Separation of Mixtures and Chemical Purity

87 soru

Soru 1Soru

A solid mixture consists of iron filings, ammonium chloride, and sodium chloride. Arrange the following laboratory steps in the correct chronological order to separate each component and isolate pure sodium chloride crystals.

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Cevap

The correct separation sequence is: 1. Remove iron filings using a magnet, 2. Sublimate ammonium chloride by heating, 3. Dissolve the residual salt in water and filter, 4. Evaporate the filtrate to collect sodium chloride crystals.
The correct order proceeds from dry physical removal to thermal transformation and finally solution-based separation. First, dry magnetization removes ferromagnetic iron filings without wetting the mixture. Second, thermal heating sublimates volatile ammonium chloride directly into gas, leaving non-volatile sodium chloride behind. Third, adding water dissolves sodium chloride, allowing filtration of insoluble residue. Finally, evaporating the filtrate recovers pure sodium chloride crystals.

Adım Adım Çözüm

1
Separate magnetic iron filings from the dry solid mixture using a magnet.
Iron filings are isolated cleanly without interference from liquids.
Magnetization requires dry solid conditions to efficiently attract ferromagnetic materials.
2
Heat the remaining mixture under an inverted funnel to separate ammonium chloride.
Ammonium chloride sublimates into gas and deposits as a pure solid on the cool funnel walls.
Ammonium chloride sublimes easily, whereas sodium chloride has a very high melting and boiling point and remains unaffected.
3
Dissolve the leftover solid residue in distilled water and pass it through filter paper.
Sodium chloride dissolves to form an aqueous filtrate while any insoluble impurities stay on the filter paper.
Water acts as a solvent for sodium chloride to separate it from insoluble solid impurities.
4
Heat the aqueous filtrate until the water completely evaporates.
Pure sodium chloride crystals remain in the evaporating dish.
Evaporation removes the liquid solvent leaving behind the dissolved solute.

Anahtar Kavram

Sequential separation of mixtures utilizing magnetization, sublimation, dissolution, filtration, and evaporation based on physical properties.
Soru 2Soru

A mixture containing four organic compounds of varying polarities—Compound P (strongly polar), Compound Q (moderately polar), Compound S (weakly polar), and Compound R (non-polar)—is separated using paper chromatography with a polar stationary phase (water bound to cellulose) and a non-polar mobile phase solvent. What is the correct sequence of these compounds when arranged in order of increasing RfR_f value (from smallest RfR_f to largest RfR_f)?

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Cevap

Compound P (strongly polar) → Compound Q (moderately polar) → Compound S (weakly polar) → Compound R (non-polar)
In paper chromatography using a polar stationary phase and non-polar mobile phase, retention factor (RfR_f) is inversely related to solute polarity. The strongly polar compound (Compound P) binds most tightly to the polar water molecules on the paper, retarding its movement and giving it the smallest RfR_f value. As compound polarity decreases (Q → S → R), solubility in the non-polar mobile phase increases, allowing the solute to travel further toward the solvent front, resulting in progressively larger RfR_f values.

Adım Adım Çözüm

1
Analyze the stationary and mobile phase characteristics.
Stationary phase is polar (water on paper), and mobile phase is non-polar solvent.
Chromatographic separation depends on partitioning between polar and non-polar phases.
2
Determine phase affinities for each compound based on polarity.
Strongly polar compounds adsorb strongly onto the polar stationary phase, while non-polar compounds dissolve preferentially in the non-polar mobile phase.
'Like dissolves like' governs partitioning between stationary and mobile phases.
3
Relate distance traveled to Retention Factor (RfR_f).
Rf=distance traveled by solutedistance traveled by solvent frontR_f = \frac{\text{distance traveled by solute}}{\text{distance traveled by solvent front}}. Solutes that travel further have larger RfR_f values.
Retention factor directly measures relative mobility.
4
Order compounds by increasing RfR_f value.
Compound P (lowest RfR_f) → Compound Q → Compound S → Compound R (highest RfR_f).
Increasing polarity corresponds to decreasing distance traveled in normal-phase paper chromatography.

Anahtar Kavram

Partitioning and Retention Factor (RfR_f) in Paper Chromatography
Soru 3Soru

A student needs to separate a solid mixture of iodine crystals and sodium chloride. Which separation technique is most appropriate to isolate pure iodine from the mixture?

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Cevap: Sublimation

Cevap

Sublimation is the correct separation technique.
Sublimation is the process where a solid changes directly into a gas without passing through a liquid state. Since iodine readily sublimates when heated while sodium chloride does not, heating the mixture drives off iodine gas, which can then be collected as pure solid crystals on a cool surface.

