Air, Water and Solubility Practice Questions — JAMB UTME — Page 2

Question 21Question →

Calculate the solubility in $\text{mol/dm}^3$ of sodium nitrate ( $\text{NaNO}_3$ ) at $25^\circ\text{C}$ , if $17.0\text{ g}$ of the salt dissolves in $100.0\text{ g}$ of water to form a saturated solution. [Relative atomic masses: $\text{Na} = 23, \text{N} = 14, \text{O} = 16$ ; density of water $= 1.0\text{ g/cm}^3$ ]

Show answer & explanation

Answer: 2

Answer

The solubility of sodium nitrate at

25^\circ\text{C}

is

2.0\text{ mol/dm}^3

.

To find solubility in

\text{mol/dm}^3

, first convert

17.0\text{ g}

of

\text{NaNO}_3

into moles by dividing by its molar mass (

85.0\text{ g/mol}

), obtaining

0.20\text{ mol}

. Next, convert

100.0\text{ g}

of water into volume, which equals

0.100\text{ dm}^3

. Dividing

0.20\text{ mol}

by

0.100\text{ dm}^3

yields

2.0\text{ mol/dm}^3

.

Step-by-Step Solution

1

Determine the molar mass of the solute (

\text{NaNO}_3

)

Molar mass of

\text{NaNO}_3 = 23 + 14 + (3 \times 16) = 85.0\text{ g/mol}

Molar mass is required to convert mass of solute to amount in moles.

2

Calculate the amount of

\text{NaNO}_3

in moles

\text{Moles} = \frac{17.0\text{ g}}{85.0\text{ g/mol}} = 0.20\text{ mol}

Solubility in

\text{mol/dm}^3

measures the amount of solute in moles per cubic decimeter of solvent.

3

Convert the mass of solvent (water) to volume in

\text{dm}^3

100.0\text{ g of water} = 100.0\text{ cm}^3 = 0.100\text{ dm}^3

Density of water is

1.0\text{ g/cm}^3

, and

1000\text{ cm}^3 = 1\text{ dm}^3

.

4

Calculate the solubility in

\text{mol/dm}^3

\text{Solubility} = \frac{0.20\text{ mol}}{0.100\text{ dm}^3} = 2.0\text{ mol/dm}^3

Dividing the moles of solute by the volume of solvent in

\text{dm}^3

gives the concentration in

\text{mol/dm}^3

.

Key Concept

Solubility expressed in molar concentration (mol/dm³)

Question 22Question →

A municipal water treatment facility processes raw river water containing fine colloidal clay particles, dissolved hydrogen sulfide, soluble iron(II) compounds, and harmful microorganisms. During the treatment process, the water is subjected to aeration, addition of potash alum, sedimentation, sand filtration, and chlorination. Which of the following statements correctly distinguishes the primary chemical roles of aeration and potash alum in this treatment sequence?

Aeration oxidizes soluble iron(II) compounds and expels dissolved foul-smelling gases, whereas potash alum induces coagulation of fine colloidal clay particles.

Aeration precipitates dissolved calcium tetraoxosulfate(VI) to remove permanent hardness, whereas potash alum sterilizes harmful microorganisms.

Aeration adjusts the pH of acidic water by dissolving atmospheric carbon(IV) oxide, whereas potash alum removes dissolved magnesium ions to soften the water.

Aeration kills pathogenic bacteria through dissolved oxygen toxicity, whereas potash alum neutralizes organic foul odors and tastes.

Show answer & explanation

Answer: Aeration oxidizes soluble iron(II) compounds and expels dissolved foul-smelling gases, whereas potash alum induces coagulation of fine colloidal clay particles.

Answer

Aeration oxidizes soluble iron(II) compounds and expels dissolved foul-smelling gases, whereas potash alum induces coagulation of fine colloidal clay particles.

In municipal water purification, aeration serves to bubble air through raw water, oxidizing dissolved iron(II) salts to insoluble iron(III) precipitates and stripping out dissolved volatile gases (such as hydrogen sulfide and carbon(IV) oxide). Potash alum (hydrated potassium aluminium tetraoxosulfate(VI)) supplies aluminium ions that coagulate fine suspended colloidal particles into larger flocs that readily settle out during sedimentation.

Step-by-Step Solution

1

Analyze the primary chemical and physical effects of aeration in municipal water treatment.

Aeration increases dissolved oxygen, which oxidizes soluble

\text{Fe}^{2+}

ions to insoluble

\text{Fe(OH)}_3

precipitates, while simultaneously purging volatile dissolved gases such as

\text{H}_2\text{S}

and excess

\text{CO}_2

.

Aeration targets volatile odorous compounds and oxidizable dissolved inorganic species.

2

Analyze the role of potash alum (

\text{KAl(SO}_4\text{)}_2 \cdot 12\text{H}_2\text{O}

) during water purification.

Potash alum dissociates to yield

\text{Al}^{3+}

ions, which neutralize the negative surface charges on microscopic colloidal clay particles, allowing them to coagulate and settle during sedimentation.

Colloidal suspensions will not settle naturally under gravity without chemical coagulation.

3

Evaluate the choices to distinguish these functions from water softening and sterilization.

Sterilization is achieved by chlorination, and water softening requires removal of

\text{Ca}^{2+}

or

\text{Mg}^{2+}

ions via chemical precipitation or ion exchange, confirming the statement regarding oxidation/gas expulsion for aeration and coagulation for alum.

