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(B) Given the formula of the metal oxide is $M_{0.98}O$.Consider $1 	ext{ mole}$ of the oxide.Moles of $M = 0.98$ and moles of $O^{2-} = 1$.Let the m

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$4.08$

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(B) Given the formula of the metal oxide is $M_{0.98}O$.Consider $1 	ext{ mole}$ of the oxide.Moles of $M = 0.98$ and moles of $O^{2-} = 1$.Let the moles of $M^{3+} = x$.Then,the moles of $M^{2+} = 0.98 - x$.Since the compound is electrically neutral,the total positive charge must equal the total negative charge:$(0.98 - x) 	imes 2 + x 	imes 3 = 1 	imes 2$$1.96 - 2x + 3x = 2$$x = 2 - 1.96 = 0.04$.Therefore,the fraction of metal existing as $M^{3+} = \frac{0.04}{0.98} 	imes 100 = 4.08 \%$.

Experimentally,it was found that a metal oxide has the formula $M_{0.98}O$. The metal $M$ is present as $M^{2+}$ and $M^{3+}$ in its oxide. The fraction of the metal which exists as $M^{3+}$ would be: .................... $\%$

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