What is the molar concentration of the cl⁻ ions in 0.54 m mncl2(aq)?

MnCl2(aq) is an ionic compound which will have the releasing of 2 Cl⁻ ions ions in water for every molecule of MnCl2 that dissolves.
MnCl2(s) --> Mn+(aq) + 2 Cl⁻(aq)
            [Cl⁻] = 0.54 mol MnCl2/1L × 2 mol Cl⁻ / 1 mol MnCl2 = 1.1 M
The answer to this question is [Cl⁻] = 1.1 M

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