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Q.
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
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 $M O =1-0.98=0.02$
Total $M^{3+}$ present in one molecule of $M O=2 \times 0.02=0.04$
That $M^{2+}$ and $M^{3+}=0.98$
Thus, $\%$ of $M^{3+}=\frac{0.04 \times 100}{0.98}=4.08 \%$