In Dumas' method of estimation of nitrogen `0.35 g` of an organic compound gave `55mL` of nitrogen collected at `300K` temperature and `715 mm` pressure. The percentage composition of nitrogen in the compound would be `: (` Aqueous tension at `300K =15 mm)`

Options

A. `16.45`
B. `17.45`
C. `14.45`
D. `15.45`

Answer

A

Explanation

According to combined gas equation ,
`(p_(1)V_(1))/(T_(1)) = (p_(2)V_(2))/(T_(2))`
Where , `p_(2)` = pressure of `N_(2)` at STP = 760 mm
`T_(2)` = Temperature of `N_(2)` at STP = 273 K
`V_(2)= ?`
Volume of `N_(2)` at STP (By gas equation)
`((rho -rho_(1))/(t + 273)) V_(1) xx (273)/(760) = V_(2)`
Where , `p_(1) = rho - rho_(1)`
`rho` = 715 mm (pressure at which `N_(2)` collected )
`rho_(1)` = aqueous tension of water = 15 mm
`T_(1) = t + 273 = 300 K`
`V_(1) = 55` mL = volume of moist nitrogen in nitrometer
`therefore " " V_(2) = ((715-15) xx 55)/(300) xx (273)/(760)`
= `46.098` mL
% of nitrogen in given compound
= `(28)/(22400) xx (V_(2))/(W) xx 100`
`= (28)/(22400) xx (46.098)/(0.35) xx 100`
= `16.45` %