Mixture problem

Question

You need a 35% alcohol solution. On hand, you have a 360 mL of a 5% alcohol mixture. You also have 80% alcohol mixture. How much of the 80% mixture will you need to add to obtain the desired solution?

You will need
240 mL of the 80% solution
to obtain
_____mL of the desired 35% solution.

Answer 1

Brooke..

 

Mixture problems, I create a chart...

 

                       %              Amt           Pure           

 

5%                   5%            360           .05(360)=18       (To get pure column, multiply % and Amt)

 

80%                 80%            x              .80x

 

Total                 35%         360+x         .35(360+x)

 

 

Once its here, 1st line plus 2nd line = 3rd line

 

18 +.80x= .35(360 + x)

 

18 +.80x =126 +.35x

 

.45x = 108

 

x= 240

 

so, you have 360ml of 5% solution, 240ml of 80%, so you would finish with 600ml(240+360) of 35% solution