Adım Adım Çözüm

1
Identify the physical properties of the components in the mixture.
Iodine is a solid that readily sublimates (transitions directly from solid to gas) upon heating, while sodium chloride has a very high melting point and does not sublimate.
Choosing an effective separation method depends on exploiting a difference in physical properties between components.
2
Select the appropriate technique based on these properties.
Heating the mixture causes iodine to vaporize directly into gas, leaving sodium chloride behind.
Sublimation allows the selective phase transition of iodine.
3
Collect the purified component.
Condensing iodine vapor on a cold surface yields pure iodine crystals.
Deposition isolates pure solid iodine.

Anahtar Kavram

Separation of mixtures by sublimation
Tahmini Süre:45s
Soru 4Soru

The presence of a soluble non-volatile impurity in a pure liquid sample depresses its boiling point and causes it to boil over a narrow temperature range.

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Cevap: False

Cevap

False. Soluble non-volatile impurities elevate the boiling point of liquids and broaden the temperature range over which boiling occurs.
Soluble non-volatile impurities elevate the boiling point of liquids rather than depressing it, and cause the liquid to boil over a broad range of temperatures instead of a narrow, sharp temperature point.

Adım Adım Çözüm

1
Analyze how non-volatile impurities affect the vapor pressure of a liquid.
Solute molecules occupy surface area, reducing the rate of solvent evaporation and lowering the vapor pressure.
Lower vapor pressure means the liquid cannot boil at its normal boiling point.
2
Determine the effect on boiling point and boiling range.
The liquid must be heated to a higher temperature to reach atmospheric pressure, causing boiling point elevation and a non-sharp boiling range.
A sharp, fixed boiling point is a criterion of purity; impure liquids boil over a range of elevated temperatures.

Anahtar Kavram

Effect of Soluble Impurities on Boiling Point and Purity Criteria
Soru 5Soru

A student performed a paper chromatography experiment to analyze the purity of an organic dye sample. On the chromatogram, the solvent front traveled 8.0 cm8.0\text{ cm} from the baseline, and a single distinct spot traveled 5.0 cm5.0\text{ cm}. Which of the following gives the correct RfR_f value and the accurate deduction about the purity of the dye sample?

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Cevap: Rf=0.625R_f = 0.625, indicating that the sample is a pure substance.

Cevap

The retention factor is Rf=0.625R_f = 0.625, which indicates that the dye sample is a pure substance because only a single spot appeared on the chromatogram.
The retention factor (RfR_f) is determined by dividing the distance moved by the solute spot (5.0 cm5.0\text{ cm}) by the distance moved by the solvent front (8.0 cm8.0\text{ cm}), giving Rf=0.625R_f = 0.625. Furthermore, a single spot on the paper chromatogram serves as a standard criterion of chemical purity, confirming that the sample is a pure substance.

Adım Adım Çözüm

1
Calculate the retention factor (RfR_f) using the formula Rf=Distance moved by soluteDistance moved by solvent frontR_f = \frac{\text{Distance moved by solute}}{\text{Distance moved by solvent front}}.
Rf=5.0 cm8.0 cm=0.625R_f = \frac{5.0\text{ cm}}{8.0\text{ cm}} = 0.625
The RfR_f value represents the relative distance traveled by the compound compared to the solvent front.
2
Interpret the number of spots observed on the developed chromatogram.
A single spot confirms the presence of only one chemical component.
Pure substances produce a single spot under specified solvent conditions, whereas mixtures resolve into two or more distinct spots.

Anahtar Kavram

Criteria of purity in chromatography: pure substances yield a single spot with a characteristic RfR_f value between 0 and 1.
Tahmini Süre:1m 0s
Soru 6Soru

Which of the following statements correctly distinguishes a chemical compound from a mixture?

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Cevap: A compound consists of elements combined in a fixed ratio by mass, whereas a mixture contains components in variable proportions.

Cevap

A compound consists of elements combined in a fixed ratio by mass, whereas a mixture contains components in variable proportions.
The correct statement identifies that chemical compounds are pure substances formed by chemical reactions where constituent elements combine in fixed, definite proportions by mass. In contrast, mixtures are formed by physical blending where the constituent components can be present in any variable proportion.

Adım Adım Çözüm

1
Analyze the fundamental definition of a chemical compound regarding composition.
Compounds are formed by chemical combination of elements in definite, fixed mass ratios (Law of Definite Proportions).
Chemical bonds form between constituent atoms in fixed stoichiometric ratios.
2
Analyze the composition and properties of a mixture.
Mixtures consist of substances physically blended together in variable mass ratios, retaining their individual identities.
No chemical bonding occurs between distinct components of a mixture.
3
Evaluate the given options against these core criteria.
The statement highlighting fixed mass proportion for compounds versus variable composition for mixtures is the accurate distinction.
Physical separation techniques apply to mixtures, sharp melting points characterize pure compounds, and compound properties differ entirely from constituent elements.

