In the crystal of $F e_{0.93} O$ , the percentage of Fe(III) will be:

15%

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85%

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30%

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78%

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Solution

The correct option is B 15%
$3 x + (0.93 - x) \times 2 = 2$
$x = 0.14$
% of $F e$ as $F e$ (III) $= \frac{0.14}{0.93} \times 100 = 15$ %

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