Distinguishing municipal water treatment for domestic supply (removal of turbidity and pathogens) from water softening avoids common conceptual confusion.

Key Concept

Distinct roles of aeration (gas removal and oxidation) and coagulation (potash alum flocculation) in municipal water treatment

Question 23Question →

The solubility of a salt $Y$ (molar mass = $160\text{ g mol}^{-1}$ ) in water is $2.5\text{ mol dm}^{-3}$ at $70^\circ\text{C}$ and $1.2\text{ mol dm}^{-3}$ at $30^\circ\text{C}$ . If $300\text{ cm}^3$ of a saturated solution of $Y$ at $70^\circ\text{C}$ is cooled to $30^\circ\text{C}$ , what mass of salt $Y$ will crystallize out of solution?

62.4\text{ g}

120.0\text{ g}

208.0\text{ g}

0.39\text{ g}

Show answer & explanation

Answer: $62.4\text{ g}$

Answer

The mass of salt

Y

that crystallizes out of solution is

62.4\text{ g}

.

The mass of salt precipitated is determined by taking the difference in molar solubility (

2.5 - 1.2 = 1.3\text{ mol dm}^{-3}

), multiplying by the volume fraction (

300/1000 = 0.3\text{ dm}^3

) to find the number of moles (

0.39\text{ mol}

), and then multiplying by the molar mass (

160\text{ g mol}^{-1}

) to obtain

62.4\text{ g}

.

Step-by-Step Solution

1

Determine the change in molar solubility upon cooling

\Delta S = 2.5\text{ mol dm}^{-3} - 1.2\text{ mol dm}^{-3} = 1.3\text{ mol dm}^{-3}

Solubility decreases as temperature drops, causing the excess solute to precipitate.

2

Calculate the amount in moles precipitated in

300\text{ cm}^3

of solution

n = 1.3\text{ mol dm}^{-3} \times \frac{300\text{ cm}^3}{1000\text{ cm}^3\text{ dm}^{-3}} = 0.39\text{ mol}

The solution volume is

300\text{ cm}^3

(

0.3\text{ dm}^3

), so the moles precipitated must be scaled from

1\text{ dm}^3

.

3

Convert the moles precipitated to mass in grams

\text{Mass} = 0.39\text{ mol} \times 160\text{ g mol}^{-1} = 62.4\text{ g}

Mass is found by multiplying the mole quantity by the given molar mass of the salt.

Key Concept

Solubility and Crystallization Calculations

Question 24Question →

A $24.6\text{ g}$ sample of hydrated magnesium tetraoxosulfate(VI), $\text{MgSO}_4 \cdot x\text{H}_2\text{O}$ , was heated strongly until all the water of crystallization was driven off, leaving behind $12.0\text{ g}$ of anhydrous $\text{MgSO}_4$ . What is the value of $x$ ? $[\text{Mg} = 24, \text{S} = 32, \text{O} = 16, \text{H} = 1]$

Show answer & explanation

Answer: 7

Answer

The value of

x

is 7.

Heating the sample evaporates

12.6\text{ g}

of water of crystallization (

24.6\text{ g} - 12.0\text{ g}

). Converting the masses to moles gives

0.1\text{ mol}

of anhydrous

\text{MgSO}_4

(

12.0\text{ g} / 120\text{ g/mol}

) and

0.7\text{ mol}

of

\text{H}_2\text{O}

(

12.6\text{ g} / 18\text{ g/mol}

). The mole ratio of water to salt is

0.7 / 0.1 = 7

, which means

x = 7

.

Step-by-Step Solution

1

Calculate the mass of water of crystallization lost during heating.

Mass of water =

24.6\text{ g} - 12.0\text{ g} = 12.6\text{ g}

.

The decrease in mass after heating represents the mass of water driven off from the hydrated salt.

2

Determine the molar masses of

\text{MgSO}_4

and

\text{H}_2\text{O}

.

Molar mass of

\text{MgSO}_4 = 24 + 32 + (4 \times 16) = 120\text{ g/mol}

; Molar mass of

\text{H}_2\text{O} = (2 \times 1) + 16 = 18\text{ g/mol}

.

Molar masses are required to convert the measured masses into chemical amounts (moles).

3

Calculate the number of moles of anhydrous salt and water.

Moles of

\text{MgSO}_4 = \frac{12.0\text{ g}}{120\text{ g/mol}} = 0.1\text{ mol}

; Moles of

\text{H}_2\text{O} = \frac{12.6\text{ g}}{18\text{ g/mol}} = 0.7\text{ mol}

.

The chemical formula stoichiometry is determined by the molar ratio of components.

4

Determine the mole ratio of water to anhydrous salt (

x

).

x = \frac{\text{Moles of } \text{H}_2\text{O}}{\text{Moles of } \text{MgSO}_4} = \frac{0.7\text{ mol}}{0.1\text{ mol}} = 7

.

The coefficient

x

is the integer ratio of moles of water of crystallization per mole of anhydrous salt.

Key Concept

Water of Crystallization Stoichiometry

Question 25Question →

In an industrial plant, dry air is cooled to $-200^\circ\text{C}$ to form liquid air after removing carbon dioxide and water vapor. When this liquid air undergoes fractional distillation in a fractionating column as its temperature is slowly raised, which statement correctly identifies the constituent that boils off first and the rationale behind its separation?