Anahtar Kavram

Distinction Between Elements, Compounds, and Mixtures
Tahmini Süre:1m 0s
Soru 7Soru

What is the correct sequential order of laboratory steps required to obtain pure, dry copper(II) sulfate crystals from a mixture of solid copper(II) sulfate and insoluble sand?

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Cevap

The correct sequence begins with dissolving the soluble component in water, filtering to remove insoluble sand, heating the filtrate to obtain a saturated solution, cooling gradually to precipitate pure crystals, and finally filtering and drying the isolated crystals.
The purification of a mixture containing a soluble salt and an insoluble impurity requires dissolving the salt in water first, followed by filtration to remove the insoluble residue. To obtain pure crystals, the filtrate must be heated only until saturated and then allowed to cool slowly, as rapid or total evaporation to dryness destroys hydrated crystals. Finally, filtering and blotting the crystals isolates pure, dry hydrated salt.

Adım Adım Çözüm

1
Dissolve the mixture in water.
Copper(II) sulfate dissolves into solution while sand remains solid.
Exploits the differential solubility of the two components.
2
Filter the suspension.
Sand is retained on the filter paper as residue; clear copper(II) sulfate solution passes through as filtrate.
Filtration physically separates an insoluble solid from a liquid.
3
Evaporate partially to crystallizing point.
A hot saturated solution is formed.
Evaporating to dryness would decompose the hydrated salt into an anhydrous powder.
4
Cool the saturated solution.
Pure copper(II) sulfate crystals precipitate from the solution.
Solubility of most solid solutes decreases as temperature falls.
5
Isolate and dry the crystals.
Pure, dry crystals of hydrated copper(II) sulfate are obtained.
Filtration collects the solid crystals, and drying with filter paper removes residual moisture without removing water of crystallization.

Anahtar Kavram

Multi-step purification of soluble and insoluble solid mixtures via selective dissolution, filtration, evaporation to saturation, and crystallization.
Soru 8Soru

A miscible liquid mixture contains three organic compounds: Liquid X (boiling point 78.3C78.3^\circ\text{C}), Liquid Y (boiling point 97.1C97.1^\circ\text{C}), and Liquid Z (boiling point 117.7C117.7^\circ\text{C}). If this mixture is subjected to fractional distillation using an efficient fractionating column, which component will be collected first as the distillate?

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Cevap: Liquid X, because it has the lowest boiling point and the highest volatility among the components.

Cevap

Liquid X will be collected first as the distillate because it has the lowest boiling point (78.3C78.3^\circ\text{C}) and highest volatility.
In fractional distillation of miscible liquids, components separate based on differences in their boiling points. Liquid X has the lowest boiling point (78.3C78.3^\circ\text{C}) among the three components, meaning it is the most volatile. As heat is applied, Liquid X vaporizes most readily and its vapors ascend to the top of the fractionating column first, where they enter the condenser and are collected as the first distillate fraction.

Adım Adım Çözüm

1
Compare the boiling points of the three miscible liquids in the mixture.
Liquid X has a boiling point of 78.3C78.3^\circ\text{C}, Liquid Y has 97.1C97.1^\circ\text{C}, and Liquid Z has 117.7C117.7^\circ\text{C}.
The boiling point determines the relative volatility of each liquid component.
2
Relate boiling point to volatility and order of vaporisation during fractional distillation.
Liquid X, having the lowest boiling point, has the highest vapor pressure at any given temperature and evaporates most readily.
Lower boiling point corresponds to weaker intermolecular forces and higher volatility.
3
Determine which component emerges first from the top of the fractionating column.
Vapors of Liquid X reach the top of the fractionating column first, condense in the Liebig condenser, and are collected as the first fraction (distillate).
Repeated condensation-vaporization cycles in the column enrich the rising vapor with the most volatile component.

Anahtar Kavram

Fractional Distillation and Boiling Point Volatility Order
Tahmini Süre:1m 0s
Soru 9Soru

Four synthesized batches (P, Q, R, and S) of an organic solvent were evaluated for chemical purity against a reference standard. The pure standard possesses a sharp boiling point of 142.0C142.0^\circ\text{C}, a fixed density of 1.04 g cm31.04\text{ g cm}^{-3}, and yields a single spot on a thin-layer chromatogram with an RfR_f value of 0.450.45. The physical data recorded for the four synthesized batches are summarized in the table below:

BatchBoiling Range (C^\circ\text{C})Density (g cm3\text{g cm}^{-3})Chromatogram Spots (RfR_f)
P141.8 – 142.21.040.45
Q138.0 – 144.51.010.45, 0.62
R142.0 – 142.11.080.45
S145.0 – 145.21.040.30

Based on the criteria of chemical purity, which batch represents the pure solvent, and what is the correct explanation regarding Batch R?

