`0.5 g` mixture of `K_(2)Cr_(2)O_(7)` and `KMnO_(4)` was treated with excess of `KI` in acidic medium. Iodine liberated required `100 cm^(3)` of `0.15N` sodium thiosulphate solution for titration. Find the per cent amount of each in the mixture.

Explanation

(a). `5e^(-)+MnO_(4)^(ɵ)toMn^(2+)(n=5)`
(b). `6e^(-)+Cr_(2)O_(7)^(2-)to2Cr^(3+)(n=6)`
(c). `2I^(ɵ)toI_(2)+2e^(-)(n=2)`
(d). `2S_(2)O_(3)^(2-)toS_(4)O_(6)^(2-)+2e^(-)(n=(2)/(2)=1)`
Let `K_(2)Cr_(2)O_(7)` and `KMnO_(4)` be a and b g, respectively.
`thereforea+b=0.5`
`m" Eq of "K_(2)Cr_(2)O_(7)+m" Eq of "KMnO_(4)=m" Eq of "KI`
`=m" Eq of "I_(2)` liberated
`=m" Eq of "Na_(2)S_(2)O_(3)`
`((a)/((294)/(6))+(b)/((158)/(5)))xx10^(3)=100xx0.15`
From equation (i) and (ii), `a=0.073,b=0.427`
`% of K_(2)Cr_(2)O_(7)=(0.073xx100)/(0.5)=14.6%`
`% of KMnO_(4)=(0.427xx100)/(0.5)=85.6%`