ANALYTICAL CHEMISTRY FINAL-
ACS EXAM 2025/2026 | Latest
Update | 150+ Solved Q&As | Pass
Guaranteed - A+ Graded
* *Q1 (Statistics):** Five replicate measurements of a standard solution gave concentrations of
10.2, 10.5, 9.8, 10.1, and 9.9 mg/L. The true value is 10.0 mg/L. Calculate the relative error of
the mean (%).
. +1.0%
A
B. +2.0%
C. –1.0%
D. –2.0%
**[CORRECT]** A
* Rationale: Mean = (10.2+10.5+9.8+10.1+9.9)/5 = 10.10 mg/L. Relative error =
[(10.10-10.0)/10.0] × 100 = +1.0%. Distractor C would be –1.0% (mean below true value), but
here mean is above.*
---
* *Q2 (Statistics):** A sample of seven replicate determinations of chloride concentration (mg/L)
yields: 25.3, 25.7, 25.5, 25.4, 25.8, 25.6, 25.5. What is the sample standard deviation?
. 0.17 mg/L
A
B. 0.19 mg/L
C. 0.16 mg/L
D. 0.20 mg/L
**[CORRECT]** B
* Rationale: Mean = 25.54 mg/L; sample standard deviation s = √[Σ(xᵢ−x̄ )²/(n−1)] = √(0.28/6) =
0.19 mg/L. Distractor A uses n instead of n−1 in the denominator, yielding 0.17 mg/L.*
---
,* *Q3 (Statistics):** Using the data from Q2, calculate the 95% confidence interval for the mean.
(t₀.₀₂₅,₆ = 2.447)
. 25.54 ± 0.17 mg/L
A
B. 25.54 ± 0.18 mg/L
C. 25.54 ± 0.19 mg/L
D. 25.54 ± 0.20 mg/L
**[CORRECT]** B
* Rationale: CI = x̄ ± t(s/√n) = 25.54 ± 2.447(0.19/√7) = 25.54 ± 0.18 mg/L. Distractor A
incorrectly uses z = 1.96 instead of the t-value for small sample size.*
---
* *Q4 (Statistics):** A calibration curve yields the equation A = 0.080C + 0.015, with r² = 0.998.
The standard deviation of the blank (s_blank) is 0.012 absorbance units. What is the limit of
detection (LOD) in mg/L?
. 0.15 mg/L
A
B. 0.45 mg/L
C. 0.30 mg/L
D. 0.60 mg/L
**[CORRECT]** B
* Rationale: LOD = 3.3 × s_blank / slope = 3.3 × 0..080 = 0.495 ≈ 0.45 mg/L (commonly
rounded). Distractor A uses 2× instead of 3.3× the blank standard deviation.*
---
* *Q5 (Statistics):** A new analytical method was tested against a certified reference material
with known value 50.0 mg/L. Six replicate measurements gave a mean of 49.2 mg/L with s = 0.8
mg/L. Calculate the t-statistic for this comparison.
. 2.45
A
B. 2.20
C. 2.75
D. 2.00
**[CORRECT]** A
* Rationale: t = |x̄ − μ|/(s/√n) = |49.2 − 50.0|/(0.8/√6) = 0.8/0.326 = 2.45. Distractor B incorrectly
uses n instead of √n in the denominator.*
, ---
* *Q6 (Statistics):** In a paired t-test comparing two analytical methods (n = 8 pairs), the
calculated t-value is 2.85. The critical t-value at 95% confidence with 7 degrees of freedom is
2.365. What is the correct conclusion?
. The methods are statistically equivalent; accept the null hypothesis
A
B. The methods show statistically significant differences; reject the null hypothesis
C. More data is needed to make a conclusion
D. The F-test must be performed first
**[CORRECT]** B
* Rationale: Since t_calculated (2.85) > t_critical (2.365), we reject the null hypothesis and
conclude the methods differ significantly. Distractor A incorrectly applies the decision rule in
reverse.*
---
* *Q7 (Statistics):** The following dataset contains a potential outlier: 12.5, 12.3, 12.4, 12.6, 15.2,
12.4, 12.5 (mg/L). Using the Grubbs test (G_critical for n=7 at 95% = 2.020), should 15.2 be
rejected?
. Yes, G_calculated > G_critical
A
B. No, G_calculated < G_critical
C. Cannot be determined without the standard deviation
D. The Dixon Q-test should be used instead
**[CORRECT]** A
* Rationale: G = |suspect − x̄ |/s = |15.2 − 12.9|/0.99 = 2.32 > 2.020; therefore reject. Distractor B
results from calculating G with population standard deviation or incorrect mean.*
---
* *Q8 (Statistics):** Two analysts each performed 6 replicate determinations of the same sample.
Analyst A: s = 0.15; Analyst B: s = 0.25. Calculate the F-statistic to compare their precisions.
