Question 257810: Number 1
A chemist is making a solution that is to be 55% chlorine. He has one solution that is 40% chlorine and another that is 65% chlorine. He wants 100 liters of the final solution. How much of each should he mix?
a. Choose a variable to represent the amount of the first solution. Then find the amount of the second solution to be added to make 100 liters. Now figure out how much chlorine is in each solution.
b. Write the equation.
c. Solve the equation.
a. Use x for the amount of the first solution.
Use (100 - x) for the amount of the second
solution. This is because when the two are
added together you want 100 liters of final
solution. For the amount of chlorine you
multiply the proper percentage and the
amount of solution. For the first it is (.40x),
and for the second it is .65(100-x) or
(65 - .65x). You want 100 liters of final
solution at 55% chlorine, so that means
.55(100) or 55.
b. Add the first two solutions, and write the
equation (simplified): .40x + 65 - .65x = 55
c. Solve
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Amount Solution
% CI
Amount Chlorine
X
40%
.40X
100-X
65%
.65(100-X)
100
55%
.55(100)
The equation comes from the last column.
Sol 1 + Sol 2 = Final solution
.40x + .65(100-x) = .55(100)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A chemist is making a solution that is to be 55% chlorine. He has one solution that is 40% chlorine and another that is 65% chlorine. He wants 100 liters of the final solution. How much of each should he mix?
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It's not that complicated.
Equation:
chlorine + chlorine = chlorine
0.40x + 0.65(100-x) = 0.55*100
Multiply thru by 100 to get:
40x + 65*100 - 65x = 55*100
-15x = -10*100
x = 66 2/3 liters (amt. of 40% solution needed in the mixture)
100 - x = 33 1/3 liters (amt. of 65% solution needed in the mixture
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Cheers,
stan H.
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