Posted by Susannah on Tuesday, February 21, 2012 at 3:54pm.
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
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.
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