Question

Two elements $A$ and $B$ which form $0.15$ moles of $A_2B$ and $AB_3$ type compounds. If both $A_2B$ and $AB_3$ weigh equally, then the atomic weight of $A$ is _____ times of atomic weight of $B$.

Answer 1

Moles of $A_2B$ = moles of $AB_3$
$\frac {W}{2+B}=\frac {W}{A+3B}$
$A + 3B = 2A + B$
$3B-2B = 2A + A$
$2B = A$
⇒ Atomic weight of $A$ is $2$ times that of $B$.

So, the answer is $2$.