`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 equivalent of `K_(2) Cr_(2) O_(7) = ( a)/( 49.0)`
No. of equivalent 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`
Equivalent 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`
`17.4 a =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`