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Grade: 11
        
 
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%
3 months ago

Answers : (3)

Khimraj
2612 Points
							
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.......................
3 months ago
Arun
18646 Points
							

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 % 

3 months ago
Arun
18646 Points
							
Dear Khimraj
 
Question is about for 60 ml. Not for 80 ml.
I think you have made a mistake
 
Regards
Arun(askIITians forum expert)
3 months ago
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