SOLUTION: A chemist has 50 g of a solution that is 10% acid if he wants to increase the strength to 20% how much pure acid should he add?

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Question 1165719: A chemist has 50 g of a solution that is 10% acid if he wants to increase the strength to 20% how much pure acid should he add?
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


(1) A typical setup using formal algebra....

50g of 10% acid, plus x g of 100% acid, equals (50+x) g of 20% acid:

.10%2850%29%2B1.00%28x%29+=+.20%2850%2Bx%29

Solve using basic algebra.

(2) A quick and easy way to solve ANY 2-part mixture problem like this....

Look at the three percentages 10, 20, and 100 on a number line and see where the 20% is in relation to the 10% and 100%.

The 20% is 1/9 of the way from 10% to 100%. (From 10 to 100 is a difference of 90; from 10 to 20 is a difference of 10; 10/90 = 1/9.)

That means 1/9 of the mixture must be the added 100% acid.

So the 50g of 10% acid is 8/9 of the mixture; 1/9 of the mixture is then 1/8 of 50g, or 6.25g.

ANSWER: 6.25 grams of 100% acid.

CHECK:
.10(50)+1.00(6.25) = 5+6.25 = 11.25
.20(50+6.25) = .20(56.25) = 11.25