In Duman's method of estimation of nitrogen, 0.35 g of an organic compound gave 55 mL of nitrogen collected at 300 K temperature and 715 mm pressure. The percentage composition of nitrogen in the compound would be:
(Aqueous tension at 300K=15 mm)
A
15.45
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B
16.45
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C
17.45
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D
14.45
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Solution
The correct option is B16.45 In Duma's method of estimation of nitrogen:- Calculation:- Volume of N2 at NTP (By gas equation) (ρ−ρ1t+273)v×273760=V ml. % of nitrogen in the given compound: 2822400×VW×100 Here, W=0.35 gm. ρ=715 mm (Pressure at which N2 collected) ρ1= aqueous tension of water =15 mm. (t+273)K=300K v ml = volume of moist nitrogen in nitrometer =55 ml. So, volume of N2 at NTP=(V)=(715−15)×55300×273760=46.098 ml. % of nitrogen =2822400×46.0980.35×100=16.45%.