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

(A) The classification of colloidal systems based on the dispersed phase and dispersion medium is as follows:
$(A)$ Solid dispersed in liquid is called a Sol $(IV)$.
$(B)$ Liquid dispersed in liquid is called an Emulsion $(I)$.
$(C)$ Gas dispersed in liquid is called a Foam $(II)$.
$(D)$ Liquid dispersed in solid is called a Gel $(III)$.
Therefore, the correct matching is $(A-IV, B-I, C-II, D-III)$.

(B) In jelly, the dispersed phase is $liquid$ and the dispersion medium is $solid$.
Thus, it is a colloidal solution of $liquid$ in $solid$.

(C) The colloidal solution of gold is prepared by dispersing solid gold particles in a liquid medium (usually water).
Since gold is a metal, it does not have an affinity for the dispersion medium.
Therefore, it cannot be prepared by simple mixing and requires special methods.
Such colloids are classified as lyophobic (solvent-hating) colloids.

(C) lyophobic colloid is a solvent-hating colloid where the dispersed phase has little or no affinity for the dispersion medium.
$Gold \ sol$ is a classic example of a lyophobic sol because gold particles have very little affinity for the water dispersion medium, making them unstable and easily coagulated.

(B) Sodium lauryl sulphate $(CH_3(CH_2)_{11}SO_4^-Na^+)$ is an anionic surfactant.
It consists of a long hydrophobic hydrocarbon chain and a hydrophilic ionic head.
In water, these molecules aggregate to form ionic micelles above the Critical Micelle Concentration $(CMC)$.

(A) Bredig's Arc (Electrical disintegration) method is used to prepare colloidal sols of metals such as gold, silver, platinum, etc.
In this method, an electric arc is struck between electrodes of the metal immersed in the dispersion medium.
The intense heat produced vaporizes the metal, which then condenses to form particles of the colloidal size.
Metal sols typically acquire a negative charge due to the preferential adsorption of anions from the dispersion medium.

(D) Antimony sulphide $(Sb_2S_3)$ sol is a negatively charged sol.
According to the Hardy-Schulze rule, the coagulating power of an electrolyte depends on the valency of the active ion (the ion with a charge opposite to that of the colloidal particles).
For a negatively charged sol, the coagulating power increases with the increase in the valency of the cation.
The cations in the given options are:
$A) Na^+$ (valency $1$)
$B) Ca^{2+}$ (valency $2$)
$C) NH_4^+$ (valency $1$)
$D) Al^{3+}$ (valency $3$)
Since $Al^{3+}$ has the highest valency $(3)$, it is the most effective coagulating agent for the negatively charged $Sb_2S_3$ sol.

(C) The flocculating value is defined as the minimum concentration of an electrolyte in millimoles per liter required to cause the coagulation of a sol.
Number of millimoles of $NaCl = \text{Molarity} \times \text{Volume in mL} = 0.5 \ M \times 10 \ mL = 5 \ \text{millimoles}$.
Since this amount is required to coagulate $1 \ L$ of the $Sb_2S_3$ sol, the flocculating value is $5 \ \text{millimoles/L}$.

(D) The Tyndall effect is the scattering of light by colloidal particles.
Statement $(I)$ is correct: It is a standard method to distinguish between a true solution (which does not show the effect) and a colloidal solution (which does).
Statement $(II)$ is correct: The scattering of light depends on the difference in refractive indices between the dispersed phase and the dispersion medium.
Statement $(III)$ is incorrect: The Tyndall effect is observed when the diameter of the dispersed particles is not much smaller than the wavelength of the light used; if particles are too small, they do not scatter light effectively.
Therefore, statements $(I)$ and $(II)$ are correct.

(C) According to the $Hardy-Schulze$ rule, the coagulating power of an ion is directly proportional to the magnitude of the charge on the ion.
$As_2S_3$ sol is a negatively charged colloid.
To coagulate a negatively charged sol, a positively charged ion is required.
The coagulating power of cations follows the order: $Al^{3+} > Ba^{2+} > Na^+$.
Among the given options, $FeCl_3$ provides $Fe^{3+}$ ions, which have the highest positive charge $(+3)$ compared to $Ca^{2+}$ $(+2)$ and $Na^+$ $(+1)$.
Therefore, $FeCl_3$ is the most effective coagulating agent.

