All practice questions

Question 261Question →

During the fractional distillation of crude oil (petroleum), different fractions are separated based on their boiling points. Arrange the following petroleum fractions in order of their distillation from first to last (lowest boiling point to highest boiling point).

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Answer

The correct order of petroleum fractions from first to last distilled is: Petroleum gas, Petrol / Gasoline, Kerosene / Paraffin, and Diesel oil / Gas oil.

Fractional distillation separates liquids based on differences in their boiling points. The fraction with the lowest boiling point (petroleum gas) is the most volatile and vaporizes first, exiting at the top of the fractionating column. As the temperature of the mixture increases, fractions with progressively higher boiling points (petrol, then kerosene, then diesel oil) vaporize and distil off sequentially.

Step-by-Step Solution

1

Identify the separation principle of fractional distillation.

Components with lower boiling points vaporize first and distil off before components with higher boiling points.

Lower boiling point liquids are more volatile and require less thermal energy to enter the vapor phase.

2

Compare the boiling points of the given petroleum fractions.

Petroleum gas (<

20^\circ\text{C}

) < Petrol (

40^\circ\text{C}-170^\circ\text{C}

) < Kerosene (

170^\circ\text{C}-250^\circ\text{C}

) < Diesel oil (

250^\circ\text{C}-350^\circ\text{C}

).

Hydrocarbon chain length increases from gas to diesel, increasing intermolecular forces and boiling points.

3

Arrange the items from lowest boiling point to highest boiling point.

Order: Petroleum gas → Petrol / Gasoline → Kerosene / Paraffin → Diesel oil / Gas oil.

This matches the sequence of collection during fractional distillation.

Key Concept

Fractional Distillation of Petroleum

Question 262Question →

Arrange the following sequential electrochemical and physical steps occurring during the reduction of alumina in the Hall-Héroult cell to extract molten aluminium metal, from initial electrolyte preparation to final anode gas emission.

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Answer

The correct sequence of steps in the Hall-Héroult cell is: (1) Dissolving alumina in molten cryolite, (2) Dissociation of alumina into

Al^{3+}

and

O^{2-}

ions, (3) Migration of

Al^{3+}

ions to the cathode, (4) Reduction of

Al^{3+}

to form molten aluminium, and (5) Oxidation of

O^{2-}

at the graphite anodes yielding carbon dioxide gas.

The Hall-Héroult process operates sequentially by first dissolving alumina (

Al_2O_3

) in molten cryolite (

Na_3AlF_6

) at around

950^\circ\text{C}

to create a conducting medium. Upon melting, alumina dissociates into mobile

Al^{3+}

and

O^{2-}

ions. Applying an electric current causes

Al^{3+}

cations to migrate to the carbon cathode at the bottom, where they are reduced to liquid aluminium metal (

Al^{3+} + 3e^- \rightarrow Al_{(l)}

). Concurrently,

O^{2-}

anions migrate to the top graphite anodes and undergo oxidation to oxygen gas (

2O^{2-} \rightarrow O_{2(g)} + 4e^-

), which reacts with the hot carbon anodes to form carbon dioxide gas (

C + O_2 \rightarrow CO_2

).

Step-by-Step Solution

1

Identify the electrolyte preparation step in the Hall-Héroult cell.

Alumina is dissolved in molten cryolite at about

950^\circ\text{C}

.

Cryolite acts as a solvent and flux to lower the high melting point of pure alumina and enhance conductivity.

2

Determine the ionization behavior of the dissolved alumina.

Alumina dissociates into mobile

Al^{3+}

cations and

O^{2-}

anions.

Liquid state ionic dissociation is necessary for current transport through the electrolyte.

3

Trace the movement of cations under the applied electric field.

Al^{3+}

cations migrate to the negatively charged carbon cathode lining at the cell floor.

Electrostatic attraction draws positive ions toward the negative electrode.

4

Determine the chemical reaction occurring at the cathode.

Al^{3+}

ions gain electrons to form molten aluminium metal (

Al^{3+} + 3e^- \rightarrow Al_{(l)}

).

Cation gain of electrons at the cathode represents the reduction process that isolates elemental aluminium.

5

Determine the chemical reaction occurring at the anode and the fate of the anode material.

O^{2-}

ions lose electrons to produce oxygen gas, which reacts with graphite anodes to produce

CO_2

gas.

Anode oxidation releases oxygen gas at high temperature, causing carbon anodes to burn away continuously.

Key Concept

Electrolytic reduction of alumina in the Hall-Héroult process

Estimated Time:1m 30s

Question 263Question →

Arrange the following crude oil (petroleum) fractions in order of increasing boiling point range, starting from the fraction with the lowest boiling point to the one with the highest boiling point.

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Answer

The correct order from lowest to highest boiling point range is Refinery gas, followed by Petrol (Gasoline), Kerosene (Paraffin), and Diesel oil (Gas oil).

In the fractional distillation of petroleum, fractions condense and exit the fractionating column at different levels depending on their boiling point ranges. Smaller alkane molecules have lower molecular masses, weaker intermolecular forces, and lower boiling points. Thus, Refinery gas (

C_1-C_4

) has the lowest boiling point, followed by Petrol (

C_5-C_{10}

), Kerosene (

C_{10}-C_{16}

), and Diesel oil (

C_{14}-C_{20}

) with the highest boiling point among the listed options.

Step-by-Step Solution

1

Determine the relationship between carbon chain length and boiling point in alkane fractions.

