# You need 660 mL of a 80% alcohol solution. On hand, you have a 40% alcohol mixture. How much of the

###### Question:

You will need

mL of the 40% solution

and

mL of pure alcohol.

Submit Question

## Answers

Let a be the amount of 40% solution and b be the amount of pure alcohol that you will use.

You want to end up with a volume of 660 mL, so

a + b = 660

For each mL of solution used, the 40% solution would contribute 0.4 mL of alcohol, and the pure alcohol will contribute 1 mL of alcohol. In the desired solution, you want a concentration of 80% alcohol, so that it contains 0.8 • 660 mL = 528 mL of alcohol, and

0.4a + b = 528

Solve the system of equations above. Subtracting the second from the first eliminates b and lets us solve for a :

(a + b) - (0.4a + b) = 660 - 528

0.6a = 132

a = 220

Then

220 + b = 660

b = 440

So you will need 220 mL of the 40% solution and 440 mL of pure alcohol.

repeat the question

step-by-step explanation:

overpriced

step-by-step explanation: