Separations of Solutions and Mixtures - College Chemistry

1. A student wants to separate a mixture of acetone (water-soluble) and toluene (water-insoluble) using a separatory funnel. Both solvents are less dense than water. After adding water and shaking, two distinct layers form. Which statement correctly describes the layer composition?
   Answer: Top layer contains toluene, bottom layer contains water with dissolved acetone When you encounter separatory funnel problems, you need to consider two key principles: density differences and solubility patterns. The denser liquid always settles to the bottom, while the less dense liquid floats on top. In this mixture, water has the highest density (1.0 g/mL), so it will form the bottom layer. Acetone, being water-soluble, will dissolve into this aqueous layer. Toluene is water-insoluble and less dense than water, so it remains as a separate organic layer floating on top of the water-acetone solution. Choice A correctly identifies this arrangement: toluene forms the top layer, while the bottom layer contains water with dissolved acetone. This follows the fundamental principle that immiscible liquids separate by density, with miscible components dissolving into their preferred phase. Choice B incorrectly reverses the density relationship, suggesting water (with acetone) is less dense than toluene, which contradicts the given information that both organic solvents are less dense than water. Choice C wrongly assumes acetone won't dissolve in water and instead remains with toluene in the organic layer. This ignores acetone's high water solubility due to its polar carbonyl group. Choice D suggests incomplete separation with all components in both layers. While real separations aren't perfectly clean, the question describes "two distinct layers," indicating complete phase separation based on solubility and density differences. Remember: in liquid-liquid extractions, always identify the densest phase first (it's your bottom layer), then determine where each component will preferentially dissolve based on polarity and solubility rules.
2. A laboratory mixture contains benzoic acid ($C_6H_5COOH$, $K_a = 6.3 \times 10^{-5}$), sodium chloride, and sand. The mixture is treated with dilute aqueous sodium hydroxide solution. Which statement best describes the result of this treatment?
   Answer: Both benzoic acid and sodium chloride dissolve, leaving only sand When you encounter acid-base chemistry problems involving mixtures, think about how each component will behave individually in the given conditions. Here, you're adding a basic solution (NaOH) to a mixture, so consider what happens to each substance. Benzoic acid ($C_6H_5COOH$) is a weak acid that will react with the strong base NaOH in an acid-base neutralization: $C_6H_5COOH + NaOH \rightarrow C_6H_5COONa + H_2O$. This forms sodium benzoate, which is highly soluble in water. Sodium chloride is an ionic compound that readily dissolves in water regardless of pH - the presence of NaOH doesn't prevent this. Sand (silicon dioxide) is insoluble in water and won't dissolve in basic conditions. Therefore, both benzoic acid (as sodium benzoate) and sodium chloride will be in solution, while sand remains undissolved. Looking at the wrong answers: Choice A incorrectly suggests sand dissolves - sand is essentially insoluble in aqueous solutions. Choice B misses that sodium chloride will also dissolve; this answer focuses only on the acid-base reaction while ignoring the other soluble component. Choice C incorrectly assumes benzoic acid won't dissolve and that only the salt dissolves, overlooking the acid-base neutralization that makes benzoic acid soluble as its sodium salt. Study tip: In separation problems, analyze each component's behavior independently. Remember that weak acids become soluble salts when neutralized by strong bases, and consider the inherent solubility of each substance in the given solvent conditions.
3. An aqueous solution contains both $Ca^{2+}$ and $Mg^{2+}$ ions. To selectively precipitate calcium while leaving magnesium in solution, which reagent should be added, and what principle governs this separation?
   Answer: Ammonium oxalate; $CaC_2O_4$ is highly insoluble while $MgC_2O_4$ is soluble When you encounter selective precipitation problems, you're dealing with the principle that different compounds have different solubilities, and you can exploit these differences to separate ions. The key is finding conditions where one ion forms a precipitate while the other remains dissolved. For calcium and magnesium separation, sodium carbonate ($Na_2CO_3$) is the ideal reagent. When added to the solution, it provides carbonate ions that react with both metals, but here's the crucial difference: calcium carbonate ($CaCO_3$) has a much lower solubility than magnesium carbonate ($MgCO_3$). The $K_{sp}$ of $CaCO_3$ is approximately $3.4 \times 10^{-9}$, while $MgCO_3$ has a $K_{sp}$ of about $6.8 \times 10^{-6}$ — nearly three orders of magnitude higher. This means $Ca^{2+}$ will precipitate first and completely, while $Mg^{2+}$ stays in solution. Option B is incorrect because both hydroxides have very low solubility, making selective precipitation difficult. Option C fails because both sulfates are actually quite soluble in water — neither would precipitate effectively. Option D is wrong because while calcium phosphates do form, the conditions required would likely precipitate magnesium as well, defeating the selectivity purpose. Remember this pattern: successful selective precipitation requires a significant difference in $K_{sp}$ values between the target compounds. Always compare solubility constants when evaluating separation strategies — the larger the difference, the cleaner your separation will be.
