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 the metal which exists as `M^(3+)` would be `:`

Options

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

Answer

B

Explanation

Metal oxide `: M_(0.98) O`
Let `M^(2+)` be x so that `M^(3+)` will be 0.98 - x .
Total charge must be zero.
`2x + 3 ( 0.98 - x ) - 2 = 0 `
`2x + 2.94 -3x -2 =0`
` -x = - 0.94 ` or ` x = 0.94`
Fraction of metal which exists as `M^(3+)`
`= 0.98 - 0.94 = 0.04`
Percentage of `M^(3+) = ( 0.04 )/( 0.98) xx 100 = 4.08 %`