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10.0 g sample of Cu_(2)O is disoolved dissolved in dil. H_(2)SO_(4) where it undergoes disproportionation quantitatively. The solution is filtered off and 8.3 g pure KI cyrstals are added to clear filtrate in order to precipitate CuI with the liberation of I_(2). The solution is again filtered and boiled till all the I_(2) is removes. Now excess of an oxidising agent is added to the rfiltrate which liberates I_(2) again. The liberated I_(2) now percentrage by mass of Cu_(2)O in the sample.

Answer» <html><body><p></p>Solution :`Cu_(2)OtoCu^(2+)+Cu^(0)` <br/> The solution after dissolution of `Cu_(2)O` in dil `H_(2)SO_(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> `Cu^(2+)` <a href="https://interviewquestions.tuteehub.com/tag/ions-1051295" style="font-weight:bold;" target="_blank" title="Click to know more about IONS">IONS</a> and `Cu`. `Cu^(2+)` ions react with KI to give `CuI_(2)` which is converted to CuI (or `CU_(2)I_(2))` and `I_(2)`. <br/> `Cu^(2+)+<a href="https://interviewquestions.tuteehub.com/tag/2i-300397" style="font-weight:bold;" target="_blank" title="Click to know more about 2I">2I</a>^(ɵ)toCuI_(2)to(Cu_(2)I_(2) or CuI)+(1)/(2)I_(2)` <br/> Millimoles of KI taken`=(8.3)/(166)xx10^(3)=50(mE of KI=166)` <br/> Now, KI left unused reacts with oxidising agent to liberate `I_(2)` again. <br/> `2I^(ɵ)overset("oxidising agent")toI_(2)overset(2S_(2)O_(3)^(2-))toS_(4)O_(6)^(2-)+I^(ɵ)` <br/> Millimoles of KI left`=` millimoles of `S_(2)O_(3)^(2-)` used <br/> `(n=(2)/(2)=1)=(n=(2)/(2)=1)` <br/> `=10xx1.0xx1` (n factor)`=10` <br/> Therefore, millimoes of KI used for `Cu^(2+)=50-10=40` <br/> Therefore, <a href="https://interviewquestions.tuteehub.com/tag/milli-560824" style="font-weight:bold;" target="_blank" title="Click to know more about MILLI">MILLI</a> moles of `Cu_(2)O=20` <br/> `{:(Coverset(+1)(u_(2))to2overset(+2)(Cu)),(mol ratio of overset(+1)(Cu_(2)):overset(+2)(Cu)=1:2):}` <br/> `(Weight)/(Mw)xx10^(3)=20` (Mw of `Cu_(2)O=143)` <br/> `(weight)/(143)xx10^(3)=20` <br/> `thereforeW_(Cu_(2)O)=2.86` <br/> `% of Cu_(2)O=(2.86xx100)/(10.0)=28.6%`</body></html>


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