4. A chromatography experiment is performed using silica gel as the stationary phase and a mixture of hexane and ethyl acetate (3:1 v/v) as the mobile phase. Three compounds are separated: compound A ($R_f = 0.85$), compound B ($R_f = 0.45$), and compound C ($R_f = 0.15$). Which statement about the relative polarities of these compounds is correct?
   Answer: Compound C is most polar, compound A is least polar When you encounter thin-layer chromatography (TLC) problems, the key relationship to remember is that $R_f$ values reveal how compounds interact with the stationary phase. In normal-phase chromatography using silica gel (a polar stationary phase), polar compounds stick more strongly to the silica and travel shorter distances, resulting in lower $R_f$ values. Looking at the $R_f$ values: compound C ($R_f = 0.15$) barely moved up the plate, indicating strong attraction to the polar silica gel, making it the most polar compound. Compound A ($R_f = 0.85$) traveled nearly to the top, showing weak interaction with silica and strong affinity for the relatively nonpolar mobile phase, making it the least polar. Compound B ($R_f = 0.45$) falls in between with intermediate polarity. Answer A incorrectly reverses the polarity relationship—it suggests the compound that traveled farthest (A) is most polar, when the opposite is true. Answer C correctly identifies A as least polar but wrongly places B as most polar, ignoring that C has the lowest $R_f$ value. Answer D misses the fundamental point that different $R_f$ values directly indicate different polarities—compounds separate precisely because they have different polarities. Answer B correctly states that compound C is most polar and compound A is least polar. Study tip: Remember the inverse relationship in normal-phase TLC: lower $R_f$ = higher polarity. Polar compounds "stick" to polar stationary phases and don't travel far.
5. An aqueous solution contains $Fe^{3+}$, $Al^{3+}$, and $Mg^{2+}$ ions. The solution is treated with aqueous ammonia ($NH_3$) until the pH reaches 9.5. Based on the $K_{sp}$ values for the hydroxides ($Fe(OH)_3: 2.8 \times 10^{-39}$, $Al(OH)_3: 3.0 \times 10^{-34}$, $Mg(OH)_2: 5.6 \times 10^{-12}$), which ions will precipitate?
   Answer: Both $Fe^{3+}$ and $Al^{3+}$ precipitate as hydroxides When you encounter precipitation problems involving multiple ions and pH changes, you need to determine which compounds will exceed their solubility limits under the given conditions. At pH 9.5, the hydroxide ion concentration is $[OH^-] = 10^{-4.5} = 3.16 \times 10^{-5}$ M. To find which hydroxides will precipitate, calculate the reaction quotient (Q) for each and compare it to the respective $K_{sp}$. For $Fe(OH)_3$: $Q = [Fe^{3+}][OH^-]^3$. Assuming typical concentrations around 0.1 M for the metal ions, $Q = (0.1)(3.16 \times 10^{-5})^3 = 3.16 \times 10^{-15}$. Since this greatly exceeds $K_{sp} = 2.8 \times 10^{-39}$, $Fe(OH)_3$ precipitates. For $Al(OH)_3$: $Q = (0.1)(3.16 \times 10^{-5})^3 = 3.16 \times 10^{-15}$, which exceeds $K_{sp} = 3.0 \times 10^{-34}$, so $Al(OH)_3$ also precipitates. For $Mg(OH)_2$: $Q = (0.1)(3.16 \times 10^{-5})^2 = 1.0 \times 10^{-10}$, which is less than $K_{sp} = 5.6 \times 10^{-12}$, so $Mg(OH)_2$ remains soluble. Answer A is wrong because $Al^{3+}$ also precipitates at this pH. Answer C is incorrect because $Mg^{2+}$ doesn't reach its precipitation threshold. Answer D shows a fundamental misunderstanding—higher $K_{sp}$ values indicate greater solubility, not precipitation tendency. Remember: smaller $K_{sp}$ values mean compounds precipitate more easily. Always calculate Q and compare to $K_{sp}$ rather than just comparing the $K_{sp}$ values alone.