Smaller hydrocarbon molecules have weaker intermolecular van der Waals forces and therefore lower boiling points.

Boiling point increases as the number of carbon atoms per molecule increases.

2

Identify the carbon chain lengths for each given fraction.

Refinery gas (

C_1-C_4

), Petrol (

C_5-C_{10}

), Kerosene (

C_{10}-C_{16}

), and Diesel oil (

C_{14}-C_{20}

).

Fractional distillation separates crude oil based on boiling point ranges governed by molecular sizes.

3

Arrange the fractions from smallest carbon number to largest carbon number.

Refinery gas

\rightarrow

Petrol

\rightarrow

Kerosene

\rightarrow

Diesel oil.

This sequence directly corresponds to increasing boiling point range.

Key Concept

Fractional Distillation and Boiling Point Trends of Petroleum Fractions

Estimated Time:45s

Question 264Question →

A laboratory technician needs to separate a dry solid mixture containing iron filings, ammonium chloride ( $NH_4Cl$ ), and fine sand. In what correct sequential order should the steps below be carried out from first to last?

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Answer

The correct sequence of steps is: first, pass a strong magnet to remove the iron filings; second, heat the remaining mixture to sublime ammonium chloride (

NH_4Cl

); third, collect the non-volatile sand residue.

The process begins with magnetization to extract ferromagnetic iron filings. Next, heating the remaining mixture causes ammonium chloride (

NH_4Cl

) to sublime directly into vapor, leaving fine sand behind as the non-volatile residue.

Step-by-Step Solution

1

Perform magnetization on the dry mixture.

Iron filings adhere to the magnet and are completely removed from the container.

Iron is ferromagnetic and can be physically isolated without affecting the other non-magnetic solids.

2

Apply heat to the remaining mixture using an inverted funnel collector.

Ammonium chloride (

NH_4Cl

) sublimes into gas and re-crystallizes as solid deposit on the cool funnel surface.

Ammonium chloride readily sublimes upon heating, separating it cleanly from sand.

3

Isolate the remaining solid in the dish.

Pure sand remains in the evaporating dish.

Sand does not sublime under normal heating conditions and is non-magnetic, making it the final isolated component.

Key Concept

Sequential separation of solid mixtures by exploiting differences in magnetic property (magnetization) and thermal volatility (sublimation).

Question 265Question →

In the biological treatment of municipal and industrial wastewater via the activated sludge process, several sequential operations are performed to purify water and manage waste. Arrange the following steps of the process in the correct chronological order from first to last.

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Answer

The correct sequence is: (1) Screening and primary sedimentation to remove coarse solids and heavy inorganic grit, (2) Aeration in tanks containing aerobic bacteria to digest dissolved organic pollutants, (3) Secondary clarification in settling basins to separate the activated biological sludge flocs from the purified effluent, and (4) Disinfection of the clarified water using chlorine or ultraviolet light prior to discharge.

The logical sequence follows the engineering progression from coarse physical removal of solids, through microbial digestion of dissolved organics during aeration, solid-liquid separation of biomass in secondary clarifiers, and final pathogenic disinfection.

Step-by-Step Solution

1

Identify the preliminary physical separation stage.

Primary screening and sedimentation occurs first.

Large debris and heavy suspended solids must be removed physically before exposing effluent to microbial action.

2

Identify the biological oxidation phase.

Aeration with aerobic microorganisms follows primary sedimentation.

Microbial breakdown requires dissolved oxygen supplied in aeration tanks to convert soluble organic waste into biomass and gas.

3

Identify the biomass separation stage.

Secondary clarification follows aeration.

The biological flocs created during aeration need quiescent conditions to settle out as activated sludge, leaving clear liquid above.

4

Identify the final pathogen neutralization stage.

Disinfection is the final operational step.

Harmful bacterial or viral pathogens are inactivated right before environmental discharge to protect aquatic ecosystems and public health.

Key Concept

Sequential physical, biological, and chemical stages of industrial biotechnology in activated sludge wastewater treatment.

Question 266Question →

Arrange the following chemical and electrochemical stages of the rusting of iron in chronological sequence, from initial anodic oxidation to the final formation of rust.

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Answer

The correct chronological sequence is: Oxidation of iron metal to

\text{Fe}^{2+}

ions

\rightarrow

Reduction of dissolved oxygen to

\text{OH}^-

ions

\rightarrow

Precipitation of

\text{Fe(OH)}_2

\rightarrow

Oxidation of

\text{Fe(OH)}_2

to hydrated

\text{Fe}_2\text{O}_3\cdot x\text{H}_2\text{O}

.

Rusting is an electrochemical process. Iron metal initially oxidizes at anodic sites to produce

\text{Fe}^{2+}

ions and electrons. The released electrons migrate through the iron to cathodic regions, where dissolved atmospheric oxygen is reduced to

\text{OH}^-

ions. These ions react in the electrolyte solution to form insoluble green

\text{Fe(OH)}_2

, which is subsequently oxidized by dissolved oxygen to form reddish-brown hydrated iron(III) oxide (rust,

\text{Fe}_2\text{O}_3\cdot x\text{H}_2\text{O}

).

Step-by-Step Solution

1

Identify the initial anodic oxidation process

\text{Fe(s)} \rightarrow \text{Fe}^{2+}\text{(aq)} + 2\text{e}^-

Rusting begins as an electrochemical reaction where iron metal acts as the anode and undergoes oxidation.

