Question

You need 675 mL of a 90% alcohol solution. On hand, you have a 25% alcohol mixture. How much of the 25% alcohol mixture and pure alcohol will you need to obtain the desired solution?

Answer 1

90 ml of the 25 percent mixture and 585 of pure alcohol

Explanation:

Firstly, you should find the quantity of alcohol in the desired mixture.

675:100*90= 675*0.9= 607.5

Firstly, define all the 25 percents mixure as x, the pure alcohol weight is y.

1. x+y= 675 (because the first and the second liquid form a desired liquid).

Then find the equation for spirit

The first mixture contains 25 percents. It is x/100*25= 0.25x

When the second one consists of pure alcohol, it contains 100 percents of spirit, so it is x.

2. 0.25x+y=607.5

Then you have a system of equations ( 1.x+y= 675 and 2. 0.25x+y= 607.5)

try 2-1 to get rid of y

x+y- (0.25x+y)= 675-607.5

0.75x= 67.5

x= 90

y= 675-x= 675-90= 585

It means that you need90 ml of the 25percents mixture and 585 0f pure alcohol