Thank you for reporting, we will resolve it shortly
Q.
The percentage of an element $M$ is 53 in its oxide of molecular formula $M_{2} O_{3}$. Its atomic mass is about
Solution:
If $m$ is the atomic mass of the element $M$, then $2\, m +48\, g$ of $M _{2} O _{3}$ contain $O=48 \,g$ $\therefore \%$ of $O$ in $M_{2} O_{3}=\frac{48}{2 m+48} \times 100=53$
$2 \,m+48=\frac{4800}{53}=90.56$
$2\, m=42.56$
$m=26.58=27$