We can solve this problem using the concept of mixtures and percentages.

Let $x$ represent the amount (in mL) of the 50% alcohol mixture to be added. We know:

1. The desired solution should be $45\%$ alcohol.
2. The total volume will be $30 + x$ mL.
3. The total alcohol content in the mixture comes from the 35% solution already present and the 50% solution to be added.

Equation:

\text{Alcohol from 35% solution} + \text{Alcohol from 50% solution} = \text{Alcohol in 45% solution}

Substituting values: $0.35(30) + 0.50(x) = 0.45(30 + x)$

Simplify:

$10.5 + 0.50x = 13.5 + 0.45x$

$0.50x - 0.45x = 13.5 - 10.5$

$0.05x = 3$

$x = \frac{3}{0.05} = 60 \, \text{mL}$

Final Answer:

You will need 60 mL of the 50% alcohol solution.

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Would you like a breakdown of similar problems or further explanation?

Related Questions:

1. How would the solution change if the desired percentage were 50% instead of 45%?
2. What is the total volume of the solution after mixing?
3. Can this method be used for mixing solutions with non-alcoholic solutes?
4. What happens if you have more than two mixtures to combine?
5. How would you handle this problem if the units were in liters?

Tip:

Always check the percentages to ensure the final mixture aligns with the desired concentration!