(C) The Tyndall effect is observed in colloidal solutions where the particles are large enough to scatter light.
$A$ sugar solution is a true solution, not a colloidal solution, because the solute particles are of molecular size $(< 1 \ nm)$ and are too small to scatter light.
Therefore, it does not show the Tyndall effect.

(D) The flocculation value is defined as the minimum concentration of an electrolyte in $mmol \ L^{-1}$ required to cause the coagulation of a colloid.
Given: Volume of colloid $= 200 \ mL$, Mass of $HCl = 0.73 \ g$.
Molar mass of $HCl = 36.5 \ g \ mol^{-1}$.
Number of moles of $HCl = \frac{0.73 \ g}{36.5 \ g \ mol^{-1}} = 0.02 \ mol = 20 \ mmol$.
Since $200 \ mL$ of the colloid requires $20 \ mmol$ of $HCl$ for coagulation, we calculate the amount required for $1000 \ mL$ (or $1 \ L$):
Flocculation value $= \frac{20 \ mmol}{200 \ mL} \times 1000 \ mL = 100 \ mmol \ L^{-1}$.
Thus, the flocculation value is $100$.

(D) The movement of colloidal particles under the influence of an applied $DC$ electric field is called $Electrophoresis$.
When an electric potential is applied across two platinum electrodes dipping in a colloidal solution, the colloidal particles move towards one or the other electrode depending on the charge they carry.

(D) According to the $Hardy-Schulze$ law, the coagulating power of an electrolyte depends on the valency of the active ion (the ion with a charge opposite to that of the colloidal particles).
For negatively charged colloids, the coagulating power of cations follows the order of their valency: $Al^{3+} > Ba^{2+} = Mg^{2+} > Na^+$.
Since $BaCl_2$ and $MgCl_2$ both provide divalent cations ($Ba^{2+}$ and $Mg^{2+}$) and $NaCl$ provides a monovalent cation $(Na^+)$, the coagulating power is $MgCl_2 = BaCl_2 > NaCl$.

(B) $Sb_2S_3$ sol is a negatively charged sol.
According to the Hardy-Schulze rule, the greater the valency of the oppositely charged ion of the electrolyte added, the faster the coagulation.
The principle is based on electrostatic forces of attraction.
Greater the valency of the flocculating ion, greater is the coagulation power.
Comparing the valency of the cations:
$Al^{3+} > Ca^{2+} > Na^{+} = NH_4^{+}$
Therefore, $Al_2(SO_4)_3$ is the most effective coagulating agent because it provides the $Al^{3+}$ ion with the highest valency.

(D) $Fe(OH)_3$ is a positively charged sol. According to the Hardy-Schulze rule, the coagulating power of an electrolyte depends on the valency of the ion carrying a charge opposite to that of the colloidal particles.
Since $Fe(OH)_3$ is positively charged, it requires an anion for coagulation.
The given electrolytes dissociate as follows:
$Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}$
$Na_3[Fe(CN)_6] \rightarrow 3Na^+ + [Fe(CN)_6]^{3-}$
$Cr(OH)_3$ and $Al(OH)_3$ are insoluble hydroxides and do not act as effective coagulating agents in this context.
Comparing the valency of the anions, $[Fe(CN)_6]^{3-}$ has a valency of $3$, while $SO_4^{2-}$ has a valency of $2$.
According to the Hardy-Schulze rule, the higher the valency of the coagulating ion, the greater is its coagulating power.
Therefore, $[Fe(CN)_6]^{3-}$ is the most effective coagulating agent among the given options.
Flocculation value = $\frac{1}{\text{Coagulating power}}$. Since $[Fe(CN)_6]^{3-}$ has the highest coagulating power, it has the lowest flocculation value, making it the best precipitating agent.

(B) The protective power of a colloid is inversely proportional to its gold number.
Smaller the gold number, greater is the protective power.
Given gold numbers:
Gelatin: $5 \times 10^{-3}$
Haemoglobin: $5 \times 10^{-2}$
Sodium acetate: $7 \times 10^{-1}$
Comparing the values: $5 \times 10^{-3} < 5 \times 10^{-2} < 7 \times 10^{-1}$.
Therefore, the order of protective power is: Gelatin $>$ haemoglobin $>$ sodium acetate.

