A: A student must make a five-fold dilution of a 2.15 M HCl solution. If the student wishes to make 75.0 mL of the diluted solution, what volume of the 2.15 M solution will he need to dilute?

B: What is the concentration of the diluted substance?

Isn't a five fold dilution just making the solution more dilute by a factor of 5?

So 2.15M/5 = 0.43M
75 mL x 0.43M = mL x 2.15 M
Solve for mL of the 2.15M solution to use.

A: To calculate the volume of the 2.15 M HCl solution needed to make a five-fold dilution, you can use the formula for preparing dilutions:

(C1)(V1) = (C2)(V2)

where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume. In this case, the initial concentration (C1) is 2.15 M, and the final concentration (C2) is the desired concentration after dilution. Since it is a five-fold dilution, the final concentration will be 2.15 M divided by 5.

Let's assume the desired volume after dilution (V2) is 75.0 mL. We can rearrange the formula to solve for V1:

V1 = (C2)(V2) / C1

Substituting the values into the formula:
V1 = (2.15 M / 5)(75.0 mL) / 2.15 M

Now we can calculate the volume of the 2.15 M solution needed for dilution:

V1 = (0.43 M)(75.0 mL) / 2.15 M
V1 = 15.0 mL

Therefore, the student will need to dilute 15.0 mL of the 2.15 M HCl solution to make a five-fold dilution with a final volume of 75.0 mL.

B: The concentration of the diluted substance can be calculated using the formula:

C2 = (C1)(V1) / V2

where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume. In this case, the initial concentration (C1) is 2.15 M, the initial volume (V1) is 15.0 mL, and the final volume (V2) is 75.0 mL.

Substituting the values into the formula:
C2 = (2.15 M)(15.0 mL) / 75.0 mL

Now we can calculate the concentration of the diluted substance:

C2 = 0.43 M

Therefore, the concentration of the diluted substance is 0.43 M.