These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. All experimental data are invented for practice. Unless a question says otherwise, paper and silica gel are polar stationary phases, and Rf = distance moved by the centre of the spot ÷ distance moved by the solvent front, both measured from the origin.
Question 1 (multiple choice · foundation)
A student pours a clear, colorless solution of glucose in water through filter paper. Which statement describes the result?
- (A) The glucose stays on the filter paper, because glucose is a solid at room temperature.
- (B) The glucose passes through with the water, because dissolved glucose molecules are much smaller than the pores in the paper.
- (C) Some glucose stays on the paper, because the glucose molecules are larger than the water molecules.
- (D) The glucose stays on the paper, because the hydrogen bonds between glucose and water break as the liquid passes through the paper.
Answer and explanation
Answer: (B). In a solution the glucose is spread out as single molecules, each less than a nanometre across. The paper’s pores are thousands of times wider, so the molecules go straight through and the filtrate is the same solution.
- (A) Pure glucose is a solid, but here it has dissolved. Filtration only traps solid that has not dissolved.
- (C) Glucose molecules are larger than water molecules, but both are far smaller than the pores.
- (D) Filtration does not break intermolecular attractions. Nothing in the paper pulls glucose away from water.
Question 2 (multiple choice · core)
On a paper chromatogram, the origin line is drawn 1.5 cm above the bottom edge of the paper. After the run, the solvent front is 12.5 cm above the origin line, and the centre of a red spot is 5.0 cm above the origin line. What is the Rf value of the red spot?
- (A) 0.40
- (B) 0.46
- (C) 0.60
- (D) 2.50
Answer and explanation
Answer: (A). Rf = 5.0 cm ÷ 12.5 cm = 0.40. Both distances are measured from the origin line.
- (B) measures both distances from the bottom edge of the paper: (5.0 + 1.5) ÷ (12.5 + 1.5) = 0.46.
- (C) uses the distance from the spot up to the solvent front: (12.5 − 5.0) ÷ 12.5 = 0.60.
- (D) inverts the ratio (12.5 ÷ 5.0). An Rf value can never be greater than 1.
Question 3 (multiple choice · core)
A mixture of three compounds of similar size is run on a silica gel TLC plate. The solvent is a mixture of hexane and ethyl acetate, which is much less polar than silica.
- toluene, C₆H₅CH₃ (no polar groups)
- benzaldehyde, C₆H₅CHO (a polar C=O group, no O–H)
- benzyl alcohol, C₆H₅CH₂OH (an O–H group)
Which list gives the compounds in order of decreasing Rf value?
- (A) benzyl alcohol > benzaldehyde > toluene
- (B) toluene > benzaldehyde > benzyl alcohol
- (C) benzaldehyde > toluene > benzyl alcohol
- (D) All three have the same Rf, because they have similar molar masses.
Answer and explanation
Answer: (B). Silica’s surface is covered in polar Si–O–H groups. Benzyl alcohol can form hydrogen bonds with them, so it is held most strongly and moves least. Benzaldehyde forms dipole–dipole attractions with the surface but cannot donate a hydrogen bond, so it is held less strongly. Toluene has only weak London dispersion attractions to the surface, so it travels furthest.
- (A) is the reverse order. It would be right on a nonpolar stationary phase, not on silica.
- (C) puts the polar benzaldehyde ahead of nonpolar toluene. Its C=O dipole makes it stick more, not less.
- (D) Similar size means similar dispersion forces, but the polar groups add extra attractions to silica.
Question 4 (multiple choice · core)
A mixture of methanol, CH₃OH, and water is distilled. The first distillate collected is rich in methanol. Which statement best explains this?
- (A) The O–H covalent bond in methanol is weaker than the O–H bonds in water, so methanol molecules break apart and leave the flask first.
- (B) A water molecule can take part in more hydrogen bonds than a methanol molecule can, so the attractions between water molecules are stronger and water has the lower vapor pressure.