Nitrogen boils off first at

-196^\circ\text{C}

because it has a lower boiling point than argon and oxygen.

Oxygen boils off first at

-183^\circ\text{C}

because

-183^\circ\text{C}

represents a higher numerical boiling point than

-196^\circ\text{C}

.

Carbon dioxide boils off first as a gas because it is the lightest component remaining in the liquid mixture.

Argon boils off first at

-186^\circ\text{C}

because unreactive noble gases always distill before diatomic gases.

Show answer & explanation

Answer: Nitrogen boils off first at $-196^\circ\text{C}$ because it has a lower boiling point than argon and oxygen.

Answer

Nitrogen boils off first at

-196^\circ\text{C}

because it has the lowest boiling point among the main liquefied atmospheric constituents.

During the fractional distillation of liquid air, components separate according to their boiling points. As the column temperature warms up from

-200^\circ\text{C}

, nitrogen (boiling point

-196^\circ\text{C}

) reaches its boiling point first, vaporizes, and is collected at the top of the fractionating column.

Step-by-Step Solution

1

Identify the main liquefied components of air after pre-purification

Purified liquid air consists primarily of liquid nitrogen, argon, and liquid oxygen. Water vapor and carbon dioxide are removed prior to liquefaction to prevent clogging the equipment.

Water freezes at 0°C and carbon dioxide sublimes at -78.5°C, making them solids well above the liquefaction temperature of air (-200°C).

2

Compare the boiling points of the liquefied components

Nitrogen boils at

-196^\circ\text{C}

(77 K), Argon boils at

-186^\circ\text{C}

(87 K), and Oxygen boils at

-183^\circ\text{C}

(90 K).

Lower boiling point values on the Celsius scale correspond to colder temperatures (e.g., -196°C is colder than -183°C).

3

Determine the distillation sequence as temperature rises from

-200^\circ\text{C}

Nitrogen reaches its boiling point first at

-196^\circ\text{C}

and turns into gas, leaving argon and oxygen behind until further warming.

In fractional distillation, the liquid with the lowest boiling point boils off first at the top of the column.

Key Concept

Fractional Distillation of Liquid Air

Estimated Time:1m 30s

Question 26Question →

A $250\text{ cm}^3$ sample of dry air is passed over excess heated phosphorus in a closed tube to remove all the oxygen gas present. Assuming oxygen constitutes $21\%$ by volume of dry air, what is the volume of the remaining gas mixture in $\text{cm}^3$ ?

Show answer & explanation

Answer: 197.5

Answer

The volume of the remaining gas mixture is

197.5\text{ cm}^3

.

Because oxygen makes up

21\%

by volume of dry air, a

250\text{ cm}^3

sample contains

0.21 \times 250\text{ cm}^3 = 52.5\text{ cm}^3

of oxygen. Heated phosphorus reacts with all the oxygen to form solid phosphorus oxide, leaving behind

250\text{ cm}^3 - 52.5\text{ cm}^3 = 197.5\text{ cm}^3

of unreacted gases.

Step-by-Step Solution

1

Calculate the volume of oxygen gas in the initial sample

52.5\text{ cm}^3

Oxygen makes up

21\%

by volume of dry air, so

0.21 \times 250\text{ cm}^3 = 52.5\text{ cm}^3

.

2

Determine the remaining gas volume after complete removal of oxygen

197.5\text{ cm}^3

Phosphorus reacts completely with oxygen, leaving the unreacted components of air:

250\text{ cm}^3 - 52.5\text{ cm}^3 = 197.5\text{ cm}^3

.

Key Concept

Percentage composition of air by volume

Question 27Question →

Smoke consists of fine solid particles dispersed in a gaseous medium. Which of the following colloidal classifications correctly describes smoke?

Aerosol

Emulsion

Foam

Sol

Show answer & explanation

Answer: Aerosol

Answer

Smoke is classified as an aerosol because it consists of solid particles dispersed in a gas.

Smoke consists of microscopic solid carbon particles suspended in air (a gas). Any colloidal dispersion where a solid or liquid is dispersed in a gas is categorized as an aerosol.

Step-by-Step Solution

1

Identify the dispersed phase and the dispersion medium of smoke.

Dispersed phase = solid (carbon/ash particles), Dispersion medium = gas (air).

Classification of colloids depends on the physical states of the dispersed phase and dispersion medium.

2

Match the phase combination (solid in gas) to the standard colloidal nomenclature.

Solid dispersed in a gas is termed a solid aerosol.

Colloids with a gaseous dispersion medium are broadly categorized as aerosols.

Key Concept

Classification of Colloidal Systems based on Dispersed Phase and Dispersion Medium

Question 28Question →

An industrial effluent contains suspended colloidal particles of clay carrying negative surface charges, with particle diameters ranging from $1\text{ nm}$ to $100\text{ nm}$ . To treat the effluent, equal volumes of four different $0.01\text{ mol dm}^{-3}$ electrolyte solutions— $Al_2(SO_4)_3$ , $CaCl_2$ , $NaCl$ , and $Na_3PO_4$ —are evaluated for their ability to induce coagulation. Which electrolyte has the highest precipitating power (lowest coagulation value) for this clay colloid, and what is the physical mechanism causing coagulation?

Al_2(SO_4)_3

, because the trivalent cation (

Al^{3+}

) carries the highest positive charge to neutralize the negative charges on the colloidal particles, causing them to aggregate into particles larger than

100\text{ nm}

.

