0.365 g of an organic compound containing nitrogen gave 56 ml of nitrogen at S.T.P. The percentage nitrogen in the given compound is

Question

0.365 g of an organic compound containing nitrogen gave 56 ml of nitrogen at S.T.P. The percentage nitrogen in the given compound is (A) 19.18 (B) 38.36 (C) 9.18 (D) 29.18

Tardigrade Admin · Accepted Answer

It is Duma's method used to find % of nitrogen %N=(28 × V a t S T P × 100/22400 × W)=(28 × 56 × 100/22400 × 0.236)=19.18 %

Q. 0.365 g of an organic compound containing nitrogen gave 56 ml of nitrogen at S.T.P. The percentage nitrogen in the given compound is

19.18

38.36

9.18

29.18

It is Duma's method used to find % of nitrogen
$% N = \frac{28 \times V a t STP \times 100}{22400 \times W} = \frac{28 \times 56 \times 100}{22400 \times 0.236} = 19.18%$

Q. 0.365 g of an organic compound containing nitrogen gave 56 ml of nitrogen at S.T.P. The percentage nitrogen in the given compound is

19.18

38.36

9.18

29.18

It is Duma's method used to find % of nitrogen %N=22400×W28×VatSTP×100​=22400×0.23628×56×100​=19.18%

It is Duma's method used to find % of nitrogen
$% N = \frac{28 \times V a t STP \times 100}{22400 \times W} = \frac{28 \times 56 \times 100}{22400 \times 0.236} = 19.18%$