A sample weighing `2.198 g` containing a mixture of `AO` and `A_(2)O_(3)` takes `0.015 "mole of" K_(2)Cr_(2)O_(7)` to oxidise the sample completely to form `AO_(4)^(-)` and `Cr^(3+)`. If `0.0187` `"mole of" AO_(4)^(-)` is formed, what is `at.wt`. of `A`?

Explanation

Let `M.wt. "of" AO` and `A_(2)O_(3)` be `m` and `n` respectively.
`therefore m=a+16`…..(1)
and `n=2a+48`……(2)
where `a` is atomic weight of `A`.
Now suppose `X` and `Y g` of `A_(2)O_(3)` are present in mixture.
Then `X+Y=2.198`.....(3)
Also Meq.of `AO+"Meq.of"A_(2)O_(3)="Meq.of" K_(2)Cr_(2)O_(7)`
`
`=0.015xx6xx1000` ....(4)`
`{(because, ,A^(2+),rarr,A^(7+)+5e ),(,,A_(2)^(3+),rarr,2A^(7+)+8e),(,6e+,Cr_(2)^(6+),rarr,2Cr^(3+)):}`
`therefore` By eqs. (4), `(5X)/(a+16)+(8Y)/(2a+48)=0.09`.....(5)
Also mole `AO_(4)^(-)` by `AO+"mole of" AO_(4)^(-)` by `A_(2)O_(3)=0.0187`
`(X)/(a+16)+(2Y)/(2a+48)=0.0187`.....(6)
`because"Moles raio of "AO:AO_(4)^(-)::1:1,A_(2)O_(3):AO_(4)^(1-)::1:2`
Solving eqs. (3), (5) and (6)
`a=100`