Separations of Solutions and Mixtures

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AP Chemistry › Separations of Solutions and Mixtures

Questions 1 - 10
1

A liquid mixture contains hexane (nonpolar, $b.p.\approx 69,^{\circ}\mathrm{C}$) and octane (nonpolar, $b.p.\approx 126,^{\circ}\mathrm{C}$). The goal is to separate and collect both liquids with good purity. Which method is most appropriate?

Decanting because the liquids will form two layers over time

Simple filtration because the liquids have different densities

Evaporation to dryness because both liquids are volatile

Fractional distillation because the liquids have different boiling points

Paper chromatography because both liquids are nonpolar

Explanation

This question tests the skill of selecting separation techniques based on physical properties of miscible liquids. Hexane and octane are both nonpolar hydrocarbons that form a homogeneous mixture, but they have significantly different boiling points (69°C vs 126°C). Fractional distillation exploits this boiling point difference by using a fractionating column that provides multiple vaporization-condensation cycles, allowing better separation of liquids with closer boiling points. The lower-boiling hexane vaporizes first and can be collected separately from the higher-boiling octane. Option E incorrectly assumes the liquids would form layers, but nonpolar liquids of similar structure are miscible and won't separate by decanting. The strategy is to identify that liquids with different boiling points can be separated by distillation, with fractional distillation preferred when high purity is needed.

2

A student has a mixture of sand and a sodium chloride solution (saltwater). The goal is to obtain dry sand and also recover solid sodium chloride. Which sequence of techniques best accomplishes both separations?

Filter to remove sand, then evaporate the filtrate to dryness to obtain NaCl(s)

Decant the liquid to remove sand, then filter the liquid to obtain NaCl(s)

Centrifuge the mixture to separate NaCl(s), then distill to remove sand

Use paper chromatography to separate sand from dissolved NaCl, then evaporate water

Distill the mixture first to collect NaCl(s), then filter to remove sand

Explanation

This question tests the skill of selecting appropriate separation techniques for a heterogeneous mixture containing both a solid and a solution. The mixture contains sand (insoluble solid) and dissolved NaCl in water, requiring two different separation methods. First, filtration removes the insoluble sand particles, which are trapped by the filter paper while the saltwater solution passes through as the filtrate. Then, evaporation of the filtrate removes water through vaporization, leaving behind solid NaCl crystals. Option B incorrectly suggests distillation would collect NaCl(s) directly, but distillation collects the volatile component (water), not the dissolved salt. The key strategy is to recognize that filtration separates solids from liquids based on particle size, while evaporation separates dissolved solids from their solvents based on volatility differences.

3

A student has a mixture of ethanol and water that appears as a single liquid phase. The student wants to increase the ethanol concentration (not necessarily to 100%) by separating based on volatility. Which technique best achieves this?

Paper chromatography because ethanol and water have different polarities

Filtration because ethanol molecules are smaller than water molecules

Decanting because ethanol and water will separate into layers when cooled

Fractional distillation because ethanol has a lower boiling point than water

Magnetic separation because water is attracted to a magnet

Explanation

This question tests the skill of separating miscible liquids based on volatility differences. Ethanol (b.p. 78°C) and water (b.p. 100°C) form a homogeneous mixture due to hydrogen bonding between molecules. Fractional distillation can increase the ethanol concentration because ethanol's lower boiling point means it vaporizes more readily than water. During distillation, the vapor phase becomes enriched in ethanol, and condensing this vapor yields a liquid with higher ethanol content than the original mixture. Option D incorrectly suggests ethanol and water would separate into layers when cooled, but these polar liquids remain miscible at all temperatures due to hydrogen bonding. The key principle is that distillation separates components based on their relative volatilities, with the more volatile component (lower boiling point) concentrating in the distillate.

4

A student has a mixture of sand and a saltwater solution in a beaker. The sand is visibly insoluble and settles to the bottom after stirring stops. Which separation method will best separate the sand from the saltwater solution?

Decanting only, because dissolved salt will settle out with the sand

Simple distillation to vaporize the water and leave sand and salt behind

Evaporation to remove water so the sand can be poured off

Filtration to trap the sand while the saltwater passes through

Paper chromatography to separate sand particles from dissolved ions

Explanation

This question tests the skill of selecting an appropriate separation technique for a heterogeneous mixture of an insoluble solid and a liquid solution based on differences in solubility and particle size. Filtration is the best method because the sand is insoluble and settles, allowing it to be trapped by the filter paper while the saltwater solution passes through as the filtrate. The principle relies on the physical barrier of the filter, which retains larger solid particles but allows dissolved ions and liquid to flow through. After filtration, the sand can be rinsed and dried if needed, effectively separating it from the solution. A tempting distractor is evaporation (choice D), which is incorrect because it would remove the water but leave the salt mixed with the sand, misunderstanding that evaporation separates solvents from solutes but not insoluble solids from dissolved ones. To separate insoluble solids from liquids, prioritize methods like filtration that exploit differences in physical state and solubility rather than phase changes.

5

A student has a mixture of two liquids that form two layers in a separatory funnel: water (bottom layer) and diethyl ether (top layer). The student wants to separate the two liquids. Which method is most appropriate?

