How many milliliters of a 2.5% bleach solution must be mixed with a 30% bleach solution to produce  300ML of an 8% bleach solution.

Let \( x \) be the amount of the 2.5% bleach solution (in milliliters), and let \( 300 - x \) be the amount of the 30% bleach solution (since the total amount of solution must be 300 mL).

We need to set up an equation based on the concentration of bleach in the final mixture. The total amount of bleach in the solution is the concentration times the volume. Therefore, the amount of bleach in each solution is:

- The amount of bleach in the 2.5% solution: \( 0.025x \)
- The amount of bleach in the 30% solution: \( 0.30(300 - x) \)
- The total amount of bleach in the final 8% solution: \( 0.08(300) \)

Now, we set up the equation:

\[
0.025x + 0.30(300 - x) = 0.08(300)
\]

### Solving for \( x \):

1. Expand both sides:
\[
0.025x + 90 - 0.30x = 24
\]

2. Combine like terms:
\[
-0.275x + 90 = 24
\]

3. Subtract 90 from both sides:
\[
-0.275x = 24 - 90
\]
\[
-0.275x = -66
\]

4. Solve for \( x \):
\[
x = \frac{-66}{-0.275}
\]
\[
x = 240 \, \text{mL}
\]

Thus, \( x = 240 \, \text{mL} \) of the 2.5% bleach solution is needed, and \( 300 - 240 = 60 \, \text{mL} \) of the 30% bleach solution.

### Conclusion:
You need to mix 240 mL of the 2.5% bleach solution with 60 mL of the 30% bleach solution to produce 300 mL of an 8% bleach solution.

Would you like a breakdown of each step or more details?

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Here are 5 related questions:

1. What is the formula for calculating the amount of solute in a solution?
2. How would the solution change if the target concentration was 10% instead of 8%?
3. What if the total amount of solution needed was 500 mL instead of 300 mL?
4. How would you adjust the equation if the 30% bleach solution concentration was different?
5. Can this type of calculation be used in diluting other chemical solutions?

**Tip:** When mixing solutions, it's always good to check that the concentrations and volumes add up to the target mixture!

How many milliliters of a 2.5% bleach solution must be mixed with a 30% bleach solution to produce 300ML of an 8% bleach solution.

Find out how to mix 240 mL of a 2.5% bleach solution with 60 mL of a 30% bleach solution to create 300 mL of an 8% bleach solution. This algebraic solution involves using the concentration equation, making it ideal for students in Grades 9-12.

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