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1.216 gm of an organic compound was reac...

1.216 gm of an organic compound was reacted under Kjeldahl's method and the ammonia evolved was absorbed in 100 ml `NH_2SO_4`. The remaining acid solution was made up to 500 ml by the addition of water. Twenty millilitres of the dilute solution required 32 ml `(N)/(10)` caustic soda solution for complete neutralisation. Calculate the percentage of nitrogen in the compound.

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`20mL` dilute unreacted acid solm. Required
`32 mL (N)/(10) NaOH` soln.
500mL dilute unreacted acid soln required
`= (32)/(20) xx 500 mL (N)/(10) NaOh` soln.
`= (32)/(20) xx (500)/(10)mL N NaOH`
`= 80 mL N NaOH`
`80 mL NaOH -= 80 mL N H_(2)SO_(4)`
Acid used for the neutralisation of `NH_(3)`
`= (100 - 80) mL N H_(2)SO_(4)`
`20 mL N H_(2)SO_(4)`
Percentage of nitrogen `= 1.4 xx N_(1) xx (V)/(W)`
`N_(1) = 1,V = 20mL` and `W = 1.216g`
So, percentage of nitrogen `=(1.4 xx 1 xx 20)/(1.216) = 23.03%`
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