`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

The redox changes are:
`{:(5e+,Mn^(7+),rarr,Mn^(2+),),(6e+,Cr_(2)^(6+),rarr,2Cr^(3+),),(,2I^(-),rarr,I_2+,2e),(,2S_(2)O_(3)^(2-),rarr,,S_(4)O_(6)^(2-)+2e):}`
Let `K_(2)Cr_(2)O_(7)` and `KMnO_(4)` be `a` and `b g` respectively.
`:. a+b=0.5`......(1)
Further Meq. of `KMnO_(4)+"Meq.of" K_(2)Cr_(2)O_(7)="Meq.of"KI="Meq.of"I_(2)"liberated"="Meq.of"Na_(2)S_(2)O_(3)`
`(a)/(294//6)xx1000+(b)/(158//5)1000=100xx0.15`.....(2)
By eqs. (1) adn (2), `a=0.073, b=0.427`
`therefore % "of" K_(2)Cr_(2)O_(7)=14.6%`
and `% "of" KMnO_(4)=85.4%`