2

Identify the cathodic reduction process

\text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} + 4\text{e}^- \rightarrow 4\text{OH}^-\text{(aq)}

Electrons released during anodic oxidation flow to cathodic sites where atmospheric oxygen dissolved in water is reduced.

3

Determine the ionic precipitation reaction

\text{Fe}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)} \rightarrow \text{Fe(OH)}_2\text{(s)}

The formed cations and anions diffuse towards each other in the aqueous electrolyte layer, forming an insoluble precipitate.

4

Determine the final oxidation step to rust

4\text{Fe(OH)}_2\text{(s)} + \text{O}_2\text{(g)} + 2x\text{H}_2\text{O(l)} \rightarrow 2(\text{Fe}_2\text{O}_3\cdot x\text{H}_2\text{O})\text{(s)}

Dissolved oxygen further oxidizes iron(II) hydroxide to hydrated iron(III) oxide, commonly known as rust.

Key Concept

Electrochemical mechanism of iron rusting

Question 267Question →

Arrange the following sequential stages involved in the industrial isolation of pure nitrogen gas from atmospheric air via fractional distillation in the correct chronological order, from ambient air intake to nitrogen gas collection.

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Answer

The correct chronological order of the process is: 1. Removal of dust, moisture, and carbon(IV) oxide; 2. Compression and Joule-Thomson expansion to produce liquid air; 3. Feeding liquid air into a fractionating column and gradual warming; 4. Selective vaporization and collection of nitrogen gas at -196 °C.

The industrial preparation of nitrogen gas relies on fractional distillation of liquid air. Ambient air must first be purified of water vapor and carbon(IV) oxide to prevent cryogenic equipment blockages caused by solid ice formation. The clean air is compressed under high pressure (around 200 atmospheres) and subjected to rapid Joule-Thomson expansion, which cools it repeatedly until it condenses into liquid air at approximately

-200\text{ }^\circ\text{C}

. When liquid air is fed into a fractionating column and warmed gradually, nitrogen gas boils off first at

-196\text{ }^\circ\text{C}

(

77\text{ K}

) due to having a lower boiling point than argon (

-186\text{ }^\circ\text{C}

) and oxygen (

-183\text{ }^\circ\text{C}

).

Step-by-Step Solution

1

Identify the initial purification stage required before gas liquefaction.

Air is scrubbed to remove dust, water vapor, and

CO_2

.

Freezing points of water (

0\text{ }^\circ\text{C}

) and carbon(IV) oxide (

-78.5\text{ }^\circ\text{C}

) are much higher than liquefaction temperature, meaning they would solidify and clog cryogenic tubes if not removed first.

2

Determine the phase change process of purified gaseous air into liquid air.

Purified air is compressed to

\sim 200\text{ atm}

and allowed to expand through a fine nozzle.

The Joule-Thomson expansion causes progressive cooling until air liquefies around

-200\text{ }^\circ\text{C}

.

3

Analyze the fractionating column feed and thermal gradient.

Liquid air enters the fractionating column and is warmed slowly.

Gradual heating drives components with lower boiling points to vaporize first.

4

Compare boiling points to establish which component distills first.

Nitrogen boils off at

-196\text{ }^\circ\text{C}

, followed by argon (

-186\text{ }^\circ\text{C}

) and oxygen (

-183\text{ }^\circ\text{C}

).

Nitrogen has the lowest boiling point, so it boils off first as a gas at the top of the column.

Key Concept

Fractional Distillation of Liquid Air for Industrial Production of Nitrogen

Question 268Question →

Given the standard reduction potentials ( $E^\circ$ ) below, arrange the metals in order of increasing reducing strength (from weakest reducing agent to strongest reducing agent):

\begin{aligned} \text{Ag}^+_{(\text{aq})} + \text{e}^- &\rightarrow \text{Ag}_{(\text{s})} \quad E^\circ = +0.80\text{ V} \\ \text{Cu}^{2+}_{(\text{aq})} + 2\text{e}^- &\rightarrow \text{Cu}_{(\text{s})} \quad E^\circ = +0.34\text{ V} \\ \text{Fe}^{2+}_{(\text{aq})} + 2\text{e}^- &\rightarrow \text{Fe}_{(\text{s})} \quad E^\circ = -0.44\text{ V} \\ \text{Zn}^{2+}_{(\text{aq})} + 2\text{e}^- &\rightarrow \text{Zn}_{(\text{s})} \quad E^\circ = -0.76\text{ V} \end{aligned}

Which sequence represents the correct order of increasing reducing strength?

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Answer

The correct order of increasing reducing strength is Silver (

\text{Ag}

), Copper (

\text{Cu}

), Iron (

\text{Fe}

), and Zinc (

\text{Zn}

).

Reducing power increases as standard reduction potential (

E^\circ

) becomes more negative, because metals with negative reduction potentials readily lose electrons. Silver has the most positive reduction potential (

+0.80\text{ V}

), followed by copper (

+0.34\text{ V}

), iron (

-0.44\text{ V}

), and zinc (

-0.76\text{ V}

). Therefore, the order from weakest to strongest reducing agent is Silver, Copper, Iron, and Zinc.

Step-by-Step Solution

1

Understand the relationship between standard reduction potential (

E^\circ

) and reducing strength.

A species with a more negative standard reduction potential has a greater tendency to undergo oxidation (lose electrons) and is therefore a stronger reducing agent. A species with a more positive

E^\circ

is a weaker reducing agent.