Na_3PO_4

, because the trivalent phosphate anion (

PO_4^{3-}

) provides the highest charge density to neutralize the colloidal dispersion.

NaCl

, because its low formula mass enables rapid diffusion into the colloidal particles, converting the colloid into a true solution of particle size under

1\text{ nm}

.

All four electrolytes possess equal precipitating power because coagulation depends strictly on the total molar concentration of solute particles rather than ionic charge.

Show answer & explanation

Answer: $Al_2(SO_4)_3$ , because the trivalent cation ( $Al^{3+}$ ) carries the highest positive charge to neutralize the negative charges on the colloidal particles, causing them to aggregate into particles larger than $100\text{ nm}$ .

Answer

The electrolyte with the highest precipitating power is

Al_2(SO_4)_3

because the trivalent cation (

Al^{3+}

) effectively neutralizes the negative surface charge on the clay particles according to the Hardy-Schulze rule.

The clay colloidal particles are negatively charged with diameters between

1\text{ nm}

and

100\text{ nm}

. According to the Hardy-Schulze rule, the coagulating power of an electrolyte is determined by the ion bearing a charge opposite to that of the colloidal particles, and it increases rapidly with the valency of the active ion. For a negative sol, cations are active. The trivalent

Al^{3+}

ion from aluminium tetraoxosulfate(VI) has a significantly higher precipitating power than divalent

Ca^{2+}

or monovalent

Na^{+}

, neutralizing the charge and aggregating particles beyond

100\text{ nm}

.

Step-by-Step Solution

1

Identify the charge carried by the colloidal clay particles.

The clay particles are negatively charged lyophobic colloidal particles in the size range

1\text{ nm}

to

100\text{ nm}

.

Coagulation requires neutralizing the electrostatic repulsion between colloidal particles using an oppositely charged ion.

2

Apply the Hardy-Schulze rule to determine which ion causes precipitation.

Positively charged cations (

Al^{3+}

,

Ca^{2+}

,

Na^{+}

) are responsible for coagulating the negative colloid.

Ions possessing a charge opposite to that of the colloidal sol are effective for coagulation.

3

Compare the valencies of the effective cations present in the electrolytes.

Al^{3+}

has a valency of

+3

,

Ca^{2+}

has

+2

, and

Na^{+}

has

+1

.

Precipitating power increases exponentially with increasing valency of the active ion (

Al^{3+} > Ca^{2+} > Na^{+}

).

4

Relate charge neutralization to particle size change.

Neutralized particles coalesce into larger aggregates (

>100\text{ nm}

) that precipitate out.

Removal of surface charge reduces electrostatic repulsion, allowing van der Waals forces to aggregate colloidal particles into suspension-sized precipitates.

Key Concept

Hardy-Schulze Rule and Coagulation of Colloidal Systems

Question 29Question →

In a municipal water treatment plant, raw water undergoes several physical and chemical processing steps. What is the primary purpose of introducing activated charcoal during the purification of water for town supply?

To remove unpleasant odors, objectionable tastes, and colored organic compounds

To destroy disease-causing bacteria and pathogenic microorganisms

To cause fine suspended colloidal clay particles to aggregate into large flocs

To precipitate calcium and magnesium ions that cause water hardness

Show answer & explanation

Answer: To remove unpleasant odors, objectionable tastes, and colored organic compounds

Answer

The primary purpose of activated charcoal in municipal water treatment is to remove unpleasant odors, objectionable tastes, and colored organic compounds.

Activated charcoal (carbon) has a micro-porous structure with an exceptionally high surface area. It physically adsorbs dissolved organic pollutants, pigments, and volatile substances that cause objectionable odors and tastes, thereby clarifying and improving the aesthetic quality of drinking water.

Step-by-Step Solution

1

Identify the chemical reagent or material specified in the question.

The specified material is activated charcoal (activated carbon).

Activated charcoal is an adsorbent material used in filtration beds during water treatment.

2

Recall the specific functional role of activated charcoal in water purification.

Activated charcoal physically adsorbs dissolved organic impurities, volatile organic compounds, pigments, and substances responsible for foul tastes and odors.

Its extremely porous structure provides a vast surface area suitable for physical adsorption.

3

Distinguish activated charcoal's role from other water treatment chemicals.

Chlorine sterilizes water, potash alum coagulates suspended particles, and slaked lime adjusts pH or precipitates hardness.

Each chemical reagent added during municipal water treatment has a distinct, specialized role.

Key Concept

Role of Activated Charcoal in Water Treatment

Question 30Question →

A water sample contains both temporary hardness due to dissolved $\text{Ca(HCO}_3)_2$ and permanent hardness due to dissolved $\text{MgSO}_4$ . Which sequence correctly arranges the operational steps required to completely soften this water sample using slaked lime followed by washing soda?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct operational sequence is: first, adding slaked lime to precipitate calcium carbonate from temporary hardness; second, filtering off the solid precipitate; third, adding washing soda to precipitate magnesium carbonate from permanent hardness; and fourth, performing a final filtration to obtain clear, soft water.

The removal process follows a logically structured chemical workflow. Temporary hardness caused by

\text{Ca(HCO}_3)_2

is treated first with a calculated amount of slaked lime (Clark's process), producing insoluble

\text{CaCO}_3

, which is filtered out. The remaining filtrate contains permanent hardness from

\text{MgSO}_4

, which is treated with washing soda (sodium trioxocarbonate(IV)) to precipitate

\text{MgCO}_3

. A second filtration removes this precipitate, leaving pure soft water.

