## Chapter 17

## Q. 17.2

**Absorbance, Transmittance, and Beer’s Law**

Find the absorbance and transmittance of a 0.002 40 M solution of a substance with a molar absorptivity of 313 M^{−1} cm^{−1} in a cell with a 2.00-cm pathlength.

## Step-by-Step

## Verified Solution

Equation 17-6 gives us the absorbance.

A = εbc = (313 M^{−1} cm^{−1})(2.00 cm)(0.002 40 M) = 1.50

Transmittance is obtained from Equation 17-5 by raising 10 to the power equal to the expression on each side of the equation:

A = log(\frac{P_0}{P} )= – log T (17-5)

\log T = −A

T = 10^{\log T} = 10^{−A} = 10^{−1.50} = 0.031 6

Just 3.16% of the incident light emerges from this solution.

* Test Yourself *The transmittance of a 0.010 M solution of a compound in a 0.100-cm-pathlength cell is T = 8.23%. Find the absorbance (A) and the molar absorptivity (ε).

(

*1.08, 1.08 × 10^{3} M^{−1} cm^{−1})*

**Answer:**If x = y, 10^{x} = 10^{y}.