College Chemistry Quiz: Beer Lambert Law
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Beer Lambert LawQuestion 1 of 10

Two solutions of the same colored compound are prepared. Solution A has twice the concentration of Solution B. If Solution B has an absorbance of 0.320 when measured in a 2.00 cm cuvette, what would be the expected absorbance of Solution A when measured in a 1.00 cm cuvette?

0.160
0.320
0.480
0.640
1.28
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College Chemistry Quiz

College Chemistry Quiz: Beer Lambert Law

Practice Beer Lambert Law in College 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 College 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

Two solutions of the same colored compound are prepared. Solution A has twice the concentration of Solution B. If Solution B has an absorbance of 0.320 when measured in a 2.00 cm cuvette, what would be the expected absorbance of Solution A when measured in a 1.00 cm cuvette?

  1. 0.160
  2. 0.320 (correct answer)
  3. 0.480
  4. 0.640
  5. 1.28
Explanation: When you encounter questions about light absorption and concentration, you're working with Beer's Law, which relates absorbance to concentration and path length: A=εbcA = \varepsilon bc, where A is absorbance, ε is the molar absorptivity constant, b is path length, and c is concentration. Let's analyze what's happening here. Solution B has absorbance 0.320 in a 2.00 cm cuvette. Solution A has twice the concentration but is measured in a 1.00 cm cuvette (half the path length). Since absorbance is directly proportional to both concentration and path length, doubling the concentration doubles the absorbance, while halving the path length halves the absorbance. These effects exactly cancel out: AA=AB×cAcB×bAbB=0.320×2×0.5=0.320A_A = A_B \times \frac{c_A}{c_B} \times \frac{b_A}{b_B} = 0.320 \times 2 \times 0.5 = 0.320 Choice A (0.160) incorrectly assumes only the path length change matters, ignoring the concentration increase. Choice C (0.480) results from adding the concentration effect (0.320 + 0.160) rather than using the multiplicative relationship. Choice D (0.640) comes from only considering the concentration doubling while ignoring the path length reduction. The correct answer is B (0.320) because the concentration doubling and path length halving produce offsetting effects. Remember this pattern: Beer's Law problems often test whether you can track multiple variables simultaneously. Always identify what's changing (concentration, path length, or both) and apply the direct proportional relationships correctly. Don't just focus on one variable while forgetting the others.

Question 2

A student measures the absorbance of a solution containing 2.5×10−4 M2.5 \times 10^{-4} \text{ M} of a dye in a 0.50 cm cuvette and obtains a value of 0.625. What is the molar absorptivity of this dye?

  1. 1.3×103 M−1cm−11.3 \times 10^3 \text{ M}^{-1}\text{cm}^{-1}
  2. 2.5×103 M−1cm−12.5 \times 10^3 \text{ M}^{-1}\text{cm}^{-1}
  3. 5.0×103 M−1cm−15.0 \times 10^3 \text{ M}^{-1}\text{cm}^{-1} (correct answer)
  4. 1.0×104 M−1cm−11.0 \times 10^4 \text{ M}^{-1}\text{cm}^{-1}
  5. 7.8×10−5 M−1cm−17.8 \times 10^{-5} \text{ M}^{-1}\text{cm}^{-1}
Explanation: When you encounter spectroscopy problems involving absorbance measurements, you're working with Beer's Law, which relates how much light a solution absorbs to its concentration and path length. The fundamental equation is A=εbcA = \varepsilon bc, where A is absorbance, ε is molar absorptivity, b is path length, and c is concentration. To find the molar absorptivity, rearrange Beer's Law to solve for ε: ε=Abc\varepsilon = \frac{A}{bc}. Substituting the given values: ε=0.625(0.50 cm)(2.5×10−4 M)=0.6251.25×10−4=5.0×103 M−1cm−1\varepsilon = \frac{0.625}{(0.50 \text{ cm})(2.5 \times 10^{-4} \text{ M})} = \frac{0.625}{1.25 \times 10^{-4}} = 5.0 \times 10^3 \text{ M}^{-1}\text{cm}^{-1}. This matches answer C. Let's examine why the other options are incorrect. Answer A (1.3×1031.3 \times 10^3) would result if you incorrectly used the concentration as 2.5×10−42.5 \times 10^{-4} but made an arithmetic error in the division. Answer B (2.5×1032.5 \times 10^3) comes from forgetting to include the path length in your calculation—dividing absorbance by concentration alone. Answer D (1.0×1041.0 \times 10^4) would arise if you accidentally used twice the correct concentration or half the correct path length. Remember that molar absorptivity is an intrinsic property of the absorbing species at a specific wavelength. Always double-check that you're using the correct units and that your final answer has the proper units of M−1cm−1\text{M}^{-1}\text{cm}^{-1}. Setting up Beer's Law problems systematically and checking unit consistency will help you avoid common calculation errors.

