How many grams of a solid mixture containing strontium chloride would one need to make 558 mL of a 0.100 M strontium chloride solution,if the solid mixture contains 58.6% strontium chloride by weight?
A)6.62 g
B)15.1 g
C)8.85 g
D)5.19 g
E)9.52 g

Question

Quizplus · Accepted Answer

To solve this problem, we first need to calculate the amount of strontium chloride (SrCl2) required to make a 0.100 M solution in 558 mL (or 0.558 L) of solution. The molar mass of SrCl2 is approximately 158.52 g/mol.The formula to calculate the mass of solute needed is:$$ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} $$Rearranging to solve for moles of solute gives:$$ \text{moles of solute} = \text{Molarity (M)} \times \text{volume of solution in liters} $$$$ \text{moles of SrCl2} = 0.100 \, \text{M} \times 0.558 \, \text{L} = 0.0558 \, \text{mol} $$Next, we convert moles of SrCl2 to grams:$$ \text{grams of SrCl2} = 0.0558 \, \text{mol} \times 158.52 \, \text{g/mol} = 8.85 \, \text{g} $$Since the mixture is only 58.6% SrCl2 by weight, we need to find out how much of the mixture is needed to get 8.85 g of SrCl2:$$ \text{mass of mixture} = \frac{\text{mass of SrCl2 needed}}{\text{percentage of SrCl2 in the mixture}} = \frac{8.85 \, \text{g}}{0.586} = 15.1 \, \text{g} $$Therefore, 15.1 g of the mixture is needed to prepare the solution.

How Many Grams of a Solid Mixture Containing Strontium Chloride