ChemistryQ501–516 of 843 questions
Page 11 of 11 · English
501
ChemistryEasyMCQMHT CET · 2025
Which among the following gases is least adsorbed on a solid at similar conditions of temperature and pressure?
A
$Cl_2$
B
$NH_3$
C
$SO_2$
D
$H_2$
Solution
(D) The extent of adsorption of a gas on a solid surface depends on the ease of liquefaction of the gas. Gases that are easily liquefiable have higher critical temperatures $(T_c)$ and are adsorbed more strongly due to stronger van der Waals forces.
Among the given gases,$H_2$ has the lowest critical temperature $(T_c \approx 33 \ K)$,whereas $Cl_2$,$NH_3$,and $SO_2$ have significantly higher critical temperatures.
Since $H_2$ is the most difficult to liquefy,it is the least adsorbed on a solid surface.
Among the given gases,$H_2$ has the lowest critical temperature $(T_c \approx 33 \ K)$,whereas $Cl_2$,$NH_3$,and $SO_2$ have significantly higher critical temperatures.
Since $H_2$ is the most difficult to liquefy,it is the least adsorbed on a solid surface.
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502
ChemistryEasyMCQMHT CET · 2025
Which of the following is $NOT$ a characteristic of chemisorption?
A
It is specific.
B
Heat released is in the range of $40-200 \ kJ/mol$.
C
Formation of multimolecular layer of adsorbate.
D
It is irreversible.
Solution
(C) Chemisorption (chemical adsorption) involves the formation of strong chemical bonds between the adsorbate and the adsorbent.
$1$. It is highly specific in nature.
$2$. The enthalpy of adsorption is high,typically in the range of $40-200 \ kJ/mol$.
$3$. It is an irreversible process.
$4$. Chemisorption results in the formation of a unimolecular layer,not a multimolecular layer.
Therefore,the formation of a multimolecular layer is a characteristic of physisorption,not chemisorption.
$1$. It is highly specific in nature.
$2$. The enthalpy of adsorption is high,typically in the range of $40-200 \ kJ/mol$.
$3$. It is an irreversible process.
$4$. Chemisorption results in the formation of a unimolecular layer,not a multimolecular layer.
Therefore,the formation of a multimolecular layer is a characteristic of physisorption,not chemisorption.
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503
ChemistryEasyMCQMHT CET · 2025
Which of the following represents the Freundlich adsorption isotherm equation for the adsorption of a gas on a solid (where $n > 1$)?
A
$x/m = k p^{1/n}$
B
$m/x = k p^{1/n}$
C
$x/m = k p^n$
D
$m/x = k p^n$
Solution
(A) The Freundlich adsorption isotherm is an empirical relationship between the amount of gas adsorbed per unit mass of adsorbent and the pressure of the gas at a constant temperature.
It is expressed by the equation: $x/m = k p^{1/n}$,where $x$ is the mass of the gas adsorbed,$m$ is the mass of the adsorbent,$p$ is the equilibrium pressure,and $k$ and $n$ are constants that depend on the nature of the adsorbent and the gas at a particular temperature.
For the adsorption process,the value of $n$ is typically greater than $1$,which implies that $1/n$ lies between $0$ and $1$.
It is expressed by the equation: $x/m = k p^{1/n}$,where $x$ is the mass of the gas adsorbed,$m$ is the mass of the adsorbent,$p$ is the equilibrium pressure,and $k$ and $n$ are constants that depend on the nature of the adsorbent and the gas at a particular temperature.
For the adsorption process,the value of $n$ is typically greater than $1$,which implies that $1/n$ lies between $0$ and $1$.
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504
ChemistryEasyMCQMHT CET · 2025
Which of the following is classified as a multimolecular colloid?
A
Starch
B
Cellulose
C
$S_8$ sulphur molecules
D
Plastics
Solution
(C) Multimolecular colloids are formed by the aggregation of a large number of atoms or smaller molecules with diameters less than $1 \ nm$.
In the case of sulphur sol,the particles consist of thousands of $S_8$ molecules.
Starch,cellulose,and plastics are examples of macromolecular colloids,as they consist of large molecules with high molecular masses.
