mols MgCl2 = grams/molar mass = ?
M MgCl2 = mols MgCl2/L = ?
Technically, there isn't any MgCl2 since all of it is ionized completely; however, I think the spirit of the question is to give the concentration of the salt, MgCl2.
You will have 2.39E-4 g Mg ^2+ there and that is 0.239 mg Mg^2+ in 2.5 L. How much is that in mg/L? That's 0.239 mg/2.5 L = approx (that's estimated) 0.096 mg/L. Since 1 mg/L = 1 ppm that is 0approx 0.096 ppm for Mg^+ and twice that for Cl^-.
Determine the concentrations of MgCl2, Mg2+, and Cl- in a solution prepared by dissolving 2.39 x 10^-4g
MgCl2 in 2.50L of water. Express all three concentrations in molarity. Also display the concentrations of ionic species in part per million (ppm).
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