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

Options

A. 0.0605
B. 0.0508
C. 0.0701
D. 0.0408

Answer

D

Explanation

Let number of `M^(3+)` in oxide is .x..
Then number of `M^(2+)` ion will be `(0.98-x)`. Oxide is neutral, therefore total charge must be zero.
`(+2)(0.98-x)+(+3)x+(-2)=0`
`1.96-2x+3x-2=0`
x=0.04
Percentage of `Mg^(3+)=("Number of "M^(3+)" ions")/("Total number of metal atoms")xx100=(0.04)/(0.98)xx100=4.08%`.