In the case of sulphur sol,the particles consist of thousands of $S_8$ molecules.
Starch,cellulose,and plastics are examples of macromolecular colloids,as they consist of large molecules with high molecular masses.
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505
ChemistryEasyMCQMHT CET · 2025
What type of colloid is fog?
A
Aerosol
B
Foam
C
Gel
D
Emulsion
Solution
(A) Fog is a type of colloid in which a liquid is dispersed in a gas.
This type of colloidal system is classified as an $Aerosol$.
This type of colloidal system is classified as an $Aerosol$.
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506
ChemistryEasyMCQMHT CET · 2025
Which of the following substances is classified as a macromolecular colloid?
A
Soap
B
Detergent
C
$S_8$ sulphur molecules
D
Nylon
Solution
(D) Macromolecular colloids are substances that have very large molecular masses and form colloidal solutions when dispersed in a suitable solvent.
Examples include naturally occurring polymers like starch,cellulose,proteins,and synthetic polymers like $Nylon$,$polythene$,and $polystyrene$.
Soap and detergents are associated colloids (micelles),while $S_8$ sulphur molecules are multimolecular colloids.
Therefore,$Nylon$ is classified as a macromolecular colloid.
Examples include naturally occurring polymers like starch,cellulose,proteins,and synthetic polymers like $Nylon$,$polythene$,and $polystyrene$.
Soap and detergents are associated colloids (micelles),while $S_8$ sulphur molecules are multimolecular colloids.
Therefore,$Nylon$ is classified as a macromolecular colloid.
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507
ChemistryMediumMCQMHT CET · 2025
Which of the following is not a negatively charged sol?
A
$As_2S_3$
B
Haemoglobin
C
Clay
D
Congo Red
Solution
(B) Colloidal particles carry an electric charge.
$As_2S_3$ (Arsenic sulfide) is a well-known negatively charged sol.
Clay is also a negatively charged sol.
Congo red is a dye that forms a negatively charged sol.
Haemoglobin is a positively charged sol because it is a protein that exists in a positively charged state at a $pH$ below its isoelectric point.
Therefore,Haemoglobin is not a negatively charged sol.
$As_2S_3$ (Arsenic sulfide) is a well-known negatively charged sol.
Clay is also a negatively charged sol.
Congo red is a dye that forms a negatively charged sol.
Haemoglobin is a positively charged sol because it is a protein that exists in a positively charged state at a $pH$ below its isoelectric point.
Therefore,Haemoglobin is not a negatively charged sol.
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508
ChemistryEasyMCQMHT CET · 2025
Identify the property of colloidal solutions where colloidal particles move freely towards respective electrodes under an applied electrical potential.
A
Brownian motion
B
Electrophoresis
C
Electroosmosis
D
Dialysis
Solution
(B) The movement of colloidal particles under an applied electric field is known as $Electrophoresis$.
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.
$Brownian \ motion$ refers to the random zig-zag movement of particles.
$Electroosmosis$ refers to the movement of the dispersion medium under an electric field.
$Dialysis$ is a process of removing impurities from a colloidal solution.
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.
$Brownian \ motion$ refers to the random zig-zag movement of particles.
$Electroosmosis$ refers to the movement of the dispersion medium under an electric field.
$Dialysis$ is a process of removing impurities from a colloidal solution.
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509
ChemistryEasyMCQMHT CET · 2025
Which of the following processes does $NOT$ result in coagulation?
A
Electrophoresis
B
Adding excess solvent
C
Mixing two oppositely charged sols
D
Boiling
Solution
(B) Coagulation is the process of aggregating colloidal particles to form a precipitate.
$1$. Electrophoresis causes particles to move towards oppositely charged electrodes and discharge,leading to coagulation.
$2$. Mixing two oppositely charged sols leads to mutual coagulation.
$3$. Boiling increases the kinetic energy of particles,causing them to collide and overcome the protective layer,leading to coagulation.
$4$. Adding excess solvent (dilution) does not cause coagulation; in fact,it often stabilizes the sol by increasing the distance between particles.
$1$. Electrophoresis causes particles to move towards oppositely charged electrodes and discharge,leading to coagulation.
