Consider a solution of an analyte with a molar absorptivity of 2000. You are measuring the absorbance in a 2cm path cell. For the same concentrations that are in this Excel sheet, calculate the apparent absorbance if the stray light amounts to 0.01%, 0.2%, and 2%, Note that this spreadsheet has one change (correction) from what I did in class. When I divided by Izero in column H, I should have use used 100+ Is as the divisor, not just 100 , because if the sample cell has no absorption, the detector will not only see the 100 units of light coming through the cell but it is also seeing the stray light at all times. \begin{tabular}{|l|l|l|l|l|l|l|l|l|} \hline \multicolumn{1}{|c|}{ ideal abs } & ideal T & ideal I & actual I & actual T & actual A \\ \hline 1.00E06 & 1.00E02 & 9.77E01 & 9.77E+01 & 9.87E+01 & 9.77E01 & 0.0099 \\ \hline 2.00E06 & 2.00E02 & 9.55E01 & 9.55E+01 & 9.65E+01 & 9.55E01 & 0.019797 \\ \hline 5.00E06 & 5.00E02 & 8.91E01 & 8.91E+01 & 9.01E+01 & 8.92E01 & 0.049476 \\ \hline 1.00E05 & 1.00E01 & 7.94E01 & 7.94E+01 & 8.04E+01 & 7.96E01 & 0.098888 \\ \hline 1.00E04 & 1.00E+00 & 1.00E01 & 1.00E+01 & 1.10E+01 & 1.09E01 & 0.962929 \\ \hline 2.00E04 & 2.00E+00 & 1.00E02 & 1.00E+00 & 2.00E+00 & 1.98E02 & 1.703291 \\ \hline 3.00E04 & 3.00E+00 & 1.00E03 & 1.00E01 & 1.10E+00 & 1.09E02 & 1.962929 \\ \hline 4.00E04 & 4.00E+00 & 1.00E04 & 1.00E02 & 1.01E+00 & 1.00E02 & 2 \\ \hline 6.00E04 & 6.00E+00 & 1.00E06 & 1.00E04 & 1.00E+00 & 9.90E03 & 2.004278 \\ \hline \end{tabular} ideal A 0.00E+005.00E051.00E041.50E042.00E042.50E043.00E043.50E044.00E044.50E04