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Cevap: Batch P is the pure solvent because it conforms to all physical criteria (narrow boiling range around 142.0C142.0^\circ\text{C}, matching density of 1.04 g cm31.04\text{ g cm}^{-3}, and a single RfR_f spot at 0.450.45), whereas Batch R is impure because its density deviates from the pure reference value.

Cevap

Batch P is the pure solvent because it satisfies all physical criteria of purity (narrow boiling range at 142.0C142.0^\circ\text{C}, density of 1.04 g cm31.04\text{ g cm}^{-3}, and a single chromatographic spot at Rf=0.45R_f = 0.45), whereas Batch R is impure due to its higher density of 1.08 g cm31.08\text{ g cm}^{-3}.
The correct option states that Batch P is the pure solvent because it meets all specified physical criteria of purity: a narrow boiling range centered at the reference temperature (141.8142.2C141.8–142.2^\circ\text{C} vs 142.0C142.0^\circ\text{C}), a matching density (1.04 g cm31.04\text{ g cm}^{-3}), and a single chromatographic spot at Rf=0.45R_f = 0.45. Batch R, despite its narrow boiling range, has an elevated density of 1.08 g cm31.08\text{ g cm}^{-3}, proving the presence of a dissolved impurity.

Adım Adım Çözüm

1
Analyze the chromatographic data for all four batches.
Batch P, R, and S each show a single spot, whereas Batch Q shows two spots (Rf=0.45R_f = 0.45 and 0.620.62).
A pure substance must yield a single spot on a chromatogram under a given solvent system. Two spots prove Batch Q is a mixture.
2
Evaluate the boiling point ranges against the reference boiling point (142.0C142.0^\circ\text{C}).
Batch P (141.8142.2C141.8 - 142.2^\circ\text{C}) and Batch R (142.0142.1C142.0 - 142.1^\circ\text{C}) boil sharply around 142.0C142.0^\circ\text{C}. Batch Q boils over a wide range (138.0144.5C138.0 - 144.5^\circ\text{C}), and Batch S boils at an elevated temperature (145.0145.2C145.0 - 145.2^\circ\text{C}).
Impure liquids boil over a wide temperature range or show boiling point elevation due to non-volatile impurities.
3
Compare the measured densities against the reference density (1.04 g cm31.04\text{ g cm}^{-3}).
Batch P and Batch S have densities of 1.04 g cm31.04\text{ g cm}^{-3}, whereas Batch R has a density of 1.08 g cm31.08\text{ g cm}^{-3}.
Chemical purity requires that all physical constants, including density and refractive index, match the reference values exactly.
4
Synthesize the findings across all criteria to identify the single pure batch.
Only Batch P meets all three physical criteria simultaneously: narrow boiling range around 142.0C142.0^\circ\text{C}, matching density of 1.04 g cm31.04\text{ g cm}^{-3}, and a single RfR_f spot at 0.450.45.
Batch R fails on density, Batch Q fails on chromatogram spot count and boiling range, and Batch S fails on boiling point and RfR_f value.

Anahtar Kavram

Criteria of Chemical Purity for Liquids
Tahmini Süre:3m 0s
Soru 10Soru

Match each distillation apparatus component or process listed on the left with its correct function or application on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Fractionating column
Liebig condenser
Simple distillation
Fractional distillation

Eşleşmeler

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Cevap

Fractionating column matches with providing a large surface area for continuous condensation/vaporisation; Liebig condenser matches with cooling hot vapor using circulating cold water; Simple distillation matches with separating liquids with widely different boiling points; Fractional distillation matches with separating miscible liquids with close boiling points.
Each item is correctly matched based on standard laboratory separation techniques: the fractionating column increases efficiency for close boiling point mixtures by providing surface area, the Liebig condenser turns vapor to liquid distillate, simple distillation separates substances with large boiling point differences, and fractional distillation resolves mixtures of miscible liquids with close boiling points.

Adım Adım Çözüm

1
Identify the primary role of structural apparatus components
The fractionating column provides surface area for multiple equilibrium stages, while the Liebig condenser serves purely to condense vapor into liquid distillate.
Apparatus functions are based on physical design principles of distillation setups.
2
Differentiate between simple and fractional distillation based on boiling point criteria
Simple distillation is suitable for large boiling point differences (>50C>50^\circ\text{C}), whereas fractional distillation is required for close boiling points (<25C<25^\circ\text{C}).
Close boiling points require repeated vaporisation-condensation steps achieved only via a fractionating column.