Step-by-Step Solution

1

Identify the chemical reaction for temporary hardness removal.

Adding

\text{Ca(OH)}_2

precipitates temporary hardness:

\text{Ca(HCO}_3)_2 + \text{Ca(OH)}_2 \rightarrow 2\text{CaCO}_3\downarrow + 2\text{H}_2\text{O}

.

Slaked lime specifically removes hydrogentrioxocarbonate(IV) salts causing temporary hardness.

2

Separate the solid calcium carbonate formed.

The solid

\text{CaCO}_3

is filtered out of the solution.

Precipitates must be removed so they do not redissolve or contaminate subsequent steps.

3

Identify the chemical reaction for permanent hardness removal.

Adding

\text{Na}_2\text{CO}_3

precipitates permanent hardness:

\text{MgSO}_4 + \text{Na}_2\text{CO}_3 \rightarrow \text{MgCO}_3\downarrow + \text{Na}_2\text{SO}_4

.

Soluble carbonate ions from washing soda precipitate divalent magnesium cations as insoluble trioxocarbonate(IV) salts.

4

Conduct final liquid-solid separation.

Filtering removes insoluble

\text{MgCO}_3

, producing soft water.

Soluble sodium tetraoxosulfate(VI) remaining in solution does not react with soap to form scum.

Key Concept

Sequential chemical removal of temporary and permanent hardness using Clark's process followed by precipitation with washing soda.

Question 31Question →

Match each municipal water purification stage or chemical additive on the left with its precise operational mechanism on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Aeration matches with expelling volatile gases and oxidizing soluble iron(II); Addition of Potash Alum matches with neutralizing negative charges on clay colloids for flocculation; Sand Filtration matches with straining out fine suspended particles remaining after sedimentation; Addition of Slaked Lime matches with adjusting pH of acidic water post-coagulation.

Aeration oxidizes soluble iron and removes dissolved gases. Potash alum supplies trivalent cations to neutralize negative charges on colloidal clay. Sand filtration physically retains fine suspended particles. Slaked lime neutralizes acidity induced by alum coagulation.

Step-by-Step Solution

1

Analyze the primary chemical and physical effects of Aeration.

Spraying raw water into air maximizes gas exchange, driving off dissolved volatile species (

H_2S

) and converting soluble

Fe^{2+}

ions to insoluble

Fe(OH)_3

precipitate.

Aeration targets volatile odors/tastes and dissolved metals rather than solid particulate filtration.

2

Analyze the coagulating action of Potash Alum (

KAl(SO_4)_2 \cdot 12H_2O

).

Trivalent

Al^{3+}

ions neutralize negative charges on suspended clay colloids, causing them to coalesce into larger settleable flocs.

Fine clay particles do not settle spontaneously due to mutual electrostatic repulsion.

3

Determine the function of Sand Filtration.

Water percolating through sand layer beds leaves behind non-settled micro-particles.

Filtration serves as a final physical straining stage after bulk sedimentation.

4

Evaluate the requirement for Slaked Lime (

Ca(OH)_2

).

Neutralizes acidity produced during alum hydrolysis, raising the pH to safe alkaline levels.

Acidic water damages metal distribution pipes through corrosive action.

Key Concept

Operational mechanisms of chemical and physical processes in town water supply treatment

Question 32Question →

Starch mucilage is classified as a colloidal system rather than a true solution. Which of the following physical properties distinguishes starch mucilage from a true solution of sodium chloride?

Ability to scatter a beam of light passing through the mixture

Rapid settling of dispersed particles under gravity upon standing

Retention of dispersed particles when passed through standard laboratory filter paper

Distinct separation of phases visible to the naked eye

Show answer & explanation

Answer: Ability to scatter a beam of light passing through the mixture

Answer

The ability to scatter a beam of light passing through the mixture (the Tyndall effect)

Colloidal systems such as starch mucilage contain particles ranging from

1\text{ nm}

to

100\text{ nm}

which scatter light (the Tyndall effect). In contrast, true solutions like sodium chloride contain ions or molecules smaller than

1\text{ nm}

that cannot scatter visible light.

Step-by-Step Solution

1

Compare particle size ranges of true solutions and colloidal systems.

True solutions have solute particles smaller than

1\text{ nm}

, whereas colloidal systems have dispersed particles in the range of

1\text{ nm}

to

100\text{ nm}

.

Particle size dictates whether light scattering occurs.

2

Identify the optical phenomenon associated with colloidal particle sizes.

Particles between

1\text{ nm}

and

100\text{ nm}

scatter light rays, making the beam's path visible (Tyndall effect).

This property clearly differentiates colloids like starch mucilage from true solutions like sodium chloride in water.

Key Concept

Tyndall Effect and Particle Size Differences in Mixtures

Estimated Time:45s

Question 33Question →

During domestic water treatment, a household adds a measured amount of calcium hypochlorite (bleaching powder) to clear well water before consumption. What is the primary chemical purpose of adding this substance?

To destroy pathogenic microorganisms present in the water

To precipitate dissolved calcium and magnesium ions that cause hardness

To coagulate fine suspended clay particles into larger settles

To neutralize organic tastes and unpleasant odors in the water

Show answer & explanation

Answer: To destroy pathogenic microorganisms present in the water

Answer

Adding calcium hypochlorite serves to sterilize the water by killing pathogenic bacteria and microorganisms.

