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

(C) Milk is an example of an emulsion,which is a colloidal system where liquid fat globules are dispersed in water.

(C) Coagulation power $\propto \frac{1}{\text{Coagulation value}}$
Lower the coagulation value,higher is the coagulating power of the electrolyte.
Given values:
$I (NaCl) = 52$
$II (BaCl_2) = 0.69$
$III (MgSO_4) = 0.22$
Comparing the values: $0.22 < 0.69 < 52$
Therefore,the order of coagulating power is: $III > II > I$.

(C) Fog is a colloidal solution of liquid in a gas,in which the liquid is the dispersed phase and the gas is the dispersion medium.

(B) The $Tyndall \ effect$ is independent of the charge on the colloidal particles.
It is an optical property that depends on the size of the colloidal particles and the wavelength of light used.
The $Tyndall \ effect$,also known as $Tyndall \ scattering$,is the scattering of light by particles in a colloid or a fine suspension.
In contrast,$electro-osmosis$,$coagulation$,and $electrophoresis$ are electrical properties that directly depend on the charge present on the colloidal particles.

(D) The $Tyndall \ effect$ is an optical phenomenon that occurs due to the scattering of light by colloidal particles.
It depends on the size of the particles and the refractive index difference between the dispersed phase and the dispersion medium,but it is independent of the charge on the particles.
In contrast,$Coagulation$,$Electrophoresis$,and $Electro-osmosis$ are all properties that depend on the electrical charge of the colloidal particles.

(B) Lyophilic sols are stabilized by adding lyophilic colloids,which protect them from precipitation. These are called protecting colloids.
Their protecting power is expressed in terms of $gold \ number$.
$Gold \ number$ is defined as the minimum amount of lyophilic colloid in milligrams that prevents the flocculation of $10 \ mL$ of gold sol by the addition of $1 \ mL$ of $10 \ \% \ NaCl$ solution.
Lesser the $gold \ number$,higher is the protecting power.

(B) According to the Hardy-Schulze rule,the coagulating power of an ion is directly proportional to the magnitude of its charge for a given sol.
Arsenic sulphide $(As_2S_3)$ sol is a negatively charged sol.
Therefore,the coagulating power depends on the charge of the cation.
The charges on the given ions are $+1$ for $Na^{+}$,$+2$ for $Ba^{2+}$,and $+3$ for $Al^{3+}$.
Thus,the increasing order of coagulating power is $Na^{+} < Ba^{2+} < Al^{3+}$.

(B) The Tyndall effect is an optical phenomenon observed in colloidal systems.
The two essential conditions for the observation of the Tyndall effect are:
$1$. The diameter of the dispersed particles is not much smaller than the wavelength of the light used.
$2$. The refractive indices of the dispersed phase and the dispersion medium differ greatly in magnitude.
Thus,conditions $2$ and $4$ are correct.

(D) $Fe(OH)_3$ sol particles are positively charged due to the adsorption of $Fe^{3+}$ ions.
During electrolysis,these positively charged sol particles migrate towards the cathode.
At the cathode,they lose their charge and undergo coagulation.
At the anode,the oxidation of water occurs,producing $O_2$ gas: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$.
Therefore,$O_2$ gas is the anode product.

(A) colloidal system is classified based on the physical state of the dispersed phase and the dispersion medium.
When a liquid is dispersed in a gas,the resulting colloid is known as an aerosol (specifically fog or cloud).
- $A$: Cloud is a liquid in gas $(L/G)$ type colloid.
- $B$: Smoke is a solid in gas $(S/G)$ type colloid.
- $C$: Milk is a liquid in liquid $(L/L)$ type colloid (emulsion).
- $D$: Dust is a solid in gas $(S/G)$ type colloid.
Therefore,the correct option is $A$.

(A) The $As_2S_3$ sol is a negatively charged sol.
According to the Hardy-Schulze rule,the coagulating power of an ion is directly proportional to the magnitude of its charge (valency) for ions carrying a charge opposite to that of the colloidal particles.
Since $As_2S_3$ is negatively charged,it is coagulated by positively charged ions (cations).
Among the given cations,$Fe^{3+}$ has the highest charge $(+3)$,followed by $Ba^{2+}$ $(+2)$.
Therefore,$Fe^{3+}$ will be the most effective in coagulating the $As_2S_3$ sol.

