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

(D) The purification of colloidal solutions involves the removal of dissolved impurities (electrolytes) from the colloidal particles.
$1$. Dialysis,Electrodialysis,and Ultrafiltration are standard methods used for the purification of colloids.
$2$. Electrophoresis is a phenomenon involving the movement of colloidal particles under the influence of an electric field,which is used to determine the charge on colloidal particles or for their coagulation,not for purification.

(B) The reaction between $AgNO_3$ and $KI$ is: $AgNO_3 + KI \rightarrow AgI(s) + KNO_3$.
For a negatively charged $AgI$ sol,the dispersion medium must contain an excess of $I^-$ ions,which get adsorbed on the surface of $AgI$ particles.
This occurs when $KI$ is in excess.
In option $B$,$50 \ mL$ of $0.2 \ M \ KI$ provides $10 \ mmol$ of $KI$,while $50 \ mL$ of $0.1 \ M \ AgNO_3$ provides $5 \ mmol$ of $AgNO_3$.
Since $KI$ is in excess,$I^-$ ions are adsorbed on $AgI$ to form a negatively charged sol.

(A) When $AgNO_3$ is added in excess to $KI$ solution,the $Ag^+$ ions from $AgNO_3$ are adsorbed on the surface of the $AgI$ precipitate formed.
$AgI + Ag^+ \rightarrow [AgI]Ag^+$
Since $Ag^+$ ions are positively charged,the resulting colloidal particles will carry a positive charge.

(A) When a beam of light is passed through a colloidal solution and observed under an ultramicroscope,the colloidal particles appear as bright spots of light against a dark background. This is due to the scattering of light by the colloidal particles,a phenomenon known as the $Tyndall$ effect. The ultramicroscope does not show the actual size or shape of the particles,but rather the light scattered by them.

(D) The color of colloidal solutions depends on the wavelength of light scattered by the dispersed particles.
This wavelength further depends on the size and nature of the particles.
For colloidal gold,as the particle size (diameter) changes,the scattering properties change,resulting in different colors (e.g.,red,purple,or blue).
Therefore,the different colors are due to the different diameters of the colloidal gold particles.

(C) The Tyndall effect is observed in colloidal solutions where the dispersed particles are large enough to scatter light.
$A$,$B$,and $D$ (Starch sol,Emulsion,and Gold sol) are all examples of colloidal systems.
$C$ (Sugar solution) is a true solution,where the solute particles are of molecular size and are too small to scatter light.
Therefore,the Tyndall effect is not observed in a sugar solution.

(B) The Tyndall effect is the scattering of light by colloidal particles.
It is more pronounced in lyophobic (hydrophobic) sols because the particles in these sols are larger and have a greater refractive index difference compared to the dispersion medium.
Lyophilic (hydrophilic) sols are more stable and the particles are often smaller or solvated,making the Tyndall effect less distinct compared to hydrophobic sols.

(C) The blue color of the sky is a result of the $Tyndall$ effect,which is the scattering of light by colloidal particles present in the atmosphere.
As sunlight enters the Earth's atmosphere,the shorter wavelengths (blue light) are scattered more strongly by the dust particles and air molecules than the longer wavelengths (red light).
This phenomenon is known as $Rayleigh$ scattering.

(C) Brownian motion is the continuous,random zig-zag movement of colloidal particles in a dispersion medium.
It is caused by the unbalanced bombardment of the particles by the molecules of the dispersion medium.
Therefore,the correct option is $C$.

(D) The phenomenon of movement of colloidal particles under an applied electric field is called $Electrophoresis$.
Since colloidal particles carry a charge,they move towards the oppositely charged electrode.
Thus,the direction of movement of the particles helps in determining the nature of the charge on the colloidal particles.

(C) The isoelectric point is defined as the $pH$ value at which the colloidal particles carry no net charge.
Since the particles are electrically neutral at this $pH$,they do not migrate towards either the cathode or the anode under the influence of an electric field during electrophoresis.
Therefore,the correct statement is that colloidal particles do not move towards any electrode.