Calcium hypochlorite (

Ca(ClO)_2

) releases hypochlorous acid and chlorine when added to water. This powerful oxidizing environment kills bacteria, viruses, and other disease-causing pathogens, fulfilling the sterilization stage of water treatment.

Step-by-Step Solution

1

Identify the chemical reagent and its functional group

Calcium hypochlorite,

Ca(ClO)_2

, acts as a chlorine-releasing disinfecting agent.

Bleaching powder hydrolyzes in water to release active chlorine ions.

2

Distinguish municipal/domestic water treatment roles

Disinfection/sterilization kills microorganisms, whereas coagulation removes turbidity and precipitation removes hardness.

Each water treatment reagent plays a specific, distinct chemical role.

Key Concept

Chemical Sterilization in Water Purification

Estimated Time:1m 0s

Question 34Question →

In a municipal water treatment plant, raw river water undergoes several sequential processing stages to ensure it is safe for domestic consumption. Arrange the following key stages of municipal water purification in the correct chronological order from first to last.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence of municipal water treatment stages from start to finish is: Screening to remove large floating debris, followed by Coagulation using alum to clump fine particles, then Sand filtration to remove tiny remaining suspended solids, and finally Chlorination to kill disease-causing germs.

The correct sequence follows the standard municipal waterworks workflow: physical removal of large debris (Screening)

\rightarrow

chemical aggregation of clay suspensions (Coagulation)

\rightarrow

mechanical straining of micro-solids (Filtration)

\rightarrow

chemical destruction of disease-causing bacteria (Chlorination).

Step-by-Step Solution

1

Identify the primary intake step.

Screening is the initial physical process to filter out large objects like leaves and sticks.

Large debris must be removed first to protect pump machinery and piping.

2

Identify the chemical clumping step.

Coagulation involves adding chemical coagulants such as alum to bind fine suspended clay particles into larger flocs.

Fine particles will not settle or filter easily without prior coagulation.

3

Identify the physical clarification step.

Filtration passes water through sand and gravel layers to remove remaining micro-suspended matter.

Water must be visually clear before chemical disinfection so pathogens are fully exposed to chlorine.

4

Identify the final disinfection step.

Chlorination is carried out last to kill bacteria and ensure biological safety for drinking.

Adding chlorine last ensures residual disinfection in the municipal piping network.

Key Concept

Sequential Stages of Municipal Water Purification

Question 35Question →

A river water sample collected for a township supply system is found to contain fine suspended clay particles, objectionable taste caused by decomposing organic matter, and pathogenic bacteria, but has negligible dissolved calcium and magnesium salts. Which combination of chemical reagents must be utilized during the treatment process to render this water clean, odorless, and biologically safe?

Potash alum, activated charcoal, and chlorine

Slaked lime, washing soda, and chlorine

Potash alum, calcium chloride, and slaked lime

Sodium carbonate, activated charcoal, and boiling water

Show answer & explanation

Answer: Potash alum, activated charcoal, and chlorine

Answer

Potash alum, activated charcoal, and chlorine form the correct combination of reagents for purifying soft, turbid, contaminated surface water for public distribution.

The correct answer combines potash alum, activated charcoal, and chlorine because each chemical directly addresses one of the identified raw water issues: potash alum coagulates colloidal clay particles into larger flocs during sedimentation; activated charcoal adsorbs dissolved organic matter causing unpleasant tastes and smells; and chlorine kills disease-causing bacteria during final disinfection.

Step-by-Step Solution

1

Analyze the specific impurities present in the raw water sample

Impurities identified: fine suspended colloidal clay (requires coagulation), organic taste/odor compounds (requires adsorption), and pathogenic microbes (requires disinfection/sterilization). Hardness minerals are absent.

Different water treatment chemicals address distinct types of physical, chemical, and biological contamination.

2

Select the appropriate chemical reagent for colloidal suspension removal

Potash alum (

KAl(SO_4)_2 \cdot 12H_2O

) provides

Al^{3+}

ions to neutralize negative charges on clay particles, causing coagulation into settleable flocs.

Physical filtration alone cannot remove extremely fine colloidal clay without prior chemical coagulation.

3

Select the reagent for taste and odor removal

Activated charcoal (carbon) adsorbs volatile organic compounds responsible for objectionable smells and tastes.

Porous activated carbon has a very large surface area optimized for organic molecule adsorption.

4

Select the chemical required for biological safety

Chlorine (or chlorine compounds) acts as a powerful oxidizing disinfectant to destroy disease-causing bacteria.

Water for public distribution must undergo chemical sterilization to eliminate waterborne pathogens.

Key Concept

Municipal water treatment requires distinct functional reagents: coagulants (potash alum) for suspended particles, adsorbents (activated carbon) for taste/odor, and disinfectants (chlorine) for sterilization. Softening reagents (slaked lime/washing soda) are only needed when hardness ions (

Ca^{2+}

,

Mg^{2+}

) are present.

Question 36Question →

Three unlabelled liquid mixtures, $P$ , $Q$ , and $R$ , are tested in a laboratory. Mixture $P$ passes completely through both filter paper and semi-permeable membranes without scattering light. Mixture $Q$ passes through filter paper but is retained by semi-permeable membranes and scatters a narrow beam of light. Mixture $R$ leaves a solid residue on filter paper upon gravity filtration. Which mixture will undergo precipitation upon the addition of an electrolyte, and what physical property accounts for this behavior?

