Which of the following will be the most effective in the coagulation of ${\rm{Fe}}{\left( {{\rm{OH}}} \right)_3}$ sol?

${\rm{M}}{{\rm{g}}_3}{\left( {{\rm{P}}{{\rm{O}}_4}} \right)_2}$

SURFACE CHEMISTRY MCQs for NEET UG CHEMISTRY — Easy, Page 2

31 Easy

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Catalysts are generally used in finely divided state because

View answer and explanation

Correct option: C

It has more surface

Option C (It has more surface) is correct because finely divided catalysts provide a larger surface area for the reaction to occur.

Solution approach

Correct Answer + Core Reason: It has more surface. A finely divided catalyst provides a larger surface area for adsorption and reaction.

Step-by-Step Solution:

Step 1: Understand the relationship between surface area and catalytic activity.

Step 2: Catalytic activity is directly proportional to the surface area of the catalyst.

Step 3: A finely divided catalyst has a larger surface area compared to a bulk catalyst.

Step 4: Therefore, a finely divided catalyst is more effective.

Why other options are wrong:
- Option A: It avoids wastage of catalyst is incorrect because the amount of catalyst used is not directly related to its state of division.
- Option B: We can see its reaction is incorrect because visibility does not affect the reaction rate.
- Option D: It has no effect on reaction rate is incorrect because the surface area significantly affects the reaction rate.

Key Concepts:
- Surface area
- Catalytic activity
- Adsorption

Memory Aids:
- Finely divided catalyst: "more surface, faster reaction".

Quick Tips:
- Remember that increasing the surface area of a catalyst increases its effectiveness.

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Which of the following forms cationic micelles above certain concentration?

View answer and explanation

Correct option: D

Cetyltrimethylammonium bromide

Option D (Cetyltrimethylammonium bromide) is correct because it contains a positively charged head group, allowing it to form cationic micelles.

Solution approach

Correct Answer + Core Reason: Cetyltrimethylammonium bromide. It has a positively charged head group and forms cationic micelles.

Step-by-Step Solution:

Step 1: Understand the structure of micelles.

Step 2: Cationic micelles are formed by surfactants with a positively charged head group.

Step 3: Analyze the given options:
- Urea is a neutral molecule and does not form micelles.
- Sodium dodecyl sulphate is an anionic surfactant and forms anionic micelles.
- Sodium acetate is a salt of a weak acid and does not typically form micelles.
- Cetyltrimethylammonium bromide is a quaternary ammonium salt with a long hydrocarbon chain, making it a cationic surfactant.

Step 4: Cetyltrimethylammonium bromide (CTAB) has a positive charge on the nitrogen atom, making it cationic.

Why other options are wrong:
- Option A: Urea is a neutral organic compound and does not form micelles.
- Option B: Sodium dodecyl sulphate is an anionic surfactant, forming anionic micelles, not cationic.
- Option C: Sodium acetate is a simple salt and does not have the necessary structure to form micelles.

Key Concepts:
- Micelles
- Surfactants
- Cationic surfactants
- Anionic surfactants

Memory Aids:
- Cationic = Positive charge

Quick Tips:
- Look for the charged head group in the surfactant molecule.

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A colloidal solution always has at least :

View answer and explanation

Correct option: D

Two phases

Option D (Two phases) is correct because a colloidal solution consists of a dispersed phase and a dispersion medium, making it a two-phase system.

Solution approach

Correct Answer + Core Reason: Two phases. A colloidal solution has a dispersed phase and a dispersion medium.

Step-by-Step Solution:

Step 1: Define a colloidal solution. A colloidal solution is a heterogeneous mixture with particles of one substance dispersed in another substance.

Step 2: Identify the phases in a colloidal solution. A colloidal solution always has two phases: the dispersed phase (the particles) and the dispersion medium (the substance in which the particles are dispersed).

Why other options are wrong:
- Option A: One phase is incorrect because a colloidal solution is a heterogeneous mixture with at least two phases.
- Option B: More than two phases is incorrect because a simple colloidal system has only two phases.
- Option C: A true solution is incorrect because a true solution is a homogeneous mixture with one phase, unlike a colloidal solution.

Key Concepts:
- Colloidal solution: A heterogeneous mixture with a dispersed phase and a dispersion medium.
- Dispersed phase: The particles in a colloidal solution.
- Dispersion medium: The substance in which the particles are dispersed.

Memory Aids:
- Colloidal = Two Phases.

Quick Tips:
- Remember that colloidal solutions are heterogeneous mixtures with two distinct phases.

