Extraction and Distillation (5C)

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MCAT Chemical and Physical Foundations of Biological Systems › Extraction and Distillation (5C)

Questions 1 - 10
1

A lab performs liquid–liquid extraction of a neutral steroid from 20 mL water into 10 mL hexane. The partition coefficient is $K = \frac{S{hex}}{S{aq}} = 10$ at this temperature. Based on the partition coefficient principle, which change would be expected to increase the fraction of steroid remaining in the aqueous layer after equilibrium is reached?

Agitate more vigorously so the steroid chemically reacts with water

Increase the hexane volume while keeping water volume constant

Decrease $K$ by switching to a more polar organic solvent than hexane

Perform the extraction at lower pressure to reduce the boiling point of hexane

Explanation

This question tests understanding of factors affecting extraction efficiency based on partition coefficients. The partition coefficient K = 10 indicates the steroid strongly prefers hexane over water. To increase the fraction remaining in water, we need to decrease the effective extraction into hexane. Choice A is correct because using a more polar organic solvent would decrease K, making the steroid less preferentially extracted and leaving more in the aqueous phase. Choice B would actually extract more steroid by increasing the organic phase volume. Choice C incorrectly relates pressure to extraction efficiency, while Choice D suggests an irrelevant chemical reaction. Remember that partition coefficients depend on the nature of both solvents and can be manipulated by changing solvent polarity.

2

A compound X is extracted between octanol and water to estimate lipophilicity. The measured partition coefficient is $K = \frac{X{oct}}{X{aq}} = 0.20$ for the neutral form. Based on this value, what conclusion can be drawn about X under these conditions?

X preferentially partitions into octanol, consistent with high lipophilicity

X must be ionic because partition coefficients are only defined for ions

X preferentially partitions into water, consistent with relatively higher hydrophilicity

X will be best separated from water by simple distillation because $K<1$

Explanation

This question tests interpretation of partition coefficient values for predicting compound properties. A partition coefficient K = 0.20 means the concentration in octanol is only 0.20 times that in water, indicating the compound prefers the aqueous phase. Since octanol represents a lipophilic environment and water represents a hydrophilic environment, this low K value indicates relatively higher hydrophilicity. Choice B correctly identifies this relationship. Choice A incorrectly interprets the K < 1 value as favoring octanol. Choice C confuses extraction with distillation, which are different separation techniques. Choice D is incorrect because partition coefficients apply to neutral molecules, not just ions. To interpret partition coefficients correctly, remember that K > 1 indicates lipophilicity while K < 1 indicates hydrophilicity.

3

A student attempts to separate a mixture of acetone (bp 56°C) and water (bp 100°C) using simple distillation at 1 atm. The mixture is heated slowly, and vapor is condensed and collected. Based on boiling point differentiation, which outcome is most consistent with simple distillation?

The distillate composition is identical to the boiling flask composition at all times

Acetone cannot be separated by distillation because it is miscible with water

Water distills first because it has stronger hydrogen bonding and thus higher vapor pressure

The distillate collected early is enriched in acetone relative to the original mixture

Explanation

This question tests understanding of simple distillation based on boiling point differences. In simple distillation, the component with the lower boiling point (higher vapor pressure) will be enriched in the vapor phase and thus in the distillate. Acetone (bp 56°C) is much more volatile than water (bp 100°C), so the initial distillate will be enriched in acetone. Choice A correctly states this outcome. Choice B is incorrect because distillate composition changes over time as the more volatile component is depleted. Choice C incorrectly relates hydrogen bonding to vapor pressure - stronger hydrogen bonding actually decreases vapor pressure. Choice D incorrectly suggests miscibility prevents separation, when in fact distillation separates based on volatility, not miscibility. Remember that lower boiling point means higher vapor pressure at a given temperature.

4

A researcher needs to separate two volatile solvents with close boiling points: Solvent A (bp 78°C) and Solvent B (bp 82°C), both at 1 atm. The lab has both simple and fractional distillation setups available. Based on the core principle of repeated vapor–liquid equilibria, what is most consistent with using fractional distillation here?