(C) The minimum concentration of an electrolyte in millimoles per litre required to cause the coagulation of a sol is called its $Flocculation \ value$ or $Coagulation \ value$.

(B) The coagulating power of an electrolyte depends on the valency of the active ion (ion carrying charge opposite to that of the colloidal particles).
According to the $Hardy-Schulze$ rule,greater the valency of the flocculating ion added,the greater is its power to cause precipitation (coagulation).
Therefore,the correct answer is $Hardy-Schulze$ rules.

(A) The coagulation value (or flocculation value) is defined as the minimum concentration of an electrolyte in $mmol/L$ required to cause the coagulation of a sol.
Therefore,it is expressed in the units of $mmol/L$.

(B) According to the Hardy-Schulze rule,the coagulating power of an electrolyte increases with the increase in the magnitude of the charge on the ion opposite to that of the colloidal particles.
Since the sol is positively charged,the coagulating power depends on the negative ion (anion).
The valency of the anions in the given electrolytes are:
$Cl^-$ (from $NaCl$): $-1$
$[Fe(CN)_6]^{4-}$ (from $K_4[Fe(CN)_6]$): $-4$
$Cl^-$ (from $ZnCl_2$): $-1$
$SO_4^{2-}$ (from $Na_2SO_4$): $-2$
The anion with the highest valency is $[Fe(CN)_6]^{4-}$,which has a charge of $-4$.
Therefore,$K_4[Fe(CN)_6]$ will have the highest coagulating power and will require the minimum concentration for coagulation.

(B) The process of destroying a sol is known as coagulation or flocculation.
Adding an electrolyte to a sol neutralizes the charge on the colloidal particles.
Once the charge is neutralized,the particles aggregate and settle down due to gravity.
Therefore,the addition of an electrolyte is the standard method to destroy a sol.

(D) According to the Hardy-Schulze rule,the coagulating power of an ion is directly proportional to its valency (charge).
Greater the charge on the ion,higher is its coagulating power.
Comparing the charges:
$PO_4^{3-}$ has a charge of $-3$.
$SO_4^{2-}$ and $SO_3^{2-}$ have a charge of $-2$.
$NO_3^-$ has a charge of $-1$.
Since $NO_3^-$ has the lowest charge (valency),it has the lowest coagulating power.

(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).
Since the arsenious sulfide sol is negatively charged,the coagulating power depends on the valency of the cation.
The higher the valency of the cation,the greater is its coagulating power.
Comparing the cations:
$Na^+$ (valency $1$)
$Mg^{2+}$ (valency $2$)
$Al^{3+}$ (valency $3$)
Since $Al^{3+}$ has the highest valency,$AlCl_3$ will have the highest coagulating power.

(A) 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).
$Fe(OH)_3$ is a positively charged sol.
Therefore,it requires negatively charged ions for coagulation.
The coagulating power increases with the increase in the magnitude of the charge on the ion.
Comparing the anions: $PO_4^{3-}$ has a charge of $-3$,while $SO_4^{2-}$ has a charge of $-2$.
Since the magnitude of the charge on $PO_4^{3-}$ is greater than that on $SO_4^{2-}$,$PO_4^{3-}$ has the highest coagulating power among the given options.

(D) The gold number is defined as the minimum amount of protective colloid in milligrams that prevents the coagulation of $10 \, mL$ of a gold sol when $1 \, mL$ of $10 \% \, NaCl$ solution is added to it.
Given: Mass of starch = $0.025 \, g = 25 \, mg$.
Since $25 \, mg$ of starch is required to protect $10 \, mL$ of gold sol,the gold number of starch is $25$.

(B) The protective power of a colloid is inversely proportional to its gold number.
Mathematically,$\text{Protective Power} \propto \frac{1}{\text{Gold Number}}$.
Given gold numbers are:
$A = 0.04$
$B = 0.004$
$C = 10$
$D = 40$
Comparing the values,the order of gold numbers is $B < A < C < D$.
Therefore,the order of protective power is $B > A > C > D$.

