What is the molar concentration of the br⁻ ions in 0.17 m cabr2(aq)?

CaBr2(aq) is an ionic compound which will have the releasing of 2 Br⁻ ions ions in water for every molecule of CaBr2 that dissolves.
CaBr2(s) --> Ca+(aq) + 2 Br⁻(aq)
            [Br⁻] = 0.17 mol CaBr2/1L × 2 mol Br⁻ / 1 mol CaBr2 = 0.34 M
The answer to this question is [Br⁻] = 0.34 M