Question
How much water must be added to an 80% acid solution to make a mixture of 50mL which is 70% acid?
Answers
Reiny
solve for x, the amount of water to be added
0(x) + .8(50-x) = .7(50)
40 - .8x = 35
-.8x = -5
x = 6.25
So they had 43.75 ml of the 80% solution, and by adding 6.25 ml of water they now had 50 ml of a 70% solution.
0(x) + .8(50-x) = .7(50)
40 - .8x = 35
-.8x = -5
x = 6.25
So they had 43.75 ml of the 80% solution, and by adding 6.25 ml of water they now had 50 ml of a 70% solution.