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

Experimentally it was found that a metal oxide has formula M_0.98 O. M

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» `5.08%` `7.01%` `4.08%` `6.05%` Solution :The FORMULA `M_0.98 O` shows that if there were 100 O-atoms present as `O^(2-)` ions, then there 98 M atoms present as `M^(2+)` and `M^(3+)` . Suppose `M^(3+) =x` , then `M^(2+)` =98-x . As the compound as a WHOLE is neutral, TOTAL charge on `M^(3+)` and `M^(2+)` =Total charge on 100 `O^(2-)` ions x (+3) + (98-x) (+2)=100 x 2 or 3x + 196 - 2x =200 or x=4 `THEREFORE` % of `M^(3+)=4/98xx100`=4.08 %