Anahtar Kavram

Simple versus Fractional Distillation and Apparatus Functions
Tahmini Süre:1m 30s
Soru 11Soru

An aqueous solution containing dissolved iodine (I2I_2) and sodium chloride (NaClNaCl) is shaken with tetrachloromethane (CCl4CCl_4, density =1.59 g/cm3= 1.59\text{ g/cm}^3) in a separating funnel and allowed to stand until two distinct liquid layers form. Which of the following correctly identifies the position and contents of the lower layer?

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Cevap: The lower layer is tetrachloromethane containing mostly dissolved iodine.

Cevap

The lower layer is tetrachloromethane containing mostly dissolved iodine.
Tetrachloromethane (CCl4CCl_4) is immiscible with water and has a higher density (1.59 g/cm31.59\text{ g/cm}^3) than water (1.00 g/cm31.00\text{ g/cm}^3), causing it to settle at the bottom of the separating funnel. According to the principle of solute partitioning, non-polar iodine dissolves preferentially in the non-polar organic solvent (CCl4CCl_4), making the lower layer tetrachloromethane enriched with iodine.

Adım Adım Çözüm

1
Determine layer placement based on density comparison.
Water has a density of 1.00 g/cm31.00\text{ g/cm}^3, whereas tetrachloromethane (CCl4CCl_4) has a density of 1.59 g/cm31.59\text{ g/cm}^3. Because CCl4CCl_4 is denser than water and immiscible with it, CCl4CCl_4 forms the bottom (lower) layer.
In a separating funnel, the denser immiscible liquid settles at the bottom under gravity.
2
Analyze solute partitioning based on solvent polarity ('like dissolves like').
Iodine (I2I_2) is non-polar and preferentially dissolves in the non-polar solvent CCl4CCl_4 (organic layer). Sodium chloride (NaClNaCl) is ionic and remains in the polar solvent H2OH_2O (aqueous layer).
Solvent extraction relies on differential solubility of solutes between two immiscible liquid phases.
3
Combine layer position and solute content to identify the correct description.
The lower layer consists of tetrachloromethane containing the extracted iodine.
The lower organic phase contains the non-polar solute.

Anahtar Kavram

Solvent Extraction and Liquid Density Layering
Soru 12Soru

A mixture of kerosene and water is poured into a separating funnel and allowed to stand undisturbed. Which of the following statements correctly explains how the mixture separates into two distinct layers?

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Cevap: Kerosene forms the upper layer because it is less dense than water.

Cevap

Kerosene forms the upper layer because it is less dense than water, allowing the denser water to be drawn off from the bottom of the funnel.
A separating funnel is designed to separate immiscible liquids based on density differences. Kerosene is non-polar and immiscible with water, and because its density is lower than that of water, it collects in the upper layer.

Adım Adım Çözüm

1
Identify the physical properties of the liquids in the mixture.
Kerosene and water are immiscible (they do not mix). Water has a density of approximately 1.0 g/cm31.0\text{ g/cm}^3, while kerosene has a density of about 0.8 g/cm30.8\text{ g/cm}^3.
Separating funnels function based on differences in liquid solubility (immiscibility) and relative density.
2
Determine the relative layer positions inside the separating funnel.
The denser liquid (water) settles at the bottom, while the less dense liquid (kerosene) floats on top.
Gravity causes the denser fluid to occupy the lower volume of the vessel.

Anahtar Kavram

Separation of immiscible liquids using a separating funnel based on density differences
Soru 13Soru

A clinical laboratory technician needs to rapidly separate dense cellular components, such as red blood cells, from liquid plasma in a blood sample. Which technique utilizes high-speed rotation to achieve this separation based on density differences?

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Cevap: Centrifugation

Cevap

Centrifugation is the correct technique for separating suspended cellular components from blood plasma based on density differences.
Centrifugation relies on spinning a fluid mixture at high speeds. The resulting outward rotational force forces denser solid components (such as red blood cells) to settle to the bottom of the tube faster than natural gravity would allow, separating them from the lighter liquid supernatant (plasma).

Adım Adım Çözüm

1
Identify the physical state and properties of the components in whole blood.
Blood is a liquid suspension consisting of denser solid/cellular components (red and white blood cells, platelets) suspended in liquid plasma.
Choosing an effective separation technique requires analyzing differences in physical properties such as particle size, state of matter, magnetic property, or density.
2
Evaluate the mechanism of centrifugation.
Centrifugation spins the mixture at extremely high speeds in a centrifuge tube.
High-speed rotation creates outward centrifugal force, accelerating the sedimentation of denser suspended particles to the bottom (forming a pellet) while the lighter liquid remains as a clear supernatant layer.
3
Compare centrifugation against alternative separation techniques.
Sublimation requires solid-to-gas phase transitions, magnetization requires magnetic components, and distillation requires thermal boiling of liquid mixtures. None of these are suitable for delicate cell suspensions.
Centrifugation is uniquely suited for separating colloidal or fine suspensions without thermal degradation.