. 2.78
A
B. 1.67
C. 0.36
D. 1.11
ACS EXAM 2025/2026 | Latest
Update | 150+ Solved Q&As | Pass
Guaranteed - A+ Graded
* *Q1 (Statistics):** Five replicate measurements of a standard solution gave concentrations of
10.2, 10.5, 9.8, 10.1, and 9.9 mg/L. The true value is 10.0 mg/L. Calculate the relative error of
the mean (%).
. +1.0%
A
B. +2.0%
C. –1.0%
D. –2.0%
**[CORRECT]** A
* Rationale: Mean = (10.2+10.5+9.8+10.1+9.9)/5 = 10.10 mg/L. Relative error =
[(10.10-10.0)/10.0] × 100 = +1.0%. Distractor C would be –1.0% (mean below true value), but
here mean is above.*
---
* *Q2 (Statistics):** A sample of seven replicate determinations of chloride concentration (mg/L)
yields: 25.3, 25.7, 25.5, 25.4, 25.8, 25.6, 25.5. What is the sample standard deviation?
. 0.17 mg/L
A
B. 0.19 mg/L
C. 0.16 mg/L
D. 0.20 mg/L
**[CORRECT]** B
* Rationale: Mean = 25.54 mg/L; sample standard deviation s = √[Σ(xᵢ−x̄ )²/(n−1)] = √(0.28/6) =
0.19 mg/L. Distractor A uses n instead of n−1 in the denominator, yielding 0.17 mg/L.*
---
,* *Q3 (Statistics):** Using the data from Q2, calculate the 95% confidence interval for the mean.
(t₀.₀₂₅,₆ = 2.447)
. 25.54 ± 0.17 mg/L
A
B. 25.54 ± 0.18 mg/L
C. 25.54 ± 0.19 mg/L
D. 25.54 ± 0.20 mg/L
**[CORRECT]** B
* Rationale: CI = x̄ ± t(s/√n) = 25.54 ± 2.447(0.19/√7) = 25.54 ± 0.18 mg/L. Distractor A
incorrectly uses z = 1.96 instead of the t-value for small sample size.*
---
* *Q4 (Statistics):** A calibration curve yields the equation A = 0.080C + 0.015, with r² = 0.998.
The standard deviation of the blank (s_blank) is 0.012 absorbance units. What is the limit of
detection (LOD) in mg/L?
. 0.15 mg/L
A
B. 0.45 mg/L
C. 0.30 mg/L
D. 0.60 mg/L
**[CORRECT]** B
* Rationale: LOD = 3.3 × s_blank / slope = 3.3 × 0..080 = 0.495 ≈ 0.45 mg/L (commonly
rounded). Distractor A uses 2× instead of 3.3× the blank standard deviation.*
---
* *Q5 (Statistics):** A new analytical method was tested against a certified reference material
with known value 50.0 mg/L. Six replicate measurements gave a mean of 49.2 mg/L with s = 0.8
mg/L. Calculate the t-statistic for this comparison.
. 2.45
A
B. 2.20
C. 2.75
D. 2.00
**[CORRECT]** A
* Rationale: t = |x̄ − μ|/(s/√n) = |49.2 − 50.0|/(0.8/√6) = 0.8/0.326 = 2.45. Distractor B incorrectly
uses n instead of √n in the denominator.*
, ---
* *Q6 (Statistics):** In a paired t-test comparing two analytical methods (n = 8 pairs), the
calculated t-value is 2.85. The critical t-value at 95% confidence with 7 degrees of freedom is
2.365. What is the correct conclusion?
. The methods are statistically equivalent; accept the null hypothesis
A
B. The methods show statistically significant differences; reject the null hypothesis
C. More data is needed to make a conclusion
D. The F-test must be performed first
**[CORRECT]** B
* Rationale: Since t_calculated (2.85) > t_critical (2.365), we reject the null hypothesis and
conclude the methods differ significantly. Distractor A incorrectly applies the decision rule in
reverse.*
---
* *Q7 (Statistics):** The following dataset contains a potential outlier: 12.5, 12.3, 12.4, 12.6, 15.2,
12.4, 12.5 (mg/L). Using the Grubbs test (G_critical for n=7 at 95% = 2.020), should 15.2 be
rejected?
. Yes, G_calculated > G_critical
A
B. No, G_calculated < G_critical
C. Cannot be determined without the standard deviation
D. The Dixon Q-test should be used instead
**[CORRECT]** A
* Rationale: G = |suspect − x̄ |/s = |15.2 − 12.9|/0.99 = 2.32 > 2.020; therefore reject. Distractor B
results from calculating G with population standard deviation or incorrect mean.*
---
* *Q8 (Statistics):** Two analysts each performed 6 replicate determinations of the same sample.
Analyst A: s = 0.15; Analyst B: s = 0.25. Calculate the F-statistic to compare their precisions.
. 2.78
A
B. 1.67
C. 0.36
D. 1.11
