In Dumas method 0.5 g of an organic compound containing nitrogen gave 112 ml of nitrogen at S.T.P. The percentage of nitrogen in the given compound is

Question

In Dumas method 0.5 g of an organic compound containing nitrogen gave 112 ml of nitrogen at S.T.P. The percentage of nitrogen in the given compound is (A) 28 (B) 38 (C) 18 (D) 48

Tardigrade Admin · Accepted Answer

% N=(28/22400)× (V o l u m e o f N2 a t S T P (. m l .)/W e i g h t o f o r g a n i c c o m p o n d)× 100 % N=(28/22400)× (112/0.5)ml× 100=28 %

Q. In Dumas method 0.5 g of an organic compound containing nitrogen gave 112 ml of nitrogen at S.T.P. The percentage of nitrogen in the given compound is

28

38

18

48

$% N = \frac{28}{22400} \times \frac{V o l u m e o f N _{2} a t STP ( m l )}{W e i g h t o f or g ani c co m p o n d} \times 100$
% $N = \frac{28}{22400} \times \frac{112}{0.5} m l \times 100 = 28$ %

Q. In Dumas method 0.5 g of an organic compound containing nitrogen gave 112 ml of nitrogen at S.T.P. The percentage of nitrogen in the given compound is

28

38

18

48

%N=2240028​×WeightoforganiccompondVolumeofN2​atSTP(ml)​×100 % N=2240028​×0.5112​ml×100=28 %

$% N = \frac{28}{22400} \times \frac{V o l u m e o f N _{2} a t STP ( m l )}{W e i g h t o f or g ani c co m p o n d} \times 100$
% $N = \frac{28}{22400} \times \frac{112}{0.5} m l \times 100 = 28$ %