Simple distillation is preferred because close boiling points increase the driving force for separation

Fractional distillation is preferred because a fractionating column provides multiple equilibration steps, improving separation for close bps

Fractional distillation works by selectively dissolving Solvent A in the column packing

Fractional distillation is unnecessary because any two liquids with different bps separate perfectly in one step

Explanation

This question tests understanding of when fractional distillation is preferred over simple distillation. Fractional distillation uses a fractionating column that provides multiple vapor-liquid equilibration steps, effectively performing many simple distillations in series. This is particularly important when boiling points are close (78°C vs 82°C), as each equilibration provides only modest enrichment. Choice A correctly identifies this principle and application. Choice B incorrectly suggests simple distillation is better for close boiling points. Choice C incorrectly describes the mechanism as dissolution rather than repeated vaporization-condensation. Choice D incorrectly claims perfect separation is unnecessary. For effective separation, use fractional distillation when boiling points differ by less than 25°C.

5

A mixture contains ethanol (bp 78°C) and a nonvolatile solute (e.g., glucose; decomposes before boiling). The solution is heated and subjected to simple distillation at 1 atm. Based on volatility differences, what outcome is most consistent with the distillation process?

Both ethanol and glucose will distill together because they are initially in the same phase

Glucose will distill before ethanol because it has more hydrogen bonding and thus higher vapor pressure

The distillate will be enriched in ethanol, while glucose remains in the boiling flask

Distillation cannot separate a solvent from a solute; chromatography is required in all cases

Explanation

This question tests understanding of distillation with volatile and nonvolatile components. Distillation separates based on volatility differences - volatile compounds vaporize and can be collected as distillate, while nonvolatile compounds remain in the boiling flask. Ethanol (bp 78°C) is volatile and will vaporize when heated, while glucose is nonvolatile and decomposes before boiling. Choice B correctly predicts that the distillate will be enriched in ethanol while glucose remains behind. Choice A incorrectly suggests glucose has higher vapor pressure despite being nonvolatile. Choice C incorrectly assumes both components distill together. Choice D incorrectly states distillation cannot separate solvent from solute. Remember that distillation is ideal for separating volatile from nonvolatile components.

6

An aqueous fermentation broth contains a volatile flavor compound F and nonvolatile salts. A purification step uses distillation under reduced pressure to avoid thermal degradation of other components. Based on the effect of external pressure on boiling, what outcome is most consistent with reducing the pressure during distillation?

Partition coefficients increase, causing salts to move into the vapor phase

Boiling points increase, improving separation by increasing intermolecular forces

Lower pressure forces nonvolatile salts to sublime into the distillate

Boiling points decrease, allowing F to distill at a lower temperature

Explanation

This question tests understanding of pressure effects on distillation. Reducing external pressure lowers the boiling points of all volatile components because molecules need less kinetic energy to escape into the vapor phase when opposing pressure is reduced. This allows distillation at lower temperatures, which is beneficial for thermally sensitive compounds. Choice A correctly identifies this effect. Choice B incorrectly states boiling points increase with reduced pressure. Choice C incorrectly invokes partition coefficients, which apply to extraction not distillation. Choice D incorrectly suggests nonvolatile salts can sublime under reduced pressure. To avoid thermal degradation during distillation, remember that vacuum distillation lowers operating temperatures by reducing boiling points.

7

A student tries to purify a volatile organic product P from an aqueous reaction mixture. They propose either (i) liquid–liquid extraction into an immiscible organic solvent or (ii) simple distillation. The key given information is: P has a high partition coefficient into the organic solvent ($K = \frac{P{org}}{P{aq}} \gg 1$) but has a boiling point similar to water. What conclusion is most consistent with these principles?