Reducing strength is inversely proportional to standard reduction potential.

2

Compare the given

E^\circ

values.

E^\circ(\text{Ag}^+/\text{Ag}) = +0.80\text{ V} > E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\text{ V} > E^\circ(\text{Fe}^{2+}/\text{Fe}) = -0.44\text{ V} > E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76\text{ V}

Listing potentials from most positive to most negative orders the elements from weakest to strongest reducing agent.

3

Arrange the metals in increasing order of reducing power.

Silver (

\text{Ag}

) < Copper (

\text{Cu}

) < Iron (

\text{Fe}

) < Zinc (

\text{Zn}

)

Silver has the highest

E^\circ

and is the weakest reducing agent, while zinc has the lowest

E^\circ

and is the strongest reducing agent.

Key Concept

Reducing Strength and Standard Electrode Potentials

Question 269Question →

A chemist needs to separate a dry solid mixture containing iron turnings, iodine crystals, and sodium chloride into its pure individual components. Arrange the following procedural steps in the correct order to achieve this separation.

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Answer

The correct sequence of steps is: First, pass a magnet over the dry mixture to isolate iron turnings; second, heat the remaining dry mixture under a cooled inverted funnel to sublime iodine crystals; third, dissolve the residual solid in distilled water and filter; fourth, evaporate the clear filtrate to dryness to obtain pure sodium chloride.

The correct order respects the physical states and properties of each component. Performing magnetization first isolates dry iron turnings cleanly. Controlled heating sublimates volatile iodine next. Dissolving the residue in water followed by evaporation isolates pure sodium chloride last.

Step-by-Step Solution

1

Remove ferromagnetic material using magnetization.

Iron turnings are completely isolated from the dry mixture.

Magnetization exploits the magnetic susceptibility of iron while the sample is dry.

2

Sublime the volatile solid component by controlled heating.

Iodine vaporizes and deposits as pure crystals on the cool funnel surface, leaving solid sodium chloride.

Iodine readily sublimates upon gentle heating, whereas sodium chloride has a very high melting point.

3

Dissolve the non-volatile residue in water and filter.

Sodium chloride dissolves completely into the aqueous filtrate.

Sodium chloride is highly soluble in water.

4

Evaporate water from the clear filtrate.

Dry crystals of pure sodium chloride are recovered.

Water vaporizes off, leaving behind the non-volatile dissolved salt.

Key Concept

Sequential separation of solid mixtures by exploiting distinct physical properties (magnetism, volatility/sublimation, and aqueous solubility).

Estimated Time:1m 30s

Question 270Question →

Arrange the following sequential transformations of nitrogen in the biological nitrogen cycle, beginning with atmospheric molecular nitrogen ( $N_2$ ) and concluding with its re-release into the atmosphere:

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Answer

The correct sequence of nitrogen cycle transformations is: Nitrogen fixation (

N_2 \rightarrow NH_4^+

), Nitrosification (

NH_4^+ \rightarrow NO_2^-

), Nitration (

NO_2^- \rightarrow NO_3^-

), Assimilation (

NO_3^- \rightarrow \text{organic plant proteins}

), and Denitrification (

NO_3^- \rightarrow N_2

).

The biological nitrogen cycle proceeds in a specific logical sequence. First, atmospheric nitrogen gas (

N_2

) must undergo nitrogen fixation to become ammonium ions (

NH_4^+

). Next, nitrification occurs in two steps: nitrosification converts ammonium (

NH_4^+

) to nitrite (

NO_2^-

) via *Nitrosomonas*, followed by nitration converting nitrite (

NO_2^-

) to nitrate (

NO_3^-

) via *Nitrobacter*. Plants then assimilate these nitrate ions (

NO_3^-

) into organic plant proteins. Finally, denitrifying bacteria reduce residual soil nitrates (

NO_3^-

) back into atmospheric nitrogen gas (

N_2

) under anaerobic conditions.

Step-by-Step Solution

1

Identify the initial entry point of atmospheric nitrogen into the biosphere.

Gaseous nitrogen (

N_2

) is fixed into ammonium ions (

NH_4^+

) by nitrogen-fixing bacteria (e.g., *Rhizobium*, *Azotobacter*).

Atmospheric

N_2

has a strong triple covalent bond (

N \equiv N

) and cannot be directly utilized by higher plants without preliminary biological fixation.

2

Determine the initial oxidation stage of nitrification (nitrosification).

Ammonium ions (

NH_4^+

) are oxidized to nitrite ions (

NO_2^-

) by *Nitrosomonas*.

Nitrification occurs in two distinct microbial steps, starting with ammonium conversion to nitrite.

3

Determine the second oxidation stage of nitrification (nitration).

Nitrite ions (

NO_2^-

) are oxidized to nitrate ions (

NO_3^-

) by *Nitrobacter*.

Nitrate (

NO_3^-

) is the most readily absorbed and utilized form of inorganic nitrogen for plants.

4

Trace the biological uptake of bioavailable soil nitrogen.

Plants absorb

NO_3^-

ions and assimilate them into plant proteins and nucleic acids.

Inorganic nitrate is reduced inside plant tissues and converted into organic amino acids.

5

Identify the pathway responsible for returning gaseous nitrogen to the atmosphere.

Denitrifying bacteria (e.g., *Pseudomonas denitrificans*) convert unabsorbed soil nitrates (

NO_3^-

) into atmospheric

N_2

gas under anaerobic conditions.

