A `2.18` g sample contains a mixture of XO and `X_(2)O_(3)`. It reacts with `0.015` moles of `K_(2)Cr_(2)O_(7)` to oxidize the sample completely to form `XO_(4)^(-) " and " Cr^(3+)` . If `0.0187` mole of `XO_(4)^(-)` is formed , what is the atomic mass of X ?

Explanation

`XO + K_(2)Cr_(2)O_(7) to Cr^(3+) + XO_(4)^(-)`
`X_(2)O_(3) + K_(2)Cr_(2)O_(7) to Cr^(3+) + XO_(4)^(-)`
Let, wt : of `K_(2)Cr_(2)O_(7) ` consumed by the mixture ` = 0.015 xx 6`
Equivalents of `XO = x/(x + 16) xx 5`
Equivalents of `X_(2)O_(3) = (2.18 -x)/(2x + 48) xx 8`
` :. x/(x + 16) xx 5 + (2.18 -x)/(2x - 14 8) xx 8 = 0.015 xx 6`
Since 1 mole of XO gives 1 mole `XO_(4)^(-)` and 1 mole of `X_(2)O_(3)` gives 2 moles of `XO_(4)^(-)`,
` :. x/(x + 16) + (2x(2.18 -x))/(2x + 48) = 0.0187`
On solving , `x = 99`