How many Br - ions are there in 2.51 grams of AlBr3

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.