Colloids, Emulsion, Gel and Their properties with application Questions in English

(A) Sodium stearate is an anionic surfactant. In water,it forms spherical micelles. The hydrophobic hydrocarbon tail,$CH_{3}(CH_{2})_{16}-$,points towards the center of the sphere to avoid contact with water,while the hydrophilic head,$-COO^{-}$,points towards the surface to interact with water.

(A) The classification of colloidal systems is as follows:
$a$. Cheese is a gel,which is a dispersion of liquid in solid $(iv)$.
$b$. Pumice stone is a solid foam,which is a dispersion of gas in solid $(iii)$.
$c$. Hair cream is an emulsion,which is a dispersion of liquid in liquid $(i)$.
$d$. Cloud is an aerosol,which is a dispersion of liquid in gas $(ii)$.
Therefore,the correct matching is $a-iv, b-iii, c-i, d-ii$.

(B) According to the Hardy-Schulze rule,the flocculating power of an ion increases with the increase in the magnitude of the charge on the ion.
For a positive sol,the coagulating ions must be negative.
The correct order of flocculating power for a positive sol is $[Fe(CN)_{6}]^{4-} > PO_{4}^{3-} > SO_{4}^{2-} > Cl^{-}$.
Therefore,the statement in option $B$ is incorrect as it presents the reverse order.

(B) According to the Hardy-Schulze rule,the coagulating power of an ion depends on its valency. The greater the valency of the flocculating ion,the greater is its power to cause precipitation.
Statement $I$ is correct: For a negative sol,the coagulating power of cations follows the order $Al^{3+} > Ba^{2+} > Na^{+}$ because the valency of the cation is directly proportional to the coagulating power.
Statement $II$ is incorrect: For a positive sol,the coagulating power depends on the valency of the anion. The order should be based on the valency of the anions $(Cl^- < SO_{4}^{2-} < PO_{4}^{3-})$. Therefore,the correct order of flocculating power for the salts should be $NaCl < Na_{2}SO_{4} < Na_{3}PO_{4}$.

(C) Statement $I$ is correct because emulsions are thermodynamically unstable systems,and oil droplets tend to coalesce and separate into two layers over time.
Statement $II$ is incorrect because adding an excess of electrolyte to an emulsion typically causes coagulation or breaking of the emulsion,not stabilization. Emulsions are stabilized by adding emulsifying agents (emulsifiers) like soaps,proteins,or gums,which form a protective film around the dispersed droplets.

(B) For the Tyndall effect to be observed,two main conditions must be met:
$1$. The diameter of the dispersed particles should not be much smaller than the wavelength of the light used.
$2$. The refractive indices of the dispersed phase and the dispersion medium must differ greatly in magnitude.
Therefore,the statement that the diameter of the dispersed particles is much smaller than the wavelength of the light used is incorrect.

(C) The cleaning action of soap is based on the formation of micelles.
Micelles are aggregates of soap molecules that form only when the concentration of soap in water reaches or exceeds the $Critical \text{ } Micelle \text{ } Concentration$ $(CMC)$.
If the concentration of soap is below the $CMC$,the soap molecules remain as individual ions or molecules and cannot effectively trap grease or dirt particles.
Therefore,using very little soap prevents micelle formation,rendering the cleaning process ineffective.

(A) . Lyophilic colloids act as protective colloids because they form a protective layer around lyophobic particles.
$B$. Emulsions are colloidal systems where both the dispersed phase and the dispersion medium are liquids (liquid-liquid colloid).
$C$. The reaction of $FeCl_3$ with hot water leads to the formation of a positively charged sol of hydrated ferric oxide $(Fe_2O_3 \cdot xH_2O)$.
$D$. The reaction of $FeCl_3$ with $NaOH$ (in excess) leads to the formation of a negatively charged sol due to the preferential adsorption of $OH^-$ ions on the surface of the precipitate.
Therefore,the correct matching is $A-II, B-I, C-IV, D-III$.

