0.2313 g of an organic substance gave 40 ml of moist nitrogen measured at `15^(@)C` and 745 mm pressure. Calcualte the % of nitrogen (Aq tension at `15^(@)C` is 12.7mm)

Explanation

Mass of the substance = 0.25 g Volume of moist dinitrogen = `30cm^(3)`
Temperature = 288K Pressure = 745 - 12.7 = 732.3 mm
Volume of dinitrogen at STP: `V_(2) = (P_(1)V_(1))/(T_(1)) xx (T_(2))/(P_(2)) = (732.3 xx 30 xx 273)/(288 xx 760) = 27.4cm^(3)`
Percentage of nitrogen in organic compound = `(28)/(22400) xx ("vol of" N_(2) "at STP")/("wt of organic compound") xx 100`
`(28)/(22400) xx (27.4)/(0.25) xx 100 = 13.6`