Question 3

A solution shows 85% transmittance when analyzed by spectrophotometry. What is the absorbance of this solution?

  1. 0.070 (correct answer)
  2. 0.15
  3. 0.85
  4. 1.2
  5. 15
Explanation: When you encounter spectrophotometry problems, you're working with the relationship between how much light passes through a solution (transmittance) and how much light the solution absorbs (absorbance). These quantities are inversely related through a logarithmic function. The key equation connecting transmittance (T) and absorbance (A) is: A=−log⁡(T)A = -\log(T), where transmittance is expressed as a decimal fraction, not a percentage. Since this solution has 85% transmittance, you first convert to decimal form: T = 0.85. Now you can calculate: A=−log⁡(0.85)=−(−0.071)=0.071A = -\log(0.85) = -(-0.071) = 0.071, which rounds to 0.070, making choice A correct. Let's examine why the other options are wrong. Choice B (0.15) might result from incorrectly using A=log⁡(T)A = \log(T) without the negative sign, or from calculation errors. Choice C (0.85) is simply the transmittance value itself—this represents a common misconception that absorbance equals transmittance. Choice D (1.2) could come from using the percentage form directly in the equation (−log⁡(85)-\log(85)) instead of converting to decimal first, or from other fundamental calculation mistakes. Remember this key relationship: high transmittance means low absorbance, and vice versa. When transmittance is above 50%, absorbance will always be less than 0.3. Practice converting between these units using the logarithmic relationship, and always convert percentages to decimals before calculating.

Question 4

When the concentration of a solution is doubled while keeping the path length constant, the transmittance changes from 50% to a new value. What is the new transmittance?

  1. 12.5%
  2. 25% (correct answer)
  3. 35.4%
  4. 70.7%
  5. 100%
Explanation: When you encounter transmittance problems, you're working with the Beer-Lambert Law, which relates how much light passes through a solution to its concentration. The key relationship is that transmittance (T) and concentration (c) follow the equation: T=10−εbcT = 10^{-\varepsilon bc}, where ε is the molar absorptivity and b is the path length. Since the path length stays constant and only concentration changes, you can use the relationship that when concentration doubles, the absorbance doubles. Starting with 50% transmittance, the absorbance is A1=−log⁡(0.50)=0.301A_1 = -\log(0.50) = 0.301. When concentration doubles, A2=2×0.301=0.602A_2 = 2 \times 0.301 = 0.602. Converting back to transmittance: T2=10−0.602=0.25=25%T_2 = 10^{-0.602} = 0.25 = 25\%. This confirms answer B is correct. Looking at the wrong answers: A (12.5%) represents what you'd get if you incorrectly applied a linear relationship, thinking doubling concentration means halving transmittance again (50% → 25% → 12.5%). C (35.4%) might come from incorrectly using natural logarithms instead of base-10 logarithms in your calculations. D (70.7%) could result from confusing the relationship direction or applying square root operations inappropriately. Remember that transmittance and concentration have an exponential, not linear, relationship. When concentration doubles, absorbance doubles, which means you square the original transmittance fraction (0.5² = 0.25). This exponential relationship is fundamental to spectroscopy problems.

Question 5

A solution has an absorbance of 1.85. What percentage of light is transmitted through this solution?

