4-15. Convert the accompanying transmittar ce data to absorbances. (a) 27.2% (b) 0.579 (c) 30.6% (d) 3.98% (e) 0.093 (f) 63.7% .16. Calculate the percent transmittance of solution that have twice the absorbance of the solutions i Problem 24-14

To convert transmittance data to absorbance, we can use Beer-Lambert law, which states that the absorbance (A) is related to the transmittance (T) by the equation:<br /><br />\[ A = -\log(T) \]<br /><br />where T is the transmittance expressed as a decimal.<br /><br />Let's calculate the absorbance for each given transmittance value:<br /><br />(a) $27.2\%$<br />\[ T = 0.272 \]<br />\[ A = -\log(0.272) \]<br />\[ A \approx 563 \]<br /><br />(b) 0.579<br />\[ T = 0.579 \]<br />\[ A = -\log(0.579) \]<br />\[ A \approx 0.241 \]<br /><br />(c) $30.6\%$<br />\[ T = 0.306 \]<br />\[ A = -\log(0.306) \]<br />\[ A \approx 0.512 \]<br /><br />(d) $3.98\%$<br />\[ T = 0.0398 \]<br />\[ A = -\log(0.0398) \]<br />\[ A \approx 1.398 \]<br /><br />(e) 0.093<br />\[ T = 0.093 \]<br />\[ A = -\log(0.093) \]<br />\[approx 1.027 \]<br /><br />(f) $63.7\%$<br />\[ T = 0.637 \]<br />\[ A = -\log(0.637) \]<br />\[ A \approx 0.198 \]<br /><br />Now, let's calculate the percent transmittance of a solution that has twice the absorbance of the solutions in Problem 24-14.<br /><br />Assuming the absorbance values from Problem 24-14 are as follows:<br />- Solution 1: 0.5<br />- Solution 2: 1.0<br />- Solution 3: 1.5<br /><br />Twice the absorbance of the highest value (1.5) is:<br />\[ 2 \times 1.5 = 0 \]<br /><br />Using the Beer-Lambert law in reverse to find the transmittance:<br />\[ T = 10^{-A} \]<br />\[ T = 10^{-3} \]<br />\[ T = 0.001 \]<br /><br />Converting this to percent transmittance:<br />\[ \text{Percent Transmittance} = T \times 100 \]<br />\[ \text{Percent Transmittance} = 0.001 \times 100 \]<br />\[ \text{Percent Transmittance} = 0.1\% \]<br /><br />Therefore, the percent transmittance of the solution with twice the absorbance of the solutions in Problem 24-14 is 0.1%.