(B) $1$. Soap solutions form associated colloids (micelles) above the Critical Micelle Concentration $(CMC)$. This statement is true.
$2$. Lyophilic colloids are reversible sols,meaning they can be easily reformed by mixing the dispersed phase with the dispersion medium after evaporation. The statement that they are irreversible is false.
$3$. Blood is a colloidal system,and its purification (removal of urea and other waste products) is indeed carried out by the process of dialysis. This statement is true.
$4$. Blood is a negative sol. Addition of excess electrolytes like $Ca^{2+}$ and $K^{+}$ causes coagulation due to the neutralization of charge. This statement is true.
Therefore,the statement that is not true is $B$.

(B) . Dialysis is a process used for the purification of colloidal solutions by removing electrolytes through a semi-permeable membrane.
$B$. Peptization is the process of converting a freshly prepared precipitate into a colloidal sol by adding a suitable electrolyte.
$C$. Flocculation (or coagulation) is the process of aggregating colloidal particles to form a precipitate.
$D$. Gold number is a measure of the protective power of a lyophilic colloid.
Therefore,the correct matching is $A-IV, B-I, C-II, D-III$.

(A) Blood is a colloidal solution in which the dispersed phase particles are negatively charged.
According to the Hardy-Schulze law,the coagulating power of an ion increases with the increase in the magnitude of the charge on the ion.
$Fe^{3+}$ ions,being highly positively charged,effectively neutralize the negative charge on the blood colloids,leading to coagulation.
Therefore,ferric chloride $(FeCl_3)$ is used to stop bleeding.

(B) The protective power of a lyophilic colloid is inversely proportional to its gold number.
$\text{Protective power} \propto \frac{1}{\text{Gold number}}$
Given gold numbers:
Gelatin: $0.005$
Haemoglobin: $0.05$
Sodium acetate: $0.7$
Since $0.005 < 0.05 < 0.7$,the order of protective power is:
$\text{Gelatin} > \text{Haemoglobin} > \text{Sodium acetate}$

(C) According to the Hardy-Schulze rule,the coagulation power of an electrolyte depends on the valency of the active ion (the ion carrying a charge opposite to that of the colloidal particles).
$As_2S_3$ sol is a negatively charged sol.
Therefore,the coagulation power depends on the valency of the cation.
The higher the valency of the cation,the greater is its coagulation power.
Comparing the cations in the given electrolytes:
$Na^+$ (from $NaCl$,valency $1$)
$Mg^{2+}$ (from $MgCl_2$,valency $2$)
$Al^{3+}$ (from $Al_2(SO_4)_3$,valency $3$)
$NH_4^+$ (from $NH_4Cl$,valency $1$)
Since $Al^{3+}$ has the highest valency $(3)$,$Al_2(SO_4)_3$ has the maximum coagulation power.

(D) The gold number of a protective colloid is defined as the minimum weight of the protective colloid in milligrams that must be added to $10 \ mL$ of a standard red gold sol so that no coagulation of the gold sol occurs when $1 \ mL$ of $10\% \ NaCl$ solution is added.
Therefore,it is a direct measure of the efficiency of the protective colloid.

(D) Lyophilic sols are solvent-loving and are formed by direct mixing of the dispersed phase and the dispersion medium. They are inherently stable and reversible.
In contrast,lyophobic sols are solvent-hating and cannot be prepared by direct mixing. They require special methods for preparation and need stabilizing agents to prevent coagulation,making them less stable.
Therefore,the statement that lyophobic sols can be formed by direct mixing is incorrect.

