Question

For the reaction, $Fe0.95)$ (molar mass:M)$\to F{e}_2{O}_3$.What is the eq. wt. of $F{e}_{0.95}O$ ?

• A. $\frac{M}{0.85}$
• B. $\frac{M}{0.95}$
• C. $\frac{M}{0.8075}$
• D. none of these

Answer 1

The correct option is A $\frac{M}{0.85}$

n factor =(3-$\frac{2}{0.95})\times 0.95$

$\Rightarrow nfactor=(3\times 0.95-2\times 1)$

$\Rightarrow nfactor=(2.85-2)$

$\Rightarrow nfactor=0.85$

Therefore, Equivalent weight $=\frac{Molecularweight}{nfactor}=\frac{M}{0.85}$

Hence, the correct option is A.