0.05g of an organic compound was treated according to Kjedahl's method. Ammonia evoled was absorbed in 50ml of 0.05 (M) `H_(2)S)_(4)`. The residual acid required 60 ml 0.5 (M) solution of NaOH for neutralisation. Find the percentage composition of N in the compound.

Explanation

Amount of `H_(2)SO_(4)` in 50 mL of 0.5 (M)`H_(2)SO_(4)`
`=(0.5xx50)/(1000)mol=(0.5xx50xx2)/(1000)g-"equiv" = 0.05g-"equiv"`
Amount of NaOH in 60 mL of 0.5 (M) NaOH
`=(0.5xx60)/(1000)mol=(0.5xx60)/(1000)g-"equiv" = 0.03g-"equiv"`
So amount of `NH_(3)` liberated by the decompositon of the compound = 0.05 - 0.03 = 0.02g - equiv = 0.02 xx 17g = 0.34g
`therefore " " "Amount of N in the compound"=0.34xx(14)/(17)=0.28g`
`therefore " " % "N in the given organic compound" = (0.28xx100)/(0.50)=56.0`