm, P_(2) = 760 mm`
`T_(1) = 300 K, T_(2) = 273 K`
`:. V_(2) = (P_(1)V_(1)T_(2))/(P_(2)T_(1)) = (700 xx 40 xx 273)/(760 xx 300) = 33.52 mL`
`%N = (28)/(22400) xx (33.52)/(0.25) xx 100 = 16.76`