Experimentally it was found that a metal oxide has formula `M_(0.98)`, Metal `M`, is present as `M_(+2)` and `M_(+3)` in its oxid. Fraction of the metal which exist as ` `M_(+3)` would be :

Options

A. `6.05%`
B. `5.08%`
C. `7.01 %`
D. `4.08%`

Answer

d

Explanation

Let number at `M^(1+) `ions is `x` then the nymber of `M^(2+)` ions will be `(0.98 - x)` .Net charge of oxide will be zero.
`:. + 2(0.98 - x) + 3(x) + (-2) 1= 0`
`1.96 + x - 2 = 0`
`x = 0.04`
Percentage of `M^(2+) = (0.04)/(0.98) xx 100 = 4.08`