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.

i do not get it   at all