Denitrification closes the global nitrogen loop by replenishing free atmospheric

N_2

.

Key Concept

Nitrogen Cycle Transformations and Microorganisms

Estimated Time:2m 0s

Question 271Question →

Arrange the following sequential processing stages involved in the treatment of raw river water for municipal supply in the correct order from initial treatment to final distribution.

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Answer

The correct sequence for municipal water purification is: Aeration → Coagulation (addition of alum) → Sedimentation → Sand Filtration → Chlorination (Disinfection).

Municipal water treatment follows a logical progression designed to clear large and dissolved impurities first before fine physical filtering and final chemical disinfection. Aeration removes volatile gases and oxidizes metals; coagulation aggregates colloidal particles using alum; sedimentation settles the heavy flocs; sand filtration removes micro-particles; and chlorination destroys disease-causing microorganisms as the final safeguard before piping to households.

Step-by-Step Solution

1

Identify the initial physical treatment step for raw water.

Aeration is performed first to remove volatile odors/gases and oxidize dissolved iron salts into insoluble forms.

Air exposure improves taste and prepares dissolved metals for precipitation.

2

Determine the step that precipitates fine colloidal particles.

Coagulation with potash alum follows aeration.

Alum causes fine, non-settling colloidal particles to aggregate into larger flocs.

3

Identify how the aggregated flocs are removed bulk-wise.

Sedimentation allows these heavy flocs to settle to the basin floor.

Gravity settling removes the bulk of suspended solids before filtration.

4

Identify the step that removes residual microscopic suspended solids.

Sand filtration passes clear water through sand and gravel beds.

Filtration traps any residual suspended particles that escaped sedimentation.

5

Identify the final biological safety step prior to distribution.

Chlorination is carried out as the final step.

Sterilization must occur after physical clarity is achieved so chlorine acts effectively on pathogenic bacteria.

Key Concept

Sequential Stages of Municipal Water Treatment

Question 272Question →

A laboratory mixture contains soluble potassium trioxonitrate(V) ( $\text{KNO}_3$ ) contaminated with insoluble clay particles. What is the correct sequential order of operational steps required to obtain pure, dry crystals of $\text{KNO}_3$ from this mixture?

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Answer

The correct sequential order of operational steps is: First, dissolve the impure mixture in warm distilled water with continuous stirring. Second, filter the mixture using filter paper and a glass funnel. Third, heat the clear filtrate gently in an evaporating dish until it reaches its saturation point. Fourth, allow the hot saturated filtrate to cool down slowly, then filter and dry the formed crystals between filter papers.

To separate and purify a soluble salt mixed with an insoluble solid, dissolution must occur first so that the salt enters the liquid phase while the insoluble component remains solid. Filtration then isolates the solid residue from the dissolved salt filtrate. The filtrate is heated only until it becomes saturated; subsequent slow cooling lowers solute solubility and allows pure crystals to precipitate cleanly without thermal degradation.

Step-by-Step Solution

1

Dissolve the mixture in distilled water.

A aqueous mixture containing dissolved

\text{KNO}_3

and suspended insoluble clay is formed.

Solubility differences allow the soluble salt to dissolve while insoluble impurities remain solid.

2

Filter the mixture through filter paper.

Insoluble clay is retained on the paper as residue, yielding clear aqueous

\text{KNO}_3

filtrate.

Filtration separates undissolved solid matter from liquid solution based on particle size.

3

Evaporate water until the solution is saturated.

A hot saturated solution of

\text{KNO}_3

is produced.

Evaporating solvent increases solute concentration until the crystallization point is achieved.

4

Cool the saturated solution slowly and dry the resulting precipitate.

Pure

\text{KNO}_3

crystals form and are isolated dry.

Solubility decreases upon cooling, causing the dissolved salt to crystallize out of solution cleanly.

Key Concept

Sequential purification of soluble salts from insoluble solids using dissolution, filtration, concentration to saturation, and crystallization.

Estimated Time:1m 30s

Question 273Question →

During the industrial extraction of iron in a blast furnace, limestone ( $\text{CaCO}_3$ ) is added to remove silica impurities ( $\text{SiO}_2$ ). Arrange the following stages of slag formation and separation in their correct chronological sequence from first to last.

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Answer

The correct chronological sequence is: Thermal decomposition of limestone (

\text{CaCO}_3

) into basic calcium oxide (

\text{CaO}

) and carbon(IV) oxide (

\text{CO}_2

)

\rightarrow

Acid-base reaction between basic calcium oxide (

\text{CaO}

) flux and acidic silicon(IV) oxide (

\text{SiO}_2

) impurity

\rightarrow

Formation of molten calcium trioxosilicate(IV) (

\text{CaSiO}_3

)

\rightarrow

Collection of molten slag as a protective layer floating above the denser liquid pig iron at the hearth.

Limestone (

\text{CaCO}_3

) decomposes thermally under high temperatures into calcium oxide (

\text{CaO}

) and carbon(IV) oxide (

\text{CO}_2

). Next, the basic

\text{CaO}

flux reacts with acidic silica (

\text{SiO}_2

) impurities in an acid-base neutralization to form molten calcium trioxosilicate(IV) (

\text{CaSiO}_3

, slag). Finally, the molten slag collects at the hearth and floats on top of the denser molten pig iron, forming a protective layer.

Step-by-Step Solution

1

Identify the initial thermal reaction of limestone in the furnace.

