433 g sample of `Crl_(3)` with percentage purity 10%, with background impurity, is completely reacted with 540 ml of `H_(2)O_(2)` solution in basic medium, where `CrI_(3)` is oxidised into `Cr_(2)O_(7)^(2-)` and `IO_(4)`, then what will be the volume strength of `H_(2)O_(2) (M_(CrI_(3)) = 433g//"mole")`?

Options

A. 28
B. 56
C. 5.6
D. 2.8

Answer

A

Explanation

Weight of pure
`Crl_(3)=433xx(10)/(100)g=43.3g`
`:.` Number of milli mole of `CrI_(3)`
`=(43.3)/(433)xx1000=100`
`:.overset(+3-1)(CrI_(3))tooverset(+6)Cr_(2)overset(-2)O_(7)+3overset(+7)(I)O_(4)^(-)`
`:.` n-factor of `CrI_(3)`
`=(6-3)+3[7-(-1)]=(6-3)+3xx8=27`
`:.` number of milli equivalents of `CrI_(3)=100xx27`
=2700 (n-factor 27)
Let the volume strength of `H_(2)O_(2)` be y, then normality of `H_(2)O_(2)=(y)/(5.6)`
Mass equivalent of `CrI_(3)` = Mass equivalent of `H_(2)O_(2)`
`2700=540xx(y)/(5.6)`
`impliesy=28`