$2$. Mixing two oppositely charged sols leads to mutual coagulation.
$3$. Boiling increases the kinetic energy of particles,causing them to collide and overcome the protective layer,leading to coagulation.
$4$. Adding excess solvent (dilution) does not cause coagulation; in fact,it often stabilizes the sol by increasing the distance between particles.
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510
ChemistryEasyMCQMHT CET · 2025
Which of the following ions exhibits the maximum power of coagulation for a positively charged sol?
A
$SO_4^{2-}$
B
$Cl^{-}$
C
$PO_4^{3-}$
D
$\left[Fe(CN)_6\right]^{4-}$
Solution
(D) According to the Hardy-Schulze rule,the coagulating power of an ion is directly proportional to the magnitude of its charge.
For a positively charged sol,the coagulating power depends on the valency of the anion.
The higher the negative charge on the anion,the greater is its coagulating power.
The charges on the given anions are:
$A: SO_4^{2-}$ (charge $-2$)
$B: Cl^{-}$ (charge $-1$)
$C: PO_4^{3-}$ (charge $-3$)
$D: [Fe(CN)_6]^{4-}$ (charge $-4$)
Since the ion $\left[Fe(CN)_6\right]^{4-}$ has the highest magnitude of negative charge $(-4)$,it exhibits the maximum power of coagulation for a positively charged sol.
For a positively charged sol,the coagulating power depends on the valency of the anion.
The higher the negative charge on the anion,the greater is its coagulating power.
The charges on the given anions are:
$A: SO_4^{2-}$ (charge $-2$)
$B: Cl^{-}$ (charge $-1$)
$C: PO_4^{3-}$ (charge $-3$)
$D: [Fe(CN)_6]^{4-}$ (charge $-4$)
Since the ion $\left[Fe(CN)_6\right]^{4-}$ has the highest magnitude of negative charge $(-4)$,it exhibits the maximum power of coagulation for a positively charged sol.
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511
ChemistryEasyMCQMHT CET · 2025
Identify the negatively charged sol from the following.
A
Eosin
B
Titanium oxide
C
Methylene blue
D
Haemoglobin
Solution
(A) Colloids are classified based on the charge on the dispersed phase particles.
$1$. Negatively charged sols include metal sulphides (e.g.,$As_2S_3$),acidic dyes (e.g.,Eosin,Congo red),and starch.
$2$. Positively charged sols include metal hydroxides (e.g.,$Fe(OH)_3$),basic dyes (e.g.,Methylene blue),and proteins like Haemoglobin.
$3$. Titanium oxide $(TiO_2)$ is generally considered a positively charged sol.
Therefore,Eosin is a negatively charged sol.
$1$. Negatively charged sols include metal sulphides (e.g.,$As_2S_3$),acidic dyes (e.g.,Eosin,Congo red),and starch.
$2$. Positively charged sols include metal hydroxides (e.g.,$Fe(OH)_3$),basic dyes (e.g.,Methylene blue),and proteins like Haemoglobin.
$3$. Titanium oxide $(TiO_2)$ is generally considered a positively charged sol.
Therefore,Eosin is a negatively charged sol.
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512
ChemistryEasyMCQMHT CET · 2025
Identify the technique used to know the binding nature of nanomaterials.
A
$SEM$
B
$TEM$
C
$XRD$
D
$FTIR$
Solution
(D) $FTIR$ (Fourier Transform Infrared Spectroscopy) is the technique used to identify the functional groups and the binding nature of nanomaterials by measuring the absorption of infrared radiation.
$SEM$ (Scanning Electron Microscopy) is used for surface morphology.
$TEM$ (Transmission Electron Microscopy) is used for internal structure and size.
$XRD$ ($X$-Ray Diffraction) is used for crystal structure analysis.
$SEM$ (Scanning Electron Microscopy) is used for surface morphology.
$TEM$ (Transmission Electron Microscopy) is used for internal structure and size.
$XRD$ ($X$-Ray Diffraction) is used for crystal structure analysis.
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513
ChemistryEasyMCQMHT CET · 2025
Which instrument is used for the determination of particle size?
