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Equivalence Relation

Given a react...

Question

Given a reaction, ${F e}_{0.95} O$ (molar mass = M) $⟶ {F e}_{2} O_{3}$ . 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 the above

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Solution

The correct option is D $\frac{M}{0.8075}$
n-factor= $(3 - \frac{2}{0.95}) 0.95 = 0.8075$
$∴ e q . w t . = \frac{M}{0.8075}$

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