https://file.aiquickdraw.com/mathbot/file/1727795751189wct6gmxe.jpg

The problem you're dealing with involves the dilution formula, which is:

\[
C_1 V_1 = C_2 V_2
\]

Where:
- \(C_1\) is the concentration of the stock solution (in mol/L),
- \(V_1\) is the volume of the stock solution (which we need to find),
- \(C_2\) is the concentration of the desired solution (in mol/L),
- \(V_2\) is the volume of the desired solution (in mL).

From the image:
- \(C_1 = 0.447 \, \text{mol/L}\),
- \(C_2 = 310 \, \text{mM} = 0.310 \, \text{mol/L}\) (because 1 mM = 0.001 mol/L),
- \(V_2 = 875 \, \text{mL}\).

We need to find \(V_1\), the volume of stock solution in mL. First, convert everything to consistent units, then solve for \(V_1\):

\[
V_1 = \frac{C_2 \times V_2}{C_1} = \frac{0.310 \times 875}{0.447} \, \text{mL}
\]

Let me calculate that for you.The volume of the aluminum chloride stock solution that the chemist should pour out is approximately \( 606.8 \, \text{mL} \), with the correct number of significant digits.

Would you like further details or have any questions about the process? 

Here are five related questions for further exploration:
1. How do you determine the correct number of significant digits in chemistry problems?
2. Why is it necessary to dilute solutions in a laboratory setting?
3. How would the calculation change if you needed a more concentrated solution?
4. What is the relationship between molarity and volume in a dilution problem?
5. Can the dilution formula be used for gases or just for solutions?

**Tip**: Always check the units in each part of your calculation to ensure consistency throughout the problem.

A chemist must prepare 875 mL of 310 mM aqueous aluminum chloride (AlCl₃) working solution. She'll do this by pouring out some 0.447 mol/L aqueous aluminum chloride stock solution into a graduated cylinder and diluting it with distilled water. Calculate the volume in mL of the aluminum chloride stock solution that the chemist should pour out. Be sure your answer has the correct number of significant digits.

Learn how to calculate the volume of stock solution needed to prepare 875 mL of a 310 mM aluminum chloride solution from a 0.447 mol/L stock. This problem involves dilution and molarity concepts, ideal for high school chemistry students.

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