AP Chemistry Quiz: Beer Lambert Law
20 questions · exam conditions
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Beer Lambert LawQuestion 1 of 20

A solution has concentration 1.0×10−5 M1.0\times10^{-5}\ \text{M} and is measured in a 1.00 cm cuvette at a wavelength where ε=3.0×104 L mol−1cm−1\varepsilon=3.0\times10^4\ \text{L mol}^{-1}\text{cm}^{-1}. What absorbance is expected?

A=0.30A=0.30
A=3.0A=3.0
A=0.03A=0.03
A=0.10A=0.10
A=0.003A=0.003
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AP Chemistry Quiz

AP Chemistry Quiz: Beer Lambert Law

Practice Beer Lambert Law in AP Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Beer Lambert Law, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A solution has concentration 1.0×10−5 M1.0\times10^{-5}\ \text{M} and is measured in a 1.00 cm cuvette at a wavelength where ε=3.0×104 L mol−1cm−1\varepsilon=3.0\times10^4\ \text{L mol}^{-1}\text{cm}^{-1}. What absorbance is expected?

  1. A=0.30A=0.30 (correct answer)
  2. A=3.0A=3.0
  3. A=0.03A=0.03
  4. A=0.10A=0.10
  5. A=0.003A=0.003

Explanation: This question tests the direct application of Beer-Lambert law with the given parameters. Using A = εℓc with c = 1.0×10⁻⁵ M, ℓ = 1.00 cm, and ε = 3.0×10⁴ L mol⁻¹ cm⁻¹: A = (3.0×10⁴)(1.00)(1.0×10⁻⁵) = 0.30. This matches the marked answer exactly. Students might choose option C (0.03) by incorrectly placing the decimal point during multiplication of scientific notation terms. When applying Beer-Lambert law, carefully track powers of 10 throughout your calculation.

Question 2

At 600 nm, a compound has ε=1.0×104 M−1cm−1\varepsilon = 1.0\times 10^4\,\text{M}^{-1}\text{cm}^{-1}. A solution of this compound has concentration c=2.0×10−5 Mc = 2.0\times 10^{-5}\,\text{M} and gives absorbance A=0.40A = 0.40. What path length ℓ\ell (in cm) was used?

  1. 0.040 cm0.040\,\text{cm}
  2. 20 cm20\,\text{cm}
  3. 0.80 cm0.80\,\text{cm}
  4. 2.0 cm2.0\,\text{cm} (correct answer)
  5. 0.20 cm0.20\,\text{cm}

Explanation: This question tests the application of the Beer-Lambert Law to find path length from absorbance, molar absorptivity, and concentration. The equation A = ε ℓ c rearranges to ℓ = A / (ε c). Inserting the values gives ℓ = 0.40 / (1.0 × 10410^4 M⁻¹ cm⁻¹ × 2.0 × 10−510^{-5} M) = 0.40 / 0.20 = 2.0 cm. This matches choice B. A tempting distractor is 0.20 cm, resulting from inverting the formula incorrectly as ℓ = (ε c) / A, a misconception in algebraic rearrangement. A transferable strategy is to solve for the unknown variable symbolically first, then substitute numbers to minimize errors in Beer-Lambert calculations.

Question 3

A solution absorbs light at 430 nm with molar absorptivity ε=150 M−1cm−1\varepsilon = 150\,\text{M}^{-1}\text{cm}^{-1}. A student uses a cuvette with ℓ=2.00 cm\ell = 2.00\,\text{cm} and measures A=0.60A = 0.60. What is the concentration of the solution?

  1. 0.0020 M0.0020\,\text{M} (correct answer)
  2. 0.0040 M0.0040\,\text{M}
  3. 0.20 M0.20\,\text{M}
  4. 0.0080 M0.0080\,\text{M}
  5. 0.0010 M0.0010\,\text{M}

Explanation: This question tests the application of the Beer-Lambert Law to calculate the concentration of a solution from absorbance, molar absorptivity, and path length. The Beer-Lambert Law states A = ε ℓ c, so concentration c = A / (ε ℓ). Substituting the values gives c = 0.60 / (150 M⁻¹ cm⁻¹ × 2.00 cm) = 0.60 / 300 = 0.0020 M. This matches choice A. A tempting distractor is 0.0040 M, which results from omitting the path length and calculating c = A / ε = 0.60 / 150 = 0.0040 M, stemming from the misconception that path length does not affect absorbance in the equation. A transferable strategy is to memorize the full Beer-Lambert equation and check that all variables are included before performing calculations.

