Two oxides of a $X$ contain $50 \%$ and $40 \%$ of non-metal respectively. If the formula of the first oxide…
Two oxides of a $X$ contain $50 \%$ and $40 \%$ of non-metal respectively. If the formula of the first oxide is $\mathrm{XO}_2$. Then the formula of second oxide is
$X_2 \mathrm{O}_3$
$X_2 \mathrm{O}_5$
$X \mathrm{O}_3$
$X_2 \mathrm{O}$
Solution
In $\mathrm{XO}_2, \mathrm{X}: \mathrm{O}:: 50: 50$
Mass of oxygen $=2 \times 16=32 \mathrm{~g}$
Hence, mass of $X=32 \mathrm{~g}$
In $\mathrm{XO}_n, \mathrm{X}: \mathrm{O}:: 40: 60$
Mass of $X$, in this case also remains $=32 \mathrm{~g}$
$40 \% \equiv 32 \mathrm{~g}$
$60 \% \equiv \frac{32}{40} \times 60=48 \mathrm{~g}$
Mass of oxygen $=48 \mathrm{~g}$
$\Rightarrow \quad n=\frac{48}{16}=3$
Formula of oxide $=\mathrm{XO}_3$