Limestone (

\text{CaCO}_3

) decomposes at around

800^\circ\text{C}-1000^\circ\text{C}

to yield calcium oxide (

\text{CaO}

) and carbon(IV) oxide (

\text{CO}_2

).

Calcium oxide (

\text{CaO}

) must first be synthesized to act as a basic flux.

2

Determine the chemical interaction between the flux and raw ore impurities.

Basic calcium oxide (

\text{CaO}

) reacts with acidic silicon(IV) oxide (

\text{SiO}_2

).

Silica is the main acidic impurity in hematite ore, requiring neutralization by the basic flux.

3

Identify the chemical product formed from this reaction.

Molten calcium trioxosilicate(IV) (

\text{CaSiO}_3

, slag) is formed via the reaction

\text{CaO(s)} + \text{SiO}_2\text{(s)} \rightarrow \text{CaSiO}_3\text{(l)}

.

Neutralization of silica forms the molten compound known as slag.

4

Describe the physical separation of slag at the hearth.

The molten slag drains down to the hearth and floats above the denser liquid iron layer.

Density differences allow slag to float on molten iron, preventing its re-oxidation by blast air.

Key Concept

Slag Formation and Impurity Removal in the Blast Furnace

Question 274Question →

Which of the following represents the correct sequential procedure for separating two immiscible liquids using a separating funnel?

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Answer

The correct procedural order begins with pouring the mixture into the closed funnel, shaking and venting pressure, allowing the layers to settle undisturbed, and finally removing the stopper to drain the lower layer.

The correct laboratory sequence requires securing the tap before filling, shaking with periodic pressure venting, resting the funnel to allow gravitational separation into distinct layers based on density, and finally removing the top stopper before draining the denser lower layer through the stopcock.

Step-by-Step Solution

1

Ensure the stopcock is closed and pour the liquid mixture into the funnel.

The mixture is safely contained within the separating funnel.

Prevents accidental loss of liquid before separation begins.

2

Stopper the funnel, invert it, and shake gently while opening the stopcock to release vapor pressure.

Vapor pressure is released safely without building up dangerous pressure inside the stoppered glass vessel.

Shaking ensures proper contact between liquid phases, while venting prevents pressure buildup.

3

Mount the funnel upright in a ring stand and allow it to stand undisturbed.

Two distinct phase layers separated by a visible interface form due to differences in density and immiscibility.

Gravity forces the denser liquid to the bottom layer and the lighter liquid to the top layer.

4

Unstopper the top of the funnel and turn the stopcock to drain the bottom layer into a container.

The lower denser layer is collected separately, completing the physical separation.

Removing the stopper prevents a negative pressure vacuum that would prevent the liquid from flowing out.

Key Concept

Standard procedural steps for separating immiscible liquids using a separating funnel

Question 275Question →

A organic mixture containing naphthalene, chlorobenzene, acetophenone, and benzoic acid is separated using silica gel column chromatography with hexane (a non-polar solvent) as the mobile phase. Arrange these four compounds in order of their elution from the column, starting with the compound that elutes first and ending with the compound that elutes last.

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Answer

The correct order of elution from first to last is: Naphthalene, Chlorobenzene, Acetophenone, Benzoic acid.

In silica gel column chromatography using a non-polar solvent like hexane, separation is governed by adsorption efficiency. Silica gel contains polar silanol groups (

-Si-OH

). Non-polar compounds (naphthalene) do not interact significantly with the polar stationary phase and dissolve readily in the non-polar eluent, causing them to move rapidly down the column and elute first. As solute polarity increases from weakly polar (chlorobenzene) to moderately polar ketone (acetophenone) to strongly polar hydrogen-bonding acid (benzoic acid), the strength of adsorption increases, delaying elution accordingly.

Step-by-Step Solution

1

Identify the nature of the stationary and mobile phases.

The stationary phase (silica gel) is polar, while the mobile phase (hexane) is non-polar.

Normal-phase column chromatography relies on adsorption differences on a polar stationary matrix.

2

Determine the polarities of the four organic solutes based on their functional groups.

Naphthalene (non-polar hydrocarbon) < Chlorobenzene (weakly polar haloarene) < Acetophenone (moderately polar ketone) < Benzoic acid (strongly polar carboxylic acid).

Carboxylic acids can hydrogen-bond, ketones have strong dipoles, haloarenes have weak dipoles, and aromatic hydrocarbons are non-polar.

3

Correlate solute polarity with retention time and elution order.

Non-polar solutes adsorb weakly to silica gel and elute first; highly polar solutes adsorb strongly and elute last.

The non-polar mobile phase preferentially dissolves and transports less polar molecules down the column more quickly.

4

Arrange the compounds in sequence from least polar (first eluted) to most polar (last eluted).

Naphthalene

\rightarrow

Chlorobenzene

\rightarrow

Acetophenone

\rightarrow

Benzoic acid.

This represents the exact elution sequence in normal-phase column chromatography.

Key Concept

Adsorption Column Chromatography Elution Order based on Solute Polarity

Estimated Time:2m 0s

Question 276Question →

Arrange the following chemical substances in order of increasing boiling point, starting from the substance with the lowest boiling point to the one with the highest boiling point based on the nature and relative strength of their intermolecular forces.

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Answer

The correct order of increasing boiling point is Methane (

CH_4

) < Hydrogen sulfide (

H_2S

) < Ammonia (

NH_3

) < Water (

H_2O

).

