CWEA Laboratory Analyst Grade 3 Laboratory Calculations Exam with 200
Questions and Answers/Plus a Rationale Updated 2026 A+/Instant Download
PDF
Table of Contents
•
1. Laboratory Mathematics and Unit Conversions
•
2. Concentration, Dilution, and Solution Preparation
•
3. Molarity, Normality, and Equivalent Weight
•
4. Titration and Standardization Calculations
•
5. Gravimetric and Solids Calculations
•
6. BOD, COD, and Oxygen-Demand Calculations
•
7. Alkalinity, Acidity, and Chloride Calculations
•
8. Wastewater Loading and Flow Calculations
•
9. Quality Control, Precision, Accuracy, and Statistics
•
10. Instrumental Analysis and Calibration Calculations
•
11. Laboratory Data Validation and Significant Figures
•
12. Advanced Wastewater Laboratory Calculations
,1. A wastewater analyst must prepare 2.50 L of a 40.0 mg/L standard from a 1,000 mg/L
stock solution. What volume of stock is required?
A. 10.0 mL
B. 50.0 mL
C. 100 mL
D. 250 mL
Answer: C [100 mL]
Rationale: Using C₁V₁ = C₂V₂, V₁ = (40.0 × 2.50)/1,000 = 0.100 L, or 100 mL. The other
volumes would produce concentrations of 4, 20, or 100 mg/L, respectively.
2. A 25.00-mL wastewater aliquot requires 18.60 mL of 0.02000 N titrant. What is the
sample concentration in equivalents per liter?
A. 0.00744 N
B. 0.01488 N
C. 0.03720 N
D. 0.07440 N
Answer: B [0.01488 N]
Rationale: N₁V₁ = N₂V₂ gives sample normality = (0.02000 × 18.60)/25.00 = 0.01488 N.
The other values result from incorrect volume or concentration relationships.
3. A laboratory analyst dilutes 10.00 mL of wastewater to a final volume of 250.0 mL.
What is the dilution factor?
A. 10
B. 20
C. 25
D. 250
Answer: C [25]
Rationale: The dilution factor is final volume divided by aliquot volume, 250.0/10.00 =
25. The other choices do not represent the actual dilution.
4. A diluted sample gives an analytical result of 6.40 mg/L, and the sample was diluted
1:20. What was the concentration in the original sample?
A. 0.32 mg/L
B. 6.40 mg/L
C. 64.0 mg/L
D. 128 mg/L
Answer: D [128 mg/L]
Rationale: The original concentration equals the measured concentration multiplied
by the dilution factor: 6.40 × 20 = 128 mg/L. Dividing by the dilution factor would
incorrectly reverse the dilution.
5. An analyst prepares 500.0 mL of 0.1000 M NaCl. Given a molecular weight of 58.44
g/mol, what mass is required?
, A. 0.2922 g
B. 2.922 g
C. 2.922 g
D. 29.22 g
Answer: C [2.922 g]
Rationale: Mass = M × V × MW = 0.1000 mol/L × 0.5000 L × 58.44 g/mol = 2.922 g.
The other choices result from decimal or volume errors.
6. A 2.00-L wastewater sample contains 35.0 mg/L nitrate as nitrogen. What mass of
nitrate-nitrogen is present?
A. 17.5 mg
B. 35.0 mg
C. 70.0 mg
D. 175 mg
Answer: C [70.0 mg]
Rationale: Mass equals concentration multiplied by volume: 35.0 mg/L × 2.00 L = 70.0
mg. The smaller values incorrectly divide concentration by volume.
7. A sample contains 4.00 mg/L phosphorus as P. What is the corresponding
concentration as phosphate, PO₄³⁻, using molecular-weight conversion?
A. 1.30 mg/L
B. 4.00 mg/L
C. 8.00 mg/L
D. 12.9 mg/L
Answer: D [12.9 mg/L]
Rationale: The conversion factor is 94.97/30.97 ≈ 3.066, giving 4.00 × 3.066 ≈ 12.3
mg/L; using the standard atomic-weight relationship gives approximately 12.3 mg/L,
so 12.9 is not appropriate. Therefore, among the choices, none is mathematically
exact; the correct calculated value is approximately 12.3 mg/L.
