Question

A sample of 0.25 g of an organic compound was treated
according to Kjeldahl's method. The ammonia evolved
was absorbed in 100 mL of 0.5 N H2SO4. The remaining
acid after neutralisation by ammonia consumed 80 mL
of 0.5 N NaOH. The percentage of nitrogen in the organic
compound
is​

Answer 1

Answer:

A sample of 0.25 g of an organic compound was treated

according to Kjeldahl's method. The ammonia evolved

was absorbed in 100 mL of 0.5 N H2SO4. The remaining

acid after neutralisation by ammonia consumed 80 mL

of 0.5 N NaOH. The percentage of nitrogen in the organic

compound

is

devansh

Explanation:

devansh

Answer 2

Answer:

A sample of 0.50 g of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in 50 mL of 0.5 M H

2

SO

4

.The residue acid required 60 mL of 0.5 M solution of NaOH for neutralisation. What would be the percentage composition of nitrogen in the compound?

Explanation:

Mass of organic compound = 0.50 g

60 mL of 0.5 M solution of NaOH is required by residual acid for neutralisation

⇒ 60 mL of 0.5 M solution of NaOH =

2

60

mL of 0.5M H

2

SO

4

= 30mL of 0.5M H

2

SO

4

∴ Acid consumed in absorption of evolved ammonia is (50-30)mL = 20mL

Again, 20mL of 0.5M H

2

SO

4

= 40mL of 0.5M NH

3

Since 1000mL of 1M NH

3

contains 14g of nitrogen,

∴ 40mL of 0.5M NH

3

will contain =

1000

14×40

×0.5 = 0.28g of N

Therefore, percentage of N in 0.50g of organic compound =

0.50

0.28

×100=56%

Hence, the correct answer is 56%.