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.

Options

A. The difference of the number of m mol of `Cu^(2+)` and `C_(2)O_(4)^(2-)` ions in the solution is 10m mol
B. The difference of the number of m mol of `Cu^(2+)` and `C_(2)O_(4)^(2-)` ions in the solution is 25.5 m mol
C. The equivalent weight of `Cu^(2+)` ions in the titration with KI is equal to the atomic weight of `Cu^(2+)`
D. The equivalent weight of KI in the titration is `M//2` (M = Molecular weight to of KI)

Answer

A, C

Explanation

`Cu^(2+)` does not react with `MnO_(4)^(-)` ion but only `C_(2)O_(4)^(2-)` reacts with `MnO_(4)^(-)` ion m.eqt of `MnO_(4)^(-) -=` m.eqts of `C_(2)O_(4)^(2-) = 20 xx (1)/(4) xx 5 = 25`
m. moles of `C_(2)O_(4)^(2-) = (25)/(2) = 12.5`
`2Cu^(2+) overset(KI)rarr Cu_(2)I_(2)+I_(2) overset(S_(2)O_(3)^(2-))rarr S_(4)O_(6)^(2-) +2I^(-)`
`Cu^(2+) +e rarr Cu^(+),n-` factor =1
m.eqts of `Cu^(2+) -=`m. equts of hypo `=25 xx (1)/(10) = 2.5`
m. moles of `Cu^(2+) = 2.5`
Difference in m. moles of `Cu^(2+)` & `C_(2)O_(4)^(2-) =12.5 - 2.5 = 10`
Ew of `Cu^(2+) = (M)/(1)` [n-factor =1]
Ew of `KI = (M)/(1) [2I^(-) rarr I_(2) +2e^(-)]` (n-factor =1)`