An element X forms two oxides. Formula of the first oxide is XO₂. The first oxide contains 50% of oxygen. If the second oxide contains 60% oxygen, the atomicity of second oxide.

${\text{First oxide (XO}}_2)$

$\begin{array}{l}\text{Let A.W of X=a}\\ \text{ X O}\\ \text{w 50gm 50gm}\\ \text{moles }\frac{50}{\text{a}}\frac{50}{16}\\ \therefore \frac{50/\text{a}}{50/16}=\frac{1}{2}\\ \text{ a = 32}\end{array}$

$\text{Second oxide }$

$\begin{array}{l}\text{ X O}\\ \text{w 40gm 60gm}\\ \text{moles }\frac{40}{32}\frac{60}{16}\\ \text{ Ratio of atoms = }\frac{40}{32}\text{ X}\frac{60}{16}=1:3\\ \therefore {\text{Atomicity of 2}}^{nd}\text{ oxide = 4}\end{array}$