Mixture

P

, because its dissolved solute particles adsorb electrolyte ions to form insoluble precipitates.

Mixture

Q

, because its dispersed particles carry surface charges and have diameters between

1\text{ nm}

and

100\text{ nm}

.

Mixture

R

, because Brownian motion maintains its suspension until electrolyte ions increase its density.

Mixture

Q

, because its solute molecules are smaller than

1\text{ nm}

and easily aggregate into coarse solids.

Show answer & explanation

Answer: Mixture $Q$ , because its dispersed particles carry surface charges and have diameters between $1\text{ nm}$ and $100\text{ nm}$ .

Answer

Mixture

Q

, because its dispersed particles carry surface charges and have diameters between

1\text{ nm}

and

100\text{ nm}

.

The experimental observations identify Mixture

Q

as a colloid because it passes through ordinary filter paper, is retained by a semi-permeable membrane, and displays light scattering (Tyndall effect). Colloidal particles range in size between

1\text{ nm}

and

100\text{ nm}

and carry surface electrical charges. Addition of an electrolyte neutralizes these charges, leading to coagulation.

Step-by-Step Solution

1

Analyze the filtration and optical properties of each mixture to classify them.

Mixture

P

is a true solution (particle size

< 1\text{ nm}

). Mixture

Q

is a colloidal system (particle size

1\text{--}100\text{ nm}

, exhibits Tyndall scattering, retained by semi-permeable membrane). Mixture

R

is a suspension (particle size

> 100\text{ nm}

, retained by filter paper).

Differentiation among true solutions, colloids, and suspensions is based on particle size and membrane permeability.

2

Determine which system undergoes coagulation upon adding an electrolyte.

Colloidal particles (Mixture

Q

) carry electric charges that stabilize them. Adding an electrolyte supplies ions of opposite charge, neutralizing the colloidal particles and causing them to coagulate/precipitate.

True solutions do not coagulate on electrolyte addition, and suspensions settle naturally by gravity.

Key Concept

Classification and properties of true solutions, colloidal systems, and suspensions

Question 37Question →

A liquid pharmaceutical preparation shows a distinct illuminated path when a beam of light passes through it in a dark room (Tyndall effect) and shows no settling of particles upon standing for several weeks. However, when poured through standard filter paper, it leaves no residue. Which of the following correctly classifies this liquid system and explains its observed behavior?

It is a true solution because its solute particles are fully dissolved, homogeneous, and smaller than

1\text{ nm}

.

It is a colloidal system because its dispersed particle diameters (

1\text{ nm}

to

100\text{ nm}

) are large enough to scatter light but small enough to pass through standard filter paper pores.

It is a suspension because its dispersed phase particles are kept suspended solely by thermal convection currents within the solvent.

It is a true solution because its particles undergo rapid sedimentation only when exposed to atmospheric air.

Show answer & explanation

Answer: It is a colloidal system because its dispersed particle diameters ( $1\text{ nm}$ to $100\text{ nm}$ ) are large enough to scatter light but small enough to pass through standard filter paper pores.

Answer

The liquid preparation is a colloidal system because its dispersed particles (diameters between

1\text{ nm}

and

100\text{ nm}

) are sufficiently large to cause the Tyndall effect by scattering light, yet small enough to pass unhindered through the pores of standard filter paper without settling.

The correct response identifies the mixture as a colloidal system. Colloidal particles range in diameter from

1\text{ nm}

to

100\text{ nm}

. This intermediate particle size makes them small enough to pass through the microscopic pores of standard laboratory filter paper while remaining large enough to scatter incident light beams (Tyndall effect) and stay suspended indefinitely via Brownian motion.

Step-by-Step Solution

1

Analyze the light scattering observation.

The observation of a visible light path (Tyndall effect) rules out true solutions, which have particle sizes smaller than

1\text{ nm}

and do not scatter light.

Particles must be comparable in size to the wavelength of light (

1\text{ nm}

to

100\text{ nm}

) to scatter light effectively.

2

Analyze the filtration and stability observations.

Passage through standard filter paper without leaving a residue and absence of settling rules out suspensions.

Suspension particles exceed

100\text{ nm}

(or

0.1\text{ }\mu\text{m}

), causing them to be trapped by filter paper pores and to settle under gravity over time.

3

Synthesize particle size characteristics to classify the system.

The mixture exhibits properties unique to a colloidal dispersion.

Colloids uniquely combine stability against gravity, ability to pass through ordinary filter paper, and distinct scattering of light.

Key Concept

Physical distinction between true solutions, colloids, and suspensions based on particle size, filtration capability, and light scattering (Tyndall effect).

Question 38Question →

During industrial water treatment, temporary hardness caused by dissolved calcium hydrogencarbonate is removed chemically via Clark's process by adding a controlled quantity of slaked lime, $\text{Ca(OH)}_2$ . Which balanced chemical equation correctly represents this water-softening reaction?

