0.2g of an organic compound on Kjeldahls analysis gives enough ammonia to neutralize 20 $c m^{3}$ of 0.1 N $H_{2} S O_{4}$ . The percentage of nitrogen in the compound is:

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Solution

The correct option is A 14
% $N_{2} = \frac{1.4 \times m e q . o f H_{2} s o_{4} u s e d t o n e u t r a l i z e N H_{3}}{m a s s o f c o m p o u n d}$
$V (H_{2} s o_{4}) = 20 m l .$
$N = 0.1 N$
% $N_{2} = \frac{1.4 \times 0.1 \times 20}{0.2}$
$= \frac{2.8}{0.2}$
$N_{2} = 14$ %

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0.5g of an organic compound on Kjeldahl’s analysis gave enough ammonia to just neutralise 10 $c m^{3}$ of 1M $H_{2} S O_{4}$ . The $%$ of N in the compound is:

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0.2g of an organic compound on Kjeldahls analysis gives enough ammonia to neutralize 20 cm3 of 0.1 N H2SO4. The percentage of nitrogen in the compound is:

The correct option is A 14%N2=1.4×meq.ofH2so4usedtoneutralizeNH3massofcompoundV(H2so4)=20ml.N=0.1N%N2=1.4×0.1×200.2=2.80.2N2=14 %

0.2g of an organic compound on Kjeldahls analysis gives enough ammonia to neutralize 20 $c m^{3}$ of 0.1 N $H_{2} S O_{4}$ . The percentage of nitrogen in the compound is:

The correct option is A 14
% $N_{2} = \frac{1.4 \times m e q . o f H_{2} s o_{4} u s e d t o n e u t r a l i z e N H_{3}}{m a s s o f c o m p o u n d}$
$V (H_{2} s o_{4}) = 20 m l .$
$N = 0.1 N$
% $N_{2} = \frac{1.4 \times 0.1 \times 20}{0.2}$
$= \frac{2.8}{0.2}$
$N_{2} = 14$ %