(D) The gold number is defined as the amount of protective colloid in milligrams that prevents the coagulation of $10 \ mL$ of a standard gold sol when $1 \ mL$ of $10 \% \ NaCl$ solution is added to it.
Given that $10^{-4} \ g$ of gelatin is required to protect $100 \ mL$ of gold sol.
Convert the mass of gelatin to milligrams: $10^{-4} \ g = 10^{-4} \times 1000 \ mg = 0.1 \ mg$.
Since $0.1 \ mg$ of gelatin protects $100 \ mL$ of gold sol, the amount of gelatin required to protect $10 \ mL$ of gold sol is:
$\frac{0.1 \ mg}{100 \ mL} \times 10 \ mL = 0.01 \ mg$.
Thus, the gold number of gelatin is $0.01$.

(D) In alkaline medium, $SnO_2$ reacts with $OH^-$ ions to form stannate ions, making the sol $[SnO_2] SnO_3^{2-}$, which is negatively charged.
In acidic medium, $SnO_2$ reacts with $H^+$ ions to form $Sn^{4+}$ ions on the surface, making the sol $[SnO_2] Sn^{4+}$, which is positively charged.
Therefore, the charges are negative and positive respectively.

(B) Silver sols are used as an eye lotion because they possess antiseptic properties and can help heal eye infections.
Hence, option $(B)$ is the correct answer.

(B) $Sb_2S_3$ is a negatively charged sol. According to the Hardy-Schulze rule, the coagulating power of an electrolyte depends on the valency of the ion carrying a charge opposite to that of the colloidal particles.
Since $Sb_2S_3$ is a negative sol, it requires a positive ion for coagulation.
The coagulating power increases with the increase in the magnitude of the charge on the cation.
The valencies of the cations are: $Na^+$ $(+1)$, $NH_4^+$ $(+1)$, and $Al^{3+}$ $(+3)$.
Since $Al^{3+}$ has the highest valency, $Al_2(SO_4)_3$ is the most effective coagulating agent.

(B) Soap molecules contain a hydrophobic (water-repelling) tail and a hydrophilic (water-attracting) head.
When soap is added to water, the hydrophobic tails dissolve in the grease or dirt particles, while the hydrophilic heads remain in the water.
This arrangement results in the formation of spherical clusters known as micelles.
Thus, a micelle is an aggregate of soap molecules surrounding a dirt or grease particle.

(B) $As_2S_3$ is a negatively charged sol.
According to the Hardy-Schulze rule, the coagulating power of an electrolyte depends on the valency of the ion carrying a charge opposite to that of the colloidal particles.
Since $As_2S_3$ is a negative sol, the positive ions (cations) are responsible for coagulation.
The coagulating power increases with the increase in the valency of the cation.
The valencies of the cations in the given options are:
$K^+$ $(KCl)$, $Mg^{2+}$ $(MgSO_4)$, $Al^{3+}$ $(AlCl_3)$.
Since $Al^{3+}$ has the highest valency $(+3)$, it is the most effective in causing the coagulation of the $As_2S_3$ sol.

(B) . Negatively charged sol: When $FeCl_3$ is added to excess $NaOH$ solution, it forms a negatively charged $Fe(OH)_3$ sol due to the adsorption of $OH^-$ ions. Thus, $A-III$.
$B$. Milk: Milk is a well-known example of an emulsion (liquid in liquid). Thus, $B-I$.
$C$. Gold number: It is a measure of the protective power of a lyophilic colloid. Thus, $C-IV$.
$D$. Colloidal antimony: It is used in the treatment of the disease kala-azar. Thus, $D-II$.
Therefore, the correct matching is $A-III, B-I, C-IV, D-II$.

(A) An emulsion is a colloidal system in which both the dispersed phase and the dispersion medium are liquids.
Milk is a classic example of an emulsion where fat droplets are dispersed in water.
While butter and vanishing cream are also emulsions, milk is the most standard example cited in textbooks for a liquid-in-liquid colloidal system.

