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0.303 g fo sample was analysed for nitro...

0.303 g fo sample was analysed for nitrogen by Kjeldahl's method. The ammonia gas evolved was absorbed in 50 ml of 0.05 M `H_(2)SO_(4)`. The excess acid required 25 ml of 0.1 M NaOH for neutralisation.

A

`11.5`

B

23

C

`12.5`

D

`14.5`

Text Solution

Verified by Experts

The correct Answer is:
A
No. of milli eq of `H_(2)SO_(4)` left after neutralisation.
= No. of m. eq of NaOH
`= 25xx0.1xx1=2.5` m eq.
No. of m eq of `H_(2)SO_(4)` taken
`= 50xx0.05xx2=5m. eq.`
No. of eq of `H_(2)SO_(4)` ultilised for neutralisations
`= 5-2.5=2.5` m.eq.
`therefore` No. of m. eq. of `NH_(3)` liberated = 2.5 m.q.
% of `N=("No. of meq. of "NH_(3)xx1.4)/("wt.of org. compound")`
`= (2.5xx1.4)/(0.303)=11.55%`.
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