SOLUTION: For a certain experiment, a student requires 300 ml of a solution that is 13% HCl (hydrochloric acid). The storeroom has only solutions that are 10% HCl and 15% HCl. How many milli
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Question 124726: For a certain experiment, a student requires 300 ml of a solution that is 13% HCl (hydrochloric acid). The storeroom has only solutions that are 10% HCl and 15% HCl. How many milliliters of each available solution should be mixed to get 300 ml of 13% HCl?
Could you please explain to me how to solve this problem? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! For a certain experiment, a student requires 300 ml of a solution that is 13% HCl (hydrochloric acid). The storeroom has only solutions that are 10% HCl and 15% HCl. How many milliliters of each available solution should be mixed to get 300 ml of 13% HCl?
:
Let x = amt of 15% HCL
the total resulting amt must be 300 ml, therefore:
(300-x) = amt of 10% HCL
:
.15x + .10(300-x) =.13(300)
:
.15x + 30 - .10x = 39
:
.15x - .10x = 39 - 30
:
.05x = 9
x =
x = 180 ml of 15% HCL
then
300 - 180 = 120 ml of 10% HCL
:
:
We can check our solution in the original equation
.15(180) + .10(120) = .13(300)
27 + 12 = 39
:
Did this make sense to you? Any questions?
