To calculate the extinction coefficient (e560) of the dye, we can use Beer-Lambert Law, which is expressed as:

$A = e \cdot C \cdot l$

Where:

• A is the absorbance (0.528),
• e is the extinction coefficient (which we need to find),
• C is the concentration (6.03 × 10⁻⁵ M),
• l is the path length of the cell (0.5 cm).

Rearranging the formula to solve for e:

$e = \frac{A}{C \cdot l}$

Substitute the known values into the equation:

$e = \frac{0.528}{(6.03 \times 10^{-5} \, \text{M}) \cdot (0.5 \, \text{cm})}$

Now, let's perform the calculation:

$e = \frac{0.528}{3.015 \times 10^{-5}}$

$e \approx 1.75 \times 10^{4} \, \text{L/mol·cm}$

So, the extinction coefficient e560 of the dye is approximately 1.75 × 10⁴ L/mol·cm.

---

Would you like further details or have any questions about this calculation?

Here are some related questions you might find interesting:

1. How does the path length affect absorbance and the extinction coefficient?
2. What is the significance of the wavelength in the context of absorbance and the extinction coefficient?
3. What other factors could influence the UV-Vis absorbance of a solution?
4. How would you determine the concentration of a solution using a UV-Vis spectrum?
5. How is the Beer-Lambert Law applied in colorimetry?

Tip: Always ensure that the units for concentration and path length are consistent with those used in the equation to avoid errors in calculations.