Answer is: there are 1.7·10²² Br⁻ ions.
m(AlBr₃) = 2.51 g; mass of aluminium bromide.
n(AlBr₃) = m(AlBr₃) ÷ M(AlBr₃).
n(AlBr₃) = 2.51 g ÷ 266.7 g g/mol.
n(AlBr₃) = 0.0094 mol; amount of substance.
In one mol of aluminium bromide there are three moles of bromine anions:
n(Br⁻) = 3 · 0.0094 mol.
n(Br⁻) = 0.028 mol.
N(Br⁻) = n(Br⁻) · Na(Avogadro constant).
N(Br⁻) = 0.028 mol · 6.022·10²³ 1/mol.
N(Br⁻) = 1.7·10²².
