Question

Experimentally it was found that a metal oxide has formula $M_{0 . 98}O$ . Metal $M$ , is present as $M^{2 +}$ and $M^{3 +}$ in its oxide. Percentage 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)

$M_{0 . 98}O\text{can be written as}M_{98}O_{100}$
Magnitude of total charge in cation = Magnitude of total charge in anion
Magnitude of total charge in anion = $100\times 2=200$
Out of total $98M$ ion, let $x$ is present in $M^{3 +}$ form then $\left(\right.98-x\left.\right)$ will be present as $M^{2 +}$ .
Now, Charge in cation $=x\times 3+\left(\right.98-x\left.\right)\times 2=x+196$
$200=x+196x=4$
$\%\text{of }M^{3 +}=\frac{\text{Number of } M^{3 +} \text{ion}}{\text{Total number of} M^{3 +} \text{ and} M^{2 +} \text{ion}}\times 100=\frac{4}{4 + 94}\times 100=4.08\%$