# An aqueous solution which is 10^-3 M absorbs 10 % of incident light in a path length of 1 cm. Calculate concentration of the solution that will absorb 90 % of the incident light in the same cell?

Feb 14, 2016

$0.022 \text{M}$

#### Explanation:

The absorbance is given by:

$A = \log \left(\frac{{I}_{0}}{I}\right)$

$I$ is the intensity of the light as measured by the spectrometer. ${I}_{0}$ is set at 100 for a solution that absorbs no light at that particular wavelength.

In the first case 10% of the light is absorbed so 90% is transmitted and $I = 90$.

So in the second case $I = 10$.

The Beer - Lambert Law gives us:

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

${\epsilon}_{\circ}$ is the molar absorption coefficient.

$c$ is the concentration in $\text{mol/l}$

$l$ is the path length of the cell.

Putting in the numbers:

$\log \left(\frac{100}{90}\right) = {\epsilon}_{\circ} \times {10}^{- 3} \times 1 \text{ } \textcolor{red}{\left(1\right)}$

$\log \left(\frac{100}{10}\right) = 1 = {\epsilon}_{\circ} \times c \times 1 \text{ } \textcolor{red}{\left(2\right)}$

Dividing $\textcolor{red}{\left(1\right)}$ by $\textcolor{red}{\left(2\right)} \Rightarrow$

$\log \left(1.11\right) = {10}^{- 3} / c$

$\therefore c = {10}^{- 3} / 0.0457 = 0.0218 \text{mol/l}$