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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» <html><body><p>`11.5`<br/>23<br/>`12.5`<br/>`14.5`</p>Solution :No. of milli eq of `H_(2)SO_(4)` <a href="https://interviewquestions.tuteehub.com/tag/left-1070879" style="font-weight:bold;" target="_blank" title="Click to know more about LEFT">LEFT</a> after <a href="https://interviewquestions.tuteehub.com/tag/neutralisation-577399" style="font-weight:bold;" target="_blank" title="Click to know more about NEUTRALISATION">NEUTRALISATION</a>.<br/>= No. of m. eq of NaOH<br/>`= 25xx0.1xx1=2.5` m eq.<br/>No. of m eq of `H_(2)SO_(4)` taken<br/>`= 50xx0.05xx2=5m. eq.`<br/>No. of eq of `H_(2)SO_(4)` ultilised for neutralisations<br/>`= 5-2.5=2.5` m.eq.<br/>`<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>` No. of m. eq. of `NH_(3)` <a href="https://interviewquestions.tuteehub.com/tag/liberated-2789268" style="font-weight:bold;" target="_blank" title="Click to know more about LIBERATED">LIBERATED</a> = 2.5 m.q.<br/>% of `N=("No. of meq. of "NH_(3)xx1.4)/("wt.of org. <a href="https://interviewquestions.tuteehub.com/tag/compound-926863" style="font-weight:bold;" target="_blank" title="Click to know more about COMPOUND">COMPOUND</a>")`<br/>`= (2.5xx1.4)/(0.303)=11.55%`.</body></html>


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