(C) The correct matches are as follows:
$A$. Negatively charged sol: $CdS$ sol $(II)$
$B$. Macromolecular colloid: Starch $(III)$
$C$. Positively charged sol: $Fe_{2}O_{3} \cdot xH_{2}O$ $(I)$
$D$. Cheese: $A$ gel $(IV)$
Therefore,the correct sequence is $A-II, B-III, C-I, D-IV$.

(C) The potential difference between the fixed layer and the diffused layer of charges on the surface of a colloidal particle is known as the Zeta potential or electrokinetic potential.

(A) The coagulating value is defined as the minimum concentration of an electrolyte in $m\,mol \, L^{-1}$ required to cause coagulation of a sol in $2 \, hours$.
$1$. Calculate the molar mass of $NaCl$: $M = 23.0 + 35.5 = 58.5 \, g \, mol^{-1}$.
$2$. Calculate the molarity of the given $42.12 \% (w/v)$ solution:
$42.12 \% (w/v)$ means $42.12 \, g$ of $NaCl$ in $100 \, mL$ of solution.
Concentration in $g \, L^{-1} = 42.12 \times 10 = 421.2 \, g \, L^{-1}$.
Concentration in $mol \, L^{-1} = \frac{421.2}{58.5} = 7.2 \, mol \, L^{-1} = 7200 \, m\,mol \, L^{-1}$.
$3$. The question asks for the coagulating value,which is the concentration required for coagulation. Given the data provided is insufficient to relate the time $(10 \, hours)$ to the standard definition $(2 \, hours)$ without a specific rate law or empirical relationship,the provided options suggest a calculation based on the concentration itself. However,based on standard textbook problems of this type,the answer is derived as $7200 \, m\,mol \, L^{-1}$. Since this is not an option,and the provided solution was 'Data insufficient',we maintain this assessment.

(D) Colloidal particles carry an electric charge,which is the primary factor for their stability.
Since all particles in a given colloidal solution carry the same type of charge,they repel each other,preventing them from coming close enough to aggregate or precipitate.
Therefore,the presence of equal and similar charges on colloidal particles provides stability to the colloidal solution.

(C) Micelle formation is a spontaneous process driven by the hydrophobic effect.
During micelle formation,the hydrophobic tails of surfactant molecules aggregate,which leads to the release of water molecules previously ordered around the tails into the bulk solvent.
This release of water molecules results in a net increase in the entropy of the system,so $\Delta S > 0$.
Additionally,the process is endothermic,meaning $\Delta H > 0$.
Therefore,statements $B$ and $C$ are correct.

(A) The colour of a colloidal solution depends on the size and shape of the dispersed particles.
As the particle size of gold sol increases,the wavelength of light scattered by the particles changes,which results in a change in the observed colour (from red to purple to blue and finally to golden).
Therefore,Assertion $A$ is true.
Reason $R$ correctly explains that the colour of a colloidal solution depends on the wavelength of light scattered by the dispersed particles,which is a direct consequence of the particle size.
Thus,both $A$ and $R$ are true,and $R$ is the correct explanation of $A$.

(C) Assertion $(A)$ is correct because dissolved substances (crystalloids) can pass through parchment paper,while colloidal particles cannot,allowing for purification via dialysis.
Reason $(R)$ is incorrect because particles in a true solution (crystalloids) can easily pass through parchment paper,whereas colloidal particles are too large to pass through it.
Therefore,$(A)$ is correct but $(R)$ is incorrect.

(A) Albumin is a protein,which is a macromolecule that forms a lyophilic sol when dispersed in a suitable solvent like water or dilute salt solutions.
Since albumin has a high affinity for the dispersion medium,it forms a lyophilic sol.

(D) The coagulation of a lyophobic sol involves the neutralization of the charge on the colloidal particles.
$(i)$ Addition of an electrolyte provides ions that neutralize the charge.
$(ii)$ Electrophoresis involves moving particles to an oppositely charged electrode,leading to discharge and coagulation.
$(iv)$ Addition of an oppositely charged sol leads to mutual coagulation.
$(iii)$ Addition of a protective colloid is used to stabilize lyophobic sols,not to coagulate them.
Therefore,the correct methods are $(i), (ii)$ and $(iv)$.

