Beer-Lambert Law - AP Chemistry

A solution has concentration $1.0\times10^{-5}\ \text{M}$ and is measured in a 1.00 cm cuvette at a wavelength where $\varepsilon=3.0\times10^4\ \text{L mol}^{-1}\text{cm}^{-1}$ . What absorbance is expected?
Answer: $A=0.30$ 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. 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.
A solution is measured at a fixed wavelength where its molar absorptivity is constant. In a $1.00\ \text{cm}$ cuvette the absorbance is $A = 0.40$ . If the concentration of the solution is doubled and the path length is halved, what absorbance is expected under the same conditions?
Answer: $A = 0.40$ This question tests understanding of how simultaneous changes in concentration and path length affect absorbance. According to Beer-Lambert law A = εℓc, if concentration doubles (c → 2c) and path length halves (ℓ → ℓ/2), the new absorbance becomes A_new = ε(ℓ/2)(2c) = εℓc = A_original. The factors of 2 and 1/2 cancel out, leaving the absorbance unchanged at 0.40. Students might incorrectly choose option B (0.80) by only considering the concentration doubling and ignoring the path length change. When multiple parameters change in Beer-Lambert law, multiply all the change factors together to determine the net effect on absorbance.
A solution has absorbance $A=0.36$ in a 1.00 cm cuvette. The student switches to a 1.50 cm cuvette with the same solution and wavelength. What absorbance is expected?
Answer: $A=0.54$ This question tests understanding of how path length affects absorbance in Beer-Lambert law. Since A = εℓc and only path length changes (from 1.00 cm to 1.50 cm), the new absorbance is A_new = A_original × (ℓ_new/ℓ_original) = 0.36 × (1.50/1.00) = 0.54. The 50% increase in path length produces a 50% increase in absorbance. Students might choose option C (0.36) by thinking path length changes don't affect absorbance. Always remember that absorbance increases proportionally with path length when other variables remain constant.
Two solutions of the same dye are measured at the same wavelength using identical 1.00 cm cuvettes. Solution 1 has concentration $1.0\times10^{-4}\ \text{M}$ and absorbance $A_1 = 0.30$ . Solution 2 has concentration $3.0\times10^{-4}\ \text{M}$ . Assuming Beer–Lambert law holds, what is the expected absorbance $A_2$ for Solution 2?
Answer: $0.90$ This question tests the proportional relationship between concentration and absorbance in Beer-Lambert law. Since both solutions use the same dye at the same wavelength with identical cuvettes, ε and ℓ are constant, making absorbance directly proportional to concentration. If concentration triples from 1.0 × 10⁻⁴ M to 3.0 × 10⁻⁴ M, absorbance also triples: A₂ = 3 × A₁ = 3 × 0.30 = 0.90. Students might incorrectly choose option D (0.30) by thinking absorbance remains constant regardless of concentration, misunderstanding the fundamental relationship in Beer-Lambert law. When concentration changes but all other factors remain constant, absorbance changes proportionally by the same factor.
A student measures absorbance for two solutions at the same wavelength in identical 1.00 cm cuvettes. Solution X has $A_X = 0.60$ and Solution Y has $A_Y = 0.20$ . If both solutions contain the same absorbing species and Beer–Lambert law applies, what is the ratio of their concentrations $c_X:c_Y$ ?
Answer: $3:1$ 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.
A student measures the absorbance of a solution at a fixed wavelength. The solution is then diluted by adding an equal volume of water (doubling the total volume). If the original absorbance was $A = 0.80$ using the same cuvette, what is the absorbance after dilution (assume Beer–Lambert law applies and $\varepsilon$ is constant)?
Answer: $0.40$ This question tests understanding of how dilution affects absorbance according to Beer-Lambert law. When a solution is diluted by adding an equal volume of water, the concentration is halved (c₂ = c₁/2). Since A = εℓc and both ε and ℓ remain constant, the absorbance is also halved: A₂ = A₁/2 = 0.80/2 = 0.40. Students might incorrectly choose option B (0.80) by thinking dilution doesn't affect absorbance, failing to recognize that concentration changes. When diluting a solution, remember that absorbance decreases proportionally to the dilution factor because concentration decreases.
A solution gives absorbance $A=0.36$ in a 1.0 cm cuvette. The concentration is $0.012\ \text{M}$ . What is the molar absorptivity $\varepsilon$ at this wavelength?
Answer: $30\ \text{M}^{-1}\text{cm}^{-1}$ This question tests calculating molar absorptivity from absorbance and concentration via the Beer-Lambert law. Using ε = A / (ℓc), ε = 0.36 / (1.0 cm × 0.012 M) = 30 M⁻¹cm⁻¹. This aligns with choice B, straightforward from the data. The 1.0 cm path length simplifies the division. A tempting distractor is choice D (300 M⁻¹cm⁻¹), arising from misreading concentration as 0.0012 M. Double-check decimal places in concentration to avoid order-of-magnitude errors.