6. A student performs recrystallization of an impure solid using hot water as the solvent. After cooling the solution to room temperature, very few crystals form. Which modification to the procedure would most likely improve crystal formation?
   Answer: Scratch the inside of the flask with a glass rod or add a seed crystal Recrystallization relies on controlled crystal formation from a supersaturated solution. When you dissolve an impure solid in hot solvent and then cool it, the decreased solubility at lower temperature should cause pure crystals to precipitate out, leaving impurities in solution. However, sometimes crystals fail to form even when the solution is supersaturated. The key issue here is nucleation - crystals need a surface or starting point to begin forming. When very few crystals appear despite proper supersaturation, the solution lacks nucleation sites. Scratching the flask's interior with a glass rod creates tiny glass particles that serve as nucleation sites, while adding a seed crystal (a small amount of pure compound) provides a perfect template for crystal growth. Both techniques give dissolved molecules a surface to organize upon and begin crystallizing. Looking at the wrong answers: (A) Using more hot water would actually make the problem worse by creating a more dilute solution that's less likely to become supersaturated upon cooling. (B) Adding cold water would dilute the solution and reduce supersaturation, working against crystal formation. (D) Higher heating temperature doesn't address the nucleation problem and could potentially decompose your compound or increase impurity solubility. Study tip: Remember that successful recrystallization requires both supersaturation AND nucleation sites. If crystals won't form from a properly prepared solution, think nucleation first - scratching or seeding almost always solves the problem. This is a common lab issue you'll encounter in organic chemistry courses.
7. A precipitate of $AgCl$ is washed with distilled water to remove soluble impurities. However, some of the $AgCl$ dissolves during washing, reducing the yield. To minimize this loss while still removing impurities, what modification should be made to the washing procedure?
   Answer: Use a dilute $HCl$ solution instead of pure water for washing This question tests your understanding of the common ion effect and solubility equilibrium. When washing precipitates, you need to balance removing impurities while minimizing dissolution of your desired product. $AgCl$ establishes an equilibrium with its ions: $AgCl_{(s)} \rightleftharpoons Ag^+_{(aq)} + Cl^-_{(aq)}$. In pure water, some $AgCl$ will dissolve according to its solubility product constant ($K_{sp}$). To minimize this dissolution, you can apply Le Chatelier's principle by adding one of the product ions to shift the equilibrium back toward the solid. Using dilute $HCl$ solution (answer B) provides $Cl^-$ ions, which creates a common ion effect. The excess $Cl^-$ shifts the equilibrium to the left, suppressing the dissolution of $AgCl$ while still allowing water to wash away soluble impurities. Answer A is wrong because hot water increases solubility, making dissolution worse. Answer C would work in principle since $Ag^+$ is also a common ion, but $AgNO_3$ is expensive and introduces new impurities ($NO_3^-$), making it impractical. Answer D incorrectly assumes temperature is the primary factor - while cold water slightly reduces solubility, the effect is minimal compared to the common ion effect, and impurity removal becomes less efficient at low temperatures. Remember: when washing ionic precipitates, use a dilute solution containing one of the precipitate's ions to suppress dissolution while maintaining washing efficiency. This is a classic application of the common ion effect in analytical chemistry.
8. An aqueous solution contains $Ba^{2+}$, $Sr^{2+}$, and $Ca^{2+}$ ions. Dilute sulfuric acid is added dropwise until no more precipitate forms. Based on the $K_{sp}$ values ($BaSO_4: 1.1 \times 10^{-10}$, $SrSO_4: 3.2 \times 10^{-7}$, $CaSO_4: 9.1 \times 10^{-6}$), which ions remain in solution after precipitation is complete?
