30% alcohol solution. 330 mL 90% alcohol mixture

Question

You need a 30% alcohol solution. you have a 330 mL of a 90% alcohol mixture. How much pure water will you need to add to obtain the desired solution?

Answer 1

Hello Clarissa,

We need a 30% alcohol solution.

We have a 330 mL, 90% alcohol mixture. This mixture is expressed as:

.90(330mL).

Let X = the amount of pure water needed to add to the 330mL,

alcohol mixture to get the desired 30% alcohol solution.

This change in the mixture is expressed as: .30(330mL + X).

We can now express the two solutions in an equation as follows:

.90(330mL) = .30(330mL + X)

Next, we solve the equation for X.

297 mL = 99mL + .30X

198mL = .30X

660 mL = X