1What is the main purpose of measuring antioxidant activity in a food sample?
Determination of total antioxidant activity of food sample
Easy
A.To identify its microbial species
B.To measure its protein content
C.To determine its water content
D.To estimate its ability to neutralize oxidants
Correct Answer: To estimate its ability to neutralize oxidants
Explanation:
Antioxidant assays estimate how effectively a food sample can neutralize oxidizing substances.
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2Which instrument is commonly used to measure absorbance in an antioxidant assay?
Determination of total antioxidant activity of food sample
Easy
A.Autoclave
B.Centrifuge
C.Spectrophotometer
D.Microscope
Correct Answer: Spectrophotometer
Explanation:
A spectrophotometer measures the absorbance of light by the reaction mixture.
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3What does a reagent blank usually contain?
Determination of total antioxidant activity of food sample
Easy
A.Only the food sample
B.The sample and standard together
C.Only distilled water
D.All reagents except the sample
Correct Answer: All reagents except the sample
Explanation:
A reagent blank accounts for absorbance produced by the reagents without the test sample.
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4Why is a standard solution used in a total antioxidant activity assay?
Determination of total antioxidant activity of food sample
Easy
A.To compare antioxidant activity
B.To increase sample volume
C.To sterilize the food extract
D.To remove the solvent
Correct Answer: To compare antioxidant activity
Explanation:
A standard provides a reference for expressing or comparing the antioxidant activity of the sample.
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5What is an antioxidant?
Determination of total antioxidant activity of food sample
Easy
A.A substance that forms starch
B.A substance that digests proteins
C.A substance that always increases oxidation
D.A substance that reduces oxidation
Correct Answer: A substance that reduces oxidation
Explanation:
Antioxidants help slow or prevent oxidation by reacting with oxidizing species.
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6What is usually prepared from a food sample before testing its antioxidant activity?
Determination of total antioxidant activity of food sample
Easy
A.A metal electrode
B.A suitable sample extract
C.A dry glass plate
D.A bacterial culture
Correct Answer: A suitable sample extract
Explanation:
The antioxidant compounds are commonly extracted from the food sample before the assay.
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7What may happen to the absorbance when more colored product is formed in an assay?
Determination of total antioxidant activity of food sample
Easy
A.The absorbance may increase
B.The absorbance always becomes zero
C.The cuvette becomes heavier
D.The wavelength disappears
Correct Answer: The absorbance may increase
Explanation:
A greater amount of colored product generally absorbs more light at the selected wavelength.
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8Why should the same wavelength be used for comparable assay readings?
Determination of total antioxidant activity of food sample
Easy
A.To ensure consistent measurements
B.To remove all antioxidants
C.To prevent pipette calibration
D.To change the sample color
Correct Answer: To ensure consistent measurements
Explanation:
Using the same wavelength makes absorbance values from different tubes directly comparable.
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9What is a common reason for testing a sample in duplicate or triplicate?
Determination of total antioxidant activity of food sample
Easy
A.To improve result reliability
B.To change the sample identity
C.To eliminate the need for reagents
D.To avoid measuring absorbance
Correct Answer: To improve result reliability
Explanation:
Repeated measurements help identify variation and provide a more reliable average result.
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10Which type of water is generally preferred for preparing assay solutions?
Determination of total antioxidant activity of food sample
Easy
A.Oily water
B.Distilled water
C.Sugary water
D.Seawater
Correct Answer: Distilled water
Explanation:
Distilled water contains fewer interfering impurities and is suitable for laboratory solutions.
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11What does TLC stand for?
Demonstration of thin layer chromatography
Easy
A.Total liquid concentration
B.Transfer layer calculation
C.Thermal laboratory culture
D.Thin-layer chromatography
Correct Answer: Thin-layer chromatography
Explanation:
TLC is the abbreviation for thin-layer chromatography.
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12What is the stationary phase in a common TLC plate?
