Calculate the molar concentration of the cl⁻ ions in 0.65 m cacl2(aq), assuming that the dissolved substance dissociates completely in solution.

The question here is solved using basic chemistry. CaCl2(aq) is an ionic compound which will have the releasing of 2 Cl⁻ ions ions in water for every molecule of CaCl2 that dissolves.
CaCl2(s) --> Ca+(aq) + 2 Cl⁻(aq)
            [Cl⁻] = 0.65 mol CaCl2/1L × 2 mol Cl⁻ / 1 mol CaCl2 = 1.3 M
The answer to this question is [Cl⁻] = 1.3 M

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