Question 4

A solution is measured at a fixed wavelength where ε\varepsilon is constant. Which change will decrease the absorbance, assuming the solute remains the same?

  1. Decrease the concentration while keeping path length constant. (correct answer)
  2. Increase the path length while keeping concentration constant.
  3. Increase both concentration and path length by the same factor.
  4. Increase the concentration while keeping path length constant.
  5. Increase the molar absorptivity while keeping concentration constant.

Explanation: This question tests understanding of what factors decrease absorbance according to Beer-Lambert law. Since A = εℓc, absorbance decreases when any of the variables (ε, ℓ, or c) decreases while others remain constant. Option A correctly states that decreasing concentration while keeping path length constant will decrease absorbance. Options B, C, and D would all increase absorbance, while option E is not practically achievable since ε is a molecular property. Students might choose option B by confusing which changes increase versus decrease absorbance. Remember that absorbance is directly proportional to concentration, path length, and molar absorptivity.

Question 5

A solution is measured at a fixed wavelength and follows Beer-Lambert law. Trial 1 uses ℓ=1.0 cm\ell=1.0\ \text{cm} and c=0.020 Mc=0.020\ \text{M} and gives A=0.50A=0.50. In Trial 2, the concentration is doubled and the path length is halved. What absorbance is expected in Trial 2?

  1. 1.001.00
  2. 0.250.25
  3. 0.500.50 (correct answer)
  4. 0.750.75
  5. 2.002.00

Explanation: This question tests the combined effect of changing concentration and path length in the Beer-Lambert law. Doubling c increases A by 2, but halving ℓ decreases A by 1/2, netting no change: A = 0.50. Alternatively, original ε = 0.50 / (1.0 × 0.020) = 25 M⁻¹cm⁻¹; new A = 25 × 0.5 × 0.040 = 0.50. This matches choice C. A tempting distractor is choice A (1.00), from only considering the concentration doubling and ignoring path halving. Evaluate net effects of multiple changes by multiplying factors in Beer-Lambert applications.

Question 6

Two solutions of the same solute are measured at the same wavelength using identical 1.00 cm cuvettes. Solution 1 has concentration 1.0×10−4 M1.0\times10^{-4}\ \text{M} and Solution 2 has concentration 3.0×10−4 M3.0\times10^{-4}\ \text{M}. Which statement best compares their absorbances?

  1. Solution 2 has one-third the absorbance of Solution 1.
  2. Solution 2 has three times the absorbance of Solution 1. (correct answer)
  3. Solution 2 has the same absorbance as Solution 1.
  4. Solution 2 has nine times the absorbance of Solution 1.
  5. Solution 2 has one-ninth the absorbance of Solution 1.

Explanation: This question tests understanding of the direct proportionality between concentration and absorbance in the Beer-Lambert law. Since A = εℓc and all other parameters (ε, ℓ, wavelength) remain constant, absorbance is directly proportional to concentration. Solution 2 has concentration 3.0×10⁻⁴ M compared to Solution 1's 1.0×10⁻⁴ M, making it exactly 3 times more concentrated. Therefore, Solution 2 will have exactly 3 times the absorbance of Solution 1. Students might incorrectly choose option D (nine times) by squaring the concentration ratio instead of using the direct proportionality. Remember that in Beer-Lambert law, absorbance scales linearly with concentration when other factors are held constant.

Question 7

A solution with concentration 1.0×10−4 M1.0\times10^{-4}\ \text{M} is measured at two different wavelengths in a 1.00 cm cuvette. At λ1\lambda_1, ε=2.0×104\varepsilon=2.0\times10^4; at λ2\lambda_2, ε=5.0×103\varepsilon=5.0\times10^3. Which statement is correct about absorbance?

  1. Absorbance at λ1\lambda_1 is twice absorbance at λ2\lambda_2.
  2. Absorbance at λ2\lambda_2 is twice absorbance at λ1\lambda_1.
  3. Absorbance at λ1\lambda_1 equals absorbance at λ2\lambda_2.
  4. Absorbance at λ1\lambda_1 is four times absorbance at λ2\lambda_2. (correct answer)
  5. Absorbance at λ2\lambda_2 is four times absorbance at λ1\lambda_1.

