Question

Given an absorbance of .493, what is the percent of alcohol in a sample of wine?

Answer 1

Ethanol and certain other alcohols react with ammonium hexanitrato cerium(IV), (NH4)2Ce(NO3)6, to yield a corresponding carboxylic acid and a cerium(III) complex. The reaction forms a deep red intermediary complex as the alcohol is oxidized. This colored species obeys Beer’s Law, meaning that the absorbance of light by the colored substance is directly proportional to the concentration of the substance in solution:

A = εlc

where A = absorbance

ε = molar absorptivity

l = path length of cell

c = concentration

A Beer’s Law plot of absorbance vs. concentration of the red complex, called a calibration curve, allows for the determination of the quantity of alcohol oxidized by ammonium hexanitrato cerium(IV). By measuring the absorbance of the complex formed, the percent alcohol in a sample can be determined from the calibration curve.

Now in this question we have A , absorbance = 0.493

standard lenght of tube = 1 cm

and ε = 26.79

So , A =εcl = 0.493 = 26.79 *c * 1

c = 0.0184 moles/L

So % v/v = ( 0.0184/46) * 100 = 0.04 %