WORd problems

Question

You need a 25% alcohol solution. On hand, you have a 300 mL of a 80% alcohol mixture. How much pure water will you need to add to obtain the desired solution?

 

You will need ?  mL of pure water

 

to obtain ? mL of the desired 25% solution.

Answer 1

(300 ml)(80%) = (x ml)(25%)

x = 960 mls final volume

You will need 960 - 300 = 660 ml pure water

to obtain 960 mls of the desired 25% solution.

 

Looking at it another way.  To go from 80% down to 25%, you must dilute it by a factor of 80/25 = 3.2

3.2 x 300 ml = 960 ml which will be the final volume.  Subtracting the 300 mls that you begin with, leaves 660 ml water to be added.

 

And a third way:

Use Allegation

HAVE......WANT......NEED....

80%........................25.....

................25%................

0%..........................55....

So, you need 25 parts of 80% and 55 parts of 0% (water). Since there are 300 ml of 80%, this becomes the 25 parts. And if 25 parts = 300 mls, then each part = 12 mls.  12 mls/part x 55 parts = 660 mls of water

Answer 2

300 * 80% + X * 0% = (300+x)*25%

 

0.8 * 300 = .25(X+300)

 

240 = .25X + 75

 

165 = 0.25X

 

16500 = 25X

 

X=660

 

So if you add 660 mL of water,

 you will get 240 ml of alchohol out of 960 Ml

 of solution which is 25%