(D) The gold number is defined as the minimum amount of protective colloid in milligrams $(mg)$ required to prevent the coagulation of $10 \, mL$ of a standard gold sol when $1 \, mL$ of $10 \% \, NaCl$ solution is added.
Given that $100 \, cm^3$ of gold sol requires $10^{-4} \, g$ of gelatin.
Therefore,$10 \, cm^3$ of gold sol requires $\frac{10^{-4}}{10} \, g = 10^{-5} \, g$ of gelatin.
Converting $g$ to $mg$: $10^{-5} \, g = 10^{-5} \times 10^3 \, mg = 10^{-2} \, mg = 0.01 \, mg$.
Thus,the gold number of gelatin is $0.01$.

(B) Lyophilic colloids act as protective colloids. When added to an emulsion,they form a protective layer around the particles of the dispersed phase,which prevents the emulsion from coagulating or breaking down. This process is known as the stabilization of the emulsion.

(C) $1$. Colloidal systems exhibit Brownian motion and Tyndall effect,which is a correct statement.
$2$. Gold number is defined as the minimum amount of protective colloid in milligrams required to prevent the coagulation of $10 \ mL$ of a standard gold sol when $1 \ mL$ of $10\% \ NaCl$ solution is added. Thus,it is a measure of the protective power of lyophilic colloids,which is a correct statement.
$3$. $A$ colloidal solution of liquid in liquid is called an emulsion,not a gel. $A$ gel is a colloidal system where a liquid is dispersed in a solid. Therefore,this statement is incorrect.
$4$. Hardy-Schulze rules state that the greater the valence of the flocculating ion added,the greater is its power to cause precipitation (coagulation). This is a correct statement.

(B) Soap molecules consist of a long hydrocarbon chain (hydrophobic) and a polar head (hydrophilic).
When soap is added to water containing grease,the hydrophobic part of the soap molecules attaches to the grease,while the hydrophilic part remains in the water.
This process breaks the large grease droplets into smaller droplets,forming a stable emulsion in water,which is then easily washed away.
This mechanism is known as emulsification.

(B) Emulsions are classified into two types: oil-in-water $(O/W)$ and water-in-oil $(W/O)$.
In oil-in-water emulsions,water is the dispersion medium (e.g.,milk).
In water-in-oil emulsions,oil is the dispersion medium (e.g.,butter,cod liver oil,and cold creams).
Therefore,butter is a classic example of a water-in-oil emulsion.

(C) Gelatin acts as a protective colloid in the preparation of ice cream. $It$ helps in stabilizing the colloidal system and prevents the growth of ice crystals,which ensures a smooth texture.

(A) Blood is a colloidal solution in which the dispersed phase particles are negatively charged.
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).
Since blood particles are negatively charged,the positively charged $Fe^{3+}$ ions from $FeCl_3$ are effective in neutralizing the charge and causing coagulation,thereby stopping the bleeding.

(A) The $Purple \ of \ Cassius$ is a colloidal solution of gold obtained by the reduction of gold chloride solution with stannous chloride $(SnCl_2)$.
It is a classic example of a gold sol used in glass coloring and ceramics.

(B) The phenomenon in which a gel shrinks and exudes liquid upon standing is known as $Syneresis$.
This occurs due to the contraction of the gel network,which squeezes out the entrapped liquid.

(C) Colloidal sols are classified based on the dispersion medium.
When water is used as the dispersion medium,the sol is called an 'aqua-sol' or 'hydrosol'.
'Aqua-dag' is a specific colloidal sol consisting of graphite dispersed in water.
Therefore,the correct option is $C$.

(B) The process of removing carbon particles from smoke is based on the principle of electrophoresis.
Smoke is a colloidal system where carbon particles are dispersed in air.
These carbon particles carry a charge.
When smoke is passed through a chamber fitted with metal plates connected to a high-voltage source,the charged carbon particles are attracted to the oppositely charged plates,get neutralized,and settle down.
This device is known as the $Cottrell$ $precipitator$.