Fractional distillation, because immiscible liquids cannot be separated by draining layers

Filtration, because the denser liquid will be trapped by the filter paper

Evaporation, because both liquids can be recovered as separate liquids by boiling off the top layer

Decanting using a separatory funnel, because the liquids are immiscible and form layers

Paper chromatography, because immiscible liquids separate based on $R_f$ values

Explanation

This question tests the skill of separating immiscible liquids using phase separation techniques based on density differences. Decanting using a separatory funnel is most appropriate because water and diethyl ether are immiscible, forming distinct layers with ether on top due to lower density, allowing the bottom layer to be drained first. This method physically separates the layers without mixing or loss. The funnel's design facilitates controlled release of each layer. Fractional distillation (choice D) is a tempting distractor but incorrect because immiscible liquids can be separated more simply by decanting, misconstruing that distillation is necessary for all liquid mixtures regardless of miscibility. For immiscible liquid mixtures, always check for layer formation and use separatory funnels for efficient, gravity-based separation.

6

A student performs paper chromatography on a black ink sample and observes that it separates into several colored spots at different heights on the paper. Which statement best explains why the spots end up at different positions?

The components have different boiling points, so higher-boiling components travel farther

The components have different solubilities and attractions to the stationary phase, so they move at different rates

The components have different densities, so denser components remain lower on the paper

The components react with the paper to form new compounds at different heights

The components have different particle sizes, so larger molecules are filtered out earlier

Explanation

This question tests the skill of understanding the principles underlying chromatographic separation of mixture components on paper. The components separate because they have different solubilities and attractions to the stationary phase (paper) versus the mobile phase (solvent), causing them to move at different rates and end up at different positions. More polar components interact strongly with the paper and travel shorter distances, while less polar ones move farther with the solvent. This partitioning results in distinct colored spots from the black ink, revealing its composition. Different boiling points (choice B) is a tempting distractor but incorrect because chromatography does not involve vaporization, misconstruing that separation relies on volatility rather than intermolecular forces. In chromatography, evaluate components' polarities and solubilities to anticipate separation patterns and interpret results effectively.

7

A mixture contains small glass beads and a solution of copper(II) sulfate in water. The student wants to separate the beads from the solution while keeping the solution unchanged. Which method is most appropriate?

Magnetic separation because $\text{Cu}^{2+}$ ions are paramagnetic and will pull the beads out

Simple distillation to collect beads as the distillate

Evaporation to remove water so the beads can be poured off from the crystals

Paper chromatography to move the beads farther than the dissolved ions

Filtration to retain the beads while the copper(II) sulfate solution passes through

Explanation

This question tests the skill of separating solids from solutions using particle size. Glass beads are larger solids, while copper(II) sulfate is dissolved, so filtration retains the beads on the paper while the solution passes through unchanged. This applies the principle of mechanical separation by pore size in filters. The solution remains intact after separation. A tempting distractor is simple distillation (choice D), which is incorrect because beads do not distill, misconstruing distillation for solid-liquid separations. For mixtures of solids in solutions, use filtration to isolate solids based on insolubility and size.

8

A mixture contains two dissolved substances in water: sodium chloride and a blue dye. The student wants to separate the dye from the salt to analyze the dye components, without evaporating all the water. Which method is most appropriate?

Simple distillation because dyes have lower boiling points than sodium chloride

Decanting because the dye settles below the salt solution over time

Filtration because dissolved dye particles are larger than dissolved ions

Paper chromatography because dye molecules partition differently while NaCl remains largely unseparated

Magnetic separation because the blue dye is colored and therefore magnetic

Explanation

This question tests the skill of using chromatography to separate dissolved components without full evaporation. The blue dye and sodium chloride have different polarities, so paper chromatography partitions the dye components differently, while NaCl may not separate much. This exploits the principle of affinity differences in stationary and mobile phases for analysis. It allows separation of the dye for further study. A tempting distractor is simple distillation (choice C), which is incorrect because dyes and salt are nonvolatile, misconstruing chromatography with boiling point separation. When analyzing dissolved mixtures, select chromatography for separations based on polarity without removing the solvent entirely.

9

A student has a homogeneous aqueous solution containing dissolved potassium chloride. The student wants to obtain solid potassium chloride. Which method is most appropriate?

Evaporation (or crystallization) to remove water and leave potassium chloride crystals

Decanting because potassium chloride settles to the bottom over time

Filtration because dissolved ions are too large to pass through filter paper

Paper chromatography because ions separate by boiling point on paper

Magnetic separation because $\text{K}^+$ is attracted to magnets

Explanation

This question tests the skill of recovering solutes from homogeneous solutions. The potassium chloride is dissolved in water, so evaporation or crystallization removes the solvent, leaving solid KCl crystals behind. This method applies the principle that nonvolatile solutes remain after solvent evaporation. It is straightforward for obtaining solids from solutions in a lab. A tempting distractor is filtration (choice A), which is incorrect because dissolved ions pass through filter paper, misconstruing filtration as effective for homogeneous solutions. When isolating solutes, consider evaporation for nonvolatile compounds to concentrate and crystallize them from solution.

10

A student has a mixture of iron filings and powdered sulfur. The student wants to separate the iron from the sulfur without using any liquids or chemical reactions. Which method is most appropriate?

Filtration because sulfur dissolves in water but iron does not

Magnetic separation because iron is attracted to a magnet and sulfur is not

Evaporation because sulfur will evaporate and leave iron behind at room temperature

Paper chromatography because iron and sulfur have different polarities

Simple distillation because iron has a higher boiling point than sulfur

Explanation

This question tests the skill of separating solid mixtures using magnetic properties without solvents or reactions. Iron filings are magnetic, while sulfur is not, so magnetic separation works by attracting iron to a magnet, leaving sulfur behind. This method exploits the principle of ferromagnetism in iron, allowing physical separation without liquids. It is ideal for dry mixtures where one component is magnetic. A tempting distractor is filtration (choice B), which is incorrect because neither component dissolves in water, misconstruing the need for solubility differences in filtration. When dealing with solid mixtures, identify unique physical properties like magnetism to choose an effective, non-chemical separation technique.

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