0.6 g of an organic compound was Kjeldhalised and NH_(3) evolved was absorbed into 50 ml of semi-normal solution of H_(2)SO_(4) . The residual acid solution was diluted with distilled water and the volume made up to 150 ml. 20 mL of this diluted solution required 35 mL of (N)/(20) NaOH solution for complete neutralisation. Calculate the % of N in the compound.

0.6 g of an organic compound was Kjeldhalised and NH_(3) evolved was a

1.	0.6 g of an organic compound was Kjeldhalised and NH_(3) evolved was absorbed into 50 ml of semi-normal solution of H_(2)SO_(4) . The residual acid solution was diluted with distilled water and the volume made up to 150 ml. 20 mL of this diluted solution required 35 mL of (N)/(20) NaOH solution for complete neutralisation. Calculate the % of N in the compound.
Answer» Solution :WEIGHT of Organic compound = 0.6 g Volume of SULPHURIC acid taken = 50 mL Strength of sulphuric acid taken=0.5 N 20 ml of diluted solution of unreacted sulphuric acid was neutralised by 35 ml of 0.05 N Sodium laydroxide `"Strength of the diluted sulphuric acid"=(35 xx 0.05)/(20) = 0.0875 N` Volume of the sulphuric acid remaining after reaction with ammonia= `V_(1)` mL Strength of `H_(2)SO_(4)` = 0.5 N Volume of the diluted `H_(2)SO_(4)` = 150 mL Strength of the diluted sulphuric acid = 0.0875 N `V_(1)= (150 xx 0.087)/(0.5) = 26.25 ml` Volume of `H_(2)SO_(4)` consumed by ammonia = 50 - 26.25 = 23.75 mL 23.75 mL of 0.5 N `H_(2)SO_(4)` - 23.75 mL of 0.5 N `NH_(3)` The amount of NITROGEN present in the 0.6 g of organic compound `=(14 g)/(1000 mL xx 1 N) xx 23.75 xx 0.5 N=0.166 g` `"Percentage of nitrogen"=(0.166)/(0.6) xx 100=27.66%`