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7 June, 23:43

A solution that is 20% acid and 80% water is mixed with a solution that is 50% acid and 50% water. If twice as much 50% acid solution is used as 20% solution, then what is the ratio of acid to water in the mixture of the solutions?

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  1. 7 June, 23:56
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    The ratio acid to water in the mixture is 2:3

    Explanation:

    Let the volume of 20% acid solution used to make the mixture = x units

    So, the volume of 50% acid solution used to make the mixture = 2x units

    Total volume of the mixture = x + 2x = 3x units

    For 20% acid solution:

    C₁ = 20%, V₁ = x

    For 50% acid solution:

    C₂ = 50%, V₂ = 2x

    For the resultant solution of sulfuric acid:

    C₃ = ?, V₃ = 3x

    Using

    C₁V₁ + C₂V₂ = C₃V₃

    20*x + 50*2x = C₃*3x

    So,

    20 + 50*2 = C₃*3

    Solving

    120 = C₃*3

    C₃ = 40 %

    Thus, for the resultant mixture,

    Acid percentage = 40%

    Water percentage = (100 - 40) % = 60%

    Ratio acid to water in the mixture = 40:60 = 2:3
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