The sequence reflects the increasing magnitude of intermolecular forces: non-polar Methane (

CH_4

) relies solely on weak London dispersion forces (lowest boiling point). Polar Hydrogen sulfide (

H_2S

) has dipole-dipole interactions but lacks hydrogen bonding because sulfur is not electronegative enough. Ammonia (

NH_3

) undergoes hydrogen bonding due to nitrogen's high electronegativity. Water (

H_2O

) forms an extensive network of strong hydrogen bonds, resulting in the highest boiling point.

Step-by-Step Solution

1

Identify the type of intermolecular forces operating in each substance

CH_4

is non-polar (London dispersion forces only);

H_2S

is polar (dipole-dipole and dispersion forces);

NH_3

is polar with hydrogen bonding;

H_2O

is polar with strong, extensive hydrogen bonding.

Boiling point depends directly on the total magnitude of attraction between molecules in the liquid state.

2

Compare non-hydrogen-bonding substances (

CH_4

vs

H_2S

)

CH_4

has the weakest intermolecular forces (dispersion only), while

H_2S

has additional permanent dipole-dipole attractions.

Permanent dipole-dipole interactions in polar molecules generally create stronger attraction than non-polar dispersion forces of comparable size.

3

Compare hydrogen-bonding substances (

NH_3

vs

H_2O

)

Both

NH_3

and

H_2O

form hydrogen bonds, but

H_2O

forms up to four hydrogen bonds per molecule in a 3D network, whereas

NH_3

is limited by its single lone pair to fewer hydrogen bonds per molecule.

Oxygen is more electronegative than nitrogen, and water has an optimal 1:1 ratio of lone pairs to hydrogen atoms for maximum hydrogen-bonding capacity.

4

Synthesize the complete sequence from lowest to highest boiling point

CH_4

<

H_2S

<

NH_3

<

H_2O

Intermolecular attraction strength increases in the order: London dispersion forces < dipole-dipole interactions < moderate hydrogen bonding < extensive hydrogen bonding.

Key Concept

Relative strengths of intermolecular forces (London dispersion, dipole-dipole, and hydrogen bonding) and their effect on physical properties like boiling point.

Estimated Time:2m 0s

Question 277Question →

Arrange the following sequential steps of a paper chromatography experiment in the correct procedural order from start to finish.

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Answer

The correct procedural order is: (1) Draw a light baseline using a lead pencil about

2\text{ cm}

from the bottom edge of the filter paper strip, (2) Apply a small, concentrated spot of the sample mixture onto the baseline and allow it to dry, (3) Suspend the paper strip in a sealed chamber, ensuring the solvent level remains strictly below the baseline, and (4) Remove the strip when the mobile phase nears the top and immediately pencil-mark the solvent front.

The correct sequence follows standard analytical chromatography protocol: first, a graphite pencil baseline is drawn at the base of the paper strip. Next, the mixture is spotted onto this origin baseline. The paper is then placed into the developing tank with the mobile phase level kept below the baseline to allow upward capillary migration. Finally, the paper is removed before the solvent reaches the top edge, and the solvent front is marked immediately for accurate distance measurements.

Step-by-Step Solution

1

Determine paper preparation step

Draw a pencil baseline near the bottom edge of the paper.

Graphite from a lead pencil is insoluble in the mobile phase and will not contaminate the chromatogram.

2

Determine sample application step

Spot the sample mixture onto the pencil baseline and dry it.

Deposition onto the origin line must occur before the paper contacts the developing solvent.

3

Determine development step

Place the paper into the chamber with the solvent line positioned below the baseline.

Maintaining the solvent level below the baseline allows the solvent to ascend by capillary action, carrying solute components upward rather than dissolving them into the bulk solvent.

4

Determine experiment termination step

Remove the paper strip and mark the solvent front.

Immediate marking of the solvent front ensures accurate measurement of mobile phase migration distance for

R_f

calculations.

Key Concept

Laboratory procedural steps and operational principles of paper chromatography

Question 278Question →

During the electrolysis of an aqueous solution containing equal concentrations of cations using inert carbon electrodes, ions migrate to the cathode where discharge occurs based on their position in the electrochemical series. Arrange the following cations in increasing order of their ease of preferential discharge at the cathode (from the least easily discharged to the most easily discharged):

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Answer

The correct sequence from least easily discharged to most easily discharged is Potassium ion (

K^+

), Zinc ion (

Zn^{2+}

), Hydrogen ion (

H^+

), and Copper(II) ion (

Cu^{2+}

).

In electrolysis at inert electrodes under equal concentrations, the primary factor determining preferential discharge of cations at the cathode is their position in the electrochemical series. Cations positioned lower in the series gain electrons more easily (have higher reduction potentials). Therefore, Potassium ion (

K^+

) is the hardest to discharge, followed by Zinc ion (

Zn^{2+}

), Hydrogen ion (

H^+

), and finally Copper(II) ion (

Cu^{2+}

), which is the most easily discharged.

Step-by-Step Solution

1

Recall the electrochemical series position for cations

The order of cations from top (most reactive / hardest to discharge) to bottom (least reactive / easiest to discharge) is

K^+

,

Zn^{2+}

,

H^+

,

Cu^{2+}

.

Ions lower in the electrochemical series have higher standard reduction potentials and accept electrons more readily at the cathode.

2

Arrange the cations in increasing order of ease of discharge

Potassium ion (

K^+

) < Zinc ion (

Zn^{2+}

) < Hydrogen ion (

H^+

) < Copper(II) ion (

Cu^{2+}

).

