Post-lab: 1. Suppose a solution was too concentrated for an accurate reading wit
ID: 932929 • Letter: P
Question
Post-lab: 1. Suppose a solution was too concentrated for an accurate reading with the colorimeter. The concentrated solution was diluted by placing 1.00 mL of the concentrated solution in 4.00 mL of water. The solution was then placed in the colorimeter and an absorbance was obtained and after a few calculations the molar concentration was calculated to be 3.5 x 10M. What was the concentration of the original stock solution before dilution? If a 0.10 M solution of a colored substance has a maximum absorbance at 500 nm and an absorbance of 0.26 at this wavelength, what will be the measured absorbance of a 0.20 M solution at 500 nm? The spectrophotometer is a device like the colorimeter. It measures the percent of light that is transmitted through the solution. The instrument then converts %T (transmittance) into absorbance using the following equation: Abs -log (T) If the absorbance of a sample is 0.85 what is the percent of light transmitted through the colored sample at this collected wavelength? 2. 3.Explanation / Answer
ANSWER:
Dear candidate you have asked Three questions. As per guidelines one question one time. Here We solve FIRST TWO QUESTIONS.
1. Use the following equation to solve the problem
M1V1 = M2V2
M1V1 = molarity and concentration of concentrated solution., M1 = ?, V1 = 1mL
M2V2 = molarity and concentration of dilute solution., M2 = 3.5X 10-6M, V2 = 1mL + 4mL = 5mL
M1 X 1 = 3.5X 10-6 X 5
M1 = 3.5X 10-6 X 5 = 17.5 X 10-6 / 1 = 1.75 X 10-5 M
2. Use the following equation
C1/C2 = A1/A2, C = concentration, A = Absorbance
C1 = 0.1, A1 = 0.26, C2 = 0.2, A2 = ?
0.1/0.2 = 0.26/A2
A2 = 0.26 X 0.2 / 0.1 = 0.52
3. Remeber the following equation
A = 2 - log%T
0.85 = 2 - log%T
1.15 = log%T
%T = antilog(1.15)
%T = 14.53
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