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.

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)