Final answer:
The concentration of the chemist's working solution of AlCl₃ after dilution is 0.726 M. This was calculated using the dilution equation M1V1 = M2V2, ensuring the final answer has the correct number of significant digits.
Step-by-step explanation:
The student is asking how to calculate the concentration of a chemist's working solution of aluminum chloride (AlCl₃) after dilution. The stock solution has a concentration of 2.94M, and the chemist added distilled water to a volume of 200. mL of this stock solution to make up a final volume of 810 mL.
To find the concentration of the working solution, we can use the dilution equation M1V1 = M2V2, where M1 and V1 are the molarity and volume of the stock solution, and M2 and V2 are the molarity and volume of the working solution. The number of moles of AlCl₃ in the concentrated stock is equal to the number of moles in the diluted working solution since the number of moles does not change upon dilution.
Using the given values:
- M1 = 2.94 M (stock concentration)
- V1 = 0.200 L (volume of stock used)
- V2 = 0.810 L (final volume)
We solve for M2:
M2 = (2.94 M x 0.200 L) / 0.810 L = 0.726 M
Thus, the concentration of the working solution is 0.726 M aluminum chloride, rounded to three significant figures to reflect the precision of the given data.