How many gallons of each of a 60% acid solution and an 80% acid solution must be mixed to produce 50 gallons of a 74% acid solution?
Accepted Solution
A:
Short way: the ratio required for the 60% and 80% to get a 74% solution is (80-74):(74-60) = 6:14 = 3:7 To make 50L, we need 50L*(3/(3+7))=15L of 60% 50L*(7/(3+7))=35L of 80%
Longer way Let x=volume of 60% solution (L), then 50-x=volume of 80% solution. We need 50L of the mixture so using CV=C1V1+C2V2, 50(0.74)=0.6x+0.8(50-x) Solve for x x=(0.8*50-50*0.74)/(0.8-0.6) =3/0.2 =15L (of 60% solution) 50-x=50-15=35L (of 80% solution)
Still longer method: Let x=volume of 60% solution y=volume of 80% solution using total volume of mixture x+y=50.....................................(1) then using CV=C1V1+C2V2 0.6x+0.8y=50(0.74) => 0.6x+0.8y=37 ....................(2) Solve system of equations (1) & (2) (1)-1.25(2) x+y-0.75x-y = 50-37*1.25 = 0.25x=3.75 x=3.75/0.25=15L x+y=50 => y=50-x = 50-15=35L So 15L of 60% and 35L of 80% as before.