Answer
a) $A(M) \approx 27\frac{g}{mol}$
b) Aluminum, $Al$
Work Step by Step
Let's denote atomic weight of metal M by $x$.
$\omega (M) =0.529= \frac{2x}{M(M_{2}O_{3})}=\frac{2x}{2x+48}$
Solving this equation for $x$ yields $x = 26.95 \frac{g}{mol}$, which corresponds to the atomic weight of aluminum, $Al$.
