For the reaction, $F e_{0.95} O (m o l a r m a s s : M) \to F e_{2} O_{3}$ . What is the equivalent weight of $F e_{0.95} O$ ?

\frac{M}{0.85}

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\frac{M}{0.95}

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\frac{M}{0.8075}

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None of these

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Solution

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

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