Chemistry - Redox Reactions Question with Solution | TestHub

For the higher oxide, oxygen constitutes 20% of the mass, while the metal accounts for 80%.

The equivalent weight ($E$) of the metal in the higher oxide is calculated as:

$E=8\times \frac{massofmetal}{massofoxygen}=8\times \frac{80}{20}=32$

The 0.48 g mass gain upon conversion indicates the amount of added oxygen.

The mass of the metal in the higher oxide is $0.8\times 4.77=3.816g$.

The mass of oxygen in the lower oxide is $4.29-3.816=0.474g$.

Therefore, the equivalent weight ($E$) of the metal in the lower oxide is:

$E_{lower}=8\times \frac{3.816}{0.474}=64.4$

Thus, the lower oxide has an equivalent weight of $E=64.4$, and the higher oxide has an equivalent weight of $E=32$. This corresponds to options (b) and (d).