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?

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Answer 1 · 2016-04-18T13:47:22

$(a)$ $2.58xx10^(3)" ""mol"^(-1)."l" "." """cm"^(-1)"$

$(b)$ $39.08%$

$(a)$

The Beer - Lambert Law states:

$A=epsiloncl$

$A$ is the absorbance

$epsilon$ is the molar absorptivity

$c$ is the molar concentration

$l$ is the path length in cm

$1"ppm"$ is equivalent to a concentration of $1"mg/l"$

The $M_r$ of $KMnO_4$ is $158.03$

$:.c=0.025/158.03=1.582xx10^(-4)"mol/l"$ .

$epsilon=A/(cl)$

$epsilon=0.408/(1.582xx10^(-4)xx1)=2.58xx10^(3)" ""mol"^(-1)."l" "." """cm"^(-1)"$

$(b)$

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

$A=log(I_0/I)$

$I_0$ is set to 100

$:.A=2-logI$

$:.logI=2-A=2-0.408=1.592$

$:.I=39.08%$

A solution containing 2525 "ppm""ppm" KMnO_4KMnO_4 gives absorbance of 0.4080.408 at a wavelength of 470470 nmnm in a 11 cmcm path length. Calculate molar absorptivity and transmittance?