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Q.
The percentage of iron present as $Fe(III)$ in $ Fe_{0.93}O_{1.0} $ is
AMUAMU 2013The Solid State
Solution:
Let $x$ atoms of $F e^{3+}$ ions are present. This means $x F e^{3+}$ ions have been replaced by $F e^{2+}$ ions. No. of $F e^{2+}$ ions $=0.93-x$ For electrical neutrality,
$2(0.93-x)+3 x=2$
$1.86+x=2$ or $x=0.14$
Fraction of $F e^{3+}=0.14, F e^{2+}=0.93-0.14=0.79$
Formula : $F e_{0.79}^{2+} F e_{0.14}^{3+} O_{1.0}^{2-}$
Total molar mass $=0.93 \times 56+1 \times 16=68.08$
$\%$ of $Fe (I I)=\frac{0.14 \times 56}{68.08} \times 100=11.5 \%$