What amount of a 70% acid solution must be mixed with a 30% solution to produce 600 mL of a 55% solution?

Question

Mathematics

Harold Patel

8 January, 19:22

What amount of a 70% acid solution must be mixed with a 30% solution to produce 600 mL of a 55% solution?

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Daniel Cowan · Answer 1

(%) (V1) + (%) (V2-V1) = (%) (V2)

(0.70) (x) + (0.30) (600mL-x) = (0.55) (600)

(0.70x) + (180) - (0.30x) = 330

(0.70-0.30) x=330-180

(0.40) x=150

x = (150) / (0.40)

x=375 mL