Experimentally it is was found that a metal oxide has formula `M_(0.95)O` .Metal `M` present as `M^(3+)` in its oxide fraction of the metal which exits as `M^(2+)` would be

Options

A. `89.47%`
B. `4.08%`
C. `6.05 %`
D. `5.08%`

Answer

a

Explanation

`M_(0.98)O`
if number of moles of `M^(2 +) = x`
`Then M^(2+) = (0..95 - x)`
Thus charge balance of `M_(0.98) O`
`(x + z)= (0.95 - x) xx 3 - (1 xx 2) = 0`
`- x = 2 - 0.95 xx 3`
`- x = 2 - 2.85 = 0.85`
`x = 2 - 2 .85 = 0.10`
Percentage of `M^(2+) = (0.10)/(0.95) = 100 = 10.52%`