The concentration of a diluted "wine" sample was found to be 0.24 % (v/v) ethanol. Assuming the wine was diluted by 75x before analysis, what was the original concentration of ethanol in the "wine"

Question

Imani Owens

Chemistry

18 October, 21:20

The concentration of a diluted "wine" sample was found to be 0.24 % (v/v) ethanol. Assuming the wine was diluted by 75x before analysis, what was the original concentration of ethanol in the "wine"

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Spears · Answer 1

The original concentration of ethanol was 18 % (v/v)

Explanation:

For the analysis of a sample, mostly dilutions are made of the original concentrated sample. The analysis results obtained from the diluted sample are then calculated for the concentrated samples.

According to the given data, the wine sample was diluted 75 times. This means that the initial concentration of ethanol was 75 times more than the final concentration i. e. 0.24 % (v/v). So mathematically:

original concentration (v/v) = final concentration (v/v) x times diluted

original concentration (v/v) = 0.24 % x 75

original concentration (v/v) = 18 %