Metal $X$ forms two oxides. Formula of the first oxide is $X O_{2}$ . It contains $50 %$ of oxygen. If the second oxide contains $60 %$ of oxygen, the formula of the second oxide is:

X_{2} O

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X O_{3}

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X_{2} O_{3}

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X_{3} O_{2}

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Solution

The correct option is C $X O_{3}$
The first oxide $X O_{2}$ contains $50 %$ oxygen.
$∴ \frac{32}{32 + x} = 0.5$
Molecular weight of the metal, $x = 32$
Now, let the molecular formula of the other metal oxide be $M O_{n}$ .
$∴ \frac{16 n}{32 + 16 n} = 0.6$
On solving, we get $n = 3$ .

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