(A) The given graphs represent the variation of molar conductivity $(\Lambda_m^c)$ with the square root of concentration $(\sqrt{c})$ for surfactants. The sharp decrease in conductivity corresponds to the Critical Micelle Concentration $(CMC)$.
As the length of the hydrophobic tail (represented by '$n$' in $R(CH_2)_nCOONa$) increases,the hydrophobicity increases,which facilitates micelle formation at lower concentrations.
Therefore,$CMC \propto \frac{1}{\text{Size of tail}}$.
Looking at the graphs,the $CMC$ values (the point of the sharp break) follow the order: $I < II < III$.
This implies the tail size follows the order: $I > II > III$.
Matching this with the given options,graph $I$ corresponds to the largest '$n$' $(16)$,graph $II$ to $12$,and graph $III$ to $8$.

(B) Purple of Cassius is a purple pigment formed by the reaction of gold salts with $tin(II)$ chloride.
It is essentially a colloidal solution of gold particles dispersed in a medium,often used in glass coloring.

(A) The Tyndall effect does not take place in a true solution because the particle size is too small to scatter light.
Therefore,the statement in option $A$ is incorrect.

(C) Option $C$ is incorrect.
The ozone layer prevents harmful ultraviolet $(UV)$ radiations from the sun from reaching the earth,not infrared radiations.
Other statements are correct: Acid rain is caused by $SO_x$ and $NO_x$; enzymes are indeed macromolecular colloids; and micelles form above the Kraft temperature $(T_k)$.

(A) $Fe(OH)_3$ sol is a positively charged sol. According to the Hardy-Schulze rule,the coagulating power of an electrolyte increases with the increase in the magnitude of the charge on the oppositely charged ion (anion in this case). The anions provided by the electrolytes are: $[Fe(CN)_6]^{4-}$,$SO_4^{2-}$,$Cl^-$,and $NO_3^-$. The magnitude of the charge is highest for $[Fe(CN)_6]^{4-}$. Therefore,$K_4[Fe(CN)_6]$ is the most effective coagulating agent.

(C) $Fe(OH)_3$ is a positively charged sol.
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 of opposite sign to that of the colloidal particle.
For a positively charged $Fe(OH)_3$ sol,the coagulating ion must be negatively charged (anion).
The anions provided by the electrolytes are:
$Na_2SO_4 \rightarrow SO_4^{2-}$
$KCl \rightarrow Cl^-$
$Mg_3(PO_4)_2 \rightarrow PO_4^{3-}$
$CH_3COONa \rightarrow CH_3COO^-$
Since $PO_4^{3-}$ has the highest negative charge $(-3)$,$Mg_3(PO_4)_2$ is the most effective coagulating agent.

(B) Statement $(I)$: False. Lyophobic colloids have little affinity for the dispersion medium and are not heavily solvated.
Statement $(II)$: False. Multimolecular colloids are formed by the aggregation of a large number of atoms or smaller molecules,and they are generally lyophobic.
Statement $(III)$: True. Starch is a lyophilic colloid. Lyophilic sols decrease the surface tension of the dispersion medium (water).
Statement $(IV)$: True. Blood is a colloidal sol of albuminoid substances. Addition of an electrolyte like $NaCl$ causes coagulation due to the neutralization of charge on the colloidal particles.
Therefore,the sequence is $(F, F, T, T)$.

(B) 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 = $0.038 \ g = 38 \ mg$.
Since $38 \ mg$ of starch prevents the coagulation of $10 \ mL$ of gold sol,the gold number of starch is $38$.

(C) The flocculation value is defined as the minimum concentration of an electrolyte in millimoles per liter $(mmol \ L^{-1})$ required to cause the coagulation of a sol.
Given mass of $HCl = 0.73 \ g$.
Molar mass of $HCl = 36.5 \ g \ mol^{-1}$.
Number of moles of $HCl = \frac{0.73 \ g}{36.5 \ g \ mol^{-1}} = 0.02 \ mol$.
Converting to millimoles: $0.02 \ mol \times 1000 = 20 \ mmol$.
Since $20 \ mmol$ of $HCl$ causes coagulation in $200 \ mL$ of the colloid,the amount required for $1000 \ mL$ $(1 \ L)$ is:
$\text{Flocculation value} = \frac{20 \ mmol}{200 \ mL} \times 1000 \ mL = 100 \ mmol \ L^{-1}$.