A student measures a solution at a wavelength where $\varepsilon$ is known to be $2.0\times10^4\ \text{L mol}^{-1}\text{cm}^{-1}$ . Using a 0.50 cm cuvette, the absorbance is 0.30. What is the concentration?
Answer: $3.0\times10^{-5}\ \text{M}$ This question tests the application of Beer-Lambert law with a shorter cuvette path length. Using A = εℓc and solving for concentration: c = A/(εℓ) = 0.30/(2.0×10⁴ × 0.50) = 0.30/(1.0×10⁴) = 3.0×10⁻⁵ M. This matches the marked answer exactly. Students might choose option B (1.5×10⁻⁵ M) by incorrectly dividing the final result by 2. When using non-standard cuvette lengths, ensure you include the correct path length value in your Beer-Lambert calculation.
A solution with concentration $0.015\ \text{M}$ is measured in a 1.0 cm cuvette and has absorbance $A=0.30$ . If the concentration is increased to $0.020\ \text{M}$ (same solute, same wavelength, same cuvette), what absorbance is expected?
Answer: $0.40$ This question tests predicting absorbance after a concentration change using Beer-Lambert proportionality. A is proportional to c, so increasing c from 0.015 M to 0.020 M (4/3 factor) increases A from 0.30 to 0.30 × 4/3 = 0.40. This aligns with choice A, with constant ℓ and ε. The same cuvette ensures no path change. A tempting distractor is choice E (0.50), from adding instead of scaling proportionally. Use ratios of concentrations for quick absorbance adjustments in similar scenarios.
A solution has $A=0.50$ in a 1.00 cm cuvette. If the concentration is tripled and the path length is halved (to 0.50 cm) at the same wavelength, what absorbance is expected?
Answer: $A=0.75$ This question tests understanding of combined effects on absorbance when both concentration and path length change. Using A = εℓc, the new absorbance is A_new = ε × (0.50 cm) × (3c) = 1.5εc compared to the original A_original = ε × (1.00 cm) × c = εc. Therefore, A_new = 1.5 × A_original = 1.5 × 0.50 = 0.75. The tripling of concentration more than compensates for the halving of path length. Students might choose option D (0.50) by thinking the effects cancel exactly. When multiple variables change, calculate the net effect by considering all changes in the Beer-Lambert equation.
A solution with concentration $2.0\times10^{-4}\ \text{M}$ is measured in a 1.00 cm cuvette and has $A=0.40$ . What absorbance is expected if the same solution is measured in a 0.25 cm cuvette at the same wavelength?
Answer: $A=0.10$ This question tests understanding of how path length affects absorbance in Beer-Lambert law. Since A = εℓc and only path length changes (from 1.00 cm to 0.25 cm), the new absorbance is A_new = A_original × (ℓ_new/ℓ_original) = 0.40 × (0.25/1.00) = 0.10. The 4-fold decrease in path length produces a 4-fold decrease in absorbance. Students might choose option C (0.40) by forgetting that path length affects absorbance. Remember that shorter cuvettes produce proportionally lower absorbances for the same solution.
A student measures two solutions of the same solute at the same wavelength in 1.00 cm cuvettes. Solution A has $A=0.20$ . Solution B has twice the concentration of Solution A. What is the expected absorbance of Solution B?
Answer: $A=0.40$ This question tests understanding of direct proportionality between concentration and absorbance in Beer-Lambert law. Since A = εℓc with constant ε, ℓ, and wavelength, absorbance is directly proportional to concentration. Solution B has twice the concentration, so it will have exactly twice the absorbance: $A_B$ = 2 × 0.20 = 0.40. Students might choose option C (0.20) by thinking concentration doesn't affect absorbance when the same solute is used. Remember that for the same solute under identical conditions, absorbance scales linearly with concentration.
A solution has absorbance $A = 0.90$ at a certain wavelength when measured in a $1.00\ \text{cm}$ cuvette. The solution is then diluted by mixing 1.00 mL of the solution with 2.00 mL of solvent (total volume 3.00 mL) and measured again in the same cuvette at the same wavelength. What absorbance is expected after dilution?
Answer: $A = 0.30$ This question tests understanding of how dilution affects absorbance according to Beer-Lambert law. When 1.00 mL of solution is mixed with 2.00 mL of solvent, the total volume becomes 3.00 mL, so the concentration is reduced by a factor of 3 (from c to c/3). Since A = εℓc and both ε and ℓ remain constant, the absorbance is also reduced by a factor of 3: A_new = 0.90/3 = 0.30. Students might incorrectly choose option C (0.60) by thinking the concentration is halved rather than reduced to one-third, confusing the volume of added solvent with the dilution factor. To solve dilution problems, always calculate the dilution factor as (initial volume)/(final volume), then apply this factor to the concentration and consequently to the absorbance.
At a certain wavelength, $\varepsilon=9.0\times10^3\ \text{L mol}^{-1}\text{cm}^{-1}$ . A student wants $A=0.90$ using a 2.00 cm cuvette. What concentration is needed?