   Answer: The relative amounts depend on the initial concentrations of each ion When you encounter precipitation problems with multiple ions, the key is understanding that lower Ksp values indicate lower solubility and greater tendency to precipitate. All three ions form sulfate precipitates with the same stoichiometry (1:1), so you can directly compare their Ksp values to determine precipitation order. BaSO₄ has the smallest Ksp (1.1 × 10⁻¹⁰), making it least soluble and first to precipitate. SrSO₄ (Ksp = 3.2 × 10⁻⁷) precipitates next, while CaSO₄ (Ksp = 9.1 × 10⁻⁶) is most soluble. As sulfuric acid is added dropwise, BaSO₄ precipitates first and most completely. However, the question asks what remains after precipitation is "complete" - meaning when no more precipitate forms upon adding more acid. At this point, the solution is saturated with respect to all three sulfates, but their different solubilities mean different amounts remain dissolved. The higher the Ksp, the more ions remain in solution at equilibrium. Choice A incorrectly assumes complete precipitation means zero ions remain - this ignores equilibrium solubility. Choice B suggests only Ca²⁺ remains, missing that Sr²⁺ also has measurable solubility. Choice C incorrectly identifies Ba²⁺ as remaining when it's actually the least soluble. Choice D correctly identifies that both Ca²⁺ and Sr²⁺ remain in solution since they have higher Ksp values than Ba²⁺. Study tip: In precipitation problems, remember that "complete" precipitation doesn't mean zero concentration - it means equilibrium is established. Always compare Ksp values to predict which ions remain most abundant in solution.
9. A mixture contains $NH_4Cl$ (sublimes at 340°C) and $NaCl$ (melts at 801°C). The mixture is heated to 400°C in a closed system with a cool condenser. What separation technique is being employed, and what is the expected outcome?
   Answer: Sublimation; $NH_4Cl$ sublimes and condenses on the cool surface, leaving $NaCl$ behind When you encounter separation technique problems, focus on the physical properties given and the conditions applied. Here, you have two compounds with very different behaviors at the heating temperature of 400°C. The key insight is understanding what happens at 400°C. $NH_4Cl$ sublimes at 340°C, meaning it transitions directly from solid to gas without melting. Since 400°C exceeds this temperature, $NH_4Cl$ will sublime completely. Meanwhile, $NaCl$ melts at 801°C, so at 400°C it remains solid since this temperature is well below its melting point. The cool condenser allows the gaseous $NH_4Cl$ to condense back to solid form on the cool surface, effectively separating it from the remaining solid $NaCl$. This is sublimation separation. Choice A is incorrect because distillation involves boiling liquids, not subliming solids, and $NaCl$ doesn't vaporize at 400°C. Choice C is wrong because we're dealing with physical phase changes, not chemical decomposition—$NH_4Cl$ sublimes but doesn't break down into different compounds. Choice D fails because fusion refers to melting, and at 400°C, $NaCl$ doesn't melt (it needs 801°C), so no melting or density-based separation occurs. Study tip: Always compare the given temperature to the phase transition temperatures of each component. When one compound undergoes a phase change while another doesn't, you can usually separate them based on that difference.
10. An aqueous solution contains $Ag^+$, $Cu^{2+}$, and $Na^+$ ions, each at 0.10 M concentration. Concentrated hydrochloric acid is slowly added to this solution. Based on the solubility products ($K_{sp}(AgCl) = 1.8 \times 10^{-10}$, $K_{sp}(CuCl_2) = 1.0 \times 10^{-6}$), what is the order of precipitation as $[Cl^-]$ increases?
    Answer: $AgCl$ precipitates first, followed by $CuCl_2$; $NaCl$ remains soluble When you encounter precipitation problems with multiple ions, you need to determine which compound will precipitate first by calculating the minimum chloride concentration required for each precipitation to begin. For precipitation to occur, the ion product must equal the $K_{sp}$. Using $K_{sp} = [M^{n+}][Cl^-]^m$ where m is the number of chloride ions: For $AgCl$: $K_{sp} = [Ag^+][Cl^-] = 1.8 \times 10^{-10}$ With $[Ag^+] = 0.10$ M: $[Cl^-] = \frac{1.8 \times 10^{-10}}{0.10} = 1.8 \times 10^{-9}$ M For $CuCl_2$: $K_{sp} = [Cu^{2+}][Cl^-]^2 = 1.0 \times 10^{-6}$ With $[Cu^{2+}] = 0.10$ M: $[Cl^-]^2 = \frac{1.0 \times 10^{-6}}{0.10} = 1.0 \times 10^{-5}$ So $[Cl^-] = 3.2 \times 10^{-3}$ M $AgCl$ precipitates first (at much lower $[Cl^-]$), then $CuCl_2$. $NaCl$ is highly soluble and won't precipitate under these conditions, making C correct. Answer A incorrectly suggests $NaCl$ will eventually precipitate, but sodium chloride is extremely soluble. Answer B reverses the precipitation order—this would happen if you incorrectly compared $K_{sp}$ values directly without accounting for the different stoichiometries and ion concentrations. Answer D is wrong because the compounds have vastly different precipitation thresholds. Study tip: Always calculate the actual concentration needed for precipitation rather than just comparing $K_{sp}$ values. Remember that compounds with different stoichiometries require different mathematical approaches, and highly soluble salts like $NaCl$ rarely precipitate from dilute solutions.