Demonstration of thin layer chromatography
Easy
A.A pool of solvent
B.A layer of silica gel
C.A tube of sample solution
D.A strip of filter paper
Correct Answer: A layer of silica gel
Explanation:
Silica gel is commonly coated onto the plate and acts as the stationary phase.
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13What is the mobile phase in TLC?
Demonstration of thin layer chromatography
Easy
A.The pencil used for marking
B.The solvent that moves upward
C.The coated plate surface
D.The sample spot at the origin
Correct Answer: The solvent that moves upward
Explanation:
The mobile phase is the solvent that travels across the stationary phase and carries sample components.
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14Where should the sample be applied on a TLC plate?
Demonstration of thin layer chromatography
Easy
A.At the top edge
B.Inside the solvent bottle
C.On a marked origin line
D.Below the lower edge
Correct Answer: On a marked origin line
Explanation:
The sample is placed on a pencil-marked origin line near the bottom of the plate.
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15Why is pencil commonly used to mark a TLC plate?
Demonstration of thin layer chromatography
Easy
A.Pencil removes the stationary phase
B.Graphite does not usually dissolve
C.Pencil ink increases separation
D.Graphite acts as the solvent
Correct Answer: Graphite does not usually dissolve
Explanation:
Graphite generally remains on the plate, whereas ink may dissolve and interfere with the chromatogram.
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16What is the solvent front in TLC?
Demonstration of thin layer chromatography
Easy
A.The thickness of the silica layer
B.The original sample spot
C.The bottom of the plate
D.The furthest point reached by solvent
Correct Answer: The furthest point reached by solvent
Explanation:
The solvent front is the highest distance traveled by the mobile phase on the plate.
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17What does an value compare?
Demonstration of thin layer chromatography
Easy
A.Sample mass with solvent volume
B.Solute distance with solvent distance
C.Spot color with spot brightness
D.Plate width with plate thickness
Correct Answer: Solute distance with solvent distance
Explanation:
The value is calculated as the distance traveled by the spot divided by the distance traveled by the solvent front.
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18Which formula represents the TLC retention factor?
Demonstration of thin layer chromatography
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
The retention factor is the solute distance divided by the solvent-front distance.
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19What does one spot on a TLC plate commonly suggest about a sample?
Demonstration of thin layer chromatography
Easy
A.It must contain no compounds
B.It may contain one major component
C.It always contains four components
D.It cannot dissolve in the solvent
Correct Answer: It may contain one major component
Explanation:
A single visible spot may indicate one major compound under the conditions used.
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20How can colorless compounds on a TLC plate often be detected?
Demonstration of thin layer chromatography
Easy
A.By weighing the plate
B.By freezing the solvent
C.By adding distilled water only
D.By ultraviolet light or a stain
Correct Answer: By ultraviolet light or a stain
Explanation:
Colorless spots are commonly visualized using ultraviolet light or a suitable chemical stain.
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21In a DPPH assay, a food extract causes the absorbance of the DPPH solution to decrease. What does this decrease primarily indicate?
Determination of total antioxidant activity of food sample
Medium
A.The extract has increased the reaction temperature
B.The extract has absorbed all assay solvent
C.The extract has reduced DPPH radicals
D.The extract has increased DPPH radicals
Correct Answer: The extract has reduced DPPH radicals
Explanation:
Antioxidants donate hydrogen atoms or electrons to DPPH radicals, converting them into a less-colored reduced form and lowering absorbance.
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22A sample has an initial DPPH absorbance of 0.800 and a final absorbance of 0.320. What is the percentage of radical-scavenging activity?
Determination of total antioxidant activity of food sample
Medium
A.40%
B.50%
C.60%
D.72%
Correct Answer: 60%
Explanation:
Using , the activity is .
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23Why is a reagent blank included when determining antioxidant activity spectrophotometrically?
