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)+6KIrarr 4K_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+7H_(2)O+3I_(2)`
`2KMnO_(4)+8H_(2)SO_(4)+10KIrarr 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`
`because" 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,`
`therefore" mass of "KMnO_(4)=(0.5-x)g`
`(x)/(49)+((0.5-x))/(31.6)=150xx0.1xx10^(-3)" (or)"`
`therefore " % of "K_(2)Cr_(2)O_(7)=(0.0732)/(0.5)xx100 rArr 14.64`