Question

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.

Answer 1

The original concentration of ethanol in the wine was 16% (v/v) after considering the 50x dilution factor and the measured concentration of 0.32% (v/v) post-dilution.

Step-by-step explanation:

To find the original concentration of ethanol in the "wine" before dilution, we can use the information provided that the dilution factor is 50x, and the concentration post-dilution is 0.32% (v/v). To calculate the original concentration, we simply multiply the diluted concentration by the dilution factor.

The original concentration (Coriginal) can be calculated using the formula:

Coriginal = Cdiluted × Dilution factor

Substituting the given values:

Coriginal = 0.32% × 50 = 16%

Therefore, the original concentration of ethanol in the wine was 16% (v/v).

Answer 2

The original concentration of ethanol is 16%(v/v)

Step-by-step 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)