Determination of total antioxidant activity of food sample
Medium
A.To increase the sample antioxidant concentration
B.To convert antioxidants into colored products
C.To measure solvent or reagent absorbance
D.To establish the sample extraction temperature
Correct Answer: To measure solvent or reagent absorbance
Explanation:
The blank accounts for absorbance contributed by reagents and solvent, allowing the sample-related absorbance change to be determined accurately.
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24A calibration curve for a standard antioxidant is used mainly to:
Determination of total antioxidant activity of food sample
Medium
A.Identify the food sample's chromatographic solvent
B.Convert absorbance into equivalent antioxidant concentration
C.Remove interfering pigments from the sample extract
D.Determine the wavelength of maximum solvent absorption
Correct Answer: Convert absorbance into equivalent antioxidant concentration
Explanation:
A standard curve relates measured absorbance or activity to known standard concentrations, allowing results to be reported as equivalents of that standard.
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25If the absorbance of a sample is higher than the reagent control in a colorimetric antioxidant assay, the most appropriate first action is to:
Determination of total antioxidant activity of food sample
Medium
A.Increase the incubation time without further checks
B.Check for sample color interference
C.Assume the sample contains no antioxidants
D.Report the result as maximum activity
Correct Answer: Check for sample color interference
Explanation:
Strongly colored food extracts can add absorbance independent of the assay reaction. A sample blank or suitable correction should be considered.
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26Why should replicate measurements be performed for a food antioxidant assay?
Determination of total antioxidant activity of food sample
Medium
A.To increase the wavelength used for measurement
B.To estimate precision and reduce random error
C.To eliminate the need for a calibration curve
D.To guarantee that all antioxidants are identified
Correct Answer: To estimate precision and reduce random error
Explanation:
Replicates reveal variation among measurements and allow a mean and dispersion estimate, improving confidence in the reported activity.
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27An antioxidant assay is performed on 2.0 g of food, and the extract is made up to 25 mL. Which change would most directly affect the reported activity per gram?
Determination of total antioxidant activity of food sample
Medium
A.Changing the cuvette optical path label
B.Changing the color of the laboratory coat
C.Changing the final extract volume
D.Changing the order of sample names in the worksheet
Correct Answer: Changing the final extract volume
Explanation:
The final volume determines the total amount represented by the measured extract concentration, so it directly affects activity calculated per gram of food.
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28Why is a fixed incubation time important in a DPPH antioxidant assay?
Determination of total antioxidant activity of food sample
Medium
A.The solvent becomes chemically inert after incubation
B.The cuvette path length changes during incubation
C.The food extract becomes a chromatographic stationary phase
D.The reaction may continue changing with time
Correct Answer: The reaction may continue changing with time
Explanation:
DPPH reduction can progress during incubation. A fixed time ensures that samples are compared at the same reaction stage.
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29A diluted extract gives an antioxidant activity within the calibration range, while the undiluted extract exceeds it. Which result is preferable for quantification?
Determination of total antioxidant activity of food sample
Medium
A.The undiluted result extrapolated beyond the curve
B.The average of both results without correction
C.The diluted result reported without concentration correction
D.The diluted result after applying its dilution factor
Correct Answer: The diluted result after applying its dilution factor
Explanation:
Measurements within the calibration range are more reliable. The original concentration is obtained by multiplying the diluted result by the dilution factor.
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30If two extracts show 80% and 40% DPPH inhibition at the same concentration, what is the most reasonable conclusion?
Determination of total antioxidant activity of food sample
Medium
A.The second extract has greater radical-scavenging activity
B.The result proves that the first extract has no pigments
D.The first extract has greater radical-scavenging activity
Correct Answer: The first extract has greater radical-scavenging activity
Explanation:
At the same concentration and under the same conditions, greater percentage inhibition indicates stronger DPPH radical-scavenging capacity.
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31In thin layer chromatography, the stationary phase is usually:
Demonstration of thin layer chromatography
Medium
A.A colored sample solution
B.A mixture of volatile solvents
C.A stream of developing vapor
D.A thin layer of silica gel
Correct Answer: A thin layer of silica gel
Explanation:
Silica gel is a polar stationary phase that interacts with analytes while the solvent carries them up the plate.
