Asked by Juan
Determine the concentrations of MgCl2 Mg2+ and Cl- in a solution prepared by dissolving 2.73*10^-4g MgCl2 in 1.00L of water. Express all the concentrations in molarity. Additionally, express the concentrations of the ionic species in ppm.
MgCl2= ____ M
Mg2+ = ____ M = _____ ppm
Cl- = ____ M = _____ ppm
MgCl2= ____ M
Mg2+ = ____ M = _____ ppm
Cl- = ____ M = _____ ppm
Answers
Answered by
DrBob222
mols MgCl2 = grams/molar mass = ? and that in 1.00 L solution is that Molarity.
So MgCl2 is ? M
Mg^2+ is ? M; i.e., the same as MgCl2 since there is 1 Mg^2+/MgCl2.
Cl^- is 2? Mi.e., twice Mg^2+ since there are 2 Cl^- per MgCl2.
To convert to ppm, Convert grams in the problem to mg MgCl2 and remember the definition that 1 ppm = 1 mg/L solution.
So MgCl2 is ? M
Mg^2+ is ? M; i.e., the same as MgCl2 since there is 1 Mg^2+/MgCl2.
Cl^- is 2? Mi.e., twice Mg^2+ since there are 2 Cl^- per MgCl2.
To convert to ppm, Convert grams in the problem to mg MgCl2 and remember the definition that 1 ppm = 1 mg/L solution.
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