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.
Fraction of the metal which exists as $M^{3+}$ would be:

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

Answer 1

The correct option is A

4.08%

Metal oxide $=M_{0.98}O$.
Let's say we have 'x' ions of M is +3 state.
$\Rightarrow 3x+2(0.98-x)=2\Rightarrow x=0.04$
$\Rightarrow \%$ of metal in +3 state $=\frac{0.04}{0.98}\times 100=4.08\%$