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Q.
The percentage of element M is 53 in its oxide of molecular formula $M_2O_3$. Its atomic mass is about
Some Basic Concepts of Chemistry
Solution:
If $m$ is the atomic mass of the element $M$, then
$2m + 48\, g$ of $M_{2}O_{3}$ contain $O=48\,g$
$\therefore $ $\%$ of $O$ in $M_{2}O_{3}=\frac{48}{2m+48}\times100=53$
(Given)
$\therefore 2m+48=\frac{4800}{53}=90.56$
or $2m=42.56$ or $m=21.28=22$