Question

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:

• A. $X_{2}O$
• B. $X{O}_3$
• C. $X_2{O}_3$
• D. $X_3{O}_2$

Answer 1

Answer: (B)

$X{O}_3$

The first oxide $X{O}_2$ contains $50\%$ oxygen.

$\therefore \frac{32}{32+x}=0.5$

Molecular weight of the metal, $x=32$

Now, let the molecular formula of the other metal oxide be $M{O}_n$.

$\therefore \frac{16n}{32+16n}=0.6$

On solving, we get $n=3$.