The problems are on the Beer-Lambert law worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Molarity is moles of solute per litre of solution, and dilution follows M₁V₁ = M₂V₂. These worked answers correct the 20 problems of the AP Chemistry worksheet on Beer-Lambert law, from molarity of salt solution to design a colorimetric analysis: each solution shows the formula, the substitution and the final answer with its unit.
Here are the Beer-Lambert law worksheet answers for AP Chemistry, explained step by step. First try the worksheet on your own, then compare, and reread the lesson if a step is unclear.
Every answer shows the relationship used, the substitution and the final result with units. When a problem asks for an explanation, the main points are given in sentences. Rounding differences in the last significant figure are acceptable.
Level: AP Chemistry · Curriculum: AP Chem Unit 3
Beer-Lambert law worksheet answers: the answers explained
Problem 1 Molarity of salt solution
n = 11.7 / 58.44 = 0.2002 molM = 0.2002 / 0.5000 = 0.400 mol/L.
Problem 2 Mass for a solution
n = 0.100 × 0.2500 = 0.0250 molm = 0.0250 × 74.55 = 1.86 g.
Problem 3 Simple dilution
M₂ = M₁V₁/V₂ = 1.00 × 25.0 / 250.0 = 0.100 mol/L.Problem 4 Calculate R_f values
1. X 2.0 cm, Y 4.4 cm, Z 6.0 cm; front 8.0 cm.2. Rf(X) = 2.0/8.0 = 0.25; Rf(Y) = 4.4/8.0 = 0.55; Rf(Z) = 6.0/8.0 = 0.75.
3. X, the spot that moved least.
Problem 5 Transmittance and absorbance
1. A = −log 0.50 = 0.301.2. T = 10−1.000 = 0.100, which is 10 percent.
Problem 6 Choose a separation method
1. Filtration.2. Simple distillation (the water evaporates and is condensed).
3. Chromatography.
4. Fractional distillation, using the difference in boiling points.
Problem 7 Ion concentrations
1. [Ca²⁺] = 0.30 M; [Cl⁻] = 2 × 0.30 = 0.60 M.2. [Al³⁺] = 2 × 0.20 = 0.40 M; [SO₄²⁻] = 3 × 0.20 = 0.60 M.
Problem 8 Lead in drinking water
1. 0.0210 / 1.50 = 0.0140 mg/L.2. 0.0140 mg/L = 14.0 µg/L.
Problem 9 Make a copper sulfate solution
n = 0.250 × 0.1000 = 0.0250 mol; m = 0.0250 × 249.70 = 6.24 g.Weigh the solid, dissolve it in a beaker with a little water, transfer to a 100.0 mL volumetric flask, rinse the beaker into the flask, and fill to the mark; invert to mix.
Problem 10 A dilution series
V₁ = M₂ × 50.0 / 0.500:0.100 M: 10.0 mL; 0.0500 M: 5.00 mL; 0.0250 M: 2.50 mL; 0.0100 M: 1.00 mL.
Problem 11 Find a concentration with Beer-Lambert
c = A/(εb) = 0.450 / (9.0 × 10³ × 1.00) = 5.0 × 10⁻⁵ mol/L.Problem 12 Change the path length
1. A is proportional to b: 0.600 and 0.150.2. T = 10−0.300 = 0.501, so 50.1 percent.
Problem 13 Use a calibration graph
1. About 3.0 × 10⁻⁵ mol/L.2. Gradient = 0.72 / (6.0 × 10⁻⁵ mol/L) = 1.2 × 10⁴ L/mol; since b = 1.00 cm, ε = 1.2 × 10⁴ L/(mol·cm).
3. With no dye there is no absorption, and the blank is set to zero.
Problem 14 Why chromatography separates
The nonpolar dye has the larger Rf.Polar silica attracts the polar dye strongly, so it spends most time on the stationary phase and moves little. The nonpolar dye is more soluble in the nonpolar solvent and is carried further.
Problem 15 Mixing two solutions
n = 0.0500 × 0.200 + 0.1500 × 0.100 = 0.0100 + 0.0150 = 0.0250 molM = 0.0250 / 0.2000 = 0.125 mol/L.
Problem 16 Volume for a titration-style reaction
n(Cl⁻) = 0.0250 × 0.100 = 2.50 × 10⁻³ mol = n(Ag⁺)V = n/M = 2.50 × 10⁻³ / 0.150 = 0.01667 L = 16.7 mL.
Problem 17 When Beer’s law fails
1. A/c = 0.12, 0.12, 0.12, 0.119, 0.113, 0.081 (units 10⁵ L/mol).2. The data are linear up to about 8 × 10⁻⁵ mol/L (A = 0.95); beyond it A/c falls.
3. The reading is outside the linear range. Dilute the sample by a known factor to bring A within the range, remeasure and multiply by the dilution factor.
Problem 18 Absorbance from percent transmittance
A = −log 0.180 = 0.745.c = 0.745 / (1.5 × 10⁴ × 1.00) = 4.96 × 10⁻⁵ mol/L.
Problem 19 Diluted sample
1. c = 0.480 / (1.20 × 10⁴ × 1.00) = 4.00 × 10⁻⁵ mol/L.2. Dilution factor 100.0/10.00 = 10, so original = 4.00 × 10⁻⁴ mol/L.
3. It selects a single wavelength (λmax) from the source so that ε is constant and the measurement is most sensitive.
Problem 20 Design a colorimetric analysis
Plan: make a set of at least five standards of known concentration by dilution; run a scan of the most concentrated standard to find λmax (the light absorbed by a blue solution is orange-red); zero the instrument with a blank; measure A for each standard and the unknown; plot A against c and draw a best-fit line through the origin; read the unknown from the line.Error: air bubbles, fingerprints on the cuvette, or an unknown outside the range of standards.
