Question

Experimentally it was found that a metal oxide has formula $M_{0.98}O$. Metal $M$, present as $M^{2+}$ and $M^{3+}$ in its oxide. Fraction of the metal which exists as $M^{3+}$ would be

• A. $7.01\%$
• B. $4.08\%$
• C. $6.05\%$
• D. $5.08\%$

Answer 1

Answer: (B)

From the valency of $M^{2+}$ and $M^{3+}$, it is clear that three $M^{2+}$ ions will be replaced by $M^{3+}$ causing a loss of one $M^{3+}$ ion.
loss of them from one molecule of $MO=1-0.98=0.02$
Total $M^{3+}$ present in one molecule of $MO=2\times 0.02=0.04$
That $M^{2+}$ and $M^{3+}=0.98$
Thus, $\%$ of $M^{3+}=\frac{0.04\times 100}{0.98}=4.08\%$