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A Sample of 0.5 gram of an organic compound was treated according to Kjeldahl's method . the ammonia evolved was absorbed in 50 ml of 0.5M H2SO4 the remaining acid after neutrailisation by Ammonia consume 80 ml of 0.5M NaOH .the percentage of Nitrogen in the organic compound is . A) 14% B) 28% C) 42% D) 56%

 
A Sample of 0.5 gram of an organic compound was treated according to Kjeldahl's  method . the ammonia evolved was absorbed in 50 ml of 0.5M H2SO4 the remaining acid after neutrailisation by Ammonia consume 80 ml of 0.5M NaOH .the percentage of Nitrogen in the organic compound is .
A) 14%
B) 28%
C) 42%
D) 56%

Grade:11

3 Answers

Khimraj
3007 Points
5 years ago
Given that, total mass of organic compound = 0.50 g
60 mL of 0.5 M solution of NaOH was required by residual acid for neutralisation.
60 mL of 0.5 M NaOH solution
60/2 mL of 0.5M
H2SO4 = 30 mL of 0.5 M H
∴Acid consumed in absorption of evolved ammonia is (50–30) mL = 20 mL
Again, 20 mL of 0.5 M H2SO4 = 40 mL of 0.5 M NH3
Also, since 1000 mL of 1 M NH3 contains 14 g of nitrogen,
∴ 40 mL of 0.5 M NH3 will contain
14*40*0.5/1000  = 0.28 g of N
Therefore, percentage of nitrogen in 0.50 g of organic compound
(0.28/0.50)*100 = 56%
Hope it clears.......................
Arun
25750 Points
5 years ago

Given that, total mass of organic compound = 0.50 g

80 mL of 0.5 M solution of NaOH was required by residual acid for neutralisation.

80 mL of 0.5 M NaOH solution

 = 80/2 ml of 0.5 M H2SO4

 = 40 mL of 0.5 M H2SO4

∴Acid consumed in absorption of evolved ammonia is (50–40) mL = 10 mL

Again, 10 mL of 0.5 MH2SO4 = 20 mL of 0.5 MNH3

Also, since 1000 mL of 1 MNH3 contains 14 g of nitrogen,

∴ 20 mL of 0.5 M NH3 will contain

14 * 20 *0.5 /1000 = 0.14 g of N

Therefore, percentage of nitrogen in 0.50 g of organic compound

0.14 /0.5= 28 % 

Arun
25750 Points
5 years ago
Dear Khimraj
 
Question is about for 60 ml. Not for 80 ml.
I think you have made a mistake
 
Regards
Arun(askIITians forum expert)

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