(D) Option $A$ is correct because lyophilic sols are stabilized by solvation and can be coagulated by adding a suitable solvent (like alcohol or acetone) that removes the solvation layer.
Option $B$ is correct because butter is a classic example of a water-in-oil $(W/O)$ emulsion.
Option $C$ is correct because if you add more of the dispersed phase to an emulsion,it does not mix uniformly and often forms a separate layer.
Option $D$ is incorrect because milk is generally considered a negatively charged colloid (due to the presence of proteins like casein which carry a negative charge at the $pH$ of milk). Therefore,the statement that milk is a positively charged colloid is false.

(D) The $As_2S_3$ sol is a negatively charged colloidal solution.
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 provided are $Ba^{2+}$,$K^{+}$,and $Al^{3+}$.
Since $Al^{3+}$ has the highest valency $(+3)$,it will be the most effective coagulating agent for the negatively charged $As_2S_3$ sol.

(C) The process of converting a freshly prepared precipitate into a colloidal solution by adding a suitable electrolyte is known as peptization. $FeCl_3$ acts as a peptizing agent for $Fe(OH)_3$ precipitate.

(B) colloid is a heterogeneous system in which one substance is dispersed as very fine particles in another substance.
$(i)$ $NaCl(aq.)$ is a true solution,not a colloid.
$(ii)$ Fog is an aerosol (liquid dispersed in gas),which is a colloid.
$(iii)$ Paint is a sol (solid dispersed in liquid),which is a colloid.
$(iv)$ Aerosols (general term) are colloids.
$(v)$ Mud is a suspension/colloidal mixture (solid dispersed in liquid),which is a colloid.
$(vi)$ Blood is a colloidal solution of proteins and other substances in water.
Therefore,$(ii), (iii), (iv), (v),$ and $(vi)$ are colloids.

(D) $Gel$ is a colloidal system in which the dispersed phase is a liquid and the dispersion medium is a solid.
$Cheese$ is a classic example of a $Gel$.
$Milk$ is an example of an emulsion (liquid in liquid).
$Smoke$ is an example of an aerosol (solid in gas).
$Gem$ $stones$ are examples of solid sols (solid in solid).

(D) $1$. Gold sol is prepared by the reduction of $AuCl_3$ with formaldehyde or stannous chloride and is known to be negatively charged. Thus,statement $A$ is correct.
$2$. Peptization is the process of converting a freshly prepared precipitate into a colloidal sol by adding a suitable electrolyte,not a method of purification. Thus,statement $B$ is incorrect.
$3$. Persistent dialysis removes the electrolyte that stabilizes the sol,leading to the coagulation of the colloidal particles. Thus,statement $C$ is correct.
$4$. Since both $A$ and $C$ are correct,the correct option is $D$.

(C) The correct option is $(C)$.
$(A)$ $As_2S_3$ sol is a negative sol and is prepared by double decomposition,not peptization.
$(B)$ Sulphur sol is a negative sol. According to the Hardy-Schulze rule,the coagulating power of an electrolyte depends on the valency of the cation. For a negative sol,the cation with the highest valency is most effective. Here,$Al^{3+}$ (from $AlCl_3$) is most effective,not $Ba^{2+}$.
$(C)$ Non-ionic detergents do not contain any ions. The structure $C_{15}H_{31}COOCH_2-C(CH_2OH)_3$ is an ester of a fatty acid with a polyol,which lacks ionic groups,making it a non-ionic detergent.
$(D)$ Soap solutions behave as electrolytes at low concentrations,but above the Critical Micelle Concentration $(CMC)$,they form micelles,which reduces the number of free ions,causing the electrical conductance to decrease or remain constant.

(C) colloidal system where a liquid is dispersed in another liquid is called an $Emulsion$,not a $Gel$. $A$ $Gel$ is a colloidal system where a liquid is dispersed in a solid. Therefore,the statement in option $C$ is false.

(B) Smoke is a colloidal system where solid particles (like carbon) are dispersed in a gaseous medium (like air).
Therefore,it is a dispersion of a solid in a gas.

