Asked by Anonymous
A chemist has one solution that is 80% acid and a second solution that is 30% acid. How many liters of each solution will the chemist need in order to make 50L of a solution that is 62% acid?
Answers
Answered by
Reiny
let amount of stronger solution to be used be x L
amount of weaker solution is 50-x L
.8x + .3(50-x) = .62(50)
.8x + 15 - .3x =31
.5x = 16
x = 16/.5 = 32
needs 32 L of the stronger and 36 of the weaker solution.
amount of weaker solution is 50-x L
.8x + .3(50-x) = .62(50)
.8x + 15 - .3x =31
.5x = 16
x = 16/.5 = 32
needs 32 L of the stronger and 36 of the weaker solution.
Answered by
MathMate
Ratio of 80% and 30% acids is
62-30:80-62
=32:18
=16:9
To make 50 L of 62%
Volume of 80% acid = 50*(16/25)=32 L
Volume of 30% acid = 50*(9/25)=18 L
62-30:80-62
=32:18
=16:9
To make 50 L of 62%
Volume of 80% acid = 50*(16/25)=32 L
Volume of 30% acid = 50*(9/25)=18 L
Answered by
Reiny
looks like I can't do a simple subtraction
of course 50-32 = 18
as MathMate had
of course 50-32 = 18
as MathMate had
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