The atomic weights of two elements A and B are 40 and 80 respectively. If $x$g of A contains $y$ atoms, how many atoms are present in $2x$g of B

Number of moles of $\text{A}=\frac{x}{40}$

Number of atoms of A =

$\frac{x}{40}\times \text{Avogadro}\text{number}=y\left(\text{say}\right)$

Number of moles of $\text{B}=\frac{2x}{80}$

Number of moles of B =

$\frac{2x}{80}\times \text{Avogadro}\text{number}$

= $\frac{2}{80}\times \frac{\text{40}y}{\text{Avogadro}\text{number}}\times \text{Avogadro}\text{number}=y$