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32A compound travels 3.0 cm while the solvent front travels 6.0 cm. What is the compound's value?
Demonstration of thin layer chromatography
Medium
A.0.25
B.0.50
C.1.50
D.2.00
Correct Answer: 0.50
Explanation:
The retention factor is .
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33Why must the sample spot be placed above the solvent level in the developing chamber?
Demonstration of thin layer chromatography
Medium
A.To prevent the solvent front from moving upward
B.To force every compound to have the same
C.To prevent the sample from dissolving into the solvent reservoir
D.To make the stationary phase completely nonpolar
Correct Answer: To prevent the sample from dissolving into the solvent reservoir
Explanation:
If the spot is submerged, the sample may dissolve directly into the solvent rather than migrate as a focused band on the plate.
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34A mixture produces three separate spots on a TLC plate. This observation most directly suggests that the mixture:
Demonstration of thin layer chromatography
Medium
A.Was applied below the solvent level
B.Contains only one component at three concentrations
C.Has failed to interact with the stationary phase
D.Contains at least three detectable components
Correct Answer: Contains at least three detectable components
Explanation:
Distinct spots generally represent components with different mobilities in the selected stationary-phase and solvent system.
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35For a silica TLC plate, which compound would generally have the lower in the same solvent system?
Demonstration of thin layer chromatography
Medium
A.A compound with a smaller spot
B.A less polar compound
C.A more polar compound
D.A compound applied in a smaller volume
Correct Answer: A more polar compound
Explanation:
Polar compounds interact more strongly with polar silica and therefore usually move more slowly, producing lower values.
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36Why is a pencil preferred over ink for marking the origin and solvent front on a TLC plate?
Demonstration of thin layer chromatography
Medium
A.Graphite reacts with every sample component
B.Graphite does not usually dissolve and migrate
C.Ink always increases silica polarity uniformly
D.Ink prevents the solvent from entering the plate
Correct Answer: Graphite does not usually dissolve and migrate
Explanation:
Ink dyes can dissolve in the mobile phase and create extra spots. Graphite generally remains at the marked position.
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37If the solvent front is not marked immediately after removing the TLC plate, the calculated value may be inaccurate because:
Demonstration of thin layer chromatography
Medium
A.All analytes return to the origin after removal
B.The solvent can continue evaporating from the plate
C.The sample spots become chemically identical
D.The silica changes into a liquid stationary phase
Correct Answer: The solvent can continue evaporating from the plate
Explanation:
The solvent front may become difficult to see or shift through evaporation, making its final distance uncertain.
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38Two compounds have identical values in one solvent system. What is the best interpretation?
Demonstration of thin layer chromatography
Medium
A.They must have different stationary phases
B.They may be identical, but identity is not proven
C.They definitely have different molecular formulas
D.They cannot be separated in any solvent system
Correct Answer: They may be identical, but identity is not proven
Explanation:
Equal values under one condition suggest similar mobility, but confirmation requires standards, another solvent system, or additional analysis.
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39A TLC spot remains near the origin even after development. Which adjustment would most likely improve its movement?
Demonstration of thin layer chromatography
Medium
A.Replace the silica plate with a dry filter paper without testing
B.Apply the sample directly into the solvent reservoir
C.Use a more polar mobile phase
D.Use a shorter development distance
Correct Answer: Use a more polar mobile phase
Explanation:
A more polar solvent can compete more effectively with the analyte for adsorption sites on silica, increasing the compound's movement.
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40Why should a TLC sample be applied as a small, concentrated spot rather than a large wet spot?
Demonstration of thin layer chromatography
Medium
A.To increase the plate thickness during development
B.To ensure the sample never contacts silica
C.To obtain sharper and better-separated bands
D.To make the solvent front stop at the origin
Correct Answer: To obtain sharper and better-separated bands
Explanation:
Small concentrated spots reduce spreading, which improves resolution and makes differences in compound mobility easier to observe.