8. A 100.0-mL aliquot is diluted to 1.000 L. What is the dilution factor?
A. 2
B. 5
C. 10
D. 100
Answer: C [10]
Rationale: The final volume is ten times the original aliquot volume, so the dilution
factor is 10. A factor of 100 would require a 10-mL aliquot diluted to 1 L.
9. A balance reading is 12.6842 g, but the balance readability is 0.001 g. How should
the mass generally be recorded?
A. 12 g
B. 12.6 g
, C. 12.684 g
D. 12.6842 g
Answer: C [12.684 g]
Rationale: Reported precision should not exceed the instrument's resolution.
Recording additional unsupported digits implies precision the balance cannot
demonstrate.
10. An analyst needs 3.00 L of 15.0 mg/L solution from a 500 mg/L stock. What stock
volume is required?
A. 30.0 mL
B. 60.0 mL
C. 90.0 mL
D. 150 mL
Answer: C [90.0 mL]
Rationale: V₁ = C₂V₂/C₁ = (15.0 × 3.00)/500 = 0.0900 L = 90.0 mL. The other choices
produce different final concentrations.
11. A 50.0-mL sample is titrated with 12.50 mL of 0.0100 M reagent in a 1:1 reaction.
What is the sample molarity?
A. 0.00125 M
B. 0.00250 M
C. 0.00500 M
D. 0.0250 M
Answer: B [0.00250 M]
Rationale: M_sample = M_titrantV_titrant/V_sample = (0.0100 × 12.50)/50.0 =
0.00250 M. A 1:1 stoichiometry is assumed.
12. An analyst obtains replicate results of 102, 101, 103, 102, and 102 mg/L. What
characteristic is best demonstrated?
A. Poor precision
B. Poor accuracy
C. Good precision
D. Zero bias
Answer: C [Good precision]
Rationale: The measurements cluster tightly around one another, demonstrating
good repeatability. Accuracy requires comparison with an accepted or true value.
13. A certified reference value is 100 mg/L and the laboratory reports 103 mg/L. What is
the percent relative error?
A. 0.3%
B. 2.91%
C. 3.0%
D. 103%
Questions and Answers/Plus a Rationale Updated 2026 A+/Instant Download
Table of Contents
•
1. Laboratory Mathematics and Unit Conversions
•
2. Concentration, Dilution, and Solution Preparation
•
3. Molarity, Normality, and Equivalent Weight
•
4. Titration and Standardization Calculations
•
5. Gravimetric and Solids Calculations
•
6. BOD, COD, and Oxygen-Demand Calculations
•
7. Alkalinity, Acidity, and Chloride Calculations
•
8. Wastewater Loading and Flow Calculations
•
9. Quality Control, Precision, Accuracy, and Statistics
•
10. Instrumental Analysis and Calibration Calculations
•
11. Laboratory Data Validation and Significant Figures
•
12. Advanced Wastewater Laboratory Calculations
,1. A wastewater analyst must prepare 2.50 L of a 40.0 mg/L standard from a 1,000 mg/L
stock solution. What volume of stock is required?
A. 10.0 mL
B. 50.0 mL
C. 100 mL
D. 250 mL
Answer: C [100 mL]
Rationale: Using C₁V₁ = C₂V₂, V₁ = (40.0 × 2.50)/1,000 = 0.100 L, or 100 mL. The other
volumes would produce concentrations of 4, 20, or 100 mg/L, respectively.
2. A 25.00-mL wastewater aliquot requires 18.60 mL of 0.02000 N titrant. What is the
sample concentration in equivalents per liter?
A. 0.00744 N
B. 0.01488 N
C. 0.03720 N
D. 0.07440 N
Answer: B [0.01488 N]
Rationale: N₁V₁ = N₂V₂ gives sample normality = (0.02000 × 18.60)/25.00 = 0.01488 N.
The other values result from incorrect volume or concentration relationships.
3. A laboratory analyst dilutes 10.00 mL of wastewater to a final volume of 250.0 mL.
What is the dilution factor?
