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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 beA. `5.08%`B. `7.01%`C. `4.08%`D. `6.05%` |
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Answer» Correct Answer - C 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 % |
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