AlBr3(aq) is an ionic compound which will have the releasing of 3 Br⁻ ions ions in water for every molecule of AlBr3 that dissolves.
AlBr3(s) --> Al+(aq) + 3 Br⁻(aq)
[Br⁻] = 0.16 mol AlBr3/1L × 3 mol Br⁻ / 1 mol AlBr3 = 0.48 M
The answer to this question is [Br⁻] = 0.48 M