(C) Statement $I:$ Colloidal particles are aggregates of molecules. Due to their large size,the number of particles in a given mass of colloid is much smaller than the number of particles in the same mass of a true solution. Since colligative properties depend on the number of particles,they are of a smaller order for colloids compared to true solutions. This statement is true.
Statement $II:$ According to the Helmholtz double layer theory,the potential difference between the fixed layer and the diffused layer of opposite charges is known as the electrokinetic potential or zeta potential. This statement is true.

(A) When $FeCl_{3}$ is added to $NaOH$ in excess,a negatively charged ferric hydroxide sol $[Fe(OH)_{3} \cdot OH^-]$ is formed.
According to the Hardy-Schulze rule,the coagulating power of an ion increases with the increase in the magnitude of its charge.
For a negatively charged sol,the coagulating power of cations follows the order: $Al^{3+} > Ca^{2+} > Na^+$.
Among the given options,$AlCl_{3}$ provides $Al^{3+}$ ions,which have the highest charge and thus the highest coagulating power,leading to the fastest rate of coagulation.

(B) . Osmosis: Solvent molecules pass through a semi-permeable membrane towards the solution side $(III)$.
$B$. Reverse osmosis: Solvent molecules pass through a semi-permeable membrane towards the solvent side $(I)$.
$C$. Electro-osmosis: The dispersion medium moves in an electric field $(IV)$.
$D$. Electrophoresis: Movement of charged colloidal particles under the influence of an applied electric potential towards oppositely charged electrodes $(II)$.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(A) In a non-polar solvent,the surfactant molecules form reverse micelles. The polar heads,which are hydrophilic,aggregate in the center to minimize contact with the non-polar solvent,while the non-polar tails,which are lipophilic,extend outwards into the non-polar solvent. This structure is represented by option $A$.

(C) colloidal system with a liquid dispersion medium is known as an emulsion (if the dispersed phase is liquid) or a sol (if the dispersed phase is solid).
Let us classify the given systems:
$1$. Gem stones: Solid sol (Solid in Solid)
$2$. Paints: Sol (Solid in Liquid)
$3$. Smoke: Aerosol (Solid in Gas)
$4$. Cheese: Gel (Liquid in Solid)
$5$. Milk: Emulsion (Liquid in Liquid)
$6$. Hair cream: Emulsion (Liquid in Liquid)
$7$. Insecticide sprays: Aerosol (Liquid in Gas)
$8$. Froth: Foam (Gas in Liquid)
$9$. Soap lather: Foam (Gas in Liquid)
The systems with liquid as the dispersion medium are: Paints,Milk,Hair cream,Froth,and Soap lather.
Total count = $5$.

(A) The coagulating power of an electrolyte is inversely proportional to its coagulating value.
$\text{Coagulating Power} (C.P.) = \frac{1}{\text{Coagulating Value} (C.V.)}$
Given:
$(C.V.)_{AlCl_3} = 0.09$
$(C.V.)_{NaCl} = 50.04$
Ratio of coagulating powers:
$\frac{(C.P.)_{AlCl_3}}{(C.P.)_{NaCl}} = \frac{(C.V.)_{NaCl}}{(C.V.)_{AlCl_3}}$
$\frac{(C.P.)_{AlCl_3}}{(C.P.)_{NaCl}} = \frac{50.04}{0.09} = 556$
Therefore,the coagulating power of $AlCl_3$ is $556$ times the coagulating power of $NaCl$. Thus,$x = 556$.

(A) Adsorption is a spontaneous process accompanied by a decrease in surface energy,which releases heat,making it an exothermic process $(\Delta H_{\text{ads}} < 0)$.
Micelle formation is a spontaneous process driven by the increase in entropy of the system,but the process itself is endothermic $(\Delta H_{\text{mic}} > 0)$.