Anahtar Kavram

Centrifugation uses centrifugal force produced by high-speed rotation to separate suspended solids from liquids or immiscible liquids of different densities.
Tahmini Süre:1m 0s
Soru 14Soru

The presence of a soluble non-volatile impurity in a pure solid sample depresses its melting point and broadens its melting temperature range.

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Cevap: True

Cevap

The statement is TRUE. Soluble non-volatile impurities depress (lower) the melting point of a pure solid and cause it to melt over a broader temperature range.
A pure solid exhibits a sharp and distinct melting point. The presence of a soluble non-volatile impurity destabilizes the crystal lattice, causing the substance to melt at a lower temperature and over a broader range.

Adım Adım Çözüm

1
Identify the melting characteristics of a pure solid.
A pure crystalline solid melts at a sharp, characteristic temperature.
Uniform intermolecular forces throughout the pure crystal lattice break simultaneously at a specific thermal energy threshold.
2
Evaluate the structural effect of adding a soluble impurity.
Impurity molecules disrupt the regular arrangement of the crystal lattice, weakening lattice stability.
Lattice disruption lowers the overall thermal energy required for the phase change from solid to liquid.
3
Determine the impact on melting point value and temperature range.
The impure solid melts below its literature melting point and across a wider range of temperatures.
Melting point depression and a broad melting range serve as definitive physical indicators of impurity.

Anahtar Kavram

Melting Point Depression and Melting Range as Criteria of Purity
Soru 15Soru

A chemist is given a dry solid mixture containing nickel powder, ammonium chloride (NH4ClNH_4Cl), fine barium sulfate (BaSO4BaSO_4) powder, and sodium chloride (NaClNaCl). What is the correct chronological sequence of laboratory procedures required to isolate each pure component from the mixture?

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Cevap

The correct sequence of separation steps is: 1. Pass a magnet over the dry solid mixture to isolate nickel powder. 2. Heat the dry residue to sublime ammonium chloride. 3. Add distilled water to dissolve sodium chloride. 4. Centrifuge the suspension to spin down fine barium sulfate powder. 5. Decant the clear supernatant liquid and evaporate to dryness to recover sodium chloride crystals.
The correct sequence begins with magnetization on the dry solid mixture to remove ferromagnetic nickel. Next, sublimation is performed on the dry residue to separate ammonium chloride, which must occur before adding water because both ammonium chloride and sodium chloride are water-soluble. Distilled water is then added to selectively dissolve sodium chloride while leaving barium sulfate in suspension. Centrifugation rapidly spins down the fine barium sulfate precipitate. Finally, decanting and evaporating the clear supernatant liquid yields pure sodium chloride crystals.

Adım Adım Çözüm

1
Extract ferromagnetic nickel powder using a bar magnet.
Dry nickel powder is separated cleanly from the solid mixture.
Nickel is ferromagnetic and is attracted by a magnetic field without requiring chemical reagents or solvent addition.
2
Apply controlled heat to the dry residue under an inverted funnel to sublime ammonium chloride (NH4ClNH_4Cl).
Ammonium chloride sublimates directly into gas upon heating and deposits as pure solid crystals on the cool funnel surface.
Sublimation must precede water dissolution because both NH4ClNH_4Cl and NaClNaCl are soluble in water; dissolving first would make thermal separation of the two salts impossible.
3
Add distilled water to the remaining mixture of NaClNaCl and BaSO4BaSO_4 and stir.
Sodium chloride dissolves into aqueous solution (NaCl(aq)NaCl_{(aq)}) while insoluble BaSO4BaSO_4 forms a fine suspension.
Water acts as a selective solvent based on solubility differences between soluble NaClNaCl and insoluble BaSO4BaSO_4.
4
Spin the suspension in a laboratory centrifuge.
Dense, fine particles of BaSO4BaSO_4 compact tightly at the bottom of the tube, separating clearly from the aqueous liquid.
Centrifugation rapidly forces fine suspended solids out of liquid phase much faster and more completely than gravity filtration.
5
Decant the supernatant solution and evaporate the solvent.
Water evaporates off, yielding pure crystalline sodium chloride.
Evaporation separates a non-volatile soluble solid solute from its volatile liquid solvent.

Anahtar Kavram

Sequential Separation of Multi-Component Mixtures
Soru 16Soru

A mixture of hexane (density =0.66 g/cm3= 0.66\text{ g/cm}^3) and an aqueous sodium chloride solution (density =1.05 g/cm3= 1.05\text{ g/cm}^3) is poured into a separating funnel. After shaking and allowing the mixture to settle into two distinct liquid layers, which of the following statements correctly identifies the bottom layer and the primary principle enabling this separation?

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Cevap: The aqueous sodium chloride solution forms the bottom layer because it is denser and immiscible with hexane.

