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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%
Given that, total mass of organic compound = 0.50 g60 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 0.5MH2SO4 = 30 mL of 0.5 M H∴Acid consumed in absorption of evolved ammonia is (50–30) mL = 20 mLAgain, 20 mL of 0.5 M H2SO4 = 40 mL of 0.5 M NH3Also, 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 NTherefore, percentage of nitrogen in 0.50 g of organic compound (0.28/0.50)*100 = 56%Hope it clears.......................
Given that, total mass of organic compound = 0.50 g80 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 mLAgain, 10 mL of 0.5 MH2SO4 = 20 mL of 0.5 MNH3Also, since 1000 mL of 1 MNH3 contains 14 g of nitrogen,∴ 20 mL of 0.5 M NH3 will contain14 * 20 *0.5 /1000 = 0.14 g of NTherefore, percentage of nitrogen in 0.50 g of organic compound0.14 /0.5= 28 %
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 %
Dear Khimraj Question is about for 60 ml. Not for 80 ml.I think you have made a mistake RegardsArun(askIITians forum expert)
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