SOLUTION: Solve Problems Using Linear Systems
11. What volume, in mililitres, of a 60% hydrochloric acid solution must be added to 100 mL of a 30% hydrochloric acid solution to make a 36% h
Algebra ->
Quadratic Equations and Parabolas
-> SOLUTION: Solve Problems Using Linear Systems
11. What volume, in mililitres, of a 60% hydrochloric acid solution must be added to 100 mL of a 30% hydrochloric acid solution to make a 36% h
Log On
Question 185643: Solve Problems Using Linear Systems
11. What volume, in mililitres, of a 60% hydrochloric acid solution must be added to 100 mL of a 30% hydrochloric acid solution to make a 36% hydrochloric acid solution?
Can you help me please? Thanks Found 2 solutions by Earlsdon, checkley77:Answer by Earlsdon(6294) (Show Source):
You can put this solution on YOUR website! Let x = the number of ml. of 60% hydrochloric acid required.
The amount of hydrochloric acid in x ml. of 60% acid solution can be expressed as 0.6x ml, and this is to be added to 100 ml. of 30% hydrochloric acid solution which, of course, contains 0.3(100) ml of hydrochloric acid.
The sum of these is to equal (100+x) ml. of 36% hydrochloric acid solution which contains 0.36(100-x) ml of hydrochloric acid.
Now you can set up the necessary equation to solve for x.
0.6x+0.3(100) = 0.36(100+x)
0.6x+30 = 36+0.36x Subtract 30 from both sides.
0.6x = 6+0.36x Now subtract 0.36x from both sides.
0.24x = 6 Finally, divide both sides by 0.24
x = 25
25 ml. of 60% hydrochloric acid solution need to be add to 100 ml. of 30% hydrochloric acid solution to obtain 125 ml. of 36% hydrochloric acid solution.
You can put this solution on YOUR website! .6X+.3*100=.36(100+X)
.6X+30=36+.36X
.6X-.36X=36-30
.24X=6
X=6/.24
X=25 ML OF 60% ACID IS NEEDED.
PROOF:
.6*25+.3*100=.36(100+25)
15+30=.36*125
45=45
(