Neither method can work because similar boiling points prevent any phase transfer during extraction

Simple distillation is likely more effective because partition coefficients determine boiling behavior

Extraction is likely more effective than simple distillation because $K\gg 1$ favors transfer into the organic phase despite similar boiling points

Chromatography is required because immiscible solvents cannot separate solutes

Explanation

This question tests choosing between extraction and distillation based on physical properties. The high partition coefficient (K >> 1) indicates the product strongly prefers the organic phase over water, making extraction highly effective. The similar boiling points of product and water would make distillation difficult and require fractional distillation for modest separation. Choice A correctly identifies that extraction is more effective given these properties. Choice B incorrectly relates partition coefficients to boiling behavior. Choice C incorrectly suggests chromatography is required. Choice D incorrectly claims similar boiling points prevent extraction, when extraction depends on solubility differences not volatility. When K >> 1, extraction provides excellent separation regardless of boiling point similarities.

8

A mixture of two liquids is separated by distillation. Liquid L has a lower boiling point than liquid H at 1 atm. The distillation is run with a fractionating column, and the first collected fraction is analyzed. Based on vapor–liquid equilibrium reasoning, what is most consistent with the composition of the first fraction?

Its composition depends primarily on the partition coefficient of L between the liquid and vapor layers

It contains only L because fractional distillation guarantees perfect separation for any mixture

It is enriched in H because the higher-boiling component condenses preferentially and thus distills first

It is enriched in L because the vapor phase is richer in the more volatile component

Explanation

This question tests understanding of vapor-liquid equilibrium in distillation. During distillation, the vapor phase is always enriched in the more volatile (lower boiling point) component compared to the liquid phase. Since liquid L has a lower boiling point than liquid H, L is more volatile and will be enriched in the vapor phase. The first fraction collected represents condensed vapor and will therefore be enriched in L. Choice B correctly identifies this enrichment. Choice A incorrectly suggests the higher-boiling component distills first. Choice C incorrectly claims perfect separation is achieved. Choice D incorrectly invokes partition coefficients, which apply to liquid-liquid systems not vapor-liquid equilibria. Remember that in distillation, lower boiling point components always concentrate in the distillate.

9

A mixture contains three volatile components with normal boiling points at 1 atm: P (60°C), Q (90°C), and R (120°C). A fractional distillation is begun and fractions are collected as the head temperature stabilizes near plateaus. Which collection order is most consistent with boiling point differentiation?

P first, then Q, then R as the temperature increases over time.

R first, then Q, then P as the temperature decreases over time.

Q first, then P, then R because the middle boiling point component equilibrates fastest.

All three co-distill together in a constant ratio because fractional distillation prevents enrichment.

Explanation

This question tests understanding of fractional distillation collection order. In fractional distillation, components are collected in order of increasing boiling point as the head temperature rises through distinct plateaus. The lowest boiling component P (60°C) distills first when the head temperature stabilizes near 60°C, followed by Q (90°C) at its plateau, and finally R (120°C). Choice B correctly identifies this P-Q-R order with increasing temperature. Choice A reverses the order, incorrectly suggesting high-boiling components distill first. Remember that fractional distillation separates based on volatility differences, with more volatile (lower boiling) components collected first.

10

A researcher performs liquid–liquid extraction of a neutral metabolite M between water and hexane. The partition coefficient is defined as $K=\frac{M{hex}}{M{aq}}$ at equilibrium. If the experiment is repeated with the same total amount of M but the hexane volume is doubled (water volume unchanged), what outcome is most consistent with the definition of $K$?

No redistribution occurs because partitioning depends only on the initial concentrations, not equilibrium.

The value of $K$ increases because more hexane is available to dissolve M.

The equilibrium concentrations adjust, but $K$ remains the same at the same temperature.

The value of $K$ decreases because dilution in hexane lowers $[M]_{hex}$.

Explanation

This question tests understanding of partition coefficients as equilibrium constants. The partition coefficient K is a thermodynamic constant that depends only on temperature, not on volumes or initial concentrations. When hexane volume is doubled, more total metabolite M will transfer to the hexane phase, but the ratio of concentrations [M]hex/[M]aq at equilibrium remains constant. Choice C correctly states that equilibrium concentrations adjust while K remains unchanged. Choice A incorrectly suggests K increases with volume, confusing the equilibrium constant with the total amount extracted. To avoid errors, remember that equilibrium constants like K are intensive properties that don't change with system size.

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