how much of each mixture will you need to obtain the desired solution

Question

You need 1125 mL of a 45% alcohol solution. On hand, you have a 5% alcohol mixture and a 80% alcohol mixture. How much of each mixture will you need to obtain the desired solution? 

 

You will need: 

____ mL of the 5% solution 

and 

____mL of the 80% solution 

Answer 1

There is a very useful procedure referred to as "alligation" that can be used to solve these kinds of problems.  Look it up and see how it's done.  My brief explanation is as follows:

You write down what you HAVE on the left, and what you WANT in the middle.  Then subtract cross-wise to find what you USE.  So, for the problem given...

 

HAVE.............WANT...........USE

5%................................... 35..

......................45%...............

80%..................................40...

 

So, you need 35 "parts" of the 5% and 40 "parts" of the 80%

Since the total volume = 1125 ml and total "parts" = 75 parts, then you get

1125/75 x 35 = 525 mls of 5%

1125/75 x 40 = 600 mls of 80%

 

I personally find this easier than applying algebra to the problem.

 

Answer 2

If we use x ml of 5% solution

we use 1125-x ml of 80% alcohol solution

 

.05x + .8(1125-x) = .45(1125)

.05x + 900 - .8x = 506.25

-.75x = -393.75

x = -393.75/(-.75)

x = 525 ml of 5% alcohol solution

1125-x = 600 ml of 80% alcohol solution