  1. 1.4% (correct answer)
  2. 15%
  3. 18.5%
  4. 81.5%
  5. 98.6%
Explanation: This question tests your understanding of the Beer-Lambert Law and the relationship between absorbance and transmittance. When light passes through a solution, some is absorbed and some is transmitted, and these quantities are mathematically related. The key relationship is: A=−log⁡TA = -\log T or T=10−AT = 10^{-A}, where A is absorbance and T is transmittance (as a decimal). With an absorbance of 1.85, you calculate: T=10−1.85=0.014T = 10^{-1.85} = 0.014. To convert to percentage, multiply by 100: 0.014 × 100 = 1.4%. Looking at the wrong answers: Choice B (15%) would correspond to an absorbance of about 0.82, showing a confusion between the exponential relationship and perhaps thinking transmittance decreases linearly with absorbance. Choice C (18.5%) represents the common trap of assuming transmittance percentage equals 100 minus absorbance (100 - 1.85 × 10 = 18.5), which incorrectly treats absorbance as if it has a simple linear relationship with transmittance. Choice D (81.5%) follows the same flawed logic but calculates 100 - 18.5, essentially getting the "opposite" of choice C. Remember that absorbance and transmittance have an inverse logarithmic relationship, not a linear one. High absorbance means very little light gets through. As a rule of thumb: absorbance of 1 = 10% transmission, absorbance of 2 = 1% transmission. This exponential decay pattern is crucial for spectroscopy calculations.

Question 6

A student dilutes a solution from 0.15 M to 0.030 M and measures the absorbance in identical cuvettes. If the original solution had an absorbance of 1.25, what should be the absorbance of the diluted solution?

  1. 0.25 (correct answer)
  2. 0.50
  3. 0.75
  4. 1.00
  5. 6.25
Explanation: When you encounter dilution problems involving absorbance measurements, you're dealing with Beer's Law, which states that absorbance is directly proportional to concentration when path length and molar absorptivity remain constant. Since identical cuvettes are used, the path length stays the same, creating a direct relationship between concentration and absorbance. To solve this, you can use the dilution relationship: A1A2=C1C2\frac{A_1}{A_2} = \frac{C_1}{C_2}, where A represents absorbance and C represents concentration. The original concentration is 0.15 M with an absorbance of 1.25, and the diluted concentration is 0.030 M. Setting up the equation: 1.25A2=0.150.030\frac{1.25}{A_2} = \frac{0.15}{0.030} Solving for A2A_2: A2=1.25×0.0300.15=0.03750.15=0.25A_2 = \frac{1.25 \times 0.030}{0.15} = \frac{0.0375}{0.15} = 0.25 Therefore, the absorbance of the diluted solution should be 0.25, making A correct. Looking at the wrong answers: B (0.50) represents only a 2.5-fold dilution rather than the actual 5-fold dilution (0.15 ÷ 0.030 = 5). C (0.75) suggests an incorrect proportional relationship that doesn't match the concentration change. D (1.00) implies the absorbance decreased by only 0.25 units, which doesn't correspond to the actual dilution factor. Remember this key strategy: in dilution problems with spectrophotometry, the dilution factor for concentration equals the dilution factor for absorbance. Calculate the dilution factor first (here, 5×), then divide the original absorbance by that same factor.

Question 7

Two solutions of the same compound are prepared with concentrations of 0.025 M and 0.075 M. If the first solution has an absorbance of 0.42 in a 1.0 cm cuvette, what is the expected absorbance of the second solution in the same cuvette?

  1. 0.14
  2. 0.42
  3. 0.84
  4. 1.26 (correct answer)
  5. 3.15
Explanation: This question tests your understanding of Beer's Law, which describes the relationship between light absorption and concentration. When you encounter problems involving absorbance and concentration changes, always think about the direct proportional relationship between these variables. Beer's Law states that A=εbcA = \varepsilon bc, where A is absorbance, ε is the molar absorptivity coefficient, b is the path length, and c is concentration. Since the same compound is used in the same cuvette, both ε and b remain constant. This means absorbance is directly proportional to concentration. You can solve this using a simple ratio: A1c1=A2c2\frac{A_1}{c_1} = \frac{A_2}{c_2}. Substituting the known values: 0.420.025=A20.075\frac{0.42}{0.025} = \frac{A_2}{0.075}. Cross-multiplying gives A2=0.42×0.0750.025=1.26A_2 = \frac{0.42 \times 0.075}{0.025} = 1.26. Looking at the incorrect answers: Choice A (0.14) represents dividing the original absorbance by 3 instead of multiplying—this reverses the concentration relationship. Choice B (0.42) assumes absorbance doesn't change with concentration, ignoring Beer's Law entirely. Choice C (0.84) results from doubling the original absorbance, which would be correct if the concentration had doubled, but 0.075 M is actually three times 0.025 M. Remember this key strategy: For Beer's Law problems, when all variables except concentration remain constant, absorbance scales directly with concentration. Set up a simple proportion rather than memorizing the full equation—it's faster and less error-prone on exams.