11. A mixture of ethyl acetate (b.p. 77°C) and ethanol (b.p. 78°C) forms an azeotropic mixture that boils at 72°C with a composition of 70% ethyl acetate and 30% ethanol. Simple distillation of this azeotropic mixture will produce:
    Answer: Azeotropic composition (70:30) in the distillate with no residue remaining When you encounter azeotrope problems, remember that azeotropes behave as single compounds during distillation—they cannot be separated by simple distillation because both components have identical vapor pressures at the azeotropic composition. This ethyl acetate-ethanol azeotrope is a minimum-boiling azeotrope (boiling point 72°C is lower than either pure component). During simple distillation of the azeotropic mixture, the vapor produced will have the exact same composition as the liquid (70% ethyl acetate, 30% ethanol) because that's the fundamental property of azeotropes. Since you're distilling the azeotropic composition itself, the entire mixture will vaporize at 72°C with no change in composition, leaving no residue. Option A is wrong because azeotropes cannot be separated into pure components by simple distillation—this would require more advanced techniques like azeotropic distillation with a third component. Option B has the same fundamental flaw, incorrectly assuming separation is possible. Option C incorrectly suggests that some material remains behind as residue; however, when you distill an azeotropic mixture at its exact azeotropic composition, the entire mixture behaves as a single compound and will completely distill over. Study tip: Remember that azeotropes are "constant-boiling mixtures." If you start with the exact azeotropic composition, simple distillation will transfer the entire mixture unchanged to the distillate. Separation only becomes relevant when you have excess of one component beyond the azeotropic ratio.
12. A mixture of ethanol ($C_2H_5OH$) and diethyl ether ($C_2H_5OC_2H_5$) needs to be separated. Both compounds have similar molecular weights (46 g/mol vs 74 g/mol) but different boiling points (78°C vs 35°C). A student considers using fractional distillation versus extraction with water. Which approach would be more effective and why?
    Answer: Water extraction, because ethanol's hydrogen bonding makes it preferentially water-soluble When you encounter separation problems in organic chemistry, focus on the fundamental principle that "like dissolves like" and consider the intermolecular forces present in each compound. Water extraction is the superior method here because ethanol contains a hydroxyl group (-OH) that can form hydrogen bonds with water molecules, making it highly water-soluble. Diethyl ether, while it has oxygen atoms, cannot form hydrogen bonds as effectively due to its ether linkage, making it much less water-soluble. This dramatic difference in water solubility allows for clean separation: ethanol dissolves into the aqueous layer while ether remains in the organic layer. Let's examine why the other options fall short. Choice A incorrectly suggests fractional distillation is better. While the 43°C boiling point difference could work, it's actually less efficient than extraction because both compounds will vaporize to some degree, requiring multiple distillation cycles and careful temperature control. Choice C is wrong because molecular weight similarity doesn't determine separation effectiveness—intermolecular forces and solubility differences are what matter. Choice D incorrectly assumes that being organic compounds with similar polarities prevents separation, ignoring the crucial hydrogen bonding capability difference. The key study tip: when evaluating separation methods, always consider hydrogen bonding first. Compounds that can hydrogen bond with water (alcohols, carboxylic acids, amines) will generally be water-extractable, while those that cannot (alkanes, ethers, esters) will prefer organic solvents. This principle makes extraction often more efficient than distillation for polar/nonpolar mixtures.
13. A student attempts to separate a mixture of acetone (boiling point 56°C) and water (boiling point 100°C) by simple distillation. The distillation is performed at standard atmospheric pressure. What is the most likely outcome of this separation attempt?
