Question

Question 12.33 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 H2SO4. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound.

Class XI Organic Chemistry : Some Basic Principles and Techniques Page 364

Answer 1

Hi

Here is your answer,

Volume of the acid taken = 50 mL of 0.5 MH₂SO₄

           = 25mL of 1.0 MH₂SO₄

Volume of alkali used for  neutralization of excess acid 

                     = 60 mL of 0.5 M NaOH

                      = 30 mL of 1.0 M NaOH

H₂SO₄ + 2NaOH --------------> Na₂SO₄ + 2H₂O

1 mole of H₂SO₄ = 2 moles of NaOH = 15 mL of 1.0 M H₂SO₄

∴ Volume of acid used by ammonia = 25 - 15 = 10 mL

                                  % of nitrogen = 1.4 × N₁ × vol. of acid used/ w

(where n = normality of acid and w = mass of organic compound taken)

               % of nitrogen = 1.4  × 2 × 10/0.5 = 56.0

Answer 2

volume of acid taken = 50ml of 0.5M H2SO4 = 25ml of 1M H2SO4
{we know, M1V1 = M2V2 , if M1=1M,M2=0.5M,V2=50ml then,V1=25ml }

volume of alkali used for the neutralisation of excess acid = 60ml of 0.5M NaOH or, 30ml of 1M NaOH {M1V1=M2V2 use and then you get volume = 30ml for 1M NaOH}

H2SO4 + 2NaOH --->Na2SO4 + 2H2O
1mole of H2SO4 = 2mole of NaOH
Hence, 30ml of 1M NaOH = 15ml of 1M H2SO4
so, volume of acid used by ammonia = 25ml - 15ml = 10ml

Now, % of nitrogen = 1.4 × Normality of acid × volume of acid/mass of organic compound
= 1.4 × 2 × 10/0.5 {normality = n-factor × molarity , normality of H2SO4 = 2 × 1 = 2 N}
= 56%

Hence, percentage of Nitrogen = 56%