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Assigning Oxidation Numbers

For the react...

Question

For the reaction, $F e_{0.95} O$ (molar mass M) $\to F e_{2} O_{3}$ , what is the equivalent weight of $F e_{0.95} O$ ?

M/0.85

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M/0.95

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M/0.80

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M/0.90

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Solution

The correct option is A M/0.85
n-factor $= (3 - \frac{2}{0.95}) \times 0.95 = 0.85.$
Equivalent weight $= \frac{M o l e c u l a r w e i g h t / A t o m i c w e i g h t}{n_{f a c t o r}}$
$= M / 0.85$

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