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%#