Mass of organic compound = 0.3 g
(i) To calculate the volume of nitrogen at N.T.P.
(Given)                                (At N.T.P.)

Applying the general gas equation


(ii) To calculate the percentage of nitrogen:
Now 22400 mL of nitrogen at N.T.P. weighs  = 28 g.

Kjeldahl’s method: It is employed largely in the analysis of food stuff and fertilisers. It is preferred to Duma’s method. It cannot be applied to the compounds which contain nitrogen atom linked either to oxygen or to other nitrogen atoms.

It consists in heating a known weight of a nitrogenous compound with concentrated sulphuric acid and a little K₂SO₄ and anhydrous CuSO₄ (or mercury). Potassium sulphate raises the boiling point of H₂SO₄ and CuSO₄ acts as a catalyst. Nitrogen present in the compound is quantitatively converted into ammonium sulphate. The resultant liquid is heated with concentrated sodium hydroxide. The ammonia gas thus liberated is absorbed in a known volume of an excess of a standard solution of acid. The acid left unused is titrated against a standard solution of alkali. Knowing the volume of acid used to neutralise ammonia, the volume of ammonia evolved and finally the percentage of nitrogen can be calculated.
Reactions involved are:

Calculations:
 Let the mass of organic compound taken = Wg
  Volume of standard  taken
                                       = V mL of  1 M solution
   After the absorption of ammonia.
  Volume of alkali required for the excess acid
                                 

         

     
Also, 1000 mL of 1 M NH₃ solution contains 14g nitrogen

Mass of the organic compound taken  = 0.5 g
Volume of  required for neutralising ammonia = 10 mL
  

Mass of organic compound taken =  0.50 g
Volume of standard H₂SO₄ taken
                       = 50 mL of 0.5 M solution
After the adsorption of ammonia,
     Volume of alkali required for the excess acid 
                                   = 60 mL of 0.5 M solution
         60 mL of 0.5 M NaOH = 30 mL of 0.5 M
                                                  H₂SO₄
ie.. the acid left unused  = 30 mL of 0.5 M
                                                 H₂SO₄