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In whipped cream, the dispersed phase and dispersion medium respectively are

View answer and explanation

Correct option: B

Gas and liquid

Option B is correct because whipped cream is a foam, which consists of a gas (air) dispersed in a liquid (cream).

Solution approach

Correct Answer + Core Reason: Gas and liquid. In whipped cream, gas (air) is dispersed in a liquid (cream).

Step-by-Step Solution:

Step 1: Define dispersed phase and dispersion medium. The dispersed phase is the substance that is present in the form of particles or droplets, and the dispersion medium is the substance in which the dispersed phase is distributed.

Step 2: Identify the dispersed phase and dispersion medium in whipped cream. In whipped cream, air (gas) is dispersed in cream (liquid). Therefore, the dispersed phase is gas, and the dispersion medium is liquid.

Why other options are wrong:
- Option A: Liquid and liquid. This describes an emulsion, not a foam like whipped cream.
- Option C: Liquid and gas. This is the reverse of what is found in whipped cream.
- Option D: Liquid and solid. This is not applicable to whipped cream.

Key Concepts:
- Colloidal solutions
- Foams
- Dispersed phase and dispersion medium

Memory Aids:
- Foam: Gas in liquid

Quick Tips:
- Remember the definitions of different types of colloids based on their dispersed phase and dispersion medium.

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Emulsions of polyvinylacetate are used in :

View answer and explanation

Correct option: B

Latex paints

Option B is correct because emulsions of polyvinylacetate are commonly used as binders in latex paints.

Solution approach

Correct Answer + Core Reason: Latex paints. Emulsions of polyvinylacetate are used in latex paints as binders.

Step-by-Step Solution:

Step 1: Understand the role of polyvinylacetate (PVA) in latex paints. Polyvinylacetate is a polymer that is commonly used as a binder in latex paints. The binder holds the pigment particles together and allows the paint to adhere to the surface.

Step 2: Explain why PVA is used in emulsion form. PVA is typically used in emulsion form because it is easier to apply and provides good film formation properties.

Step 3: Consider the other options.
- Polishes: Polishes may contain other polymers or waxes, but PVA emulsions are not typically the primary component.
- Fire works: Fire works use different chemical compounds for color and explosive effects.
- Rayons: Rayon is a manufactured fiber made from regenerated cellulose.

Why other options are wrong:
- Option A: Polishes. Polishes may contain other polymers or waxes, but PVA emulsions are not typically the primary component.
- Option C: Fire works. Fire works use different chemical compounds for color and explosive effects.
- Option D: Rayons. Rayon is a manufactured fiber made from regenerated cellulose.

Key Concepts:
- Polymers
- Emulsions
- Latex paints
- Binders

Memory Aids:
- PVA = Paints

Quick Tips:
- Remember the common applications of different polymers.

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At CMC (critical micelle concentration), the surfactant molecules undergo

View answer and explanation

Correct option: B

Micelle formation

Option B (Micelle formation) is correct because at the critical micelle concentration (CMC), surfactant molecules aggregate to form micelles.

Solution approach

Correct Answer + Core Reason: Micelle formation. At CMC, surfactant molecules aggregate to form micelles.

Step-by-Step Solution:

Step 1: Define CMC. The critical micelle concentration (CMC) is the concentration above which micelles spontaneously form.

Step 2: Explain micelle formation. At concentrations below the CMC, surfactant molecules exist as individual molecules. Above the CMC, they aggregate to form micelles.

Why other options are wrong:
- Option A: Dissociation is incorrect because surfactant molecules associate to form micelles, not dissociate.
- Option C: Both (1) and (2) is incorrect because only micelle formation occurs at CMC.
- Option D: None of these is incorrect because micelle formation is the correct answer.

Key Concepts:
- Critical micelle concentration (CMC): The concentration at which micelles form.
- Micelle: An aggregate of surfactant molecules.

Memory Aids:
- CMC = Micelle Formation.

Quick Tips:
- Remember that micelle formation is the defining characteristic of CMC.

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Freundlich adsorption isotherm is

View answer and explanation

Correct option: A

\frac{x}{m} = k{p^{1/n}}

Option A (

\frac{x}{m} = k{p^{1/n}}

) is correct because it represents the Freundlich adsorption isotherm equation, which relates the amount of gas adsorbed to the pressure.

Solution approach

Correct Answer + Core Reason: $\frac{x}{m} = k{p^{1/n}}$ . This is the standard equation for the Freundlich adsorption isotherm.

