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0.5 g mixture of K(2)Cr(2)O(7) and KMnO(...

`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.

Text Solution

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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")/(5)=(158)/(5)=31.6`
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
`=(100xx0.15)/(1000)=0.015`
Equivalents of `K_(2)Cr_(2)O_(7)+` Equivalents of `KMnO_(4)`
`-=` Equivalents of iodine
`-=` Equivalents of `Na_(2)S_(2)O_(3)`
`(a)/(49.0)+((0.5-a))/(31.6)=0.015`
17.4a=1.274
a= 0.0732
% of `K_(2)Cr_(2)O_(7)=(0.0732xx100)/(0.5)=14.64`
% of `KMnO_(4)=85.36`
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