Question

Suppose a solution of Co(NO3)2 has an extinction coefficient of 5.1L/mol-cm at 505 nm. On graph paper, 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-cm cell. On the same graph plot, the percent transmittance, or %T, of each solution versus concentration

Answer 1

From Beer's law,
Absorbance = extinction coefficient (5.1L/mol-cm) x concentration x path length (1 cm)

Percent transmittance = 100 x 10^-A

So for concentrations of 0.020, 0.040, 0.060, 0.080 and 0.100 M, the corresponding absorbance values are 0.102, 0.204, 0.306, 0.408 and 0.510, and the corresponding percentage tranmittances are 79.1%, 62.5%, 49.4%, 39.1% and 30.9%.   

Answer 2

From Beer's law,
Absorbance = extinction coefficient (5.1L/mol-cm) x concentration x path length (1 cm)

Percent transmittance = 100 x 10^-A

So for concentrations of 0.020, 0.040, 0.060, 0.080 and 0.100 M, the corresponding absorbance values are 0.102, 0.204, 0.306, 0.408 and 0.510, and the corresponding percentage tranmittances are 79.1%, 62.5%, 49.4%, 39.1% and 30.9%.