Beers law - converting from %T to absorbance and too many other question for one post

1. Prior to analyzing the samples provided by Q you must demonstrate (for yourself and others) that you, your instruments, and your company will produce reliable results. One of the ways you decide to do this is by making a Beer’s law plot using pure M. You obtain the following data:

Sample Identification Code Concentration of M (mol/L) %T A
Q5000 4.00 x 10-4 17.9
Q5001 3.20 x 10-4 25.0
Q5002 2.40 x 10-4 35.7
Q5003 1.60 x 10-4 50.2
Q5004 8.000 x 10-5 70.8

Convert %T to absorbance and prepare a Beer’s law plot using these data.

2. You are now ready to begin analyzing the drug. You are sent five bottles of the drug from batch 021015. You analyze these samples and obtain the following data:

Sample Identification Code %T A
Q021015-01 43.7
Q021015-02 44.1
Q021015-03 43.8
Q021015-04 44.1
Q021015-05 43.8

What is the concentration of M in these samples?

3. Company Q has reported that batch 021015 has an M concentration of 3.00 x 10-4 mol/L. Your analysis is better than theirs; therefore, you assume that your analysis is correct. What was their percent error? Remember from Chemistry 1, that formula is:

% error = theoretical value – actual value x 100%
theoretical value

4. By law, company Q must have an M concentration of 3.00 x 10-4 mol/L +/- 5%. (In other words, the M concentration must be between 2.85 x 10-4 mol/L and 3.15 x 10-4 mol/L.) Does Batch 021015 meet legal requirements?

A = 2 - log10 %T  will convert between %T and A for the first question