Free Printable WorksheetScienceGrade 11Science CurriculumAP Chem Units 1, 3, 8, 9 and Science Practices 1-6; NGSS SEPs; CCSS HSS-ID, HSA-CED.A, WHST.11-12.7-9

11.126 Beer Lambert Law Worksheet

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Worksheet Overview & Skills Covered

Core academic competencies and curriculum alignment

This Chemistry Science worksheet provides structured practice on 11.126 Beer Lambert Law. Students are guided through targeted exercises that build foundational understanding, computational accuracy, and critical thinking skills.

Beer-Lambert Law: Use absorbance, path length, molar absorptivity, and concentration relationships to determine unknown concentration. Aligned to NGSS, CCSS, AP Chemistry/Pre-AP Chemistry, and representative state chemistry framework crosswalks AP Chem Units 1, 3, 8, 9 and Science Practices 1-6; NGSS SEPs; CCSS HSS-ID, HSA-CED.A, WHST.11-12.7-9 - Beer-Lambert Law Chemistry Standards Crosswalk: National: AP Chem Units 1, 3, 8, 9 and Science Practices 1-6; NGSS SEPs; CCSS HSS-ID, HSA-CED.A, WHST.11-12.7-9. State frameworks: TX TEKS 112.43(c)(1)-(4); MA HS-PS1-11(MA); FL/GA/VA inquiry and data-analysis standards; CA/NY/NJ science practices. Multiple-choice items can present numeric data and require correct setup, units, significant figures, and answer selection for beer-lambert law.

Whether you are an educator preparing classroom handouts, a tutor reinforcing weekly topics, or a parent supporting homeschool learning, this printable resource gives students the deliberate practice required to achieve subject fluency and master state standards.

Grade

Chemistry

Subject

Science

Domain

Science Curriculum

Formats

PDF & DOCX

Sample Practice Problems

Preview the types of problems and questions covered in this worksheet

Below are sample questions representative of the exercises included on this worksheet and related 11.126 Beer Lambert Law skill assessments:

Problem 1Difficulty: Mastery

Three replicate raw absorbances are 0.930, 0.940, and 0.920. The blank absorbance is 0.030. If ε = 9.00 x 10³ L mol^-1 cm^-1 and b = 1.00 cm, what concentration is calculated from the average corrected absorbance?

A1.03 x 10^-4 M

B9.00 x 10^-5 M

C3.00 x 10^-5 M

D1.00 x 10^-4 M

Solution & Explanation:The average raw absorbance is 0.930. Corrected A = 0.930 - 0.030 = 0.900. c = 0.900/(9.00 x 10³) = 1.00 x 10^-4 M.

Problem 2Difficulty: Mastery

A final colored solution is prepared by diluting 10.00 mL of original sample to 25.00 mL after adding reagent. Beer-Lambert analysis of the final solution gives c = 8.00 x 10^-5 M. What concentration was in the original sample, assuming the colored complex represents the analyte concentration?

A3.20 x 10^-5 M

B8.00 x 10^-5 M

C2.00 x 10^-4 M

D5.00 x 10^-5 M

Solution & Explanation:The dilution factor is 25.00/10.00 = 2.50. Original concentration = (8.00 x 10^-5)(2.50) = 2.00 x 10^-4 M.

Problem 3Difficulty: Mastery

A calibration equation is A = 0.00250C + 0.015, where C is in µM. A diluted unknown gives A = 0.265 and was prepared with a dilution factor of 6.00. What was the original concentration?

A100 µM

B600 µM

C112 µM

D1.50 µM

Solution & Explanation:For the diluted unknown, C = (0.265 - 0.015)/0.00250 = 100 µM. Multiplying by the dilution factor 6.00 gives 600 µM.

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QWhat grade level and standards does this worksheet align with?

This worksheet is designed specifically for Chemistry Science students. It aligns with educational standard AP Chem Units 1, 3, 8, 9 and Science Practices 1-6; NGSS SEPs; CCSS HSS-ID, HSA-CED.A, WHST.11-12.7-9 (National: AP Chem Units 1, 3, 8, 9 and Science Practices 1-6; NGSS SEPs; CCSS HSS-ID, HSA-CED.A, WHST.11-12.7-9. State frameworks: TX TEKS 112.43(c)(1)-(4); MA HS-PS1-11(MA); FL/GA/VA inquiry and data-analysis standards; CA/NY/NJ science practices.).