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16. A 2.3 L sample of a 4.9 M solution is diluted to 52 L. What is the molarity of the diluted solution?

17. To what volume should you dilute 65.4 mL of a 12.0 M acetic acid solution to obtain a 1.55 M acetic acid solution?

Answer :

Let's tackle these two problems step-by-step:

### Problem 16: Determine the molarity of the diluted solution.

Given:
- Initial volume ([tex]\(V_1\)[/tex]) = 2.3 L
- Initial molarity ([tex]\(M_1\)[/tex]) = 4.9 M
- Final volume ([tex]\(V_2\)[/tex]) = 52 L

Solution:

To find the final molarity ([tex]\(M_2\)[/tex]) of the diluted solution, we use the dilution formula:

[tex]\[ M_1 \times V_1 = M_2 \times V_2 \][/tex]

Rearrange the formula to solve for [tex]\(M_2\)[/tex]:

[tex]\[ M_2 = \frac{M_1 \times V_1}{V_2} \][/tex]

Substitute the known values:

[tex]\[ M_2 = \frac{4.9 \text{ M} \times 2.3 \text{ L}}{52 \text{ L}} \][/tex]

Calculating this gives:

[tex]\[ M_2 \approx 0.217 \text{ M} \][/tex]

So, the molarity of the diluted solution is approximately 0.217 M.

### Problem 17: Determine the volume for dilution.

Given:
- Initial volume ([tex]\(V_1\)[/tex]) = 65.4 mL
- Initial molarity ([tex]\(M_1\)[/tex]) = 12.0 M
- Final molarity ([tex]\(M_2\)[/tex]) = 1.55 M

Solution:

We will use the same dilution formula:

[tex]\[ M_1 \times V_1 = M_2 \times V_2 \][/tex]

Rearrange to solve for [tex]\(V_2\)[/tex] (the final volume):

[tex]\[ V_2 = \frac{M_1 \times V_1}{M_2} \][/tex]

Substitute the known values:

[tex]\[ V_2 = \frac{12.0 \text{ M} \times 65.4 \text{ mL}}{1.55 \text{ M}} \][/tex]

Calculating this gives:

[tex]\[ V_2 \approx 506.32 \text{ mL} \][/tex]

So, you should dilute the solution to approximately 506.32 mL to achieve a 1.55 M acetic acid solution.

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