Cevap

The aqueous sodium chloride solution forms the bottom layer because it is denser and immiscible with hexane.
A separating funnel is used to separate two immiscible liquids based on their density differences. Hexane (a non-polar organic liquid) and an aqueous sodium chloride solution (a polar liquid) do not mix. Since the aqueous solution has a greater density (1.05 g/cm31.05\text{ g/cm}^3) than hexane (0.66 g/cm30.66\text{ g/cm}^3), it sinks to form the lower layer, which can then be drained through the tap.

Adım Adım Çözüm

1
Identify the physical properties required for separating funnel extraction.
Separating funnels require two immiscible liquids that separate into distinct phases.
Hexane is a non-polar organic solvent, while water/aqueous solution is polar, making them immiscible.
2
Compare the densities of the two immiscible liquids.
Aqueous NaCl solution (1.05 g/cm31.05\text{ g/cm}^3) is denser than hexane (0.66 g/cm30.66\text{ g/cm}^3).
Gravity causes the liquid with the higher density to sink to the bottom of the funnel.
3
Determine the bottom layer.
The denser aqueous NaCl solution settles at the bottom, while the less dense hexane forms the top layer.
Layer order in a separating funnel is strictly governed by density difference and immiscibility.

Anahtar Kavram

Separating Funnel Principles: Immiscibility and Density
Tahmini Süre:1m 0s
Soru 17Soru

Match each industrial process on the left with its corresponding primary separation technique on the right.

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Öğeler

Production of oxygen gas from atmospheric air
Refining of crude oil into petrol, kerosene, and diesel fractions
Removal of suspended particulate impurities during municipal water purification
Recovery of sodium hydrogentrioxocarbonate(IV) in the Solvay process

Eşleşmeler

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Cevap

Production of oxygen gas matches liquefaction followed by fractional distillation; Refining of crude oil matches fractional distillation based on hydrocarbon boiling point differences; Removal of suspended particulate impurities matches coagulation with alum followed by sand filtration; Recovery of sodium hydrogentrioxocarbonate(IV) matches filtration of precipitated solid crystals.
Industrial separation methods rely on specific physical property differences: atmospheric gases and crude oil fractions are separated by fractional distillation due to boiling point differences; suspended particles in water are settled via coagulation and filtered; and solid precipitate crystals in the Solvay process are separated from liquid by filtration.

Adım Adım Çözüm

1
Analyze the industrial separation of atmospheric air into components.
Air consists of gases (N2N_2, ArAr, O2O_2) with very low boiling points. Liquefying the mixture under high pressure and low temperature followed by fractional distillation enables their separation.
Fractional distillation separates miscible substances with distinct boiling points.
2
Analyze the refining process of petroleum (crude oil).
Crude oil is a complex liquid mixture of hydrocarbons separated into fractions (gasoline, kerosene, diesel) inside a fractionating column based on boiling point ranges.
Smaller hydrocarbon molecules vaporize and rise higher up the column than larger molecules with higher boiling points.
3
Examine the municipal water clarification process.
Alum (potassium aluminium sulfate) causes fine suspended mud particles to coagulate into heavy flocs, which are removed as the water passes through sand beds.
Coagulation aggregates fine particles so that mechanical filtration can physically trap them.
4
Determine how solid sodium hydrogentrioxocarbonate(IV) is isolated in the Solvay process.
NaHCO3NaHCO_3 has lower solubility than ammonium chloride in cold water, precipitating out as solid crystals that are separated from the solution by filtration.
Filtration is used to separate insoluble or precipitated solid products from liquid reaction mixtures.

Anahtar Kavram

Industrial Applications of Separation Methods
Soru 18Soru

Match each mixture or suspension with the most appropriate physical separation technique used to separate its components.

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Öğeler

A dry solid mixture of ammonium chloride and sodium chloride
A mixture of magnetic iron filings and non-magnetic sulfur powder
A suspension of red blood cells in liquid blood plasma

Eşleşmeler

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Cevap

Ammonium chloride and sodium chloride mixture matches Sublimation; Iron filings and sulfur powder mixture matches Magnetization; Blood cell suspension in plasma matches Centrifugation.
Each physical mixture is matched to its correct separation technique based on specific properties: Sublimation separates ammonium chloride because it vaporizes directly from solid state; Magnetization extracts iron filings due to magnetic attraction; Centrifugation separates blood cells from plasma by density differences under centrifugal force.

