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 $(98-x)$ will be present as $M^{2+}$ .
Now, Charge in cation $=x\times 3+(98-x)\times 2=x+196$
$200=x+196x=4$
$\%\text{of }{M}^{3+}=\frac{\text{Number of }{M}^{3+}\text{ion}}{\text{Total number of}{M}^{3+}\text{ and}{M}^{2+}\text{ion}}\times 100=\frac{4}{4+94}\times 100=4.08\%$