Explanation: This question tests understanding of how molar absorptivity affects absorbance at different wavelengths. At λ₁: A₁ = (2.0×10⁴)(1.00)(1.0×10⁻⁴) = 2.0. At λ₂: A₂ = (5.0×10³)(1.00)(1.0×10⁻⁴) = 0.5. The ratio A₁/A₂ = 2.0/0.5 = 4, so absorbance at λ₁ is four times that at λ₂. This occurs because the molar absorptivity at λ₁ is four times larger than at λ₂. Students might choose option A by inverting the ratio. When comparing absorbances at different wavelengths, always calculate both values and determine their ratio carefully.

Question 8

A solution is measured in a 1.00 cm cuvette and has absorbance A=0.66A=0.66. The molar absorptivity is ε=3.3×103 L mol−1cm−1\varepsilon=3.3\times10^3\ \text{L mol}^{-1}\text{cm}^{-1}. What is the concentration?

  1. 5.0×10−4 M5.0\times10^{-4}\ \text{M}
  2. 1.0×10−4 M1.0\times10^{-4}\ \text{M}
  3. 5.0×10−5 M5.0\times10^{-5}\ \text{M}
  4. 2.0×10−4 M2.0\times10^{-4}\ \text{M} (correct answer)
  5. 2.0×10−5 M2.0\times10^{-5}\ \text{M}

Explanation: This question tests the application of Beer-Lambert law to determine concentration from absorbance measurements. Using A = εℓc and solving for concentration: c = A/(εℓ) = 0.66/(3.3×10³ × 1.00) = 0.66/(3.3×10³) = 2.0×10⁻⁴ M. This matches the marked answer exactly. Students might choose option B (5.0×10⁻⁴ M) by incorrectly estimating the division result. When working with Beer-Lambert calculations, always perform the division carefully to avoid computational errors.

Question 9

A solution of a colored compound is measured at 400 nm. In a 1.00 cm cuvette, A=0.10A=0.10 for a solution with c=2.0×10−5 Mc=2.0\times10^{-5}\ \text{M}. What is ε\varepsilon at 400 nm?

  1. ε=5.0×104 L mol−1cm−1\varepsilon=5.0\times10^4\ \text{L mol}^{-1}\text{cm}^{-1}
  2. ε=1.0×103 L mol−1cm−1\varepsilon=1.0\times10^3\ \text{L mol}^{-1}\text{cm}^{-1}
  3. ε=5.0×103 L mol−1cm−1\varepsilon=5.0\times10^3\ \text{L mol}^{-1}\text{cm}^{-1} (correct answer)
  4. ε=2.0×103 L mol−1cm−1\varepsilon=2.0\times10^3\ \text{L mol}^{-1}\text{cm}^{-1}
  5. ε=2.0×104 L mol−1cm−1\varepsilon=2.0\times10^4\ \text{L mol}^{-1}\text{cm}^{-1}

Explanation: This question tests the calculation of molar absorptivity from Beer-Lambert law data with very dilute solutions. Using A = εℓc and solving for ε: ε = A/(ℓc) = 0.10/(1.00 × 2.0×10⁻⁵) = 0.10/(2.0×10⁻⁵) = 5.0×10³ L mol⁻¹ cm⁻¹. This matches the marked answer exactly. Students might choose option B (2.0×10⁴) by incorrectly handling the very small concentration value in their calculation. When working with dilute solutions, pay careful attention to powers of 10 in scientific notation.

Question 10

Two solutions of different solutes are measured at the same wavelength using 1.00 cm cuvettes. Solution P has ε=1.0×104\varepsilon=1.0\times10^4 and c=1.0×10−4 Mc=1.0\times10^{-4}\ \text{M}. Solution Q has ε=5.0×103\varepsilon=5.0\times10^3 and c=2.0×10−4 Mc=2.0\times10^{-4}\ \text{M}. Which statement best compares absorbances?