Adım Adım Çözüm

1
Analyze the solid mixture of ammonium chloride and sodium chloride.
Ammonium chloride is a sublimating salt while sodium chloride is not.
Sublimation is used to separate a mixture containing a component that changes directly from solid to gas on heating.
2
Analyze the mixture of iron filings and sulfur powder.
Iron is attracted to a magnet while sulfur is non-magnetic.
Magnetization relies on differences in magnetic properties to separate magnetic substances from non-magnetic ones.
3
Analyze the blood cell suspension in plasma.
Cells and liquid plasma have different densities.
Centrifugation accelerates the sedimentation of suspended particles in liquids using rapid rotation based on density differences.

Anahtar Kavram

Selection of physical separation methods based on sublimation, magnetic property, and density differences
Soru 19Soru

An organic solute XX dissolved in an aqueous solution is to be extracted using ethoxyethane (diethyl ether, density=0.71 g/cm3\text{density} = 0.71\text{ g/cm}^3, boiling point=34.6C\text{boiling point} = 34.6^\circ\text{C}). Which of the following statements correctly identifies the essential properties of ethoxyethane for this process and the location of the organic layer in the separating funnel?

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Cevap: Ethoxyethane must be immiscible with water, have a higher solubility for solute XX than water, and form the upper layer because its density is less than that of water.

Cevap

Ethoxyethane must be immiscible with water, have a higher solubility for solute XX than water, and form the upper layer because its density is less than that of water.
Solvent extraction using a separating funnel relies on three key principles: the organic solvent must be immiscible with water, the solute must be preferentially more soluble in the organic solvent than in water, and the relative density determines layer placement. Since ethoxyethane has a density (0.71 g/cm30.71\text{ g/cm}^3) lower than that of water (1.00 g/cm31.00\text{ g/cm}^3), it forms the less dense upper layer.

Adım Adım Çözüm

1
Identify the requirement for liquid-liquid separation in a separating funnel.
The extraction solvent must be immiscible with the original solvent (water) so that two distinct liquid phases form.
If the two liquids are miscible, they form a single homogeneous solution and cannot be physically separated using a tap funnel.
2
Determine the solubility criterion (distribution ratio) for effective extraction.
The target solute XX must be significantly more soluble in the extracting organic solvent than in water.
Solvent extraction relies on preferential partitioning (distribution) of the solute into the organic layer upon shaking.
3
Evaluate the layer positioning based on physical properties.
Because ethoxyethane has a lower density (0.71 g/cm30.71\text{ g/cm}^3) than water (1.00 g/cm31.00\text{ g/cm}^3), it floats on top to form the upper layer.
Layering in a separating funnel is determined exclusively by relative density, not by boiling point or vapor pressure.

Anahtar Kavram

Principles of Solvent Extraction and Density Layering in a Separating Funnel
Soru 20Soru

A student is provided with a crude sample of rock salt containing sodium chloride (NaClNaCl) contaminated with insoluble calcium carbonate (CaCO3CaCO_3) and sand (SiO2SiO_2). Arrange the following laboratory procedures in the correct sequence to obtain pure, dry crystals of sodium chloride from this mixture.

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Cevap

The correct sequence of procedures is: First, grind the sample and dissolve it in distilled water. Next, filter the mixture to separate the insoluble solids from the NaClNaCl filtrate. Then, heat the filtrate to its crystallization point to obtain a saturated solution. After that, cool the solution slowly to allow crystal formation. Finally, filter and dry the crystals between filter papers.
The correct order follows the standard laboratory procedure for recovering a pure salt from a mixture containing insoluble solids: dissolution \rightarrow filtration \rightarrow evaporation to crystallization point \rightarrow cooling \rightarrow final filtration and drying.

Adım Adım Çözüm

1
Dissolution of soluble component
Sodium chloride dissolves in distilled water while sand and calcium carbonate remain undissolved solids.
Solubility differences allow selective dissolution of NaClNaCl from insoluble impurities.
2
Gravity filtration
Insoluble sand and CaCO3CaCO_3 are retained on the filter paper as residue; aqueous NaClNaCl passes through as filtrate.
Filtration separates insoluble solids from aqueous solutions.
3
Controlled evaporation to saturation point
Water evaporates until the NaClNaCl solution reaches saturation (crystallization point).
Heating to dryness should be avoided when pure crystals are desired, as rapid evaporation can trap impurities or spur spitting.
4
Controlled cooling and crystallization
Well-defined crystals of sodium chloride precipitate out of the cooling solution.
Solubility decreases with lowering temperature, allowing solute molecules to arrange into crystal lattices.
5
Isolation and drying of crystals
Pure, dry crystals of NaClNaCl are collected.
Filtering removes the remaining mother liquor containing residual soluble contaminants, and filter paper absorbs surface moisture.

Anahtar Kavram

Multi-step purification of a soluble salt from insoluble impurities using dissolution, filtration, evaporation to saturation, and crystallization.
Tahmini Süre:2m 0s
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Separation of Mixtures and Chemical Purity Alıştırma Soruları — JAMB UTME | Examkin