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

Options

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

Answer

B

Explanation

Let number of `M^(3+)` ion will be (0.98-x) . Oxide is neutral, therefore total change must be zero.
`(+2)(0.98-x)+(+3)x+(-2)1=0`
`1.96-2x+3x-2=0`
x=0.04
Percentage of `M^(3+)=("Number of" M^(3+) "ions")/("Total number of metal atoms")xx100`
`=(0.04)/(0.98)xx100`
`=4.08 %`