1.

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 :

Answer»

`7.01 %`
4.08%
6.05%
5.08%

SOLUTION :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 %`


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