0.25 g of an organic compound on Kjeldahl's analysis gave enough ammonia to just neutralize 10 text cm3 of 0.5 text M text H2SO4 . The percentage of nitrogen in the compound is

Question

0.25 g of an organic compound on Kjeldahl's analysis gave enough ammonia to just neutralize 10 text cm3 of 0.5 text M text H2SO4 . The percentage of nitrogen in the compound is (A) 28 (B) 56 (C) 14 (D) 112

Tardigrade Admin · Accepted Answer

From Kjeldahl's method, Per cent of nitrogen =(1.4× N× V/W) =(1.4× 0.5× 2× 10/0.25) =56%

Q. $0.25 g$ of an organic compound on Kjeldahl's analysis gave enough ammonia to just neutralize $10 c m^{3}$ of $0.5 M H_{2} S O_{4}$ . The percentage of nitrogen in the compound is

28

56

14

112

42

From Kjeldahl's method,
Per cent of nitrogen
$= \frac{1.4 \times N \times V}{W}$ $= \frac{1.4 \times 0.5 \times 2 \times 10}{0.25}$ $= 56$

Q. 0.25g of an organic compound on Kjeldahl's analysis gave enough ammonia to just neutralize 10cm3 of 0.5MH2​SO4​ . The percentage of nitrogen in the compound is

28

56

14

112

42

From Kjeldahl's method, Per cent of nitrogen =W1.4×N×V​ =0.251.4×0.5×2×10​ =56

Q. $0.25 g$ of an organic compound on Kjeldahl's analysis gave enough ammonia to just neutralize $10 c m^{3}$ of $0.5 M H_{2} S O_{4}$ . The percentage of nitrogen in the compound is

From Kjeldahl's method,
Per cent of nitrogen
$= \frac{1.4 \times N \times V}{W}$ $= \frac{1.4 \times 0.5 \times 2 \times 10}{0.25}$ $= 56$