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CHM2046L Final Exam Study Guide - Spring 2026 University of Florida | General Chemistry 2 Lab Review & Practice

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CHM2046L Final Exam Study Guide - Spring 2026 University of Florida | General Chemistry 2 Lab Review & Practice

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CHM2046L Final Exam Study Guide - Spring
2026 University of Florida | General
Chemistry 2 Lab Review & Practice



Topic 1: Spectrophotometry & Beer-Lambert Law (Questions 1-20)

Question 1
Which of the following is the correct mathematical expression for the Beer-Lambert law?
A) A = εdc
B) A = ε / (dc)
C) A = dc / ε
D) A = εd / c

Answer: A

• Rationale for A: This is the standard form of the Beer-Lambert law, where A is absorbance, ε is
the molar absorptivity (L mol⁻¹ cm⁻¹), d is the path length (cm), and c is the concentration
(mol/L). The law states that absorbance is directly proportional to both concentration and path
length .

• Rationale for B: This would suggest an inverse relationship between absorbance and
concentration or path length, which is incorrect.

• Rationale for C: This incorrectly places concentration and path length in the numerator and ε in
the denominator, reversing their relationships.

• Rationale for D: This suggests absorbance decreases with concentration or path length, which
contradicts the direct proportionality described by the Beer-Lambert law.

Question 2
A solution of a colored compound has an absorbance of 0.451 at a concentration of 2.0 × 10⁻⁵ M
measured in a 1.00 cm cuvette. What is the molar absorptivity (ε) of the compound?
A) 1.13 × 10³ L mol⁻¹ cm⁻¹
B) 2.26 × 10⁴ L mol⁻¹ cm⁻¹
C) 4.52 × 10⁴ L mol⁻¹ cm⁻¹
D) 9.04 × 10⁴ L mol⁻¹ cm⁻¹

Answer: B

, • Rationale for B: Using the Beer-Lambert law, A = εdc. Rearranged to solve for ε gives ε = A / (dc).
Substituting the values: ε = 0.451 / (1.00 cm × 2.0 × 10⁻⁵ M) = 2.255 × 10⁴ L mol⁻¹ cm⁻¹, which
rounds to 2.26 × 10⁴ L mol⁻¹ cm⁻¹ .

• Rationale for A: This answer might come from incorrectly multiplying A and c instead of
dividing.

• Rationale for C: This could result from forgetting to account for the path length (d) in the
calculation.

• Rationale for D: This might arise from inverting the concentration term (dividing by 2.0 × 10⁵
instead of 10⁻⁵).

Question 3
You generate a Beer-Lambert calibration curve and obtain the linear regression equation y = 22066x +
0.003 with an R² value of 0.999. What does the y-intercept represent in this context?
A) The molar absorptivity of the compound
B) The absorbance of a solution with zero concentration of the colored analyte
C) The concentration of the most dilute standard
D) The path length of the cuvette

Answer: B

• Rationale for B: In a calibration curve of Absorbance (y) vs. Concentration (x), the y-intercept is
the theoretical absorbance value when the concentration is zero. Ideally, it should be zero,
meaning no light is absorbed when no analyte is present. A non-zero intercept indicates a
systematic error, such as not properly blanking the spectrophotometer .

• Rationale for A: The slope (22066) represents the product of ε and d (εd), not the intercept.

• Rationale for C: The x-axis represents concentration, so the intercept's value has no relation to
the standards' concentrations.

• Rationale for D: Path length is a constant physical property of the cuvette and is not
represented by the y-intercept.

Question 4
A student measures the absorbance of an unknown solution and obtains A = 1.25. Using the calibration
curve equation y = 22066x + 0.003, what is the concentration of the unknown solution?
A) 2.50 × 10⁻⁵ M
B) 5.65 × 10⁻⁵ M
C) 6.77 × 10⁻⁵ M
D) 1.25 × 10⁻⁵ M

Answer: B

• Rationale for B: To find the concentration, we solve for x (concentration) by plugging y
(absorbance) into the equation. 1.25 = 22066x + 0.003 -> 22066x = 1.247 -> x = 1. =
5.65 × 10⁻⁵ M .

