Lesson Using effective methodology to solve many-steps dilution problems

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Using effective methodology to solve many-steps dilution problems

Problem 1

Ruth has a beaker containing a solution of 800 mL of acid and 200 mL of water.
She thinks the solution is a little strong, so she drains 300 mL from the beaker, adds 300 mL of water, and stirs the solution.
Ruth thinks the solution is still too strong, so again she drains 400 mL from the beaker, and adds 400 mL of water.
Ruth again thinks the solution is still too strong, so again she drains 100 mL from the beaker,
and adds 100 mL of water. How many milliliters of water are now in the beaker?

Solution

        In this problem, after each step, Ruth keeps the total volume of the mixture in the beaker unchangeable, 1000 mL.

        Therefore, in my solution, I will calculate and will watch/track for the amount and the concentration of the acid, only, in the beaker.

        I will calculate the amount of water in the beaker after the last step, but will not calculate/watch/track it at the intermediate steps.

        It will allow me to reduce the volume of calculations and will allow me to concentrate my attention on one component, only
        (which is acid). It will diminish the volume of calculations and will diminish possible errors.

(0)  Starting amount of the acid is 800 mL;  the starting concentration is   = 0.8.



(1)  First step is to drain 300 mL of the mixture from the beaker.

     With it,  0.8*300 = 240 mL of the acid goes out.

     The amount of the acid remained in the beaker is 800-240 = 560 mL.

     The amount of water is added to keep the total volume of the mixture in the beaker 1000 mL.

     The concentration after step 1  is   = 0.56.



(2)  Second step is to drain 400 mL of the mixture from the beaker.

     With it,  0.56*400 = 224 mL of the acid goes out.

     The amount of the acid remained in the beaker is 560-224 = 336 mL.

     The amount of water is added to keep the total volume of the mixture in the beaker 1000 mL.

     The concentration after step 1  is   = 0.336.



(3)  Third step is to drain 100 mL of the mixture from the beaker.

     With it,  0.336*100 = 33.6 mL of the acid goes out.

     The amount of the acid remained in the beaker is 336-33.6 = 302.4 mL.

     The amount of water is added to keep the total volume of the mixture in the beaker 1000 mL.

     
     Hence, the amount of water in the beaker is  1000 - 302.4 = 697.6 mL.


ANSWER.  The amount of water in the beaker after step 3 is  697.6 mL.

Of course, the most important element of this solution is the METHODOLOGY,
which EXCLUDES unnecessary calculations.

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