0.35 g of an organic substance was Kjeldahlised and the ammonia obtained was passed into 100 ml of N/5 `H_(2)SO_(4)`. The excess acid required 154 ml of N/10 `NaOH` for neutralisation. Calculate the percentage of nitrogen in the compound.

Explanation

Volume of `N//5 H_(2)SO_(4)` solution taken = 100 mL
The volume of `N//5 H_(2)SO_(4)` neutralised by N/10
NaOH can be obtained as follows :
`N_("acid")xxV_("acid")=N_("alkali")xxV_("alkali")`
`(N)/(5)xx V_("acid")=(N)/(10)xx154mL`
`V_("acid")=(154)/(10)xxmL`
= 77 nL.
Therefore,
Volume of `N//% H_(2)SO_(4)` used for
neutralisting ammonia `=(100-77)mL`
`=23mL`
Then,
`"Percentage of nitrogen in the sample "="1.4"xx"Normality volume of acid"`
`("used for neutralising "NH_(3))/("Mass of the compound taken")`
Percentage of nitrogen in the sample
`=(1.4xx1//5xx23)/(0.35)=18.4`