0.5 g of an organic compound on Kjeldahl’s analysis gave enough ammonia to just neutralize 10 cm3 of 1 M H2SO4. The percentage of nitrogen in the compound is

Question

0.5 g of an organic compound on Kjeldahl’s analysis gave enough ammonia to just neutralize 10 cm3 of 1 M H2SO4. The percentage of nitrogen in the compound is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

1 M of 10 ml H2SO4 = 1 M of 20 ml of NH3 1000 ml of 1 M ammonia contain = 14 g of nitrogen 20 ml of 1 M ammonia contain =(14×20/1000)g of nitrogen =(14×20/1000) × (100/0.5) =56.0 %

Q. $0.5$ g of an organic compound on Kjeldahl’s analysis gave enough ammonia to just neutralize $10$ cm $^{3}$ of 1 M $H_{2} S O_{4}$ . The percentage of nitrogen in the compound is

$1$ M of $10$ ml $H_{2} S O_{4} = 1$ M of $20$ ml of $N H_{3}$
$1000$ ml of $1$ M ammonia contain $= 14$ g of nitrogen
$20$ ml of $1$ M ammonia contain $= \frac{14 \times 20}{1000}$ g of nitrogen
$= \frac{14 \times 20}{1000} \times \frac{100}{0.5} = 56.0%$

Q. 0.5 g of an organic compound on Kjeldahl’s analysis gave enough ammonia to just neutralize 10 cm3 of 1 M H2​SO4​. The percentage of nitrogen in the compound is

1 M of 10 ml H2​SO4​=1 M of 20 ml of NH3​ 1000 ml of 1 M ammonia contain =14 g of nitrogen 20 ml of 1 M ammonia contain =100014×20​g of nitrogen =100014×20​×0.5100​=56.0%

Q. $0.5$ g of an organic compound on Kjeldahl’s analysis gave enough ammonia to just neutralize $10$ cm $^{3}$ of 1 M $H_{2} S O_{4}$ . The percentage of nitrogen in the compound is

$1$ M of $10$ ml $H_{2} S O_{4} = 1$ M of $20$ ml of $N H_{3}$
$1000$ ml of $1$ M ammonia contain $= 14$ g of nitrogen
$20$ ml of $1$ M ammonia contain $= \frac{14 \times 20}{1000}$ g of nitrogen
$= \frac{14 \times 20}{1000} \times \frac{100}{0.5} = 56.0%$