(A) According to the $Hardy-Schulze$ law, the coagulating power of an ion depends on the magnitude of the charge carried by the ion.
$Fe(OH)_{3}$ sol is a positively charged sol.
Therefore, the ion with the highest negative charge will be most effective for its flocculation.
Comparing the charges of the given ions:
$PO_{4}^{3-}$ has a charge of $-3$.
$SO_{4}^{2-}$ has a charge of $-2$.
$SO_{3}^{2-}$ has a charge of $-2$.
$NO_{3}^{-}$ has a charge of $-1$.
Since $PO_{4}^{3-}$ carries the highest negative charge, it is the most effective ion for the flocculation of $Fe(OH)_{3}$ sol.

(A) Option $(a)$ is false because colloidal sols are not homogeneous.
Colloidal sols are heterogeneous mixtures of dispersed phase and dispersion medium.

(C) Fog is a type of colloid where a liquid is dispersed in a gas.
Therefore, the dispersed phase is $liquid$ and the dispersion medium is $gas$.

(B) $As_{2}S_{3}$ is a negatively charged sol. 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).
For a negatively charged sol, the coagulating power increases with the increase in the magnitude of the positive charge on the cation.
The charges on the given ions are: $Na^{+}$ $(+1)$, $Ba^{2+}$ $(+2)$, and $Al^{3+}$ $(+3)$.
Therefore, the order of coagulating power is $Na^{+} < Ba^{2+} < Al^{3+}$.

(D) According to the Hardy-Schulze law, the coagulating power of an electrolyte depends on the valency of the active ion (the ion with a charge opposite to that of the colloidal particles).
For a negatively charged $AgI$ sol, the coagulating power depends on the positive ions: $Na^{+}$, $Ba^{2+}$, and $Al^{3+}$.
The coagulating power increases with the increase in the magnitude of the charge: $Na^{+} < Ba^{2+} < Al^{3+}$.
Since the amount of electrolyte required is inversely proportional to its coagulating power, the order of the amount required is: $NaNO_{3} > Ba(NO_{3})_{2} > Al(NO_{3})_{3}$.

(C) $1$. Nanomaterials are classified based on the number of dimensions that are outside the nanoscale range $(< 100 \text{ nm})$.
$2$. $A$ material with all three dimensions in the nanoscale range $(< 100 \text{ nm})$ is called a zero-dimensional $(0D)$ nanomaterial, such as nanoparticles or nanoshells.
$3$. Nanotubes and nanowires are one-dimensional $(1D)$ nanomaterials (two dimensions are $< 100 \text{ nm}$).
$4$. Thin films are two-dimensional $(2D)$ nanomaterials (one dimension is $< 100 \text{ nm}$).
$5$. Therefore, nanoshells are the correct answer.

(D) Nanoparticles are defined as materials having at least one dimension in the range of $1-100 \text{ nm}$. However, in the context of classification based on the number of dimensions outside the nanoscale, nanoparticles are typically considered to be $0$-dimensional structures where all three dimensions are in the nanoscale range $(< 100 \text{ nm})$.

(B) Nanostructures are classified based on the number of dimensions that are outside the nanometer range $(1-100 \text{ nm})$.
$(1)$ Zero-dimensional $(0D)$ structures have all three dimensions in the nanometer range (e.g., nanoparticles).
$(2)$ One-dimensional $(1D)$ structures have two dimensions in the nanometer range, while the third dimension is much larger (e.g., nanotubes, nanowires).
$(3)$ Two-dimensional $(2D)$ structures have one dimension in the nanometer range (e.g., thin films, graphene).
$(4)$ Three-dimensional $(3D)$ structures have all dimensions outside the nanometer range (e.g., bulk materials).
Therefore, nanotubes are classified as one-dimensional nanostructures.

(C) thin film is a layer of material ranging from fractions of a nanometer to several micrometers in thickness. In a thin film, only one dimension (the thickness) is in the nanometer range, while the other two dimensions are significantly larger. Therefore, it is classified as a two-dimensional nanostructure.

(C) $1$. Nanomaterials possess unique properties due to their small size $(1-100 \text{ nm})$.
$2$. They are used in manufacturing scratch-proof glasses, advanced electronic devices, and self-cleaning surfaces (e.g., lotus effect).
$3$. Conventional water treatment relies on traditional methods like filtration, sedimentation, and chlorination, not on nanotechnology.
$4$. Therefore, conventional water treatment is the correct answer.