(C) Blood is a colloidal system which is negatively charged.
According to the Hardy-Schulze rule,the coagulating power of an ion is directly proportional to its valency.
$Fe^{3+}$ ions,being highly positively charged,effectively neutralize the negative charge on the blood colloids,leading to coagulation and stopping the bleeding.

(D) Lyophilic sols act as protective colloids for lyophobic sols. When a lyophilic sol is added to a lyophobic sol,the lyophilic particles form a protective film or layer around the lyophobic particles,thereby preventing their coagulation by electrolytes.

(C) The Tyndall effect is observed when the diameter of the dispersed particles is not much smaller than the wavelength of the light used.
When the particle size is large enough (comparable to the wavelength of light),the light hits the particles and scatters in all directions,making the path of the light visible.
Therefore,both Assertion $A$ and Reason $R$ are correct,and $R$ provides the correct explanation for $A$.

(A) The coagulating value is defined as the minimum concentration of an electrolyte in millimoles per liter $(mmol/L)$ required to cause coagulation of a sol in $2$ hours.
Given:
Volume of $NaCl$ solution = $20 \ mL$
Molarity of $NaCl$ solution = $0.5 \ M$
Volume of $As_2S_3$ sol = $200 \ mL$
Number of millimoles of $NaCl$ = $Molarity \times Volume \ (in \ mL) = 0.5 \times 20 = 10 \ mmol$.
These $10 \ mmol$ of $NaCl$ are required for $200 \ mL$ of sol.
For $1000 \ mL$ $(1 \ L)$ of sol,the required millimoles = $(10 / 200) \times 1000 = 50 \ mmol/L$.
Therefore,the coagulating value of $NaCl$ is $50$.

(D) Pumice stone is a type of colloid known as a solid sol.
In this system,the dispersed phase is $Gas$ and the dispersion medium is $Solid$.

(B) $Statement-1$ is True: Micelles are indeed formed by surfactant molecules only above the critical micellar concentration $(CMC)$ and the Kraft temperature $(T_k)$.
$Statement-2$ is True: At the $CMC$,individual surfactant ions aggregate to form large,bulky micellar particles. These large particles have lower mobility compared to individual ions,which leads to a sharp decrease in the molar conductivity of the solution.
Conclusion: While both statements are true,the decrease in conductivity is a consequence of micelle formation,not the reason why micelles form. Therefore,$Statement-2$ is not the correct explanation for $Statement-1$.

(A) The Critical Micelle Concentration $(CMC)$ of a surfactant decreases as the length of the hydrophobic hydrocarbon chain increases.
This is because a longer hydrophobic chain makes the surfactant molecule more hydrophobic,which decreases its solubility in water and increases its tendency to aggregate into micelles to minimize contact with water.
Comparing the chain lengths:
Option $A$: $16$ carbons
Option $B$: $12$ carbons
Option $C$: $7$ carbons
Option $D$: $12$ carbons
Since option $A$ has the longest hydrocarbon chain ($16$ carbons),it will have the lowest $CMC$ value.

(C) According to the Hardy-Schulze rule,the coagulating power of an electrolyte depends on the valency of the active ion (the ion carrying a charge opposite to that of the colloidal particles).
$Sb_2S_3$ sol is a negatively charged sol. Therefore,it is coagulated by the positively charged ions of the added electrolytes.
The positively charged ions in the given electrolytes are:
$Na^+$ (from $Na_2SO_4$),
$Ca^{2+}$ (from $CaCl_2$),
$Al^{3+}$ (from $Al_2(SO_4)_3$),
$NH_4^+$ (from $NH_4Cl$).
According to the Hardy-Schulze rule,the coagulating power increases with the increase in the magnitude of the charge on the ion. The order of coagulating power is $Al^{3+} > Ca^{2+} > Na^+ = NH_4^+$.
Since $Al^{3+}$ has the highest charge $(+3)$,$Al_2(SO_4)_3$ is the most effective coagulating agent for $Sb_2S_3$ sol.
Hence,option $C$ is correct.