Answer: $5.0\times10^{-5}\ \text{M}$ 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.
A solution has $A=0.15$ when measured in a 1.00 cm cuvette. If the solution is diluted to one-third its original concentration and measured again in the same cuvette at the same wavelength, what absorbance is expected?
Answer: $A=0.05$ This question tests understanding of how dilution affects absorbance according to Beer-Lambert law. When a solution is diluted to one-third its original concentration, the new concentration becomes c_new = c_original/3. Since A = εℓc and only concentration changes, the new absorbance is A_new = A_original/3 = 0.15/3 = 0.05. The direct proportionality between concentration and absorbance means dilution by a factor of 3 reduces absorbance by the same factor. Students might choose option D (0.15) by forgetting that dilution changes the concentration and thus the absorbance. Always account for concentration changes when calculating absorbance after dilution.
A solution has absorbance $A=0.45$ in a 1.50 cm cuvette. The molar absorptivity is $\varepsilon=3.0\times10^3\ \text{L mol}^{-1}\text{cm}^{-1}$ . What is the concentration?
Answer: $1.0\times10^{-4}\ \text{M}$ This question tests the ability to calculate concentration using Beer-Lambert law with non-standard path lengths. Using A = εℓc and solving for concentration: c = A/(εℓ) = 0.45/(3.0×10³ × 1.50) = 0.45/(4.5×10³) = 1.0×10⁻⁴ M. This matches the marked answer perfectly. Students might choose option C (1.5×10⁻⁴ M) by forgetting to account for the 1.50 cm path length in their calculation. When working with non-standard cuvette lengths, always include the path length in your Beer-Lambert calculations.
A student uses a 1.00 cm cuvette to measure a solution and records $A=0.20$ . The student then prepares a new solution with twice the concentration of the original and measures it in the same cuvette at the same wavelength. What absorbance is expected for the new solution?
Answer: $A=0.40$ This question tests understanding of the direct proportionality between concentration and absorbance in Beer-Lambert law. Since A = εℓc with constant ε, ℓ, and wavelength, absorbance is directly proportional to concentration. If concentration doubles, absorbance also doubles: A_new = 2 × 0.20 = 0.40. The linear relationship means that any change in concentration produces a proportional change in absorbance. Students might incorrectly choose option C (0.20) by thinking concentration changes don't affect absorbance. Remember that Beer-Lambert law establishes a direct linear relationship between concentration and absorbance.
A solution has concentration $c=6.0\times10^{-5}\ \text{M}$ and gives $A=0.30$ in a 1.00 cm cuvette. What absorbance is expected for a solution of the same solute at $c=2.0\times10^{-5}\ \text{M}$ in the same cuvette at the same wavelength?
Answer: $A=0.10$ This question tests understanding of direct proportionality between concentration and absorbance in Beer-Lambert law. Since A = εℓc with constant ε, ℓ, and wavelength, absorbance is directly proportional to concentration. The ratio of concentrations is (2.0×10⁻⁵)/(6.0×10⁻⁵) = 1/3, so the absorbance ratio should also be 1/3: A_new = 0.30/3 = 0.10. Students might choose option D (0.30) by forgetting that different concentrations produce proportionally different absorbances. When comparing solutions of different concentrations, always apply the concentration ratio to predict the absorbance ratio.
A student uses Beer-Lambert law, $A=\varepsilon \ell c$ , at a fixed wavelength. If the concentration of a solution is doubled while $\varepsilon$ and $\ell$ remain constant, which change in absorbance is expected?
Answer: The absorbance doubles. This question tests understanding the proportionality of absorbance to concentration in the Beer-Lambert Law when ε and ℓ are constant. The law A = ε ℓ c shows absorbance is directly proportional to concentration. Doubling c while keeping ε and ℓ unchanged therefore doubles A. This corresponds to choice D. A tempting distractor is that absorbance is cut in half, which might come from confusing concentration with transmittance or inverting the relationship, a misconception about the direct proportionality in the equation. A transferable strategy is to consider how each variable affects A independently by holding others constant and observing the linear relationships.
A student measures a solution with $A=0.35$ using a 1.00 cm cuvette. The molar absorptivity is $\varepsilon=7.0\times10^3\ \text{L mol}^{-1}\text{cm}^{-1}$ . What is the concentration?
Answer: $5.0\times10^{-5}\ \text{M}$ This question tests the application of Beer-Lambert law to calculate concentration from absorbance data. Using A = εℓc and solving for concentration: c = A/(εℓ) = 0.35/(7.0×10³ × 1.00) = 0.35/(7.0×10³) = 5.0×10⁻⁵ M. This matches the marked answer exactly. Students might choose option E (7.0×10⁻⁵ M) by confusing the molar absorptivity coefficient with the concentration result. When solving for concentration, clearly identify the unknown variable and systematically substitute the known values.

Start this quiz