Question

The
percentage transmittance of an aqueous solution
of unknown compound is A20% at 25 c and
300 nm for a C= 2x105 M solution in a d4 cm
cell. Calculate the absorbance and the molar
extinction cofficient.​

Answer 1

Here is an example of directly using the Beer's Law Equation (Absorbance = e L c) when you were given the molar absorptivity constant (or molar extinction coefficient). In this equation, e is the molar extinction coefficient. L is the path length of the cell holder. c is the concentration of the solution.

Answer 2

The absorbance of the solution is 0.699 and the molar extinction coefficient of the unknown compound at 300 nm is 8.738 x 10^-6 M^-1 cm^-1.

Given:

Percentage transmittance (T) = 20%

Wavelength (λ) = 300 nm

Path length (l) = 0.4 cm (since d = 4 cm and the solution is in a 0.1 cm cuvette)

Concentration (C) = 2 x 10^5 M

To find:

Absorbance (A)

Molar extinction coefficient (ε)

Solution:

Calculating Absorbance:

Beer's Law states that absorbance is proportional to the concentration of the solution and the path length of the sample. The formula for absorbance is:

A = -log(T)

Where T is the percentage transmittance. Substituting the given values, we get:

A = -log(0.2) = 0.699

Therefore, the absorbance of the solution is 0.699.

Calculating Molar Extinction Coefficient:

The molar extinction coefficient (ε) is a measure of how strongly a particular substance absorbs light at a given wavelength. It is defined as:

ε = A / (l * C)

Substituting the given values, we get:

ε = 0.699 / (0.4 * 2 x 10^5) = 8.738 x 10^-6 M^-1 cm^-1

Therefore, the molar extinction coefficient of the unknown compound at 300 nm is 8.738 x 10^-6 M^-1 cm^-1.

In summary, the absorbance of the solution is 0.699 and the molar extinction coefficient of the unknown compound at 300 nm is 8.738 x 10^-6 M^-1 cm^-1.

#SPJ3