SOLUTION: a chemist has two soltions of H. One has a 40% concentration and the other has a 25% concentration. How many liters of each solution must be mixed to obtain 135 liters of a 37% sol

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Question 990480: a chemist has two soltions of H. One has a 40% concentration and the other has a 25% concentration. How many liters of each solution must be mixed to obtain 135 liters of a 37% solution?
Answer by ikleyn(52817) About Me  (Show Source):
You can put this solution on YOUR website!
.
1. Never heard or saw this term  "soltion of H".
I understand that you wanted to write  "solutions"  instead of "soltions",  but again I never heard or saw the term  "solution of H".  I suspect that such thing do not exist.

2. OK,  actually,  it doesn't matter.  We will talk about,  let say,  an abstract solution.

Then you have the system of two linear equations in two unknowns  x  and  y  that are two unknowns volumes of the solutions in units of volume,  let say,  in liters

system%28x+%2B+y+=+135%2C%0D%0A0.4%2Ax+%2B+0.25%2Ay+=+0.37%2A%28x%2By%29%29,

or

system%28x+%2B+y+=+135%2C%0D%0A0.4%2Ax+%2B+0.25%2Ay+=+0.37%2A135%29,

or

system%28x+%2B+y+=+135%2C%0D%0A0.4%2Ax+%2B+0.25%2Ay+=+49.95%29.

The first equation is the conservation law for the volume of liquid.
The second equation is the conservation law for the volume of solute.

This is how to construct  (to make,  to write)  the system of equations.  It is the most important part in solving problems on mixtures and on solutions.
The rest is just techniques  (which is important, though).

Try to solve the system yourself.  You may use the substitution method
(express   y = 135 -x  from the first equation,
substitute it into the second equation,
you will get the single equation for the unknown  x,
then solve this equation
and back substitute the found value of  x  into the first equation of the original system to get the solution for  y).

Good luck!