0.35 g of an organic susbtance was Kjeldahilsed and the ammonia obtained was passed into 100ml of M/`10H_(2)SO_(4)` The excess acid required 154 ml of `M//10 NaOH` for neurtralisation, calculate the % of nitrogen in the compound.

Explanation

Vol. of M/ 10 `H_2 SO_4` taken `= 100` mL Let us first calculate the volume of excess M/10 `H_2 SO_4` which was neutralised by `154` mL of M/ 10 `Na OH`
154 mL of `( M)/(10) Na OH -= (154)/( 2)` mL of `(M)/( 10) H_2 SO_4`
`therefore` Volume of `(M)/( 10) H_2 SO_4` left unused `= 77` mL
Volume of `(M )/( 10) H_2 SO_4` used for neutralisation of `NH_3`
`= 100-77 = 23` mL
Now,
23 mL of `(M )/( 10) H_2 SO_4 = 2 xx 23` mL of `( M)/( 10) NH_3`
`= 46` mL of ` (M )/( 10) NH_3`
Now 1000 mL of M `NH_3` contain nitrogen `= 14` g
Now 1000 mL of 1 M `NH_3` contain nitrogen `=14` g
`46` mL of `(M )/ ( 10) NH_3` contain nitrogen
`= (14)/( 1000 ) xx ( 46 xx 1)/( 10) `
`therefore` Percentage of nitrogen `= (14 xx 46 xx 100)/( 1000 xx 10 x 0.35)`
`= 18.4 %`