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A sample containing 0.4775 of (NH_(4))_(2)C_(2)O_(4) and inert material was dissolved in water and made strongly alkaline with KOH which converted NH_(4)^(o+) to NH_(3) The liberated NH_(3) was distilled of H_(2)SO_(4) was back titrated with 11.3 " mL of " 0.1214 M NaOH. Calculate (a) % of (NH_(4))_(2)C_(2)O_(4)=124.10 And atomic weight of N=14.0078.

Answer» <html><body><p></p>Solution :`(NH_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>))_(2)C_(2)O+(4)+2KOHtoK_(2)C_(2)O_(4)+2NH_(3)+2H_(2)O` <br/> Total `H_(2)SO_(4)` <a href="https://interviewquestions.tuteehub.com/tag/used-2318798" style="font-weight:bold;" target="_blank" title="Click to know more about USED">USED</a> `=50xx0.05035xx2NH_(2)SO_(4)` <br/> `=5.035m" Eq of "H_(2)SO_(4)` <br/> Excess of `H_(2)SO_(4)=11.3xx0.124M NaOH` <br/> `=1.372 m" Eq of " NaOH` <br/> `=1.372 m" Eq of "H_(2)SO_(4)` <br/> `H_(2)SO_(4)` used `=5.035-1.372` <br/> `=3.663 m" Eq of "H_(2)SO_(4)` <br/> `=3.663 <a href="https://interviewquestions.tuteehub.com/tag/meq-1093952" style="font-weight:bold;" target="_blank" title="Click to know more about MEQ">MEQ</a> NH_(3)` <br/> `=3.663 m" Eq of "(NH_(4))_(2)C_(2)O_(4)` <br/> Ew of `(NH_(4))_(2)C_(2)O_(4)=(124.1)/(2)=62.06g` <br/> Weight of `(NH_(4))_(2)C_(2)O_(4)=3.663xx10^(-3)xx62.05=0.2273g` <br/> `124.1 g of (NH_(4))_(2)C_(2)O_(4)` <a href="https://interviewquestions.tuteehub.com/tag/contains-11473" style="font-weight:bold;" target="_blank" title="Click to know more about CONTAINS">CONTAINS</a> `14.0078 g of N`. <br/> `0.2273 g of (NH_(4))_(2)C_(2)O_(4)=(14.0078xx0.2273)/(124.1)=0.02565` g <br/> `% of N=(0.02565)/(0.4775)xx100=5.373%` <br/> `% of (NH_(4))_(2)C_(2)O_(4)=(0.2273)/(0.4775)xx100=5.373%`</body></html>


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