Question 1
Isotopically labeled internal standards are used in LC-MS to mitigate matrix effects because:
Correct Answer:
They behave chemically like the analyte but are distinguishable by mass, compensating for matrix effects and losses during sample prep, enabling accurate quantitation.
Explanation:
In LC-MS, matrix effects from co-eluting substances can alter ionization, leading to inaccurate quantitation. Isotopically labeled internal standards help because they behave chemically like the target analyte, so they undergo the same extraction, cleanup, chromatography, and ionization processes. The key difference is their mass, created by the isotope label, which makes them distinguishable in the mass spectrometer. This means both analyte and internal standard are subjected to the same matrix effects and losses, so taking the ratio of their signals corrects for those variations and yields accurate results. They aren’t intended to replace the analyte, and while they often co-elute closely due to nearly identical chemistry, exact identical retention time isn’t guaranteed. They’re used primarily to compensate for matrix effects and sample prep losses, not just to calibrate the detector.
Question 2
How can TLC determine purity?
Correct Answer:
Pure compounds usually produce one spot.
Explanation:
In TLC, purity is judged by the number of distinct spots the sample produces on the plate. A pure substance is just one chemical species, so under a given solvent system it should appear as a single spot with a consistent Rf value. If impurities are present, they are other compounds that travel differently, creating additional spots at different positions. Therefore, a single spot strongly suggests purity, while multiple spots indicate impurities. Remember, color alone isn’t a reliable measure—many substances are colorless, and spot intensity reflects concentration rather than purity.
Question 3
Which statement correctly distinguishes partition chromatography from adsorption chromatography, with appropriate examples?
Correct Answer:
Partition chromatography involves distribution of solute between a liquid stationary phase and the mobile phase; example: reversed-phase HPLC with aqueous-organic mobile phase and C18 stationary phase.
Explanation:
Partition chromatography is based on the solute distributing between two immiscible liquid phases. In practice this means a liquid stationary phase paired with a separate mobile phase; how strongly the solute prefers the stationary liquid versus the mobile liquid sets how long it is retained. In reversed‑phase HPLC, the stationary phase is nonpolar (for example, C18 groups bonded to silica) and the mobile phase is polar (water with an organic modifier). The analyte partitions between the nonpolar stationary phase and the polar mobile phase, so more hydrophobic compounds spend more time on the stationary phase and elute later. This liquid–liquid distribution is what defines partition chromatography. Adsorption chromatography, by contrast, relies on the solute sticking to a solid surface. A typical example is normal‑phase silica chromatography, where polarity drives adsorption to the polar silica surface. So the correct statement captures that partition chromatography involves distribution between a liquid stationary phase and the mobile phase, with reversed‑phase HPLC illustrating this idea.
Question 4
Which parameter is inversely related to peak broadening and related to the plate count?
Correct Answer:
N
Explanation:
Peak sharpness reflects how efficiently a column separates solute molecules, and this efficiency is captured by the plate count. A higher number of theoretical plates means the peak is narrower and less broadened, indicating better separation efficiency. The plate count N is linked to tR and the peak width W (for a Gaussian peak, N ≈ 16 (tR/W)²). Since W increases with peak broadening, increasing broadening lowers N, making N inversely related to how broad the peak becomes. That’s why this parameter best reflects the trade-off between peak width and column efficiency. W is directly the measure of how broad the peak is, so it correlates with broadening itself rather than inversely. Retention time tR tells you when the peak elutes, not how broad it is. The mobile-phase residence time tM isn’t a measure of peak broadening.
Question 5
If two adjacent peaks exhibit Rs of 2.0, what does this indicate?
Correct Answer:
They are baseline separated.
Explanation:
Resolution between two adjacent peaks is a measure of how well they are separated in chromatography. It combines how far apart their retention times are and how wide the peaks are. An Rs value of 2.0 indicates good, baseline separation: there is a clear baseline between the two peaks, so they do not overlap and you can quantify each peak independently. If peaks were partially overlapping, the Rs would be lower, not 2.0. If they had the same retention time, they would co-elute and the separation would be effectively nonexistent. If they were not detected, there would be no peaks to measure at all. So, a value of 2.0 means the peaks are baseline separated.
Question 1
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Prepare with the Chromatography Practice Test practice quiz. This question bank includes 10 questions covering chromatography, normal-phase, reversed-phase, plate, and hplc. Use it to review important concepts, identify knowledge gaps, and build confidence for the related exam, course, or assessment.

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Chromatography Practice Test

This practice set contains 10 questions from the matching question bank and focuses on chromatography, normal-phase, reversed-phase, plate, and hplc. Work through each question carefully, review the provided solutions, and revisit topics that need more study before your next attempt.

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