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What fraction of Fe exists as Fe(III) in $Fe_{0.96}O$? (Consider $Fe_{0.96}O$ to be made up of Fe(II) and Fe(III) only)
Detailed Solution
$Fe_{0.96}O=Fe_{96}O_{100}$. Let x Fe be $Fe^{2+}$; then $(96-x)$ are $Fe^{3+}$. Charge balance: $2x+3(96-x)=200 \Rightarrow x=88$. $Fe^{3+}=96-88=8$. Fraction of $Fe^{3+}=\frac{8}{96}=\frac{1}{12}$.