(D) The correct option is $D$ $(I: CH_3OH, II: KCl, III: CH_3(CH_2)_{11}OSO_3^{-}Na^{+})$.
$1$. $CH_3OH$ is a solute that decreases the surface tension of water as it disrupts the hydrogen bonding network of water,leading to a gradual decrease in surface tension with increasing concentration (Sketch $I$).
$2$. $KCl$ is an inorganic electrolyte. When added to water,it increases the intermolecular forces of attraction,leading to a slight increase in surface tension with increasing concentration (Sketch $II$).
$3$. $CH_3(CH_2)_{11}OSO_3^{-}Na^{+}$ is a surfactant (soap/detergent). It significantly lowers the surface tension of water until the critical micelle concentration $(CMC)$ is reached,after which the surface tension remains nearly constant or increases slightly (Sketch $III$).

(A) $(I)$ Lyophobic colloids have no affinity between the dispersed phase and the dispersion medium,so they cannot be prepared by simple mixing; special methods are required.
$(II)$ Emulsions are defined as colloidal systems where both the dispersed phase and the dispersion medium are liquids.
$(III)$ Micelles are formed only above a specific concentration called the Critical Micelle Concentration $(CMC)$ and above a specific temperature called the Kraft temperature $(T_k)$.
$(IV)$ The Tyndall effect occurs only when there is a significant difference in the refractive indices of the dispersed phase and the dispersion medium.
Therefore,statements $(I)$ and $(II)$ are correct.

(C) Lyophobic colloids are inherently unstable and are stabilized primarily by the presence of an electric charge on their surface.
$(A)$ The preferential adsorption of common ions from the solution creates a charged surface,which leads to electrostatic repulsion between particles,preventing aggregation.
$(D)$ The potential difference between the fixed layer and the diffused layer (known as the Zeta potential) provides the necessary repulsive force to maintain stability.
Therefore,both $(A)$ and $(D)$ are correct reasons for the stability of lyophobic colloidal particles.

(A) Sodium stearate is a surfactant that behaves as a strong electrolyte at low concentrations.
As the concentration increases,the molar conductivity $\left(\Lambda_m\right)$ decreases due to the interionic attractions,similar to other strong electrolytes.
However,once the concentration reaches the Critical Micelle Concentration $(CMC)$,the individual ions aggregate to form large,bulky micelles.
These micelles have a much lower mobility compared to individual ions,leading to a sharp decrease in the rate of change of molar conductivity with respect to $\sqrt{c}$ after the $CMC$ point.
This results in a plot where the slope changes significantly at the $CMC$,which is best represented by the graph in option $A$.

(D) The process of precipitating a colloidal sol by an electrolyte is called coagulation or flocculation,not peptization. Peptization is the process of converting a precipitate into a colloidal sol. Hence,$(A)$ is false.
$(B)$ Colloidal particles have very high molar masses,so for a given mass concentration,the molar concentration is very low. Consequently,the depression in freezing point $(\Delta T_f)$ is very small for colloidal solutions compared to true solutions. Therefore,the freezing point of a colloidal solution is higher than that of a true solution. Hence,$(B)$ is true.
$(C)$ Surfactants form micelles only above the critical micelle concentration $(CMC)$. The $CMC$ is a characteristic property that depends on the nature of the surfactant and the temperature. Hence,$(C)$ is true.
$(D)$ Micelles are associated colloids,whereas macromolecular colloids are formed by large molecules like proteins or polymers. Hence,$(D)$ is false.

(B) Statement $(i)$ is correct: Peptization is the process of converting a freshly prepared precipitate into a colloidal sol by adding a suitable electrolyte.
Statement $(ii)$ is incorrect: Lyophilic sols are generally organic in nature (e.g.,starch,proteins,gums).
Statement $(iii)$ is correct: Lyophilic sols are highly solvated because the dispersed phase particles have a strong affinity for the dispersion medium.
Statement $(iv)$ is incorrect: Lyophobic sols are irreversible in nature and require special methods for preparation,whereas lyophilic sols are reversible.
Therefore,statements $(i)$ and $(iii)$ are correct.

