The non-stoichimetric compound $F e_{0.94} O$ is formed when $x$ % of $F e^{2 +}$ ions are replaced by 2/3 $F e^{3 +}$ ions, then $x$ is:

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Solution

The correct option is A 18
Number of $F e^{3 +}$ ions replacing $x F e^{2 +}$ ions is equal to $\frac{2 x}{3}$
$∴$ Vacancies of cations $= x - \frac{2}{3} x = \frac{x}{3}$
But $\frac{x}{3} = 1 - 0.94 = 0.06 or x = 0.06 \times 3 = 0.18$ i.e. $18$ %.

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