(A) The process of converting a freshly prepared precipitate into a colloidal sol by adding a small amount of an electrolyte is known as $Peptization$.
The electrolyte added is called a $Peptizing \text{ } agent$.
This occurs due to the preferential adsorption of ions from the electrolyte on the surface of the precipitate particles,leading to electrostatic repulsion and dispersion.

(C) The micelles are formed by the association of dispersed particles above a certain concentration,which is known as the critical micelle concentration $(CMC)$.
Above the $CMC$,the surfactant molecules associate to form a larger aggregate known as a micelle.

(D) The Tyndall effect is the scattering of light by colloidal particles.
For the Tyndall effect to be observed,two primary 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 significantly in magnitude.
Therefore,statements $(i)$ and $(ii)$ are correct.

(A) An emulsifier is a substance that stabilizes an emulsion by preventing the dispersed droplets from coalescing.
Soap acts as an emulsifier because its molecules have a long hydrocarbon tail that is hydrophobic (oil-loving) and a polar head that is hydrophilic (water-loving).
When soap is added to a mixture of oil and water,the hydrocarbon tails dissolve in the oil droplets,while the polar heads remain in the water.
This creates a stable emulsion where the oil droplets are dispersed in water and prevented from merging due to electrostatic repulsion between the negatively charged heads.
Therefore,$Soap$ is the correct answer.

(D) According to the Hardy-Schulze rule,the flocculating power of an ion is directly proportional to its valency (charge).
Therefore,the greater the valency of the flocculating ion,the greater is its flocculating power.

(D) The $As_2S_3$ sol is a negatively charged sol.
According to the Hardy-Schulze rule,the coagulating power of an electrolyte depends on the valency of the cation or anion that neutralizes the charge on the colloidal particles.
For a negatively charged sol,the coagulating power is determined by the cation.
The higher the valency of the cation,the greater is its coagulating power.
Comparing the cations in the given options:
$A$: $Zn^{2+}$
$B$: $Na^+$
$C$: $Zn^{2+}$
$D$: $Al^{3+}$
Since $Al^{3+}$ has the highest valency $(+3)$,it will have the maximum coagulating power for the negatively charged $As_2S_3$ sol.

(B) Fog is a type of colloidal system known as a liquid aerosol. In this system,a liquid (dispersed phase) is dispersed in a gas (dispersion medium).

(A) Lyophobic colloids are inherently unstable and tend to aggregate.
They are stabilised primarily by the presence of an electric charge on the colloidal particles.
Due to the preferential adsorption of ions from the dispersion medium,the particles acquire a similar charge.
This results in electrostatic repulsion between the particles,which prevents them from coming close enough to aggregate,thereby providing stability to the colloidal system.

(D) 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).
Since $As_2S_3$ sol is negatively charged,the coagulating power depends on the valency of the cation.
The cations present in the given options are:
$A) \, Zn^{2+}$
$B) \, Na^+$
$C) \, Zn^{2+}$
$D) \, Al^{3+}$
According to the Hardy-Schulze rule,the higher the valency of the flocculating ion,the greater is its coagulating power.
Comparing the valencies: $Al^{3+} (3) > Zn^{2+} (2) > Na^+ (1)$.
Therefore,$AlCl_3$ has the maximum coagulating power.

(B) When the dispersed phase is $Liquid$ and the dispersion medium is $Solid$,the resulting colloidal system is classified as a $Gel$.
Examples of $Gel$ include $Butter$,$Jelly$,and $Cheese$.

(D) Rubber is a natural polymer with a high molecular mass. When dissolved in a suitable solvent,it forms a macromolecular colloid. Therefore,the solution of rubber is an example of a macromolecular colloid.

(A) $Fe(OH)_3$ sol 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.
For a positively charged sol,the coagulating power is determined by the negative ion (anion). The order of coagulating power is: $PO_4^{3-} > SO_4^{2-} > Cl^-$.
Since the flocculation value is inversely proportional to the coagulating power,the electrolyte with the lowest charge on the anion will have the maximum flocculation value.
Comparing the anions: $Cl^-$ (charge $-1$),$SO_4^{2-}$ (charge $-2$),$PO_4^{3-}$ (charge $-3$).
Therefore,$NaCl$ has the lowest coagulating power and the maximum flocculation value.