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. 17.36
B. 18.2
C. 16.76
D. 15.76

Answer

C

Explanation

Mass of the substance taken = `0.25` g
Volume of nitrogen collected = 40 ml
Atmospheric pressure = 725 mm
Room temperature = 300 k
Aqueous tension at 300 k = 25 mm
Actual pressure of the gas = (725-25) mm Hg
= 700 mm
To convert the volume at experimental conditions to volume at STP.
Experimental value `" " ` At STP
`P_(1) = 700 mm " " P_(2) = 760 mm`
`V_(1) = 40 ml " " V_(2) = ? `
`T_(1) = 300 k " " T_(2) = 273 k`
Substituting these values in the gase-eq.
`(P_(2)V_(2))/(T_(2)) = (P_(1)V_(1))/(T_(1))`
we get , `(760 xx V_(2))/(273) = (700 xx 40)/(300)`
`V_(2) = (700 xx 40)/(300) xx (273)/(760) = 3353` ml
To convert volume at STP into mass
22400 ml of nitrogen at STP weigh = 28 g
`therefore 33.53` ml of nitrogen at STP will
weigh = `(28xx 3353)/(22400 xx 0.25) `
To calculate percentage of nitrogen
= `(28 xx 3353)/(22400 xx 0.25) xx 100`
= `16.76`%