Question

You need a 40% alcohol solution. On hand, you have a 540 mL of a 10% alcohol mixture. How much pure alcohol will you need to add to obtain the desired solution?

Answer 1

270 mL of pure alcohol is needed.

Explanation:

In 540 mL of the mixture, you have 540*10% = 54 mL of pure alcohol.

If you add x mL pure alcohol, then you have (540 + x) mL solution, in it there is (54 + x) mL of pure alcohol

so (54 + x)/(540 + x) = 40%

Then

54 + x = 0.4(540 + x)

54 + x = 216 + 0.4 x

0.6x = 216 - 54 = 162

x = 162/0.6 = 270 mL