10.0 g sample of `Cu_(2)O` is 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 removed. Now excess of an oxidising agent is added to the filtrate which liberates `I_(2)` again. The liberated `I_(2)` now requires 10 mL of 0.1M `Na_(2)S_(2)O_(3)`.The percentage by mass of `Cu_(2)O` in the sample is

Explanation

`Cu_(2)OtoCu^(2+)+Cu^(0)`
The solution after dissolution of `Cu_(2)O` in dil `H_(2)SO_(4)` contains `Cu^(2+)` ions 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)`.
`Cu^(2+)+2I^(ɵ)toCuI_(2)to(Cu_(2)I_(2) or CuI)+(1)/(2)I_(2)`
Millimoles of KI taken`=(8.3)/(166)xx10^(3)=50(mE of KI=166)`
Now, KI left unused reacts with oxidising agent to liberate `I_(2)` again.
`2I^(ɵ)overset("oxidising agent")toI_(2)overset(2S_(2)O_(3)^(2-))toS_(4)O_(6)^(2-)+I^(ɵ)`
Millimoles of KI left`=` millimoles of `S_(2)O_(3)^(2-)` used
`(n=(2)/(2)=1)=(n=(2)/(2)=1)`
`=10xx1.0xx1` (n factor)`=10`
Therefore, millimoes of KI used for `Cu^(2+)=50-10=40`
Therefore, milli moles of `Cu_(2)O=20`
`{:(Coverset(+1)(u_(2))to2overset(+2)(Cu)),(mol ratio of overset(+1)(Cu_(2)):overset(+2)(Cu)=1:2):}`
`(Weight)/(Mw)xx10^(3)=20` (Mw of `Cu_(2)O=143)`
`(weight)/(143)xx10^(3)=20`
`thereforeW_(Cu_(2)O)=2.86`
`% of Cu_(2)O=(2.86xx100)/(10.0)=28.6%`