You need 560 mL of a 65% alcohol solution. On hand, you have a 80% alcohol mixture. How much of the 80% alcohol mixture and pure water will you need to obtain the desired solution?

Answer 1

Call w the amount of water in mL and m the amount of mixture in mL

We need 560mL in total so therefore:

w + m = 560

The second equation we get by realizing that:

Alcohol in mixture + Alcohol in water = Alcohol in mixture combined with water

The amount of alcohol in the mixture is 80% of the mixture. Therefore the amount of alcohol in the mixture can be written as 0.80m

The amount of alcohol in the water is 0%. Therefore, the amount of alcohol in the water can be written as 0w.

The amount of alcohol in total after we mix the mixture with the water is 65% of the combined water and mixture. Thus, 0.65*560.

Our second equation is:

0.8m + 0w = 0.65*560

0.8m = 0.65*560

m = 455

And m + w = 560. So, w = 105