1.

0.6 g of an organic compound was Kjeldhalised and NH_(3) evolved was absorbed into 50 ml of semi-normal solution of H_(2)SO_(4) . The residual acid solution was diluted with distilled water and the volume made up to 150 ml. 20 mL of this diluted solution required 35 mL of (N)/(20) NaOH solution for complete neutralisation. Calculate the % of N in the compound.

Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/weight-1451304" style="font-weight:bold;" target="_blank" title="Click to know more about WEIGHT">WEIGHT</a> of Organic compound = 0.6 g <br/> Volume of <a href="https://interviewquestions.tuteehub.com/tag/sulphuric-655420" style="font-weight:bold;" target="_blank" title="Click to know more about SULPHURIC">SULPHURIC</a> acid taken = 50 mL <br/> Strength of sulphuric acid taken=0.5 N <br/> 20 ml of diluted solution of unreacted sulphuric acid was neutralised by 35 ml of 0.05 N Sodium laydroxide <br/> `"Strength of the diluted sulphuric acid"=(35 xx 0.05)/(20) = 0.0875 N` <br/> Volume of the sulphuric acid remaining after reaction with ammonia= `V_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)` mL <br/> Strength of `H_(2)SO_(4)` = 0.5 N <br/> Volume of the diluted `H_(2)SO_(4)` = <a href="https://interviewquestions.tuteehub.com/tag/150-275254" style="font-weight:bold;" target="_blank" title="Click to know more about 150">150</a> mL <br/> Strength of the diluted sulphuric acid = 0.0875 N <br/> `V_(1)= (150 xx 0.087)/(0.5) = 26.25 ml` <br/> Volume of `H_(2)SO_(4)` consumed by ammonia = 50 - 26.25 = 23.75 mL <br/> 23.75 mL of 0.5 N `H_(2)SO_(4)` - 23.75 mL of 0.5 N `NH_(3)` <br/> The amount of <a href="https://interviewquestions.tuteehub.com/tag/nitrogen-1118291" style="font-weight:bold;" target="_blank" title="Click to know more about NITROGEN">NITROGEN</a> present in the 0.6 g of organic compound <br/> `=(14 g)/(1000 mL xx 1 N) xx 23.75 xx 0.5 N=0.166 g` <br/> `"Percentage of nitrogen"=(0.166)/(0.6) xx 100=27.66%`</body></html>


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