Question

Suppose a solution of Co(NO3)2 has an extinction coefficient of 5.1 L/mol-cm at 505nm. On the graph provided, plot a graph of A versus C (mol/L) for solutions of 0.020, 0.040, 0.060, 0.080, and 0.100 M Co(NO3)2 in a 1.00 cm cell. On the same graph, plot the percent transmittance, %T, of each solution versus concentration.

Answer 1

Absorbance = molar absortivity x path length x concentration

with,

molar absorptivity =5.1 L/mol-cm

path length = 1 cm

we get,

absorbance = 5.1 x conc.

Also,

log(%T) = 2 - Absorbance

Calculate all the values and plot the graphs,

Abs	Conc	%T
0.102	0.02	79.068
0.204	0.04	62.517
0.306	0.06	49.431
0.408	0.08	39.084
0.51	0.1	30.903

Plot Abs vs Conc 0.6 0.5 0.4 2 0.3 8 0.2 0.1 -Series1 0.02 0.04 0.06 0.08 0.1 Conc. (mol/L)

plot %T vs conc 100 80 60 40 20 -Series1 0.1 0.04 0D. 0.02 Conc. (moVL)