(C) The process of converting a freshly prepared precipitate into a colloidal sol by adding a suitable electrolyte is known as peptization.
In this process,the precipitate particles adsorb ions from the electrolyte,which causes them to break down and disperse into the colloidal state.

(B) Silver nanoparticles $(AgNPs)$ are widely known for their potent antimicrobial properties. They are frequently used to coat filter materials in water purification systems because they effectively kill bacteria and prevent biofilm formation,acting as an efficient bacterial disinfectant.

(D) The size ranges are as follows:
- Water and Glucose are at the molecular scale,typically ranging from $0.1$ to $1 \ nm$.
- Viruses typically range from $20$ to $300 \ nm$.
- Bacteria are much larger,typically ranging from $0.2$ to $10 \ \mu m$ ($200$ to $10,000 \ nm$).
Therefore,among the given options,bacteria are the largest.

(C) Zinc oxide $(ZnO)$ and Titanium dioxide $(TiO_2)$ nanoparticles are commonly used in sunscreen lotions.
These compounds protect the skin against harmful $UV$ (ultraviolet) rays by absorbing or reflecting the radiation,thereby preventing skin damage.

(A) Nanowires are one-dimensional nanostructures in which two dimensions are in the nanoscale $(< 100 \ nm)$.
Quantum dots are zero-dimensional structures in which all three dimensions are in the nanoscale $(< 100 \ nm)$.
Microcapsules are generally larger than the nanoscale range.
Nanorings are typically considered zero-dimensional or specific nanostructures depending on their geometry,but nanowires are the standard example of having two dimensions in the nanoscale.

(A) The Scanning Electron Microscope $(SEM)$ is the instrument used to examine the surface structure of materials at high magnification.
It works by scanning a focused beam of electrons across the surface of a sample.
The electrons interact with the atoms on the surface,producing signals that are used to create an image of the surface's topography,morphology,and composition.

(A) One-dimensional nanostructures are materials that have two dimensions in the nanometer range and one dimension that is much larger, such as $Nanowires$.
$Nanoparticles$ are zero-dimensional, $Thin films$ are two-dimensional, and $Quantum dots$ are zero-dimensional nanostructures.

(C) Nanomaterials are classified based on their dimensions.
$0$-dimensional nanostructures have all three dimensions in the nanoscale (e.g.,nanoparticles).
$1$-dimensional nanostructures have two dimensions in the nanoscale (e.g.,nanowires,nanotubes).
$2$-dimensional nanostructures have one dimension in the nanoscale (e.g.,thin films,coatings).
Therefore,thin films are an example of a $2$-dimensional nanostructure.

(B) $(1)$ Nanorods: $1D$ structure: Elongated in one dimension (length),confined in the other two (width and height).
$(2)$ Nanoparticles: $0D$ structure: No elongation in any dimension,confined in all directions (usually spherical).
$(3)$ Thin films: $2D$ structure: Thin in one dimension (thickness) but extends in the other two (length and width).
$(4)$ Fibres: $1D$ structure: Elongated in one dimension (length),confined in the other two.

(B) One-dimensional nanomaterial: In one-dimensional nanomaterial,one dimension is outside the nanoscale range,while two dimensions are within the nanoscale range. This can be visualized as a wire. Examples of $1D$ nanomaterials include nanotubes,nanorods,and nanowires.

(A) $UV-visible$ spectroscopy is commonly used for the preliminary confirmation of the synthesis of nanoparticles,especially metal nanoparticles,due to the phenomenon of Surface Plasmon Resonance $(SPR)$.
$SPR$ causes a characteristic absorption peak in the $UV-visible$ region,which serves as a quick and simple indicator of nanoparticle formation.