(C) First,convert the volume to liters:
$1 \ mm^3 = (10^{-1} \ cm)^3 = 10^{-3} \ cm^3 = 10^{-3} \ mL = 10^{-6} \ L$.
Calculate the total number of detergent molecules in $1 \ mm^3$:
$\text{Number of molecules} = \text{Molarity} \times \text{Volume in L} \times N_A$
$= 10^{-3} \ mol/L \times 10^{-6} \ L \times 6 \times 10^{23} \ mol^{-1} = 6 \times 10^{14}$ molecules.
Given that there are $10^{13}$ micelles in this volume,the number of detergent molecules per micelle is:
$\text{Number of ions per micelle} = \frac{\text{Total molecules}}{\text{Number of micelles}} = \frac{6 \times 10^{14}}{10^{13}} = 60$.

(C) Lyophobic colloids are those in which the dispersed phase has little or no affinity for the dispersion medium.
Among the given options,$Gold$ $sol$ is a lyophobic colloidal solution.
Starch and protein solutions are lyophilic colloids,and polymer solutions in organic solvents are also typically lyophilic.

(D) Volume of gold present in solution = $\frac{\text{Mass of gold}}{\text{Density of gold}} = \frac{1.9 \times 10^{-4} \ g}{19 \ g/cm^3} = 1.0 \times 10^{-5} \ cm^3$.
For spherical particles of gold with radius $r = 10 \ nm = 10 \times 10^{-7} \ cm = 10^{-6} \ cm$,the volume of each particle = $\frac{4}{3} \pi r^3 = \frac{4}{3} \times 3.14 \times (10^{-6} \ cm)^3 = 4.186 \times 10^{-18} \ cm^3$.
Number of gold particles present in $1 \ L$ $(1000 \ cm^3)$ = $\frac{\text{Total volume of gold}}{\text{Volume of one particle}} = \frac{1.0 \times 10^{-5} \ cm^3}{4.186 \times 10^{-18} \ cm^3} \approx 2.389 \times 10^{12} \text{ particles}$.
Since $1 \ L = 1000 \ cm^3 = 10^6 \ mm^3$,the number of particles per $mm^3$ = $\frac{2.389 \times 10^{12}}{10^6} \approx 2.4 \times 10^6 \text{ particles/mm}^3$.

(A) According to the Hardy-Schulze law,the flocculating value is inversely proportional to the valency of the coagulating ion,i.e.,$\text{Flocculating value} \propto \frac{1}{z}$,where $z$ is the valency of the coagulating ion.
$Fe(OH)_3$ is a positively charged sol,so it is coagulated by negative ions.
The negative ions present in the given electrolytes are:
$A) Cl^-$ (valency = $1$)
$B) S^{2-}$ (valency = $2$)
$C) PO_4^{3-}$ (valency = $3$)
$D) SO_4^{2-}$ (valency = $2$)
Since $Cl^-$ has the lowest valency $(z=1)$,it will have the maximum flocculation value. Therefore,$NaCl$ has the maximum flocculation value.

(D) The reaction between $FeCl_3$ and excess $NaOH$ produces a negatively charged ferric hydroxide sol: $FeCl_3 + 3NaOH \to Fe(OH)_3(s) + 3NaCl$.
Due to the adsorption of $OH^-$ ions from the excess $NaOH$,the sol is negatively charged: $[Fe(OH)_3]OH^-$.
According to the Hardy-Schulze rule,the coagulating power of an ion is directly proportional to its valency. For a negatively charged sol,the coagulating power increases with the valency of the cation.
The valencies of the given cations are: $Ba^{2+}$ $(+2)$ and $Al^{3+}$ $(+3)$.
Since $Al^{3+}$ has the highest valency among the cations,it has the maximum coagulating power and,therefore,the minimum coagulating value.