  1. Solution P has half the absorbance of Solution Q.
  2. Solution Q has twice the absorbance of Solution P.
  3. Solution P and Solution Q have the same absorbance. (correct answer)
  4. Solution P has twice the absorbance of Solution Q.
  5. Solution Q has half the absorbance of Solution P.

Explanation: This question tests understanding of how different combinations of ε and c affect absorbance in Beer-Lambert law. For Solution P: A = εℓc = (1.0×10⁴)(1.00)(1.0×10⁻⁴) = 1.0. For Solution Q: A = εℓc = (5.0×10³)(1.00)(2.0×10⁻⁴) = 1.0. Both solutions have the same absorbance because the product εc is identical in both cases. Students might choose option A or B by only comparing individual parameters rather than their combined effect. When comparing absorbances, always calculate the complete Beer-Lambert expression for each solution.

Question 11

A solution is measured in a 1.00 cm cuvette and has A=0.48A=0.48. The molar absorptivity is ε=1.6×104 L mol−1cm−1\varepsilon=1.6\times10^4\ \text{L mol}^{-1}\text{cm}^{-1}. What is the concentration?

  1. 3.0×10−4 M3.0\times10^{-4}\ \text{M}
  2. 7.5×10−4 M7.5\times10^{-4}\ \text{M}
  3. 7.5×10−5 M7.5\times10^{-5}\ \text{M}
  4. 3.0×10−5 M3.0\times10^{-5}\ \text{M} (correct answer)
  5. 4.8×10−5 M4.8\times10^{-5}\ \text{M}

Explanation: This question tests the application of Beer-Lambert law to determine concentration from absorbance data. Using A = εℓc and solving for concentration: c = A/(εℓ) = 0.48/(1.6×10⁴ × 1.00) = 0.48/(1.6×10⁴) = 3.0×10⁻⁵ M. This matches the marked answer exactly. Students might choose option B (7.5×10⁻⁵ M) by making computational errors in the division. When working with Beer-Lambert calculations, double-check your arithmetic, especially when dividing by large numbers in scientific notation.

Question 12

A solution has A=1.20A=1.20 in a 2.00 cm cuvette. If the same solution is measured in a 1.00 cm cuvette at the same wavelength, what absorbance is expected?

  1. A=2.40A=2.40
  2. A=0.60A=0.60 (correct answer)
  3. A=1.20A=1.20
  4. A=0.30A=0.30
  5. A=1.80A=1.80

Explanation: This question tests understanding of how path length affects absorbance when moving between different cuvettes. Since A = εℓc and only path length changes (from 2.00 cm to 1.00 cm), the new absorbance is A_new = A_original × (ℓ_new/ℓ_original) = 1.20 × (1.00/2.00) = 0.60. Halving the path length results in halving the absorbance. Students might choose option A (2.40) by incorrectly thinking shorter cuvettes increase absorbance. Remember that absorbance decreases proportionally when path length decreases.

Question 13

A student measures a solution and obtains A=0.27A=0.27 using a 1.00 cm cuvette. The concentration is 9.0×10−5 M9.0\times10^{-5}\ \text{M}. What is ε\varepsilon?

  1. ε=3.0×103 L mol−1cm−1\varepsilon=3.0\times10^3\ \text{L mol}^{-1}\text{cm}^{-1} (correct answer)
  2. ε=3.0×104 L mol−1cm−1\varepsilon=3.0\times10^4\ \text{L mol}^{-1}\text{cm}^{-1}
  3. ε=2.4×103 L mol−1cm−1\varepsilon=2.4\times10^3\ \text{L mol}^{-1}\text{cm}^{-1}
  4. ε=2.4×104 L mol−1cm−1\varepsilon=2.4\times10^4\ \text{L mol}^{-1}\text{cm}^{-1}
  5. ε=9.0×103 L mol−1cm−1\varepsilon=9.0\times10^3\ \text{L mol}^{-1}\text{cm}^{-1}

Explanation: This question tests the calculation of molar absorptivity from Beer-Lambert law experimental data. Using A = εℓc and solving for ε: ε = A/(ℓc) = 0.27/(1.00 × 9.0×10⁻⁵) = 0.27/(9.0×10⁻⁵) = 3.0×10³ L mol⁻¹ cm⁻¹. This matches the marked answer exactly. Students might choose option B (3.0×10⁴) by incorrectly handling the powers of 10 during the division. When calculating molar absorptivity from dilute solutions, pay careful attention to scientific notation arithmetic.

