# A solution containing 25 "ppm" KMnO_4 gives absorbance of 0.408 at a wavelength of 470 nm in a 1 cm path length. Calculate molar absorptivity and transmittance?

Apr 18, 2016

$\left(a\right)$ $2.58 \times {10}^{3} \text{ ""mol"^(-1)."l" "." """cm"^(-1)}$

$\left(b\right)$ 39.08%

#### Explanation:

$\left(a\right)$

The Beer - Lambert Law states:

$A = \epsilon c l$

$A$ is the absorbance

$\epsilon$ is the molar absorptivity

$c$ is the molar concentration

$l$ is the path length in cm

$1 \text{ppm}$ is equivalent to a concentration of $1 \text{mg/l}$

The ${M}_{r}$ of $K M n {O}_{4}$ is $158.03$

$\therefore c = \frac{0.025}{158.03} = 1.582 \times {10}^{- 4} \text{mol/l}$.

$\epsilon = \frac{A}{c l}$

$\epsilon = \frac{0.408}{1.582 \times {10}^{- 4} \times 1} = 2.58 \times {10}^{3} \text{ ""mol"^(-1)."l" "." """cm"^(-1)}$

$\left(b\right)$

Absorbance in terms of % transmittance $I$ is given by:

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

${I}_{0}$ is set to 100

$\therefore A = 2 - \log I$

$\therefore \log I = 2 - A = 2 - 0.408 = 1.592$

:.I=39.08%