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 :

6.05%

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5.08%

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7.01%

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4.08%

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Solution

The correct option is C 4.08%
Average oxidation no. of M $= + \frac{200}{98}$ (lies between 2 and 3)
Let % of $M^{2 +}$ be $α$ and of $M^{3 +}$ be $b$
or, $\frac{2 a + 3 (100 - a)}{100} = 2.04 (∵ a + b = 100)$
$∴ 2 a + 300 - 3 a = \frac{200}{98}$
$∴ + a = 300 - 2.04 \times 100$
$= 300 - 204 = 96$
Thus $M^{2 +} = 96 %$
$M^{3 +} = 4 %$

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Similar questions

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:

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.

%

of the metal which exists as

M^{3 +}

would be :

Experimentally it was found that a metal oxide has a formula of M0.98 . Metal M is present as M^+2 and M^+3 in its oxide. Fraction of the metal which exists as M^+3 would be ?

Options -----

(A) 4.08 percentage

(B) 6.05 percentage

(D) 7.01 percentage

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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. Fraction of the metal which exists as M3+ would be :

The correct option is C 4.08%Average oxidation no. of M=+20098 (lies between 2 and 3)Let % of M2+ be α and of M3+ be bor, 2a+3(100−a)100=2.04(∵a+b=100)∴2a+300−3a=20098∴+a=300−2.04×100 =300−204=96Thus M2+=96%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. Fraction of the metal which exists as $M^{3 +}$ would be :