Question

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You need a 50% alcohol solution. On hand, you have a 240 mL of a 10% alcohol mixture. You also have 70% alcohol mixture. How much of the 70% mixture will you need to add to obtain the desired solution?

You will need

------------mL of the 70% solution

Answer 1

Let x be the amount of 70% alcohol mixture needed in mL.

To obtain a 50% alcohol solution, we need to find the amount of alcohol in the resulting mixture and set it equal to 50% of the total volume of the resulting mixture.

The amount of alcohol in the 240 mL of 10% alcohol mixture is:

0.1 × 240 mL = 24 mL

When we add x mL of the 70% alcohol mixture, the resulting total volume will be:

240 mL + x mL

The amount of alcohol in x mL of the 70% alcohol mixture is:

0.7x mL

The total amount of alcohol in the resulting mixture is:

24 mL + 0.7x mL

Since we want a 50% alcohol solution, we set up the equation:

0.5(240 mL + x mL) = 24 mL + 0.7x mL

Simplifying and solving for x, we get:

120 mL + 0.5x mL = 24 mL + 0.7x mL

96 mL = 0.2x mL

x = 480 mL

Therefore, you need to add 480 mL of the 70% alcohol mixture to obtain a 50% alcohol solution.