    Answer: Partial separation with acetone-rich distillate followed by water-rich residue When you encounter distillation problems, focus on the relationship between boiling points and the nature of the liquid mixture. Simple distillation works best when components have significantly different boiling points (typically >25°C difference) and don't interact strongly with each other. Acetone and water have boiling points differing by 44°C, which seems favorable for separation. However, these compounds can form hydrogen bonds with each other, creating a non-ideal mixture that doesn't behave like two completely independent liquids. During distillation, the vapor will be enriched in the more volatile component (acetone) but won't be pure acetone. As distillation proceeds, you'll collect an acetone-rich distillate initially, then progressively more water-rich fractions, leaving a water-rich residue. This makes answer B correct. Answer A is wrong because the hydrogen bonding between acetone and water prevents complete separation by simple distillation - you won't get pure components. Answer C incorrectly suggests azeotrope formation. While acetone and water do interact, they don't form a true azeotropic mixture that would prevent any separation at standard conditions. Answer D is incorrect because neither compound decomposes at their respective boiling points under normal atmospheric pressure - both are thermally stable at these temperatures. Remember that "partial separation" is often the realistic outcome in distillation problems involving polar compounds that can hydrogen bond. Complete separation typically requires fractional distillation or other advanced techniques when dealing with interacting molecules.
14. A mixture contains 25.0 g of sodium chloride, 15.0 g of sand (silicon dioxide), and 10.0 g of iron filings. Which sequence of separation techniques would most efficiently isolate each component in pure form?
    Answer: Magnetic separation, dissolution in water, filtration, evaporation When approaching mixture separation problems, you need to consider the unique physical properties of each component and design a sequence that efficiently exploits these differences without interfering with subsequent steps. The correct approach (B) starts with magnetic separation to remove the iron filings, since iron is ferromagnetic while sodium chloride and sand are not. This immediately isolates one pure component. Next, you dissolve the remaining mixture in water - sodium chloride is highly soluble while sand is insoluble. Filtration then separates the dissolved salt solution from the insoluble sand, giving you pure sand. Finally, evaporation removes the water from the filtrate, yielding pure sodium chloride crystals. Option A fails because it attempts magnetic separation after the other steps, but by then the iron filings would already be mixed with wet sand after filtration, making magnetic separation much less effective. Option C suggests starting with filtration, but there's nothing to filter initially since all components are dry solids. Option D begins with evaporation, which makes no sense since there's no liquid present in the original mixture to evaporate. The key insight is that magnetic separation works best on dry materials and should be done first when possible. Once you add water to dissolve the salt, subsequent magnetic separation becomes messy and inefficient. Study tip: Always start separation sequences by removing components with the most distinctive properties (like magnetism) first, then move to solubility-based separations. This prevents complications from wet materials interfering with physical separation methods.
15. A mixture contains naphthalene ($C_{10}H_8$, sublimes at 80°C) and sodium chloride (decomposes above 800°C). The mixture is heated to 100°C under reduced pressure. What separation technique is being employed, and what is the expected result?
    Answer: Sublimation; naphthalene vaporizes and resolidifies, leaving pure $NaCl$ behind When you encounter separation problems involving different physical properties, focus on the key temperatures and phase changes described. This question tests your understanding of sublimation as a purification technique. At 100°C under reduced pressure, naphthalene will sublime - it transitions directly from solid to gas without becoming liquid, since its sublimation point is 80°C. The vaporized naphthalene can then be collected on a cool surface where it resolidifies, leaving behind pure sodium chloride in the original container. This is a classic sublimation separation where one component has a much lower sublimation temperature than the decomposition temperature of the other. Answer A is incorrect because distillation involves liquid-to-vapor transitions, but naphthalene sublimates (solid-to-vapor) and $NaCl$ doesn't vaporize at 100°C - it would need over 800°C. Answer C is wrong because crystallization involves dissolving substances and allowing them to reform crystals, which isn't happening here with just heat application. Answer D incorrectly suggests naphthalene decomposes, but 100°C is far too low to break down this stable aromatic compound - it simply changes phase. Remember that sublimation works best when there's a large temperature gap between the sublimation point of one substance and the melting/decomposition point of another. Look for keywords like "reduced pressure" (which lowers sublimation temperatures) and check whether the heating temperature falls between the relevant transition points of the compounds involved.

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