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3 years ago

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

1 answer:

5 0

(%)(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

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