Step-by-Step Solution:

Step 1: Define Freundlich adsorption isotherm: It is an empirical equation describing the adsorption of gases onto solid surfaces.

Step 2: State the equation: The Freundlich adsorption isotherm is given by $\frac{x}{m} = k{p^{1/n}}$ , where:
- $x$ is the mass of the gas adsorbed
- $m$ is the mass of the adsorbent
- $p$ is the pressure of the gas
- $k$ and $n$ are empirical constants

Step 3: Relate to the question: The question asks for the Freundlich adsorption isotherm, which is $\frac{x}{m} = k{p^{1/n}}$ .

Why other options are wrong:
- Option B: $x = mk{p^{1/n}}$ is incorrect because it is a rearrangement of the correct equation, but not the standard form.
- Option C: $x/m = k{p^{ - n}}$ is incorrect because the exponent of $p$ should be $1/n$ , not $-n$ .
- Option D: All of these is incorrect because only option A is correct.

Key Concepts:
- Freundlich Adsorption Isotherm
- Adsorption Isotherms

Memory Aids:
- Freundlich: Fraction ( $\frac{x}{m}$ ) = constant * pressure^(1/n).

Quick Tips:
- Remember the form of the Freundlich adsorption isotherm equation.

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The migration of colloidal particles under the influence of an electric field is known as

View answer and explanation

Correct option: C

Electrophoresis

Option C (Electrophoresis) is correct because it describes the movement of colloidal particles under an electric field.

Solution approach

Correct Answer + Core Reason: Electrophoresis. It is the movement of colloidal particles under the influence of an electric field.

Step-by-Step Solution:

Step 1: Define electrophoresis.

Step 2: Electrophoresis is the migration of charged colloidal particles through a fluid under the influence of an electric field.

Step 3: This movement is due to the electrostatic interaction between the charged particles and the electric field.

Why other options are wrong:
- Option A: Brownian movement is the random motion of particles in a fluid due to collisions with other molecules.
- Option B: Dialysis is the separation of smaller molecules from larger ones through a semipermeable membrane.
- Option D: Electro-osmosis is the movement of a liquid through a membrane under the influence of an electric field.

Key Concepts:
- Electrophoresis
- Colloids
- Electric field

Memory Aids:
- Electrophoresis: "movement by electricity".

Quick Tips:
- Remember the definitions of different colloidal phenomena.

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Emulsion is a colloidal solution of

View answer and explanation

Correct option: A

Liquid in liquid

Option A is correct because an emulsion is a colloidal solution where both the dispersed phase and the dispersion medium are liquids.

Solution approach

Correct Answer + Core Reason: Liquid in liquid. An emulsion is a colloid where both the dispersed phase and the dispersion medium are liquids.

Step-by-Step Solution:

Step 1: Define emulsion. An emulsion is a colloidal dispersion of two or more immiscible liquids, where one liquid is dispersed as droplets in the other liquid.

Step 2: Identify the dispersed phase and dispersion medium in an emulsion. Both the dispersed phase (the liquid that is present in the form of droplets) and the dispersion medium (the liquid in which the droplets are dispersed) are liquids.

Step 3: Examples of emulsions. Common examples of emulsions include milk (fat dispersed in water), mayonnaise (oil dispersed in vinegar), and many cosmetic creams and lotions.

Why other options are wrong:
- Option B: Liquid in solid. This describes a gel, not an emulsion.
- Option C: Gas in liquid. This describes a foam, not an emulsion.
- Option D: Solid in gas. This is not a common type of colloid.

Key Concepts:
- Colloidal solutions
- Emulsions
- Dispersed phase and dispersion medium

Memory Aids:
- Emulsion: Think of \"immiscible liquids\" to remember it's liquid in liquid.

Quick Tips:
- Remember the definitions of different types of colloids based on their dispersed phase and dispersion medium.

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An increase in the concentration of adsorbate at the surface relative to its concentration in bulk phase is called :

View answer and explanation

Correct option: A

Adsorption

Option A (Adsorption) is correct because adsorption is the process where the concentration of a substance increases at the surface of a material compared to its concentration in the bulk phase.

Solution approach

Correct Answer + Core Reason: Adsorption. Adsorption is the accumulation of adsorbate at the surface.

Step-by-Step Solution:

Step 1: Define the terms: Adsorption, Absorption, Enthalpy.

Step 2: Adsorption is a surface phenomenon where molecules adhere to the surface of a solid or liquid.

Step 3: Absorption is a bulk phenomenon where molecules penetrate into the volume of a solid or liquid.

