Chemist needs to mix 20% acid solution with a 70% acid solution to obtain the 8 liters 50% many liters do you

Question

A chemist needs to mix a 20% acid solution with a 70% acid solution to obtain leave you a message asking how many liters each of the solution must be

Answer 1

We are mixing two acids.

 

x = liters of 20% acid solution

y = liters of 70% acid solution

 

x + y = 8    This represents the total number of liters

and

.2x + .7y = .5(8)   This represents the concentration of the solutions

 

Since x+y=8, y = 8-x

 

We can use substitution.

.2x + .7(8-x) = 4

.2x + 5.6 - .7x = 4

Combining like terms

-0.5x = - 1.6

Dividing by -0.5

x = 3.2

 

There are 3.2 liters of the 20% solution to be mixed with 4.8 liters of the 70% solution.

Answer 2

X is the amount of 20% acid solution

Y is the amount of 70% acid solution

 

X + Y = 8 liters ---> Y = 8 - x

 

0.2X + 0.7Y = 0.5*8 = 4

 

0.2*X + 0.7*(8-x) = 4

 

0.2X + 5.6 - 0.7X = 4

 

-0.5X = -1.6

 

5x = 16

 

x=16/5

 

y = 8 - x = 8 - 16/5 = 40/5 - 16/5 = 24/5

 

As a check, 

The amount of acid found in these two amounts are then

 

16/5 * 20%  + 24/5 * 70% = 16/5 * 20/100 + 24/5*70/100

 

                                          = 320/500 + 1680/500

 

                                          = 2000/500 

 

                                          = 4

 

                                           = 50% * 8

 

 

 

  The answers are checked, verified, and proven to be correct.                                                                                 

 

x=16/5 = 3 and 1/5  = 3.2 liters of the 20% acid solution

 

AND

y = 24/5 = 4 and 4/5 = 4.8 liters of the 70% acid solution