3. using the data at right:\na. calculate the absorbance of each sample. you will need to express the %t…

3. using the data at right:\na. calculate the absorbance of each sample. you will need to express the %t values as decimals before calculating the logarithms. (e.g., for t = 25%, a = -log 0.25 = 0.602)\nb. use the graph paper on the back of this page (or a spreadsheet) to plot absorbance vs. concentration. make a proper graph, with labeled axes, uniform scales, etc. use a ruler to draw the best - fit line.\nc. use the graph to find the concentration of the unknown sample. (draw dotted lines on the graph showing how you found your answer.)\n\nconcentration (ppm)\nunknown 500 400 300 200 100 0\npercent transmission 60.3 22.3 30.1 40.7 54.9 74.1 100.0\nabsorbance

3. using the data at right:\na. calculate the absorbance of each sample. you will need to express the %t values as decimals before calculating the logarithms. (e.g., for t = 25%, a = -log 0.25 = 0.602)\nb. use the graph paper on the back of this page (or a spreadsheet) to plot absorbance vs. concentration. make a proper graph, with labeled axes, uniform scales, etc. use a ruler to draw the best - fit line.\nc. use the graph to find the concentration of the unknown sample. (draw dotted lines on the graph showing how you found your answer.)\n\nconcentration (ppm)\nunknown 500 400 300 200 100 0\npercent transmission 60.3 22.3 30.1 40.7 54.9 74.1 100.0\nabsorbance

Answer

Explanation:

Step1: Recall the formula for absorbance

The formula for absorbance $A$ in terms of percent - transmission ($%T$) is $A = -\log\left(\frac{%T}{100}\right)$.

Step2: Calculate absorbance for each sample

For $%T = 100.0$: $A = -\log\left(\frac{100.0}{100}\right)=-\log(1) = 0$ For $%T = 74.1$: $A=-\log\left(\frac{74.1}{100}\right)=-\log(0.741)\approx0.13$ For $%T = 54.9$: $A = -\log\left(\frac{54.9}{100}\right)=-\log(0.549)\approx0.26$ For $%T = 40.7$: $A=-\log\left(\frac{40.7}{100}\right)=-\log(0.407)\approx0.39$ For $%T = 30.1$: $A=-\log\left(\frac{30.1}{100}\right)=-\log(0.301)\approx0.52$ For $%T = 22.3$: $A=-\log\left(\frac{22.3}{100}\right)=-\log(0.223)\approx0.65$ For $%T = 60.3$: $A=-\log\left(\frac{60.3}{100}\right)=-\log(0.603)\approx0.22$

Answer:

CONCENTRATION (ppm) PERCENT TRANSMISSION ABSORBANCE
0 100.0 0
100 74.1 0.13
200 54.9 0.26
300 40.7 0.39
400 30.1 0.52
500 22.3 0.65
unknown 60.3 0.22