0.4 g of an organic compound containing N was Kjedahlised and `NH_(3)` obtained was passed into 50 mL (N/2) `H_(2)SO_(4)` solution. Volumn of the acid solution was increased to 150 mL by adding distilled water. 20 mL of this acid solution required 31 mL (N/20) NaOH for complete neutralisation. Calculate the percentage of N.

Explanation

20 mL of (partially neutralised) diluted acid solution `-=31mL(1)/(20)(N)NaOH` solution.
`therefore` Strenght of (partially neutralised) diluted acid solution
`=31xx(1)/(20)xx(1)/(20)(N)=(31)/(400)(N)`
`therefore` Amount of `H_(2)SO_(4)` present in 150mL (partially neutralised) diluted acid solution `=(31xx150)/(400xx1000)` g - equiv.
Now, 50 mL of `(1)/(2)(N)H_(2)SO_(4)` solution contains `=(1xx50)/(2xx1000)` g - equiv. `H_(2)SO_(4)`.
`therefore NH_(3)` produced by decomposition 0.4 g of the organic compound `=((50)/(2000)-(31xx150)/(400000))` g - equivalent
=0.013375 g - equivalent
`= 0.013375 xx 17` g
Now, `0.013375xx17gNH_(3)=(14)/(17)xx0.013375xx17g N`.
`therefore` % of nitrogen in the organic compound
`=(14xx0.013375)/(0.4)xx100=46.81`