Analysis shows that metal oxide has formnla `M_(0.96)O_(1.00)` What fractions of the metal exist as `M^(2+)` and `M^(3+)`

Explanation

Let number of `M^(2+)` ions =x
Number of `M^(3+)` ions= 0.98 - x
Total positive charge = total negative charge
`Xxx(+2)+(0.98-x)3=(+2)1`
2x+2.94-3x=+2
-x=+2-0.94
-x=-0.94
x=0.94
Number of `M^(2+)` ions = 0.94
Number of `M^(3+)` ions = 0.98 - 0.94 = 0.04
% of `M^(2+) ions =0.94/0.98xx10=95.9%`
% of `M^(3+) ions = 0.04/0.98xx100=4.08%`