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3 March, 15:59

qizlet, The concentration of a diluted "wine" sample was found to be 0.32 % (v/v) ethanol. Assuming the wine was diluted by 50x before analysis, what was the original concentration of ethanol in the "wine"? Enter your answer with 2 significant figures.

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  1. 3 March, 19:16
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    The original concentration of ethanol is 16% (v/v)

    Explanation:

    concentration before dilution, C1 = ?

    Concentration after dilution, C2 = 0.32% (v/v)

    Volume before dilution, V1 = x

    Volume after dilution, V2 = 50x

    From the dilution principle

    C1V1 = C2V2

    (C1) * x = (0.32) * 50x

    C1 = (16x) / x

    C1 = 16% (v/v)
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