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tino4ka555 [31]
3 years ago
10

Calculate the molar concentration of the Cl⁻ ions in 0.92 M MnCl2(aq).

Chemistry
1 answer:
scZoUnD [109]3 years ago
4 0
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⁻] = 0.92 mol MnCl2/1L × 2 mol Cl⁻ / 1 mol MnCl2 = 1.8 M
The answer to this question is [Cl⁻] = 1.8 M
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