Question 14

A compound has ε=1.6×104 L mol−1cm−1\varepsilon=1.6\times10^4\ \text{L mol}^{-1}\text{cm}^{-1} at 500 nm. A solution in a 1.00 cm cuvette shows A=0.32A=0.32. What is the concentration?

  1. 2.0×10−5 M2.0\times10^{-5}\ \text{M} (correct answer)
  2. 5.0×10−5 M5.0\times10^{-5}\ \text{M}
  3. 2.0×10−4 M2.0\times10^{-4}\ \text{M}
  4. 5.0×10−4 M5.0\times10^{-4}\ \text{M}
  5. 3.2×10−5 M3.2\times10^{-5}\ \text{M}

Explanation: This question tests the application of Beer-Lambert law to determine concentration from absorbance data. Using A = εℓc and solving for concentration: c = A/(εℓ) = 0.32/(1.6×10⁴ × 1.00) = 2.0×10⁻⁵ M. This matches the marked answer exactly. Students might incorrectly choose option B (5.0×10⁻⁵ M) by making arithmetic errors with the scientific notation. When using Beer-Lambert law, always substitute values carefully and double-check your powers of 10.

Question 15

A solution gives absorbance A=0.25A=0.25 in a 1.00 cm cuvette at a wavelength where ε=1.0×104 L mol−1cm−1\varepsilon=1.0\times10^4\ \text{L mol}^{-1}\text{cm}^{-1}. What absorbance is expected if the same solution is measured in a 2.00 cm cuvette at the same wavelength?

  1. A=0.50A=0.50 (correct answer)
  2. A=0.125A=0.125
  3. A=0.25A=0.25
  4. A=2.0A=2.0
  5. A=0.0625A=0.0625

Explanation: This question tests how path length affects absorbance according to the Beer-Lambert law. First, we determine the concentration from the initial measurement: c = A/(εℓ) = 0.25/(1.0×10⁴ × 1.00) = 2.5×10⁻⁵ M. When the same solution is measured in a 2.00 cm cuvette, the new absorbance is A = εℓc = (1.0×10⁴)(2.00)(2.5×10⁻⁵) = 0.50. Since path length doubled while concentration remained constant, the absorbance also doubled. Students might incorrectly choose option C (0.25) by thinking absorbance doesn't change with path length. Always remember that absorbance is directly proportional to both concentration and path length.

Question 16

At a certain wavelength, ε=9.0×103 L mol−1cm−1\varepsilon=9.0\times10^3\ \text{L mol}^{-1}\text{cm}^{-1}. A student wants A=0.90A=0.90 using a 2.00 cm cuvette. What concentration is needed?

  1. 1.0×10−4 M1.0\times10^{-4}\ \text{M}
  2. 5.0×10−5 M5.0\times10^{-5}\ \text{M} (correct answer)
  3. 2.0×10−4 M2.0\times10^{-4}\ \text{M}
  4. 1.8×10−4 M1.8\times10^{-4}\ \text{M}
  5. 9.0×10−5 M9.0\times10^{-5}\ \text{M}

Explanation: This question tests the calculation of required concentration for a desired absorbance using Beer-Lambert law with a 2.00 cm cuvette. Using A = εℓc and solving for concentration: c = A/(εℓ) = 0.90/(9.0×10³ × 2.00) = 0.90/(1.8×10⁴) = 5.0×10⁻⁵ M. This matches the marked answer exactly. Students might choose option A (1.0×10⁻⁴ M) by forgetting to include the 2.00 cm path length in their calculation. Always account for the actual cuvette path length when calculating required concentrations.

Question 17

A solution has concentration c=1.0×10−4 Mc = 1.0\times10^{-4}\ \text{M} and is measured in a 1.00 cm cuvette at two different wavelengths. At 520 nm, ε520=2.0×103 L mol−1cm−1\varepsilon_{520} = 2.0\times10^{3}\ \text{L mol}^{-1}\text{cm}^{-1}. At 600 nm, ε600=5.0×103 L mol−1cm−1\varepsilon_{600} = 5.0\times10^{3}\ \text{L mol}^{-1}\text{cm}^{-1}. Which statement correctly compares the absorbances A520A_{520} and A600A_{600}?

