SOLUTION: A laboratory has 60 cubic centimeters (cm^3) of a solution that is 40% HCl acid. How many cubic centimeters of a 15% solution of HCl acid should be mixed with the 60 cm^3 of 40% ac

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Question 1207820: A laboratory has 60 cubic centimeters (cm^3) of a solution that is 40% HCl acid. How many cubic centimeters of a 15% solution of HCl acid should be mixed with the 60 cm^3 of 40% acid to obtain a solution of 25% HCl? How much of the 25% solution is there?
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39616) (Show Source):
You can put this solution on YOUR website!
To ignore denstities and choice between w/w, w/v;

60 volume units of 40 % solution;
Solution available to mix, 15%;
Mix result wanted, 25%.

v, amount of the 15% solution to add and mix:

--------this equation without explanation; based on thinking in decimal fractions; solve for v.

..... maybe a little explanation.

Amount of resulting solute:

Amount of resulting mixture:

As a rational expression, if 25% solution is wanted, this is 0.25.
amount of solute as numerator and amount of final mixture as denominator
must be equal to 0.25.

Answer by greenestamps(13198) (Show Source):
You can put this solution on YOUR website!

First, a standard formal algebraic solution....

You are mixing 60cc of 40% acid with x cc of 15% acid to obtain (60+x) cc of 25% acid:

Perhaps multiply by 100 to get rid of the decimals....

ANSWER: 90cc

Next, a solution by an informal method that can be used to solve any 2-part mixture problem like this....

Look at the three percentages 15, 25, and 40 (on a number line, if it helps) and observe/calculate that 25 is 10/25 = 2/5 of the way from 15 to 40.

That means 2/5 of the mixture is the higher percentage acid -- i.e., the mixture is 2 parts 40% acid and 3 parts 15% acid.

Use a proportion to find that, since 2 parts of the mixture is 60cc, 3 parts of the mixture is 90cc.

ANSWER (again, of course): 90 cc