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13 April, 22:25

You need 180 mL of a 25% alcohol solution. On hand, you have a 30% alcohol mixture. How much of the 30% alcohol mixture and pure water will you need to obtain the desired solution?

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  1. 13 April, 22:55
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    150 mL of the 30% alcohol mixture and 30 mL of the pure water will we need to obtain the desired solution.

    Explanation:

    Let the volume of 30% alcohol used to make the mixture = x L

    For 25% alcohol:

    C₁ = 25%, V₁ = 180 mL

    For 30% alcohol:

    C₂ = 30%, V₂ = x L

    Using

    C₁V₁ = C₂V₂

    25*180 = 30*x

    So,

    x = 150 mL

    Pure water = 180 mL - 150 mL = 30 mL

    150 mL of the 30% alcohol mixture and 30 mL of the pure water will we need to obtain the desired solution.
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