Question

A chemist makes 980.mL of aluminum chloride (AICI₃) working solution by adding distilled water to 280 mL of a 2.29 M stock solution of aluminum chloride in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.

Answer 1

The concentration of the chemist's working solution of aluminum chloride (AlCl₃) is 0.652 M when a 2.29 M stock solution is diluted with distilled water to a final volume of 980 mL.

Step-by-step explanation:

To calculate the concentration of the chemist's working solution of aluminum chloride (AlCl₃), we use the principle of dilution, which states that the amount of solute remains constant when a solution is diluted. The formula for this is M1V1 = M2V2, where M1 is the molarity of the initial solution, V1 is the volume of the initial solution, M2 is the molarity of the final solution, and V2 is the total volume of the final solution.

Starting with a 2.29 M stock solution and diluting it to a final volume of 980 mL (0.980 L), we have:

M1 = 2.29 M (stock concentration)
V1 = 280 mL (0.280 L)
V2 = 980 mL (0.980 L)

Using the dilution formula:
2.29 M × 0.280 L = M2 × 0.980 L

M2 = (2.29 × 0.280) ÷ 0.980
M2 = 0.652 M (rounded to three significant digits)

Therefore, the concentration of the working solution is 0.652 M.