A solution of `0.2 g` of a compound containing `Cu^(2+)` and `C_(2)O_(4)^(2-)` ions on titration with `0.02M KMnO_(4)` in presence of `H_(2)SO_(4)` consumes `22.6mL` oxidant. The resulting solution is neutralized by `Na_(2)CO_(3)`, acidified with dilute `CH_(3)COOH` and titrated with excess of `KI`. The liberated `I_(2)` required `11.3 mL "of" 0.05M Na_(2)S_(2)O_(3)` for complete reduction. Find out mole ratio of `Cu^(2+)` and `C_(2)O_(4)^(2+)` in compound.

Explanation

`KMnO_(4)` reacts only with oxalate
`2MnO_(4)^(ɵ)+5C_(2)O_(4)^(2-)+16H^(o+)to2Mn^(2+)+10CO_(2)+8H_(2)O`
m" Eq of "`KMnO_(4)=MxxVxxn` factor
`=0.02xx22.6xx5`
m" Eq of "oxalate`=0.02xx22.6xx5`
millimoles of oxalate`=(0.02xx22.6xx5)/(2)`
(n factor of oxalate is 2)
KI reacts only with `Cu^(2+)`
`2Cu^(2+)+4I^(ɵ)toCu_(2)I_(2)+I_(2)`
`I_(2)+2S_(2)O_(3)^(2-)to2I^(ɵ)+S_(4)O_(6)^(2-)`
m" Eq of "thiosulphate `=MxxVxxn` factor
`=0.05xx11.3xx1`
m" Eq of "iodine`=0.05xx11.3xx1`
mmoles of `Cu^(2+)=0.05xx11.3xx1`
(n-factor for `Cu^(2+)` is 1)
`("mmoles" of Cu^(2+))/("mmoles of oxalate")=(0.05xx11.3xx1)/((0.02xx22.6xx5)/(2))=(1)/(2)`