**A student has 2.0 mM blue#1 dye solution and needs 100.0 mL of 4.2 times 10^-4
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Question
**A student has 2.0 mM blue#1 dye solution and needs 100.0 mL of 4.2 times 10^-4 M blue dye #1 solution. How much of the 2.0 mM solution should the student add to the 100.0 mL volumetric flask? A) 1.2 mL B) 21 mL C) 37 mL D) 79 mL ** A student measures the absorbance of solution of red dye #40. Which of the following is a possible lambda max for the red dye? A) 400 nm B) 500 nm C) 600 nm D) 700 nm ** A student obtained the calibration curve for the absorbance of red dye vs the concentration of the red dye solution. If an unknown red dye solution has an absorbance of 0.38, what is its concentration? A) 8.4 times 10^-4 M B) 1.4 times 10^-5 M C) 3.8 times 10^-6 M D) 7.6 times 10^-2 MExplanation / Answer
16) M1V1 = M2V2
2*10-4 mol/L * x L = 4.2*10-4 mol/L * 0.1L
V1 = 0.21L = 0.21*1000 = 21mL, option b)
17) Option d) 700nm is the answer, this is the maximum possible length for the red color in
visible spectrum.
18) In the graph, from 0.38 on Y-axis (absorbance) draw a straight line onto the line y = 98022x+0.0042
and look for the corresponding value onto the X-axis.
3.8*10-6M is the answer, option c)
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