Experimentally it was found that a metal oxide has formula M_0.98O. Metal M, is present as M²⁺ and M³⁺ in its oxide. Fraction of the metal which exists as M³⁺ would be

Let the fraction of metal which exists as M³⁺ be x. Then the fraction of metal exists as M²⁺ = (0.98 $-$ x)

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\mathrm{x}+2(0.98-\mathrm{x})=2\\ \hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em} \mathrm{x}+1.96=2\\ \hspace{0.25em} \mathrm{x}=0.04\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} \mathrm{\%}\text{ of }{\mathrm{M}}^{3+}=\frac{0.04}{0.98}\times 100=4.08\mathrm{\%}\end{array}$