When 50.0 mL of a 0.3000 M AgNO₃ solution is added to 50.0 mL of a solution of MgCl₂, an AgCl precipitate forms immediately. The precipitate is then filtered from the solution, dried, and weighed. If the recovered AgCl is found to have a mass of 0.1183 g, what was the concentration of magnesium ions in the original MgCl₂ solution A) 0.300 M B) 8.25 * 10^-3 M C) 1.65 * 10^-2 M D) 2.06 * 10^-5 M E) 4.13 * 10^-3 M

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Quizplus · Accepted Answer

The mass of AgCl (143.32 g/mol) formed is 0.1183 g. Using the molar mass to find moles: $0.1183 \, \text{g} \div 143.32 \, \text{g/mol} = 8.25 \times 10^{-4} \, \text{moles}$. Since the reaction is 1:1 for Ag+ and Cl-, the moles of Cl- originally present are the same. The total volume after mixing is 100.0 mL or 0.1 L. Concentration of Cl- (and thus Mg2+ since it's from MgCl2) is $8.25 \times 10^{-4} \, \text{moles} \div 0.1 \, \text{L} = 8.25 \times 10^{-3} \, \text{M}$.

When 500 ML of a 0