  1. A520=A600A_{520} = A_{600} because concentration is the same
  2. A520>A600A_{520} > A_{600} because longer wavelength means higher absorbance
  3. A520<A600A_{520} < A_{600} because molar absorptivity is larger at 600 nm (correct answer)
  4. A520>A600A_{520} > A_{600} because molar absorptivity is smaller at 600 nm
  5. A520<A600A_{520} < A_{600} because path length is shorter at 520 nm

Explanation: This question tests the application of Beer-Lambert law at different wavelengths where molar absorptivity varies. At 520 nm: A₅₂₀ = ε₅₂₀ℓc = (2.0 × 10³)(1.00)(1.0 × 10⁻⁴) = 0.20. At 600 nm: A₆₀₀ = ε₆₀₀ℓc = (5.0 × 10³)(1.00)(1.0 × 10⁻⁴) = 0.50. Since ε₆₀₀ > ε₅₂₀, we have A₆₀₀ > A₅₂₀, making A₅₂₀ < A₆₀₀. Students might incorrectly choose option A by thinking concentration alone determines absorbance, ignoring that molar absorptivity varies with wavelength. When comparing absorbances at different wavelengths, always consider how molar absorptivity changes—it's a wavelength-dependent property.

Question 18

A solution has A=0.40A=0.40 in a 1.00 cm cuvette. The student changes to a different solute that has half the molar absorptivity at the same wavelength, but uses the same concentration and cuvette. What absorbance is expected?

  1. A=0.80A=0.80
  2. A=0.20A=0.20 (correct answer)
  3. A=0.40A=0.40
  4. A=0.10A=0.10
  5. A=0.60A=0.60

Explanation: This question tests understanding of how changing molar absorptivity affects absorbance while keeping other variables constant. Since A = εℓc and only ε changes (decreases by half) while c and ℓ remain constant, the new absorbance becomes A_new = (ε/2) × ℓ × c = 0.40/2 = 0.20. When the molar absorptivity is halved, the absorbance is also halved due to direct proportionality. Students might choose option A (0.80) by thinking smaller molar absorptivity increases absorbance. Remember that absorbance decreases when molar absorptivity decreases.

Question 19

A solution is analyzed at a wavelength where ε\varepsilon is constant. Using a 1.00 cm cuvette, the absorbance is A=0.25A = 0.25. If the same solution is placed in a 2.00 cm cuvette and measured at the same wavelength, what absorbance is expected?

  1. 0.500.50 (correct answer)
  2. 0.1250.125
  3. 0.750.75
  4. 2.002.00
  5. 0.250.25

Explanation: This question tests how path length affects absorbance in Beer-Lambert law. According to A = εℓc, absorbance is directly proportional to path length when concentration and molar absorptivity remain constant. Doubling the path length from 1.00 cm to 2.00 cm doubles the absorbance: A₂ = 2 × A₁ = 2 × 0.25 = 0.50. Students might incorrectly choose option C (0.25) by thinking absorbance remains unchanged with different cuvette sizes, not recognizing path length's role in the equation. When changing cuvette size, remember that absorbance scales linearly with path length—a longer path means more light absorption.

Question 20

A student measures absorbance for two solutions at the same wavelength in identical 1.00 cm cuvettes. Solution X has AX=0.60A_X = 0.60 and Solution Y has AY=0.20A_Y = 0.20. If both solutions contain the same absorbing species and Beer–Lambert law applies, what is the ratio of their concentrations cX:cYc_X:c_Y?

  1. 1:31:3
  2. 2:12:1
  3. 3:13:1 (correct answer)
  4. 1:21:2
  5. 1:11:1

Explanation: This question tests understanding of the proportional relationship between absorbance and concentration. Since both solutions are measured under identical conditions (same wavelength, cuvette, and species), the ratio of absorbances equals the ratio of concentrations: A_X/A_Y = c_X/c_Y. Therefore, c_X/c_Y = 0.60/0.20 = 3/1, giving a ratio of 3:1. Students might incorrectly choose option D (1:2) by inverting the ratio or misinterpreting which solution has higher concentration. When comparing absorbances of the same species under identical conditions, higher absorbance always indicates proportionally higher concentration.