# A solution of concentration 5 x 10^-5 M has absorbance of 0.75 in a cell of 1 cm length. Calculate molar absorptivity and transmittance of the cell?

Feb 14, 2016

$\left(a\right)$

$1.5 \times {10}^{4} \text{L/mol/cm}$.

$\left(b\right)$

17.8%

#### Explanation:

$\left(a\right)$

The Beer - Lambert Law gives us:

$A = {\epsilon}_{\circ} c l$

$\therefore {\epsilon}_{\circ} = \frac{A}{c l}$

$\therefore {\epsilon}_{\circ} = \frac{0.75}{5 \times {10}^{- 5} \times 1}$

${\epsilon}_{\circ} = 1.5 \times {10}^{4} \text{L/mol/cm}$

$\left(b\right)$

$\log \left({I}_{0} / I\right) = 0.75$

Since ${I}_{0} = 100 \Rightarrow$

$2 - \log I = 0.75$

I=17.8% which is the transmittance.