\text{Ca(HCO}_3)_2\text{(aq)} + \text{Ca(OH)}_2\text{(aq)} \rightarrow 2\text{CaCO}_3\text{(s)} + 2\text{H}_2\text{O(l)}

\text{Ca(HCO}_3)_2\text{(aq)} + \text{Ca(OH)}_2\text{(aq)} \rightarrow \text{CaCO}_3\text{(s)} + \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}

\text{CaSO}_4\text{(aq)} + \text{Ca(OH)}_2\text{(aq)} \rightarrow 2\text{CaCO}_3\text{(s)} + \text{H}_2\text{SO}_4\text{(aq)}

\text{Ca(HCO}_3)_2\text{(aq)} + \text{CaO(s)} \rightarrow \text{CaCO}_3\text{(s)} + \text{H}_2\text{O(l)}

Show answer & explanation

Answer: $\text{Ca(HCO}_3)_2\text{(aq)} + \text{Ca(OH)}_2\text{(aq)} \rightarrow 2\text{CaCO}_3\text{(s)} + 2\text{H}_2\text{O(l)}$

Answer

\text{Ca(HCO}_3)_2\text{(aq)} + \text{Ca(OH)}_2\text{(aq)} \rightarrow 2\text{CaCO}_3\text{(s)} + 2\text{H}_2\text{O(l)}

The equation

\text{Ca(HCO}_3)_2\text{(aq)} + \text{Ca(OH)}_2\text{(aq)} \rightarrow 2\text{CaCO}_3\text{(s)} + 2\text{H}_2\text{O(l)}

represents Clark's process. Slaked lime provides hydroxide ions that neutralize the hydrogencarbonate ions in temporary hard water, precipitating all calcium as insoluble calcium trioxocarbonate(IV), leaving soft water.

Step-by-Step Solution

1

Identify the chemical cause of temporary hardness in the water sample

Temporary hardness is caused by dissolved calcium hydrogencarbonate,

\text{Ca(HCO}_3)_2\text{(aq)}

.

Hydrogencarbonate salts of calcium and magnesium decompose or precipitate when treated with appropriate alkaline agents.

2

Apply the principles of Clark's process for temporary hardness removal

Clark's process involves adding a calculated amount of slaked lime,

\text{Ca(OH)}_2\text{(aq)}

, which supplies hydroxide ions to convert hydrogencarbonate ions into insoluble trioxocarbonate(IV) ions.

Adding excess lime would re-introduce calcium ions and cause artificial hardness, so a stoichiometric amount is used.

3

Balance the precipitation chemical equation

\text{Ca(HCO}_3)_2\text{(aq)} + \text{Ca(OH)}_2\text{(aq)} \rightarrow 2\text{CaCO}_3\text{(s)} + 2\text{H}_2\text{O(l)}

One mole of calcium hydrogencarbonate reacts with one mole of calcium hydroxide to produce two moles of insoluble calcium trioxocarbonate(IV) and two moles of water.

Key Concept

Clark's process (slaked lime precipitation) for removal of temporary water hardness

Question 39Question →

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.

Drag items to arrange them in the correct order

Show answer & explanation

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 40Question →

Match each chemical species or substance involved in water chemistry with its primary role or behavior regarding water hardness.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Magnesium hydrogencarbonate causes temporary hardness that decomposes on boiling; magnesium tetraoxosulfate(VI) causes permanent hardness unaffected by boiling; sodium trioxocarbonate(IV) removes all hardness via chemical precipitation; sodium permutit removes all hardness via ion exchange.

Magnesium hydrogencarbonate is a soluble hydrogencarbonate salt causing temporary hardness removed by boiling. Magnesium tetraoxosulfate(VI) is a soluble sulfate salt causing permanent hardness that cannot be removed by boiling. Sodium trioxocarbonate(IV) removes all hardness types by precipitating calcium/magnesium ions as insoluble carbonates. Permutit softened water by exchanging hardness cations for soluble sodium ions.

Step-by-Step Solution

1

Identify the cause of temporary hardness.

Magnesium hydrogencarbonate,

\text{Mg(HCO}_3)_2

, dissolves in water to cause temporary hardness. Upon boiling, hydrogencarbonate ions decompose to insoluble carbonate precipitates:

\text{Mg(HCO}_3)_2(aq) \rightarrow \text{MgCO}_3(s) + \text{H}_2\text{O}(l) + \text{CO}_2(g)

.

Temporary hardness is specifically caused by hydrogen carbonate salts of calcium and magnesium, which are thermally unstable.

2

Identify the cause of permanent hardness.

Magnesium tetraoxosulfate(VI),

\text{MgSO}_4

, causes permanent hardness because sulfate salts do not undergo thermal decomposition upon boiling.

Permanent hardness is caused by soluble tetraoxosulfate(VI) or chloride salts of calcium and magnesium.

3

Analyze the action of washing soda (

\text{Na}_2\text{CO}_3

).

Adding washing soda introduces

\text{CO}_3^{2-}

ions, which combine with dissolved

\text{Mg}^{2+}

or

\text{Ca}^{2+}

ions to precipitate them as

\text{MgCO}_3

or

\text{CaCO}_3

.

Precipitation of divalent metallic cations as insoluble carbonates removes both temporary and permanent hardness.

4

Analyze the action of sodium permutit (zeolite).

Permutit acts as an ion exchanger:

\text{Na}_2\text{Z}(s) + \text{Mg}^{2+}(aq) \rightarrow \text{MgZ}(s) + 2\text{Na}^+(aq)

.

Hardness-causing divalent cations are bound to the zeolite matrix while harmless sodium ions are released into solution.

Key Concept

Classification of water hardness causes (hydrogencarbonate vs sulfate salts) and chemical removal mechanisms (boiling, precipitation, ion exchange).

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