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 0.15 M sodium thiosulphate solution for titration. The percent amount of `KMnO_(4)` in the mixture is
(Atomic weight – K = 39, Cr = 52, Mn=55, Na=23, S=32)

Options

A. `85.36%`
B. `82.34%`
C. `54.22%`
D. `34.56%`

Answer

A

Explanation

Let the weight of `K_(2)Cr_(2)O_(4)` be .a. gm
Then the weight of `KMnO_(4)=0.5` -a gm
Now E quivalents of so dium th iosulphate
`implies((0.5-a))/(158//5)+(a)/(294//6)=(0.15)xx1xx(100)/(1000)`
`implies(5(0.5-a))/(158)+(6a)/(294)=0.015`
`implies(2.5-5a)/(158)+(6a)/(294)=0.015`
`implies735-1470a+948a=696.78`
`implies522a=38.22`
`impliesa=0.0732`
`implies:.` weight of `K_(2)Cr_(2)O_(7)=0.0732g`
& weight of `KMnO_(4)=0.5-0.0732=0.42682`
% of `K_(2)Cr_(2)O_(7)=(0.0732)/(0.5)xx100`
`=14.64%`
Percent of `KMnO_(4)=(0.4268)/(0.5)xx100`
`=85.36%`