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.

0.303 g fo sample was analysed for nitrogen by Kjeldahl's method. The

1.	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.
Answer» `11.5` 23 `12.5` `14.5` Solution :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%`.