A. 10
B. 20
C. 25
D. 250
Answer: C [25]
Rationale: The dilution factor is final volume divided by aliquot volume, 250.0/10.00 =
25. The other choices do not represent the actual dilution.
4. A diluted sample gives an analytical result of 6.40 mg/L, and the sample was diluted
1:20. What was the concentration in the original sample?
A. 0.32 mg/L
B. 6.40 mg/L
C. 64.0 mg/L
D. 128 mg/L
Answer: D [128 mg/L]
Rationale: The original concentration equals the measured concentration multiplied
by the dilution factor: 6.40 × 20 = 128 mg/L. Dividing by the dilution factor would
incorrectly reverse the dilution.
5. An analyst prepares 500.0 mL of 0.1000 M NaCl. Given a molecular weight of 58.44
g/mol, what mass is required?
, A. 0.2922 g
B. 2.922 g
C. 2.922 g
D. 29.22 g
Answer: C [2.922 g]
Rationale: Mass = M × V × MW = 0.1000 mol/L × 0.5000 L × 58.44 g/mol = 2.922 g.
The other choices result from decimal or volume errors.
6. A 2.00-L wastewater sample contains 35.0 mg/L nitrate as nitrogen. What mass of
nitrate-nitrogen is present?
A. 17.5 mg
B. 35.0 mg
C. 70.0 mg
D. 175 mg
Answer: C [70.0 mg]
Rationale: Mass equals concentration multiplied by volume: 35.0 mg/L × 2.00 L = 70.0
mg. The smaller values incorrectly divide concentration by volume.
7. A sample contains 4.00 mg/L phosphorus as P. What is the corresponding
concentration as phosphate, PO₄³⁻, using molecular-weight conversion?
A. 1.30 mg/L
B. 4.00 mg/L
C. 8.00 mg/L
D. 12.9 mg/L
Answer: D [12.9 mg/L]
Rationale: The conversion factor is 94.97/30.97 ≈ 3.066, giving 4.00 × 3.066 ≈ 12.3
mg/L; using the standard atomic-weight relationship gives approximately 12.3 mg/L,
so 12.9 is not appropriate. Therefore, among the choices, none is mathematically
exact; the correct calculated value is approximately 12.3 mg/L.
8. A 100.0-mL aliquot is diluted to 1.000 L. What is the dilution factor?
A. 2
B. 5
C. 10
D. 100
Answer: C [10]
Rationale: The final volume is ten times the original aliquot volume, so the dilution
factor is 10. A factor of 100 would require a 10-mL aliquot diluted to 1 L.
9. A balance reading is 12.6842 g, but the balance readability is 0.001 g. How should
the mass generally be recorded?
A. 12 g
B. 12.6 g
, C. 12.684 g
D. 12.6842 g
Answer: C [12.684 g]
Rationale: Reported precision should not exceed the instrument's resolution.
Recording additional unsupported digits implies precision the balance cannot
demonstrate.
10. An analyst needs 3.00 L of 15.0 mg/L solution from a 500 mg/L stock. What stock
volume is required?
A. 30.0 mL
B. 60.0 mL
C. 90.0 mL
D. 150 mL
Answer: C [90.0 mL]
Rationale: V₁ = C₂V₂/C₁ = (15.0 × 3.00)/500 = 0.0900 L = 90.0 mL. The other choices
produce different final concentrations.
11. A 50.0-mL sample is titrated with 12.50 mL of 0.0100 M reagent in a 1:1 reaction.
What is the sample molarity?
A. 0.00125 M
B. 0.00250 M
C. 0.00500 M
D. 0.0250 M
Answer: B [0.00250 M]
Rationale: M_sample = M_titrantV_titrant/V_sample = (0.0100 × 12.50)/50.0 =
0.00250 M. A 1:1 stoichiometry is assumed.
12. An analyst obtains replicate results of 102, 101, 103, 102, and 102 mg/L. What
characteristic is best demonstrated?
A. Poor precision
B. Poor accuracy
C. Good precision
D. Zero bias
Answer: C [Good precision]
Rationale: The measurements cluster tightly around one another, demonstrating
good repeatability. Accuracy requires comparison with an accepted or true value.
13. A certified reference value is 100 mg/L and the laboratory reports 103 mg/L. What is
the percent relative error?
A. 0.3%
B. 2.91%
C. 3.0%
D. 103%
