`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

Let .a. g of `K_(2)Cr_(2)O_(7)` be present in the mixture
Mass of `KMnO_(4) = (0.5 - a) ` g
Eq. Mass of `K_(2)Cr_(2)O_(7) = "Mol mass "/6 = (294)/6 = 49.0`
Eq. mass of `KMnO_(4) = " Mol mass "/5 = (158)/5 = 31.6`
No. of equivalents of `K_(2)Cr_(2)O_(7) = a/(49.0) `
No. of equivalents of `K_(2)Cr_(2)O_(7) = a/(49.0)`
No. of equivalents of `KMnO_(4) = ((0.5 -a))/(31.6) `
No . of equivalents of `Na_(2)S_(2)O_(3)` in 100 `cm^(3)" of " 0.15` N Solution
` = (100 xx 0.15)/1000 = 0.015`
Equivalents of `K_(2)Cr_(2)O_(7) + ` Equivalents of `KMnO_(4)`
` -= ` Equivalents of iondine
`-=` Equivalennts of `Na_(2)S_(2)O_(3)`
`rArr a/(49.0) + ((0.5 -a))/(31.6) = 0.015`
`rArr 17.4a = 1.274`
` a = 0.0732`
% of `K_(2)Cr_(2)O_(7) = (0.0732 xx 100)/(0.5) = 14.64`
% of `KMnO_(4) = 85.36`.