A
$SEM$
B
$TEM$
C
$FTIR$
D
$UV$ visible spectroscopy
Solution
(B) Both $SEM$ (Scanning Electron Microscopy) and $TEM$ (Transmission Electron Microscopy) are used for the determination of particle size and morphology. However,$TEM$ is generally considered the standard technique for precise particle size measurement at the nanoscale. Given the options,both $A$ and $B$ are technically correct,but $TEM$ is the most common answer in this context.
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514
ChemistryEasyMCQMHT CET · 2025
Which of the following nanomaterials has all three dimensions less than $100 \ nm$?
A
Nanoshells
B
Nanotubes
C
Thin films
D
Fibres
Solution
(A) Nanomaterials are classified based on the number of dimensions that are outside the nanoscale range $(< 100 \ nm)$.
$1$. $0-D$ nanomaterials: All three dimensions are in the nanoscale range (e.g.,nanoparticles,quantum dots,nanoshells).
$2$. $1-D$ nanomaterials: Two dimensions are in the nanoscale range (e.g.,nanotubes,nanowires,fibres).
$3$. $2-D$ nanomaterials: One dimension is in the nanoscale range (e.g.,thin films,graphene).
Since nanoshells have all three dimensions less than $100 \ nm$,they are classified as $0-D$ nanomaterials.
$1$. $0-D$ nanomaterials: All three dimensions are in the nanoscale range (e.g.,nanoparticles,quantum dots,nanoshells).
$2$. $1-D$ nanomaterials: Two dimensions are in the nanoscale range (e.g.,nanotubes,nanowires,fibres).
$3$. $2-D$ nanomaterials: One dimension is in the nanoscale range (e.g.,thin films,graphene).
Since nanoshells have all three dimensions less than $100 \ nm$,they are classified as $0-D$ nanomaterials.
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515
ChemistryMediumMCQMHT CET · 2025
Match column $I$ (process) with column $II$ (application).
| Column $I$ | Column $II$ |
|---|---|
| $A$. Dialysis | $a$. Cleansing action of soap |
| $B$. Peptization | $b$. Coagulation |
| $C$. Emulsification | $c$. Colloidal solution preparation |
| $D$. Electrophoresis | $d$. Purification of colloidal solution |
A
$A-d, B-c, C-a, D-b$
B
$A-d, B-c, C-b, D-a$
C
$A-c, B-b, C-d, D-a$
D
$A-b, B-c, C-a, D-d$
Solution
(A) . 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.
$C$. Emulsification is the process of stabilizing an emulsion,which is the principle behind the cleansing action of soap.
$D$. Electrophoresis is the movement of colloidal particles under an applied electric field,which leads to their coagulation at the electrodes.
Therefore,the correct matching is $A-d, B-c, C-a, D-b$.
$B$. Peptization is the process of converting a freshly prepared precipitate into a colloidal sol by adding a suitable electrolyte.
$C$. Emulsification is the process of stabilizing an emulsion,which is the principle behind the cleansing action of soap.
$D$. Electrophoresis is the movement of colloidal particles under an applied electric field,which leads to their coagulation at the electrodes.
Therefore,the correct matching is $A-d, B-c, C-a, D-b$.
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516
ChemistryEasyMCQMHT CET · 2025
Identify the sequential reactions involved in the Sol-Gel process.
A
Polycondensation and hydrolysis
B
Hydrolysis and polycondensation
C
Polycondensation and oxidation
D
Oxidation and hydrolysis
Solution
(B) The Sol-Gel process is a chemical solution deposition technique used to produce solid materials from small molecules.
It involves two primary sequential steps:
$1$. Hydrolysis: The precursor (usually a metal alkoxide) reacts with water to form hydroxyl groups.
$2$. Polycondensation: These hydroxyl groups then undergo condensation reactions to form a network of metal-oxygen-metal bonds $(M-O-M)$,leading to the formation of a gel.
It involves two primary sequential steps:
$1$. Hydrolysis: The precursor (usually a metal alkoxide) reacts with water to form hydroxyl groups.
$2$. Polycondensation: These hydroxyl groups then undergo condensation reactions to form a network of metal-oxygen-metal bonds $(M-O-M)$,leading to the formation of a gel.
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