Question 892649

What quantity of 75 per cent acid solution must be mixed with a 20 per cent solution to produce 440 mL of a 50 per cent solution?
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Let amount of 75% solution to be mixed, be S
Then amount of 20% solution is: 440 – S
Thus, we have: .75(S) + .2(440 – S) = .5(440)
.75S + 88 - .2S = 220
.75S - .2S = 220 – 88
.55S = 132
S, or amount of 75% solution to be mixed = {{{132/.55}}}, or {{{highlight_green(240)}}} mL
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