# The absorbance of 2 x 10^-4 M solution was found to be 0.123 when placed in a cell of 1 cm length. Calculate the transmittance, percent transmittance and molar absorptivity of a cell of length 2 cm?

Feb 12, 2016

$T = 0.568$
%T=56.8%
$\epsilon = 615 {M}^{- 1} \cdot c {m}^{- 1}$

#### Explanation:

$C = 2 \times {10}^{- 4} M$
$A = 0.123$
$l = 2 c m$
?T, ?%T, ?epsilon

The relationship between transmittance $T$ and absorbance is:

$A = - \log T \implies T = {10}^{- A}$

Since the length of the cell was increased to $l = 2 c m$, the absorbance will double and therefore, ${A}_{2} = 2 \times 0.123 = 0.246$

$\implies T = {10}^{- 0.246} = 0.568$

Therefore, %T=100xxT=100xx0.568=56.8%

In order to solve for the molar absorptivity $\epsilon$, we will use the Beer-Lambert law:

$A = \epsilon \times l \times C \implies \epsilon = \frac{A}{l \times C}$

$\epsilon = \frac{0.123}{1 c m \times 2 \times {10}^{- 4} M} = 615 {M}^{- 1} \cdot c {m}^{- 1}$

Here is a video that explains Beer-Lambert law and it use experimentally:
AP Chemistry Investigation #1: Beer-Lambert Law.