Question

a chemistry lab assistant wants to prepare 10 liters of 48% acid solution. in stock he has 80% and 40% acid solutions. how many liters of each in stock solution should he mix to prepare a 10-liter mixture?

Answer 1

L = no. of liters of 80% solution
.80L = value of 80% solution
10 - L = no. of liters of 40% solution
.40(10 - L) = value of 40% solution
.48(10) = value of mixture

.80L + .40(10 - L) = .48(10)
Solve for L

L = no. of liters of 80% solution
10 - L = no. of liters of 40% solution