let the amount of the 20% solution used be x ml
then the amount of the 50% solution used is 300-x m.
solve for x ....
.2x + .5(300-x) = .3(300)
times 10
2x + 5(300-x) = 3(300)
take it from there
A chemist has a solution that is 20% HIC and a solution that is 50% HCI. How much of each solution should be mixed to attain 300ml of a solution that is 30% HCI?
2 answers
It may help to set up a table for these kinds of problems:
............Amount.....%....HCl
Solution 1....x......0.2....0.2x
Solution 2...300-x...0.5....0.5(300-x)
Mixture......300.....0.3....0.3(300)
Equation:
0.2x + 0.5(300-x) = 0.3(300)
Solve for x. (Be sure to list the amount of both solutions!)
I hope this helps.
............Amount.....%....HCl
Solution 1....x......0.2....0.2x
Solution 2...300-x...0.5....0.5(300-x)
Mixture......300.....0.3....0.3(300)
Equation:
0.2x + 0.5(300-x) = 0.3(300)
Solve for x. (Be sure to list the amount of both solutions!)
I hope this helps.