0.5 gmixture of `K_(2)Cr_(2)O_(7)` and `KMnO_(4)` was treated with excess of KI in acidic medium. Iodine liberated required `150 cm^(3)` of 0.10 N solution of thiosulphate solution for titration.
Find trhe percentage of `K_(2)Cr_(2)O_(7)` in the mixture :

Options

A. 14.64
B. 34.2
C. 65.69
D. 50

Answer

A

Explanation

Reactions of `K_(2)Cr_(2)O_(7)` and `KMnO_(4)` with KI may be given as:
`K_(2)Cr_(2)O_(7) +7H_(2)SO_(4)+6KI rarr 4K_(2)SO_(4) + Cr_(2)(SO_(4))_(3) +7H_(2)O + 3I_(2)`
`2KMnO_(4)+8H_(2)SO_(4)+10KI rarr 6K_(2)SO_(4) +2MnSO_(4) +5I_(2)`
Thus, equivalent wt of `K_(2)Cr_(2)O_(7) = (294)/(6) = 49`
equivalent wt of `KMnO_(4) = (158)/(5) = 31.6`
`:'` meq of `K_(2)Cr_(2)O_(7) +` meq of `KMnO_(4)`
= meq of `I_(2) =` meq of hypo
Let the mass of `K_(2)Cr_(2)O_(7) = xg`
`:.` mass of `KMnO_(4) = (0.5 -x)g`
`(x)/(49) +((0.5 -x))/(31.6) = 150 xx 0.1 xx 10^(-3)` (or)
`:. %` of `K_(2)Cr_(2)O_(7) = (0.0732)/(0.5) xx 100 rArr 14.64`