- (C) Methanol is nonpolar, so its molecules are held together only by London dispersion forces.
- (D) Methanol has the larger molar mass, so it has stronger London dispersion forces and therefore the higher vapor pressure.
Answer and explanation
Answer: (B). Both liquids form hydrogen bonds. Each water molecule has two O–H hydrogens and two lone pairs, while each methanol molecule has only one O–H hydrogen and a nonpolar CH₃ group. Water’s hydrogen-bond network is more extensive, so water has the lower vapor pressure and methanol dominates the first vapor.
- (A) Distillation does not break covalent bonds. The methanol in the distillate is intact CH₃OH.
- (C) Methanol is polar and forms hydrogen bonds through its O–H group.
- (D) has two errors: stronger intermolecular forces would lower the vapor pressure, and in these small molecules hydrogen bonding outweighs the small difference in dispersion forces.
Question 5 (constructed response · core)
A student separates a mixture of a blue dye and a yellow dye on a column packed with silica gel, using a solvent of low polarity. She collects the liquid leaving the column in small fractions. The blue dye leaves the column in the fractions collected between 12 mL and 20 mL. The yellow dye leaves between 35 mL and 46 mL.
(a) Identify which dye is more polar. Justify your answer in terms of intermolecular interactions. (b) The student runs the same mixture on a silica TLC plate with the same solvent. Predict which dye has the higher Rf value. Explain. (c) She repeats the column with a more polar solvent. Predict whether the yellow dye will leave the column after a smaller or a larger volume than before. Explain.
Worked solution and suggested Marlbridge rubric
(a) The yellow dye is more polar. It took longer to pass through the column (35–46 mL against 12–20 mL), so it spent more time held by the silica. Silica is polar, with surface Si–O–H groups, so the dye with the stronger dipole–dipole or hydrogen-bonding attractions to it is held back more. That is the more polar dye.
(b) The blue dye has the higher Rf. The same stationary phase and solvent are used, so the dye that moves fastest through the column (weaker attraction to silica) also moves furthest up the plate.
(c) Smaller volume. A more polar solvent is attracted more strongly to the silica surface and to the polar dye, so it competes better with the stationary phase. The yellow dye spends less time held on the silica and is carried out sooner.
| Point | What earns it |
|---|---|
| 1 | Yellow is more polar, linked to its slower passage through the column |
| 1 | Explains that stronger intermolecular attractions to the polar silica hold the dye back |
| 1 | Blue has the higher Rf, with a reason linking column order to TLC order |
| 1 | Smaller volume, explained by the polar solvent competing with the stationary phase for the dye |
Do not award the point in (a) for “yellow is more polar” without a reason based on attraction to the stationary phase.
Question 6 (constructed response · core)
A bottle holds a cloudy green liquid. It is water containing undissolved chalk powder (calcium carbonate) and a green color. A student wants to answer the question: “Is the green color one dye or a mixture of dyes?”
(a) Describe how to remove the chalk, and explain why this method works for the chalk. (b) After step (a) the liquid is clear but still green. Explain why step (a) did not remove the color. (c) Describe a paper chromatography procedure that would answer the student’s question. Include where the sample is placed, how the solvent is added, and what result would show that the green is a mixture. (d) Another student suggests distilling the clear green liquid to separate the dyes. Evaluate this suggestion.
Worked solution and suggested Marlbridge rubric
(a) Filter the liquid. The chalk has not dissolved, so it exists as solid grains much larger than the pores in the filter paper. The grains stay on the paper while the water and everything dissolved in it pass through.
(b) The dye or dyes are dissolved: they are spread through the water as single molecules, far smaller than the pores, so they pass through the filter.
