Calculating Molar Extinction Coefficient using Beer-Lambert Law

This problem requires calculating the molar extinction coefficient ($\epsilon$) using the Beer-Lambert Law, given concentration ($c$), path length ($l$), and light absorption.

1. Determine Absorbance (A)

The problem states that $90\%$ of the incident light is absorbed. This means the transmittance ($T$) is $100\% - 90\% = 10\%$. We convert transmittance to a decimal:

• $T = 10\% = 0.10$

Absorbance ($A$) is calculated using the formula $A = -\log_{10}(T)$:

• $A = -\log_{10}(0.10) = -(-1) = 1.00$

2. Convert Units

Ensure all units are consistent. The Beer-Lambert Law typically uses concentration in molarity ($M$) and path length in centimeters ($cm$).

• Concentration ($c$): $10 \text{ mM} = 10 \times 10^{-3} \text{ M} = 0.01 \text{ M}$
• Path Length ($l$): $10 \text{ mm} = 1 \text{ cm}$

3. Apply Beer-Lambert Law

The Beer-Lambert Law is given by $A = \epsilon c l$. To find the molar extinction coefficient ($\epsilon$), we rearrange the formula:

• $\epsilon = \frac{A}{c l}$

4. Calculate Molar Extinction Coefficient ($\epsilon$)

Substitute the known values into the rearranged formula:

• $A = 1.00$
• $c = 0.01 \text{ M}$
• $l = 1 \text{ cm}$

Calculation:

• $\epsilon = \frac{1.00}{(0.01 \text{ M}) \times (1 \text{ cm})} = \frac{1.00}{0.01} \text{ M}^{-1}\text{cm}^{-1} = 100 \text{ M}^{-1}\text{cm}^{-1}$

Rounding to two decimal places gives $100.00 \text{ M}^{-1}\text{cm}^{-1}$. This value falls within the given range of 98 to 102.