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41A food extract gives an absorbance of 0.620 at 695 nm. The ascorbic-acid calibration equation is , where is in . If the extract was diluted 20-fold before measurement and prepared from 0.50 g of food to a final volume of 25 mL, what is the antioxidant activity in mg ascorbic-acid equivalents per gram of food?
Determination of total antioxidant activity of food sample
Hard
A.24.0 mg AAE/g
B.60.0 mg AAE/g
C.48.0 mg AAE/g
D.50.0 mg AAE/g
Correct Answer: 50.0 mg AAE/g
Explanation:
The diluted extract concentration is . The original concentration is . Thus, the 25 mL extract contains 25 mg AAE, corresponding to mg AAE/g.
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42In a phosphomolybdenum total antioxidant assay, the sample absorbance is 0.410, the reagent blank is 0.080, and the sample matrix blank without molybdate reagent is 0.050. The standard curve is . Which concentration should be used to calculate the sample antioxidant equivalent?
Determination of total antioxidant activity of food sample
Hard
A.38.0
B.33.0
C.35.0
D.30.0
Correct Answer: 35.0
Explanation:
Corrected absorbance is . Substitution into the calibration equation gives only if the reagent blank and matrix blank are both independently additive to the uncorrected reading. Here the matrix blank already contains the reagent background, so subtracting the reagent blank twice is incorrect; use , giving .
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43A calibration curve is linear from 10 to 80 . A diluted sample produces an absorbance corresponding to 135 . Which action gives the most defensible result?
Determination of total antioxidant activity of food sample
Hard
A.Report 135 directly
B.Extrapolate the calibration line to 135
C.Subtract the highest standard from the sample signal
D.Dilute the sample further and repeat the measurement
Correct Answer: Dilute the sample further and repeat the measurement
Explanation:
The response lies outside the validated linear range. Further dilution places the sample within the calibration interval, after which the measured concentration is multiplied by the additional dilution factor.
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44Two food extracts have identical total antioxidant-equivalent values by the phosphomolybdenum assay, but one has a much higher ferric-reducing power. What is the best interpretation?
Determination of total antioxidant activity of food sample
Hard
A.The assays respond differently to antioxidant compounds
B.The phosphomolybdenum assay has no quantitative value
Correct Answer: The assays respond differently to antioxidant compounds
Explanation:
Different assays use different reaction mechanisms and conditions. Equal equivalent values do not prove identical composition, and differing results can reflect compound-specific response, kinetics, or matrix effects.
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45A sample is tested at two dilutions. The 10-fold dilution gives 0.88 absorbance, whereas the 20-fold dilution gives 0.47. The calibration range is 0.10–0.80 absorbance. Which result should be accepted for quantification?
Determination of total antioxidant activity of food sample
Hard
A.The 10-fold result because it is less diluted
B.The average because both readings are reproducible
C.Neither result because dilution changes antioxidant chemistry
D.The 20-fold result because it lies within range
Correct Answer: The 20-fold result because it lies within range
Explanation:
The 10-fold sample exceeds the validated absorbance range and may be nonlinear. The 20-fold sample falls within the range and should be back-calculated using its dilution factor.
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46A food sample is extracted twice. The first extraction recovers 82% of the final measured antioxidant activity, and the second recovers the remaining 18%. If only the first extract is assayed, what is the principal consequence?
Determination of total antioxidant activity of food sample
Hard
A.The antioxidant concentration is overestimated
B.The reagent blank becomes negligible
C.The calibration slope becomes steeper
D.The antioxidant concentration is underestimated
Correct Answer: The antioxidant concentration is underestimated
Explanation:
Antioxidants remaining in the second extract are omitted from the analytical total. Unless the procedure defines a single extraction, incomplete recovery produces a low result.
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47A sample extract is intensely colored at the assay wavelength. The reaction mixture has an absorbance of 0.900, while a sample blank containing extract and solvent but no assay reagent has an absorbance of 0.300. What is the primary correction?
