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Question

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 B 16.45
In Duma's method of estimation of nitrogen:-
Calculation:- Volume of N2 at NTP (By gas equation)
(ρρ1t+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)=(71515)×55300×273760=46.098 ml.
% of nitrogen =2822400×46.0980.35×100=16.45%.

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