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0.1688 g when analysed by the Dumas meth...

0.1688 g when analysed by the Dumas method yield 31.7 mL of moist nitrogen measured at `14^(@)C` and 758 mm mercury pressure. Determine the % of N in the substance (Aqueous tension at `14^(@)C =12 mm` of Hg).

Text Solution

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Weight of Organic compound =0.168 g
`"Volume of moist nitrogen" (V_(1))=31.7 ml=31.7 xx 10^(-3) L`
`"Temperature" (T_(1)) = 14^(@)C = 14 + 273 = 287 K`
Pressure of Moist nitrogen (P) =758 mm Hg
Aqueous tension at `14^(@)C` = 12 mm of Hg
`:.` Pressure of dry nitrogen=(`P-P^(1)`)=758-12=746 mm of Hg
`(P_(1)V_(1))/(T_(1))=(P_(0)V_(0))/(T_(0))`
`V_(0)=(746 xx 31.7 xx 10^(-3))/(287) xx (273)/(760)`
`V_(0)=29.8 xx 10^(-3) L`
`"Percentage of nitrogen"=((28)/(22.4) xx (V_(0))/(W)) xx 100`
`=(28)/(22.4) xx (29.58 xx 10^(-3))/(0.1688) xx 100=21.90%`
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