Determination of total antioxidant activity of food sample
Hard
A.Ignore the blank because color is sample-specific
B.Divide the reaction value by the sample blank
C.Add the blank absorbance to the reaction value
D.Subtract the sample blank from the reaction value
Correct Answer: Subtract the sample blank from the reaction value
Explanation:
The sample blank measures absorbance from the extract itself. Corrected reaction absorbance is , preventing sample color from being interpreted as antioxidant-derived signal.
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48For a standard curve, concentrations are 0, 20, 40, 60, and 80 , with absorbances 0.02, 0.19, 0.37, 0.55, and 0.73. The unknown gives 0.46. Which concentration is the most appropriate estimate by linear interpolation?
Determination of total antioxidant activity of food sample
Hard
A.40.0
B.50.0
C.45.0
D.55.0
Correct Answer: 50.0
Explanation:
The response is halfway between 0.37 at 40 and 0.55 at 60 . Linear interpolation therefore gives 50 .
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49A standard curve has slope 0.015 absorbance units per and an intercept of 0.025. Replicate sample readings are 0.410, 0.414, and 0.620. Which statistical decision is most appropriate before calculating antioxidant activity?
Determination of total antioxidant activity of food sample
Hard
A.Investigate the high reading as a possible outlier
B.Discard the lowest reading automatically
C.Use only the middle reading as the result
D.Average all readings without inspection
Correct Answer: Investigate the high reading as a possible outlier
Explanation:
The first two measurements agree closely, whereas 0.620 is markedly discordant. It should be checked for pipetting, dilution, contamination, or transcription error before exclusion or averaging.
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50A result is reported as 12.6 mg Trolox equivalents per gram using a 0.200 g sample extracted to 10.0 mL. If the analyst accidentally records the sample mass as 0.0200 g during back-calculation, how will the reported value change?
Determination of total antioxidant activity of food sample
Hard
A.It will be ten times too high
B.It will be unchanged
C.It will be one hundred times too high
D.It will be ten times too low
Correct Answer: It will be ten times too high
Explanation:
Activity per gram is calculated by dividing the measured antioxidant amount by sample mass. Using a mass one-tenth of the true mass makes the calculated value ten times larger.
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51A TLC plate has a solvent-front distance of 8.0 cm. Compound X travels 2.4 cm and compound Y travels 5.6 cm. Which conclusion is justified?
Demonstration of thin layer chromatography
Hard
A.The of X is 0.70 and Y is 0.30
B.The of X is 0.30 and Y is 0.70
C.Y must have twice the molecular mass of X
D.X is always more polar than Y
Correct Answer: The of X is 0.30 and Y is 0.70
Explanation:
equals distance traveled by the compound divided by distance traveled by the solvent front. Thus, and .
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52On a silica TLC plate developed with a relatively nonpolar solvent, an antioxidant spot remains near the origin while a second spot travels close to the solvent front. Which interpretation is most likely?
Demonstration of thin layer chromatography
Hard
A.The front spot is necessarily more polar
B.The origin spot interacts more strongly with silica
C.The solvent front was measured from the origin incorrectly
D.The origin spot has the lower molecular mass
Correct Answer: The origin spot interacts more strongly with silica
Explanation:
Silica is polar and retains compounds through adsorption and hydrogen bonding. A spot near the origin generally has stronger stationary-phase interactions under the selected solvent conditions.
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53A sample spot is applied below the solvent level in the developing chamber. What chromatographic artifact is most expected?
Demonstration of thin layer chromatography
Hard
A.The stationary phase will become chemically nonpolar
B.The sample will dissolve directly into the solvent reservoir
C.The solvent front will move without capillary action
D.The spot will remain sharply fixed at the origin
Correct Answer: The sample will dissolve directly into the solvent reservoir
Explanation:
If the applied spot is submerged, the sample can dissolve into the mobile-phase reservoir rather than migrate as a discrete band, causing severe loss and distorted separation.
