Question

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

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

Answer 1

Answer: (B)

$4.08 \%$

Answer 2

Let the fraction of metal which exists as $M^{3+}$ be $x$.

Then the fraction of metal as $M^{2+} = (0.98 - x)$
$\therefore 3x + 2(0.98 -x ) = 2$
$x + 1.96 =2$
$x = 0.04$
$\therefore \%$ of $M^{3+}=\frac{0.04}{0.98}\times 100$
$=4.08\%$