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When \mathrm{Fe}_{0.93} \mathrm{O} is heated in presence of oxygen, it converts to \mathrm{Fe}_2 \mathrm{O}_3.The number of correct statement/s from the following is

A. The equivalent weight of \mathrm{Fe}_{0.93} \mathrm{O}$ is $\frac{\text { Molecular weight }}{0.79}

B. The number of moles of \mathrm{Fe}^{2+} and \mathrm{Fe}^{3+}in 1 mole of \mathrm{Fe}_{0.93} \mathrm{O}  is 0.79 and 0.14 respectively

C.\mathrm{Fe}_{0.93} \mathrm{O} is metal deficient with lattice comprising of cubic closed packed arrangement of \mathrm{O}^{2-} ions

D.The \% composition of \mathrm{Fe}^{2+}and \mathrm{Fe}^{3+}in \mathrm{Fe}_{0.93} \mathrm{O} is 85 \% and 15 \% respectively

Option: 1

4


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

{Fe}_{0.93} \mathrm{O}
2 \mathrm{x}+(0.93-\mathrm{x}) 3=2
-\mathrm{x}+3 \times 0.93=2
\mathrm{x}=0.79
0.79=\text { no. of } \mathrm{Fe}^{2+} \text { ion }
0.14=\text { no. of } \mathrm{Fe}^{3+} \text { ion }
\mathrm{nf}=0.79

\text{Equivalent wt} =\frac{\text { Molecular weight }}{0.79}
Due to presence of \mathrm{Fe}^{3+} in \mathrm{FeO} lattice, Metal deficiency occurs.
\% \text { Composition :- } \mathrm{Fe}^{2+} \text { ions }=\frac{0.79}{0.93} \times 100
85 \%
\mathrm{Fe}^{3+} \text { ion }=\frac{0.14}{0.93} \times 100
15 \%

Posted by

Ajit Kumar Dubey

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