We calculate the molar concentration [Cl⁻] using stoichiometry. 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⁻] = 0.68 mol MnCl2/1L × 2 mol Cl⁻ / 1 mol MnCl2 = 1.4 M
The answer to this question is [Cl⁻] = 1.4 M