Ease of discharge increases down the electrochemical series.

Key Concept

Position of cations in the electrochemical series determines their relative ease of discharge at the cathode.

Question 279Question →

In the industrial manufacture of tetraoxosulfate(VI) acid via the Contact Process, several crucial chemical and physical steps are carried out in a specific sequence to maximize yield and prevent efficiency loss. Arrange the following steps of the Contact Process in the correct sequential order from first to last.

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Answer

The correct sequence of the Contact Process is: Combustion/roasting to produce

SO_2

gas

\rightarrow

Purification of

SO_2

gas to remove catalyst poisons

\rightarrow

Catalytic conversion of

SO_2

to

SO_3

over

V_2O_5

catalyst

\rightarrow

Absorption of

SO_3

gas in concentrated

H_2SO_4

to form oleum

\rightarrow

Controlled dilution of oleum with water to yield

H_2SO_4

.

The Contact Process must follow a rigorous order: first generating raw

SO_2

gas, purifying it to prevent catalyst poisoning by impurities like

As_2O_3

, catalytically oxidizing

SO_2

to

SO_3

over

V_2O_5

, absorbing

SO_3

in 98%

H_2SO_4

to form oleum (

H_2S_2O_7

), and finally diluting oleum with water to produce concentrated

H_2SO_4

.

Step-by-Step Solution

1

Identify the initial feedstock generation stage.

Combustion of sulfur or roasting of sulfide ores produces

SO_2

gas.

Sulfur(IV) oxide is the essential chemical precursor required for the process.

2

Determine the necessary gas purification stage prior to catalysis.

Passing the

SO_2

and air mixture through scrubbers and precipitators removes dust particles and arsenic(III) oxide (

As_2O_3

).

Arsenic compounds act as catalyst poisons, permanently deactivating the vanadium(V) oxide catalyst if not removed first.

3

Identify the catalytic oxidation step.

Purified

SO_2

reacts with

O_2

over a

V_2O_5

catalyst at 450 °C and 1–2 atm to form

SO_3

.

This exothermic equilibrium reaction converts sulfur(IV) oxide to sulfur(VI) oxide.

4

Determine the absorption stage for sulfur(VI) oxide.

SO_3

gas is absorbed into concentrated (98%)

H_2SO_4

to form oleum (

H_2S_2O_7

).

Direct hydration of

SO_3

with water is extremely exothermic and produces a fine acid fog that cannot be easily condensed industrially.

5

Identify the final dilution/hydration step.

Oleum (

H_2S_2O_7

) is diluted with a calculated volume of water to form concentrated

H_2SO_4

.

Reacting oleum with water yields high-purity tetraoxosulfate(VI) acid safely and efficiently.

Key Concept

Sequential chemical and industrial stages of the Contact Process for tetraoxosulfate(VI) acid production

Estimated Time:2m 0s

Question 280Question →

During the laboratory preparation and isolation of nitrogen gas from atmospheric air, a student must pass atmospheric air through several reagents in a specific sequence to remove impurities. What is the correct order of steps to isolate nitrogen gas from atmospheric air, starting from the initial removal of carbon(IV) oxide to the final collection of the gas?

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Answer

The correct sequence to isolate nitrogen gas from atmospheric air is: 1. Pass air through concentrated caustic alkali solution (

KOH

/

NaOH

) to remove

CO_2

. 2. Pass the gas through concentrated

H_2SO_4

to remove water vapor. 3. Pass dry gas over red-hot copper turnings to remove

O_2

. 4. Collect the remaining nitrogen gas over water.

Atmospheric air consists primarily of

N_2

(78%),

O_2

(21%),

CO_2

(0.03%), water vapor, and noble gases. To isolate nitrogen, impurities are removed according to chemical reactivity:

CO_2

is removed first by neutralisation with an alkali (

KOH

), moisture is absorbed by a dehydrating agent (concentrated

H_2SO_4

), oxygen is removed by reduction of red-hot copper turnings to

CuO

, and the remaining nitrogen gas (mixed with trace noble gases like argon) is collected over water.

Step-by-Step Solution

1

Identify the acidic gas impurity in air and select its removal agent.

Carbon(IV) oxide (

CO_2

) is acidic and must be scrubbed first using concentrated

KOH

or

NaOH

solution:

2KOH_{(aq)} + CO_{2(g)} \rightarrow K_2CO_{3(aq)} + H_2O_{(l)}

.

Removing

CO_2

first prevents it from contaminating subsequent drying and heating apparatus.

2

Dry the remaining gas mixture.

Passing the remaining gases (

O_2

,

N_2

, water vapor, noble gases) through concentrated

H_2SO_4

absorbs moisture.

Gas must be thoroughly dried before passing over hot copper turnings to prevent thermal shock and unwanted reactions.

3

Remove oxygen gas chemically.

Passing dry air over red-hot copper turnings removes

O_2

:

2Cu_{(s)} + O_{2(g)} \rightarrow 2CuO_{(s)}

.

Hot copper chemically binds oxygen, leaving only unreactive nitrogen and traces of noble gases.

4

Collect the purified nitrogen gas.

Nitrogen gas is collected over water.

Nitrogen has very low solubility in water, making water displacement ideal for gas collection.

Key Concept

Laboratory Isolation of Nitrogen from Atmospheric Air

Estimated Time:1m 30s

Answer

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