What amount of a 80% acid solution must be mixed with a 55% solution to produce 400 mL of a 60% solution?

Answer 1

.8x + (400-x).55 = 400(.6)

.25x = 400(.6)- 400(.55)

x = 400(.05/.25 = 80 mL of 80%

400-80 = 320 ml of 55%

60-55 =5

80-60=20

20/5 = 4 = 320/80

you need 4 times as much of 55 as 80%