A) 8.3%
B) 9.6%
C) 11.5%
D) 17.7%
E) (e) None of these above
Correct Answer: E
Solution :
(e)Percentage of Fe (lll) in \[F{{e}_{0.93}}{{O}_{1.0}}\] Formula \[F{{e}_{0.93}}{{O}_{1.00}}\] shows that Fe : 0 = 0.93 : 1.00 or 93 : 100 If there are 100 oxygen atoms then Fe atoms = 93 Charge on oxide ion \[=100\times (-2)=-200\] Suppose \[F{{e}^{2+}}=x\], then \[F{{e}^{3+}}=93-x\] Total positive charge \[=(2\times x)+3(93-x)\] \[2x+279-3x=279-x\] Positive charge = negative charge \[279-x=200\] \[x=79=F{{e}^{2+}}\] \[F{{e}^{3+}}=93-79=14\] % of \[F{{e}^{3+}}=\frac{14\times 100}{93}=15.05%\] Hence, None of these options is correct.
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