How many liters of 70% alcohol solution must be mixed to obtain 15 liters of 50% alcohol solution? ___ liters of 70% solution ___ liters of 20% solution

Question

How many liters of 70% alcohol solution must be mixed to obtain 15 liters of 50% alcohol solution?

___ liters of 70% solution

___ liters of 20% solution

Answer 1

Sorry about my comment, I misread the problem. Here is the answer.

 

the formula is C1V1 + C2V2 = C3V3 where C = concentration and V = volume

 

to simplify the process, we will change the % figures to decimals; 20% = 0.20, 50% = 0.50, 70% = 0.70

 

let x = volume of 70% solution, then 15L - x = volume of 20% solution

 

our equation becomes 0.7x + 0.2(15L - x) = 15L•0.5

 

expanding, we get 0.7x + 3L -0.2x = 7.5L

 

combining, we get 0.5x = 4.5L

 

solving, x = 9L; then 15L - x = 6L

 

Proof: 0.7(9L) + 0.2(6L) = 0.5(15L); 6.3L +1.2L = 7.5L