Question 103959: A chemist has 300 grams of 20% hydrochloric acid solution. He wishes to drain some off and replace it with an 80% solution so as to obtain a 25% solution. How many grams must be drain and replace it with 80% solution.
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! Let x = the number of grams of 20% hydrochloric acid solution to be withdrawn and replaced.
The chemist has 300 grams to start so after draining, there will be (300-x) grams of 20% acid solution. This can be written as:
(0.2)(300-x)
Now the chemist needs to add x grams of 80% acid solution. This can be written as: (0.8)(x)
When added together, the chemist will have 300 grams of 25% acid solution. This can be written as:
(0.25)(300) So, putting it all together, we can write the necessary equation to find x.
(0.2)(300-x)+(0.8)(x) = (0.25)(300) Simplifying this, we get:
60-0.2x+0.8x = 75 Combining like-terms:
60+0.6x = 75 Subtracting 60 from both sides.
0.6x = 15 Finally, dividing both sides by 0.6, we get:
x = 25
The chemist will need to drain 25 grams of the 20% hydrochloric acid solution from the 300 grams and replace it with 25 grams of 80% hydrochloric acid solution to obtain 300 grams of 25% hydrochloric acid solution.
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