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Solution:

Here, the concentration (strength) units are the same, as they should be. However, you'll need to change ounces to the metric equivalent. One ounce equals 30 mL; therefore, 8 oz. equals 240 mL.

[tex]
\[
\begin{align*}
x \times 70\% &= 240 \, \text{mL} \times 40\% \\
70x &= 9600 \\
x &= 137 \, \text{mL}
\end{align*}
\]
[/tex]

137 mL of the 70% isopropyl alcohol stock is needed to make 8 oz. (240 mL) of 40% isopropyl alcohol. You would add enough water to make up to 240 mL (that means about 103 mL of water will be added).

Answer :

To solve this problem, we need to find out how much of a 70% isopropyl alcohol solution is necessary to make 240 mL (or 8 ounces) of a 40% isopropyl alcohol solution. Let's go step-by-step:

1. Understand the goal: We're trying to create a diluted solution with a 40% concentration using a stronger solution that has a 70% concentration.

2. Set up the equation based on concentrations:
- We know that the mass of the pure alcohol in the initial 70% solution is equal to the mass of the pure alcohol in the final 40% solution.
- This can be expressed with the equation for the concentration of a solution:
[tex]\[
\text{(Volume of 70% solution)} \times 70\% = 240 \, \text{mL} \times 40\%
\][/tex]
- Let [tex]\( x \)[/tex] be the volume in mL of the 70% solution needed.

3. Rewrite the equation:
[tex]\[
x \times 0.70 = 240 \times 0.40
\][/tex]

4. Solve for [tex]\( x \)[/tex]:
- Calculate the right-hand side:
[tex]\[
240 \times 0.40 = 96
\][/tex]
- Now use this in the equation:
[tex]\[
0.70x = 96
\][/tex]
- Solve for [tex]\( x \)[/tex] by dividing both sides by 0.70:
[tex]\[
x = \frac{96}{0.70} \approx 137.14 \, \text{mL}
\][/tex]

5. Calculate the additional water needed:
- Since the final solution must total 240 mL, find how much more liquid (as water) we need to add.
- If we're using about 137.14 mL of the 70% solution, then:
[tex]\[
\text{Water added} = 240 - 137.14 \approx 102.86 \, \text{mL}
\][/tex]

Therefore, approximately 137.14 mL of the 70% isopropyl alcohol is needed, and about 102.86 mL of water should be added to achieve the final desired solution of 240 mL at 40% concentration.

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