SOLUTION: A chemist is making 200 L of a solution that is 62% acid. He is mixing an 80% acid solution with a 30% acid solution. How much of the 80% acid solution will he use?

Question 633555: A chemist is making 200 L of a solution that is 62% acid. He is mixing an 80% acid solution with a 30% acid solution. How much of the 80% acid solution will he use?

Answer by Maths68(1474) (Show Source): You can put this solution on YOUR website!
Mixture
========
Amount=200L
Concentration=62%=0.62

Solution A
=========
Amount = x
Concentration =80% =0.8

Solution B
=========
Amount = 200-x
Concentration =30% = 0.3

[Amount Solution A * Concentration A] + [Amount Solution B * Concentration of B] = [Amount of Mixture *Concentration of Mixture]
(x)(0.8)+(200-x)(0.3)=(200)(0.62)
0.8x+60-0.3x=124
0.5x=124-60
0.5x=64
0.5x/0.5=64/0.5
x=128

Chemist has to mix 128L of the 80% solution with 72L of 30% solution to get a 200L mixture of 62% acid solution..

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