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54A TLC plate develops with a curved solvent front, and compounds near the edges show different values from identical compounds in the center. Which procedural change best addresses the problem?
Demonstration of thin layer chromatography
Hard
A.Use a level chamber and pre-equilibrate it
B.Increase the spotting volume at the plate edges
C.Scratch deeper channels across the stationary phase
D.Measure all distances from the plate midpoint
Correct Answer: Use a level chamber and pre-equilibrate it
Explanation:
Uneven chamber saturation, an unlevel plate, or excessive evaporation can produce a curved front. Chamber equilibration and a level setup improve solvent uniformity.
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55A sample produces one visible spot in solvent system A but three well-separated spots in solvent system B. What is the strongest conclusion?
Demonstration of thin layer chromatography
Hard
A.Solvent B chemically synthesized two new compounds
B.Solvent A permanently destroyed the stationary phase
C.The sample is pure because one system gives one spot
D.The sample contains components resolved more effectively by B
Correct Answer: The sample contains components resolved more effectively by B
Explanation:
A single spot can represent co-migrating compounds. The second solvent system provides better selectivity, revealing components that were unresolved in solvent A.
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56The solvent front reaches the top edge of a TLC plate before the plate is removed. Why are the resulting values less reliable?
Demonstration of thin layer chromatography
Hard
A.All compounds become permanently fluorescent
B.The denominator approaches the plate length limit
C.The silica layer becomes completely soluble
D.The solvent front becomes impossible to identify
Correct Answer: The solvent front becomes impossible to identify
Explanation:
The solvent-front position must be marked immediately when development stops. If the front reaches or runs off the plate, its true distance cannot be measured accurately, invalidating calculations.
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57A compound gives in a solvent system. Which adjustment is most likely to improve separation from the solvent front?
Demonstration of thin layer chromatography
Hard
A.Apply a larger sample spot
B.Use a more polar mobile phase
C.Use a less polar mobile phase
D.Increase the chamber temperature sharply
Correct Answer: Use a less polar mobile phase
Explanation:
On silica, a compound with an near 1 is moving too readily with the mobile phase. Decreasing solvent polarity generally increases retention and lowers the .
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58A TLC spot appears as a broad streak rather than a compact band. Which combination of causes is most plausible?
Demonstration of thin layer chromatography
Hard
A.Excess sample and incomplete drying between applications
B.A perfectly saturated chamber and a tiny sample
C.A short origin distance and a fresh silica surface
D.A low analyte concentration and immediate development
Correct Answer: Excess sample and incomplete drying between applications
Explanation:
Overloading and wet reapplication allow the sample to spread laterally and dissolve during development. Small, concentrated applications with drying between additions reduce streaking.
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59Two antioxidant standards have values of 0.42 and 0.44, while an unknown gives one spot at 0.43 under the same conditions. What is the strongest defensible claim?
Demonstration of thin layer chromatography
Hard
A.The unknown contains neither standard under any condition
B.The unknown contains both standards in equal amounts
C.The unknown has exactly the same molecular structure as both standards
D.The unknown may contain a compound matching either standard
Correct Answer: The unknown may contain a compound matching either standard
Explanation:
Similar values support tentative identity but do not prove it. Co-spotting, altered solvent systems, spectral detection, or an independent method is needed for stronger identification.
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60A TLC plate is viewed under UV light, but the antioxidant compound is not visible. The solvent front and a reference standard are visible. Which explanation is most likely?
Demonstration of thin layer chromatography
Hard
A.The solvent front converted the compound into silica
B.The compound must have an greater than one
C.The compound may lack a UV-active chromophore
D.The reference standard proves the sample was absent
Correct Answer: The compound may lack a UV-active chromophore
Explanation:
UV visualization depends on absorption by the analyte or fluorescence quenching. A compound can be present yet invisible under UV and may require iodine vapor or a specific spraying reagent.
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