Question

You need 910 mL of a 5% alcohol solution. On hand, you have a 65% alcohol mixture. How much of the 65% alcohol mixture and pure water will you need to obtain the desired solution?

Answer 1

Step 1: Calculate the total volume of the desired 5% alcohol solution.

910 mL = 0.0910 L

Step 2: Calculate the amount of 5% alcohol solution needed.

(0.0910 L) * (0.05) = 0.00455 L

Step 3: Calculate the amount of pure water needed.

(0.0910 L) - (0.00455 L) = 0.08645 L

Step 4: Calculate the amount of 65% alcohol mixture required.

(0.00455 L) / (0.65) = 0.007 L

Step 5: Calculate the amount of pure water in the 65% alcohol mixture.

(0.007 L) * (1 - 0.65) = 0.00245 L

Step 6: Calculate the total amount of pure water required.

(0.08645 L) + (0.00245 L) = 0.0889 L

To obtain the desired 910 mL of 5% alcohol solution, you will need 0.007 L of 65% alcohol mixture and 0.0889 L of pure water.