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.

Explanation

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%`