, • Rationale for A: This might come from misplacing the decimal or using an incorrect formula,
such as not subtracting the intercept.

• Rationale for C: This answer could result from adding 0.003 to 1.25 instead of subtracting it.

• Rationale for D: This is a simple order-of-magnitude guess and does not use the provided
calibration equation.

Question 5
A solution appears green. What wavelength of visible light is it primarily absorbing?
A) 400 nm (violet)
B) 480 nm (blue)
C) 560 nm (green)
D) 680 nm (red)

Answer: D

• Rationale for D: The color we see is the complementary color of the light being absorbed. A
solution appears green because it is absorbing light from the red region of the spectrum
(approximately 630-750 nm), which is the complement of green .

• Rationale for A & B: Violet and blue light are complements of yellow and orange, respectively. If
a solution absorbed these, it would appear yellow or orange.

• Rationale for C: If a solution absorbed green light (560 nm), it would appear red, which is the
complementary color to green.

Question 6
A student prepares a series of standard solutions for a Beer-Lambert calibration curve. The R² value of
the linear regression is 0.95. What does this indicate?
A) 95% of the variation in absorbance is explained by concentration.
B) The calibration curve is perfect with no error.
C) The data do not follow Beer-Lambert law at all.
D) The cuvette path length is incorrect.

Answer: A

• Rationale for A: R², the coefficient of determination, quantifies how well the regression line fits
the data. An R² of 0.95 means that 95% of the variance in the dependent variable (absorbance)
can be predicted from the independent variable (concentration). While acceptable for some
analyses, a value above 0.990 is generally expected for a high-quality calibration curve in
analytical chemistry .

• Rationale for B: A perfect curve would have an R² value of 1.000, indicating no deviation from
the line.

• Rationale for C: An R² of 0.95 still indicates a strong linear correlation, but it's not perfect.

, • Rationale for D: An incorrect path length would systematically affect the slope of the curve, but
the data could still be perfectly linear (R² = 1.000) if all measurements were made with the same
incorrect cuvette.

Question 7
You have a stock solution of 6.00 × 10⁻⁵ M dye. What volume of this stock solution is needed to prepare
20.00 mL of a 2.00 × 10⁻⁵ M dye solution?
A) 3.33 mL
B) 6.67 mL
C) 10.0 mL
D) 15.0 mL

Answer: B

• Rationale for B: Use the dilution equation M₁V₁ = M₂V₂. M₁ = 6.00 × 10⁻⁵ M, M₂ = 2.00 × 10⁻⁵ M,
V₂ = 20.00 mL. Solving for V₁: V₁ = (M₂V₂) / M₁ = (2.00 × 10⁻⁵ M * 20.00 mL) / (6.00 × 10⁻⁵ M) =
6.67 mL .

• Rationale for A: This might result from incorrectly using M₁V₁ = M₂V₂ as V₁ = (M₁V₂)/M₂.

• Rationale for C: This would be correct if you were making a 3.00 × 10⁻⁵ M solution (half the
stock concentration).

• Rationale for D: This is the volume needed to make a 4.50 × 10⁻⁵ M solution.

Question 8
What is the primary purpose of using a "blank" solution in spectrophotometry?
A) To calibrate the instrument to zero absorbance.
B) To increase the absorbance of the sample.
C) To dilute the sample.
D) To clean the cuvette.

Answer: A

• Rationale for A: A blank solution contains all components of the sample matrix except the
analyte of interest. By placing the blank in the spectrophotometer and setting its absorbance to
zero, we account for any light absorbed by the solvent, the cuvette, or other non-analyte
species. This ensures that subsequent absorbance readings are due only to the analyte .

• Rationale for B: The blank does not affect the sample's intrinsic properties.

• Rationale for C: Dilution is a separate preparatory step, not the function of a blank.

• Rationale for D: Cuvettes are cleaned with solvents, not by running a blank.

Question 9
A solution has 50% transmittance (%T). What is its absorbance?
A) 0.301
B) 0.699

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