Question 252173: a chemist has one solution that is 80% acid and another that is 30% acid. how much of each solution is needed to make a 9 200l solution that is 62% acid?
Answer by drk(1908)   (Show Source):
You can put this solution on YOUR website! All mixture problems are just variations of the standard RTD problems. We need 3 columns labeled: % , L, %L. Next we need three rows labeled A, B, and Mix. The A and B stand for the percentages of what you need. We set up the following table based on the information given. The table doesn't show up very well. BIG SPACE means place a big space between the terms. The % numbers should line up under the percentage sign (%). The liter numbers should line up under the liter sign (L). The %L terms should line up under the %L sign.
% BIG SPACE L BIG SPACE %L
A BIG SPACE 80 BIG SPACE X BIG SPACE 80X
B BIG SPACE 30 BIG SPACE 9200-X BIG SPACE 276000 - 30X
Mix BIG SPACE 62 BIG SPACE 9200 BIG SPACE 570400
Where did all of this stuff come from? 80 = % acid in one solution; 30 = % acid in another solution; 9200 the total number of liters in the mixture; 62 = % needed in the mixture. Now for the variable. We don't know the exact number of liters for A or B, so we call one of them X. The other will always be
total liters - X. Finally column 3 (%L): I got these number by multiplying each % and L information.
Next step is to add the first 2 terms of the %L column and set equal to the last term. So, we get
80X + 276000 - 30X = 570400
50X = 294400
X = 5888 L. So we need 5,888 L of the 80% solution.
9200-X = 3112L. We need 3,112L of the 30% solution.
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