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Experimentally it was found that a metal...

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 :

A

`4.08%`

B

`6.05%`

C

`5.08%`

D

`7.01%`

Text Solution

Verified by Experts

The correct Answer is:
A
Metal oxide `= M_(0.98)O`
If `'x'` ons of `M` are in `+3` state, then
`3x + (0.98 -x) xx 2 = 2`
` x = 0.04`
So, the percentage of metal in `+3` state would be
`(0.04)/(0.98) xx 100 = 4.08%`
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