The straight-line plot of absorbance vs concentration follows:

Answer: Beer-Lambert Law (A = epsilon b c) Beer-Lambert law states absorbance is proportional to molar absorptivity, path length, and concentration; a linear A vs c calibration curve confirms it.

What is the molar absorptivity (epsilon) of CuSO4 at 635 nm?

Answer: 2.10 M^-1 cm^-1 Beer-Lambert: A = epsilon b c. With b=1.00 cm, slope of A vs c equals epsilon = 2.10 M^-1 cm^-1.

What is the concentration of the unknown CuSO4 solution?

Answer: 0.250 M The slope is 2.10 M^-1 (e.g., 0.840/0.400). Using A = slope c, c = 0.525 / 2.10 = 0.250 M.

AP Chemistry · Topic 3.13

Beer-Lambert Law Practice

Part of Properties of Substances and Mixtures.(SAP-8.C)

Practice questions

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Sample questions

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Sample 1difficulty 1/5
$Calibration data for a colored complex are plotted as absorbance versus concentration.$
$The straight-line plot of absorbance vs concentration follows:$
- A
  $Henry's Law$
- B
  $Beer-Lambert Law (A = epsilon * b * c)$
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- C
  $Raoult's Law$
- D
  $Boyle's Law$
Why
$Beer-Lambert law states absorbance is proportional to molar absorptivity, path length, and concentration; a linear A vs c calibration curve confirms it.$
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Sample 2difficulty 1/5
$A student prepares five aqueous CuSO4 standard solutions by dilution from a 0.500 M stock and measures their absorbance at 635 nm in a 1.00 cm cuvette. The blank is distilled water. Concentrations and absorbances are: 0.050 M (A=0.105), 0.100 M (A=0.210), 0.200 M (A=0.420), 0.300 M (A=0.630), 0.400 M (A=0.840). An unknown CuSO4 solution gives A=0.525 in the same cuvette.$
$What is the molar absorptivity (epsilon) of CuSO4 at 635 nm?$
- A
  $2.10 M^-1 cm^-1$
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- B
  $21.0 M^-1 cm^-1$
- C
  $0.477 M^-1 cm^-1$
- D
  $0.210 M^-1 cm^-1$
Why
$Beer-Lambert: A = epsilon * b * c. With b=1.00 cm, slope of A vs c equals epsilon = 2.10 M^-1 cm^-1.$
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Sample 3difficulty 2/5
$A student prepares five aqueous CuSO4 standard solutions by dilution from a 0.500 M stock and measures their absorbance at 635 nm in a 1.00 cm cuvette. The blank is distilled water. Concentrations and absorbances are: 0.050 M (A=0.105), 0.100 M (A=0.210), 0.200 M (A=0.420), 0.300 M (A=0.630), 0.400 M (A=0.840). An unknown CuSO4 solution gives A=0.525 in the same cuvette.$
$What is the concentration of the unknown CuSO4 solution?$
- A
  $0.105 M$
- B
  $1.10 M$
- C
  $0.525 M$
- D
  $0.250 M$
  check_circle
Why
$The slope is 2.10 M^-1 (e.g., 0.840/0.400). Using A = slope * c, c = 0.525 / 2.10 = 0.250 M.$
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Sample 4difficulty 2/5
$Absorbance spectrum of an unknown shows a peak at the dashed wavelength.$
$For most accurate Beer-Lambert quantitation, measurements should be made at:$
- A
  $lambda_max (the wavelength of maximum absorbance)$
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- B
  $Any wavelength$
- C
  $The shortest wavelength shown$
- D
  $The longest wavelength shown$
Why
$Working at lambda_max maximizes signal-to-noise and minimizes errors from small wavelength uncertainties because the curve is locally flat at the peak.$
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Sample 5difficulty 2/5
$Calibration of Cu2+ shows a slope giving A = 2.0 * [Cu2+].$
$A solution with absorbance 0.40 has concentration:$
- A
  $0.40 M$
- B
  $0.10 M$
- C
  $0.80 M$
- D
  $0.20 M$
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Why
$From A = 2.0 * c, c = 0.40/2.0 = 0.20 M, matching the dashed grid line.$
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