Question 8

A solution of copper(II) sulfate has an absorbance of 0.450 at 635 nm in a 1.00 cm cuvette. If the molar absorptivity of CuSO4CuSO_4 at this wavelength is 12.0 M−1cm−112.0 \text{ M}^{-1}\text{cm}^{-1}, what is the concentration of the solution?

  1. 0.0375 M (correct answer)
  2. 0.0450 M
  3. 0.540 M
  4. 5.40 M
  5. 13.5 M
Explanation: This question tests your understanding of Beer's Law, the fundamental relationship in spectrophotometry that connects light absorption to solution concentration. When you see absorbance, molar absorptivity, and path length given, you should immediately think of the Beer's Law equation: A=εbcA = \varepsilon bc, where A is absorbance, ε is molar absorptivity, b is path length, and c is concentration. To find the concentration, rearrange Beer's Law to solve for c: c=Aεbc = \frac{A}{\varepsilon b}. Substituting the given values: c=0.450(12.0 M−1cm−1)(1.00 cm)=0.45012.0=0.0375 Mc = \frac{0.450}{(12.0 \text{ M}^{-1}\text{cm}^{-1})(1.00 \text{ cm})} = \frac{0.450}{12.0} = 0.0375 \text{ M} Looking at the wrong answers: Choice B (0.0450 M) comes from incorrectly dividing absorbance by path length only, ignoring molar absorptivity entirely. Choice C (0.540 M) results from multiplying absorbance by molar absorptivity instead of dividing, showing a fundamental misunderstanding of the Beer's Law relationship. Choice D (5.40 M) comes from multiplying all three values together rather than using the correct algebraic manipulation. The correct answer is A (0.0375 M). Study tip: Always write out Beer's Law and clearly identify each variable before solving. The most common mistakes involve algebraic errors or forgetting to include all variables in the calculation. Practice rearranging A=εbcA = \varepsilon bc for each variable until it becomes automatic.

Question 9

A colored solution in a 1.0 cm cuvette has an absorbance of 0.48. The same solution is then analyzed in a 2.5 cm cuvette. Assuming the solution follows Beer's Law, what absorbance would be measured?

  1. 0.19
  2. 0.48
  3. 0.96
  4. 1.2 (correct answer)
  5. 3.0
Explanation: When you encounter questions about absorbance and path length, you're dealing with Beer's Law, which describes how light absorption relates to solution concentration and the distance light travels through the sample. Beer's Law states that A=εbcA = \varepsilon bc, where A is absorbance, ε is the molar absorptivity coefficient, b is the path length (cuvette width), and c is concentration. Since the same solution is used in both measurements, ε and c remain constant. This means absorbance is directly proportional to path length. You can solve this using a simple ratio. If the path length increases from 1.0 cm to 2.5 cm, that's a factor of 2.5. Since absorbance is directly proportional to path length, the new absorbance will be: 0.48×2.5=1.20.48 \times 2.5 = 1.2 Looking at the wrong answers: A) 0.19 incorrectly divides the original absorbance by 2.5, suggesting an inverse relationship between absorbance and path length. B) 0.48 assumes absorbance stays the same regardless of path length, ignoring Beer's Law entirely. C) 0.96 multiplies by 2 instead of 2.5, perhaps from misreading the new path length or making an arithmetic error. Remember that in Beer's Law problems, absorbance scales linearly with path length when all other variables are held constant. Always check whether you're dealing with a longer or shorter path length, and adjust absorbance proportionally in the same direction.

Question 10

Use the data in the table to determine the molar absorptivity of the compound at the given wavelength. Based on the table shown, what is the molar absorptivity in M−1cm−1\text{M}^{-1}\text{cm}^{-1}?

  1. 1.8×1031.8 \times 10^3
  2. 2.4×1032.4 \times 10^3
  3. 3.6×1033.6 \times 10^3 (correct answer)
  4. 4.5×1034.5 \times 10^3
  5. 7.2×1037.2 \times 10^3
Explanation: The molar absorptivity is calculated as ε = A/(bc) for each data point. Using any row: For 1.0×10⁻⁴ M: ε = 0.36/(1.0×10⁻⁴ × 1.0) = 3.6×10³ M⁻¹cm⁻¹. This can be verified with other data points. Choice A uses only half the absorbance value. Choice B results from calculation errors. Choice D comes from using wrong concentration units. Choice E doubles the correct value.