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
To solve this problem, we need to determine what fraction of the metal \( M \) exists as \( M^{3+} \) in the metal oxide \( M_{0.98}O \). The oxide contains both \( M^{2+} \) and \( M^{3+} \).
Let the number of moles of \( M^{2+} \) be \( x \) and the number of moles of \( M^{3+} \) be \( y \).
The total moles of the metal \( M \) present is given by:
$$ x + y = 0.98 $$
(as indicated by the formula \( M_{0.98}O \)).
For the compound to maintain electrical neutrality, the total positive charge must equal the negative charge of the oxygen, which is \( -2 \) per oxygen. Thus, the charge balance equation is:
$$ 2x + 3y = 2 \cdot 1 = 2 $$
Now, we have two equations:
$$ x + y = 0.98 $$
$$ 2x + 3y = 2 $$
We can solve these equations simultaneously. First, express \( x \) from the first equation:
$$ x = 0.98 - y $$
Substitute \( x \) in the second equation:
$$ 2(0.98 - y) + 3y = 2 $$
Expand and simplify the equation:
$$ 1.96 - 2y + 3y = 2 $$
$$ 1.96 + y = 2 $$
$$ y = 2 - 1.96 = 0.04 $$
So, \( y = 0.04 \). This means 0.04 moles of \( M \) are present as \( M^{3+} \). Hence, the fraction of the metal existing as \( M^{3+} \) is:
$$ \text{Fraction of } M^{3+} = \frac{0.04}{0.98} $$
Calculate the percentage:
$$ \text{Percentage of } M^{3+} = \left(\frac{0.04}{0.98}\right) \times 100\% = 4.08\% $$
Therefore, the fraction of the metal existing as \( M^{3+} \) is 4.08%, which matches the provided correct answer.
