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

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Solution

Average oxidation number of M $= + \frac{200}{98}$ (lies between 2 and 3).
Let $%$ of $M^{2 +}$ be $α$ and of $M^{3 +}$ be $b$ .
or, $\frac{2 \times a + (100 - a) \times 3}{100} = 2.04 (∵ a + b = 100)$
$∴ 2 a + 300 - 3 a = \frac{200}{98}$
$∴ + a = 300 - 2.04 \times 100$ $= 300 - 204 = 96$
Thus, $M^{2 +}$ = $96 %$ and $M^{3 +}$ = $4 %$ .

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Q. 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 :

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:

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Experimentally it was found that a metal oxide has formula M0.98O. Metal M is present as M2+ and M3+ in its oxide. % of the metal which exists as M3+ would be :

Average oxidation number of M=+20098 (lies between 2 and 3).Let % of M2+ be α and of M3+ be b.or, 2×a+(100−a)×3100=2.04(∵a+b=100)∴2a+300−3a=20098∴+a=300−2.04×100 =300−204=96Thus, M2+ = 96% and M3+ = 4%.

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