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3 November, 12:58

How many mL of 50% (v/v) alcohol should be mixed with 10% alcohol to obtain 50mL of 35% (v/v) alcohol? Round to the hundredths place.

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  1. 3 November, 13:24
    0
    The volume of 50% alcohol used to make the mixture = 31.25 L

    The volume of 10% alcohol used to make the mixture = 18.75 L

    Explanation:

    Let the volume of 50% alcohol used to make the mixture = x L

    Let the volume of 10% alcohol used to make the mixture = y L

    Total volume of the mixture = x + y = 50 L ... (1)

    For 50% alcohol:

    C₁ = 50%, V₁ = x L

    For 10% alcohol:

    C₂ = 10%, V₂ = y L

    For the resultant alcohol solution:

    C₃ = 35%, V₃ = 50 L

    Using

    C₁V₁ + C₂V₂ = C₃V₃

    50*x + 10*y = 35*50

    So,

    5x + y = 175 ... (2)

    Solving 1 and 2 we get,

    x = 31.25 L

    y = 18.75 L
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