(C) $1$. Nanomaterials are classified based on the number of dimensions that are in the nanoscale range (less than $100 \text{ nm}$).
$2$. Fibers have two dimensions in the nanoscale range.
$3$. Thin films have one dimension in the nanoscale range.
$4$. Quantum dots are zero-dimensional nanomaterials, meaning all three dimensions are in the nanoscale range (less than $100 \text{ nm}$).
$5$. Nanotubes have two dimensions in the nanoscale range.
$6$. Therefore, quantum dots are the correct answer.

(C) The nano scale is generally defined as the range of dimensions where materials exhibit unique physical and chemical properties due to their small size. By international convention, the nano scale is defined as the range of $1 \text{ nm}$ to $100 \text{ nm}$.

(C) $1$. Nanomaterials are classified based on the number of dimensions that are in the nanoscale range $(< 100 \text{ nm})$.
$2$. Quantum dots have three dimensions in the nanoscale range ($0D$ nanomaterials).
$3$. Nanowires and nanotubes have two dimensions in the nanoscale range ($1D$ nanomaterials).
$4$. Thin films or coatings have only one dimension in the nanoscale range ($2D$ nanomaterials).
$5$. Therefore, thin films are the correct answer.

(D) $1$. Emulsions are colloidal systems where both the dispersed phase and the dispersion medium are liquids.
$2$. In water-in-oil $(W/O)$ emulsions, water is the dispersed phase and oil is the dispersion medium.
$3$. Examples of $W/O$ emulsions include butter, cold cream, and cod liver oil.
$4$. Milk and vanishing cream are examples of oil-in-water $(O/W)$ emulsions, where water is the dispersion medium.
$5$. Therefore, butter is the correct example of a water-in-oil emulsion.

(B) $1$. An emulsion is a colloidal system where both dispersed phase and dispersion medium are liquids.
$2$. In an oil-in-water $(O/W)$ emulsion, oil is the dispersed phase and water is the dispersion medium.
$3$. Milk and vanishing cream are examples of $O/W$ emulsions.
$4$. Butter is a water-in-oil $(W/O)$ emulsion, where water is the dispersed phase and oil (fat) is the dispersion medium.
$5$. Therefore, butter is not an example of an oil-in-water emulsion.

(D) Step $1$: Analyze the definitions of the given colloidal phenomena.
Step $2$: Brownian Motion is the random, zigzag movement of colloidal particles due to collisions with molecules of the dispersion medium, which corresponds to $(III)$.
Step $3$: Tyndall Effect is the scattering of light by colloidal particles, which corresponds to $(IV)$.
Step $4$: Electrophoresis is the movement of charged colloidal particles under the influence of an electric field, which corresponds to $(II)$.
Step $5$: Dialysis is the process of removing impurities from colloidal sols using a semipermeable membrane, which corresponds to $(I)$.
Step $6$: Matching the pairs: $(A)$-$(III)$, $(B)$-$(IV)$, $(C)$-$(II)$, $(D)$-$(I)$. This matches option $(D)$.

(A) $1$. 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 sol).
$2$. The ferric oxide sol is positively charged, so the coagulating power depends on the valency of the anion.
$3$. The anions provided by the electrolytes are: $NO_3^-$ (valency $1$), $SO_4^{2-}$ (valency $2$), $PO_4^{3-}$ (valency $3$), and $[Fe(CN)_6]^{4-}$ (valency $4$).
$4$. According to the Hardy-Schulze rule, the coagulating power increases with the increase in the valency of the oppositely charged ion.
$5$. Therefore, the ion with the lowest valency $(NO_3^-)$ will have the minimum coagulating power.

(C) The colour of colloidal solutions depends on the wavelength of light scattered by the dispersed particles. The wavelength of light further depends on the size and nature of the particles. Therefore, the scattering of light is responsible for the development of colour in colloids.

(B) According to the Hardy-Schulze rule, the coagulating power of an ion increases with the increase in the magnitude of the charge on the ion used for coagulation.
For a positively charged sol, the coagulating power of anions follows the order: $[Fe(CN)_6]^{4-} > PO_4^{3-} > SO_4^{2-} > Cl^-$.
Since $[Fe(CN)_6]^{4-}$ has the highest negative charge $(-4)$, it has the greatest coagulating power.