(D) Lyophobic sols are prepared by chemical methods such as double decomposition,oxidation,reduction,and hydrolysis.
For example:
$1$. Double decomposition: $As_2O_3 + 3H_2S \rightarrow As_2S_3 (\text{sol}) + 3H_2O$
$2$. Oxidation: $2H_2S + SO_2 \rightarrow 3S (\text{sol}) + 2H_2O$
$3$. Reduction: $2AuCl_3 + 3HCHO + 3H_2O \rightarrow 2Au (\text{sol}) + 3HCOOH + 6HCl$
$4$. Hydrolysis: $FeCl_3 + 3H_2O \rightarrow Fe(OH)_3 (\text{sol}) + 3HCl$
Since all these methods are used,the correct option is $D$.

(C) The movement of colloidal particles towards the anode indicates that the sol is negatively charged.
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 the sol is negative,the coagulating power depends on the charge of the cation ($Na^+$,$Ba^{2+}$,$Al^{3+}$).
Greater the charge on the cation,higher is its coagulating power.
Thus,the order of coagulating power is $Al^{3+} > Ba^{2+} > Na^+$,which corresponds to $AlCl_3 > BaCl_2 > NaCl$.

(A) Since the sol particles migrate towards the cathode,they are positively charged.
According to the Hardy-Schulze law,the coagulating power of an ion is directly proportional to its valency.
For a positively charged sol,the effective coagulating ions are the anions: $Cl^-$,$SO_4^{2-}$,and $PO_4^{3-}$.
The coagulating power order is $PO_4^{3-} > SO_4^{2-} > Cl^-$.
Since the coagulating value is inversely proportional to the coagulating power,the order of coagulating values is $NaCl > Na_2SO_4 > Na_3PO_4$.

(A) The protective power of a colloid is inversely proportional to its gold number.
Lower gold number indicates higher protective power.
Given gold numbers: $A = 0.50$,$B = 0.01$,$C = 0.10$,$D = 0.005$.
Arranging in increasing order of gold number: $D (0.005) < B (0.01) < C (0.10) < A (0.50)$.
Therefore,the order of protective power is $A < C < B < D$.
Thus,the correct option is $A$ or $C$ (as both represent the same sequence).

(A) Colloidal antimony is used in the treatment of the disease kala azar.

(C) Electrodialysis involves the movement of ions towards oppositely charged electrodes under the influence of an electric field.
Urea $(NH_2CONH_2)$ is a covalent compound and does not dissociate into ions in an aqueous solution.
Since it does not form ions,it cannot be moved by an electric field and thus cannot be removed by electrodialysis.
In contrast,$NaCl$,$K_2SO_4$,and $CaCl_2$ are ionic compounds that dissociate into ions ($Na^+$,$Cl^-$,$K^+$,$SO_4^{2-}$,$Ca^{2+}$) and can be easily removed by this process.

(A) The sky appears blue due to the scattering of sunlight by colloidal particles (dust and air molecules) present in the atmosphere.
According to Rayleigh scattering,the intensity of scattered light is inversely proportional to the fourth power of the wavelength $(I \propto 1/\lambda^4)$.
Since blue light has a shorter wavelength compared to red light,it is scattered more effectively by these particles,making the sky appear blue.

(B) The assertion is correct because in a micelle,the hydrophilic $-COO^{-}$ heads are oriented towards the water at the surface,while the hydrophobic hydrocarbon tails point inward.
The reason is also correct because the addition of soap (sodium stearate) acts as a surfactant,which lowers the surface tension of water.
However,the reduction of surface tension is a property of surfactants,not the direct explanation for why the $-COO^{-}$ groups are at the surface of the micelle. The orientation is due to the amphiphilic nature of the molecules.

(A) The colour of colloidal solutions depends on the size of the colloidal particles.
For gold sol,the colour varies with the particle size; very fine gold sol is red in colour.
This colour arises due to the scattering of light by the colloidal gold particles.