The ammonia evolved from 2g of a compound in Kjeldahl's estimation of nitrogen neutralizes 10 mL of 2 M $H_{2} S O_{4}$ solution. The weight percentage of nitrogen in the compound is

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Solution

The correct option is B
28

$2 N H_{2} + H_{2} S O_{4} - \to (N H_{4})_{2} S O_{4}$
10ml
2ml
millimole of $H_{2} S O_{4} = \frac{mmol of N H_{3}}{2} = 20$
Mmol $N H_{3}$ =mmol of N=40
$= \frac{40 \times 14}{1000} = \frac{560}{1000} = 0.56 g$
$W_{n}$
% of N $= \frac{0.56}{2} \times 100 = 28$

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The ammonia evolved from 2g of a compound in Kjeldahl's estimation of nitrogen neutralizes 10 mL of 2 M $H_{2} S O_{4}$ solution. The weight percentage of nitrogen in the compound is

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The ammonia evolved from 2g of a compound in Kjeldahl's estimation of nitrogen neutralizes 10 mL of 2 M H2SO4 solution. The weight percentage of nitrogen in the compound is

The correct option is B 28 2NH2+H2SO4 −→(NH4)2SO4 10ml 2ml millimole of H2SO4=mmol of NH32=20 Mmol NH3=mmol of N=40 =40×141000=5601000=0.56g Wn % of N=0.562×100=28

The ammonia evolved from 2g of a compound in Kjeldahl's estimation of nitrogen neutralizes 10 mL of 2 M $H_{2} S O_{4}$ solution. The weight percentage of nitrogen in the compound is

The correct option is B
28

$2 N H_{2} + H_{2} S O_{4} - \to (N H_{4})_{2} S O_{4}$
10ml
2ml
millimole of $H_{2} S O_{4} = \frac{mmol of N H_{3}}{2} = 20$
Mmol $N H_{3}$ =mmol of N=40
$= \frac{40 \times 14}{1000} = \frac{560}{1000} = 0.56 g$
$W_{n}$
% of N $= \frac{0.56}{2} \times 100 = 28$