(c) Draw a pencil origin line about 1–2 cm above the bottom of a strip of chromatography paper. Put a small, concentrated spot of the green filtrate on the line and let it dry. Stand the paper in a covered container with solvent below the origin line, so the spot does not dissolve into the pool of solvent. Remove the paper when the solvent front is near the top and mark the front. If the green spot separates into two or more spots at different heights (for example, one blue and one yellow), the green is a mixture. A single spot suggests one dye, although two dyes with identical Rf values in that solvent would also give one spot, so a second solvent would make the test stronger.
(d) The suggestion will not work. Distillation separates components by vapor pressure. Water has a far higher vapor pressure than the dyes, which are not volatile. Water would distil over and the dyes would be left together as a residue in the flask, still mixed. Distillation can remove the water, but it cannot separate the dyes from each other.
| Point | What earns it |
|---|---|
| 1 | Filtration, explained by the chalk being undissolved solid larger than the pores |
| 1 | Color passes through because the dye is dissolved as individual molecules |
| 1 | Pencil origin line above the solvent level, with the spot on the line |
| 1 | Two or more separated spots = mixture |
| 1 | Distillation rejected: dyes are nonvolatile, so they stay together while water distils |
Accept thin-layer chromatography in (c). A sketch with labels can earn the procedure point in (c).
Question 7 (argumentation · stretch)
Two compounds, A and B, are spotted side by side on two identical silica TLC plates. Plate 1 is developed with hexane. Plate 2 is developed with a 1 : 1 mixture of hexane and ethyl acetate, which is more polar than hexane alone. On both plates the solvent front moves 6.0 cm from the origin.
| Plate | Distance moved by A (cm) | Distance moved by B (cm) |
|---|---|---|
| 1 (hexane) | 0.6 | 2.1 |
| 2 (hexane/ethyl acetate) | 2.7 | 5.1 |
(a) Calculate the Rf value of each compound on each plate. (b) A student claims: “Compound A became less polar in the second solvent, because its Rf value increased.” Evaluate this claim. (c) Which solvent gives the better separation of A and B? Justify your answer with the data.
Worked solution and suggested Marlbridge rubric
(a) Plate 1: A = 0.6 ÷ 6.0 = 0.10; B = 2.1 ÷ 6.0 = 0.35. Plate 2: A = 2.7 ÷ 6.0 = 0.45; B = 5.1 ÷ 6.0 = 0.85.
(b) The claim is incorrect. Polarity is a property of A’s molecular structure, and that structure is the same in both solvents. What changed is the mobile phase. The more polar solvent is attracted more strongly to the silica surface and to A, so it competes better with the stationary phase for A. A therefore spends less time held on the silica, and its Rf rises. The evidence that A’s relative polarity is unchanged: A still has a lower Rf than B on both plates, so A is still the more polar of the two.
(c) Plate 2 (hexane/ethyl acetate). The spots are 2.4 cm apart on plate 2 (5.1 − 2.7) but only 1.5 cm apart on plate 1 (2.1 − 0.6); in Rf terms the difference is 0.40 against 0.25. On plate 1, A has barely left the origin (Rf 0.10).
| Point | What earns it |
|---|---|
| 1 | All four Rf values correct |
| 1 | States the claim is wrong because A’s structure, and so its polarity, has not changed |
| 1 | Explains the higher Rf by the more polar mobile phase competing with the silica for A |
| 1 | Uses the unchanged order (A below B on both plates) as evidence |
| 1 | Chooses plate 2 with a quantitative comparison of the separations |
How did you do?
- Q1 or Q6(a)–(b) wrong: re-read “Why a filter cannot separate a solution” in the study guide.
- Q2, Q3 or Q7(a) wrong: practise Rf calculations and polarity ranking with Figure 1 and Worked example 1.
- Q4 or Q6(d) wrong: revisit the distillation section and Worked example 2: compare intermolecular forces, then vapor pressures.
- Q5 or Q7(b)–(c) incomplete: your answer needs the competition between the mobile phase and the stationary phase, not only “it is more polar”.
Then tick off the topic checklist.