Step 4: Enthalpy is a thermodynamic property related to the heat content of a system.

Step 5: The question describes a surface phenomenon where the concentration of a substance increases at the surface. This is the definition of adsorption.

Why other options are wrong:
- Option B: Enthalpy is a thermodynamic property and not a process related to concentration changes at a surface.
- Option C: Absorption is a bulk phenomenon, not a surface phenomenon as described in the question.
- Option D: Option A is correct, so "None of these" is incorrect.

Key Concepts:
- Adsorption
- Absorption
- Surface Chemistry

Memory Aids:
- Adsorption: "At the surface"
- Absorption: "Into the bulk"

Quick Tips:
- Differentiate between adsorption (surface) and absorption (bulk).

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Adsorption is accompanied by the evolution of heat. So, according to Le-Chatelier principle the amount of substance adsorbed should

View answer and explanation

Correct option: A

Increase with decrease in temperature

Option A (Increase with decrease in temperature) is correct because adsorption is an exothermic process, and according to Le Chatelier's principle, decreasing the temperature favors the exothermic direction, thus increasing adsorption.

Solution approach

Correct Answer + Core Reason: Increase with decrease in temperature. Adsorption is exothermic, so lower temperature favors adsorption.

Step-by-Step Solution:
Step 1: Adsorption is generally an exothermic process, meaning heat is released during adsorption ( $\Delta H < 0$ ).
Step 2: According to Le Chatelier's principle, if a system at equilibrium is subjected to a change in temperature, pressure, or concentration, the system will adjust itself to counteract the change and restore a new equilibrium.
Step 3: In this case, since adsorption is exothermic, decreasing the temperature will favor the forward reaction (adsorption) to release more heat and counteract the change.
Step 4: Therefore, the amount of substance adsorbed should increase with a decrease in temperature.

Why other options are wrong:
- Option B (Increase with increase in temperature): This is incorrect because increasing the temperature would favor the reverse reaction (desorption), not adsorption.
- Option C (Decrease with decrease in temperature): This is incorrect because decreasing the temperature favors the exothermic process of adsorption.
- Option D (Decrease with increase in temperature): This is correct for the reverse process (desorption) but not for adsorption itself.

Key Concepts:
- Adsorption
- Exothermic Process
- Le Chatelier's Principle

Memory Aids:
- Exothermic = Heat released = Favored by low temperature

Quick Tips:
- Remember that exothermic processes are favored by lower temperatures and endothermic processes are favored by higher temperatures.

42 Easy

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Which is a homogeneous system?

View answer and explanation

Correct option: A

A solution of sugar in water

Option A (A solution of sugar in water) is correct because a homogeneous system has uniform composition and properties throughout, which is true for a sugar solution.

Solution approach

Correct Answer + Core Reason: A solution of sugar in water. A homogeneous system has uniform composition throughout.

Step-by-Step Solution:

Step 1: Define a homogeneous system. A homogeneous system is a mixture that has uniform composition and properties throughout.

Step 2: Consider the given options. A solution of sugar in water is a homogeneous mixture because the sugar molecules are evenly distributed throughout the water.

Why other options are wrong:
- Option B: Concrete is a heterogeneous mixture of cement, sand, and gravel.
- Option C: Muddy water is a heterogeneous mixture of water and soil particles.
- Option D: Bread is a heterogeneous mixture of flour, water, yeast, and other ingredients.

Key Concepts:
- Homogeneous system: A mixture with uniform composition and properties.
- Heterogeneous system: A mixture with non-uniform composition and properties.

Memory Aids:
- Homogeneous = Uniform.

Quick Tips:
- Remember that solutions are typically homogeneous, while suspensions and colloids are heterogeneous.

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The substance that gets adsorbed on the surface of solid is called

View answer and explanation

Correct option: A

Adsorbate

Option A (Adsorbate) is correct because the substance that gets adsorbed on the surface of a solid is, by definition, the adsorbate.

Solution approach

Correct Answer + Core Reason: Adsorbate. The substance being adsorbed is called the adsorbate.

Step-by-Step Solution:

Step 1: Define adsorption. Adsorption is the adhesion of atoms, ions, or molecules from a gas, liquid, or dissolved solid to a surface.

Step 2: Identify the components. In adsorption, there are two key components: the adsorbate and the adsorbent.

Step 3: Define adsorbate. The adsorbate is the substance that is being adsorbed onto the surface.

Step 4: Define adsorbent. The adsorbent is the solid surface onto which the adsorbate is adsorbed.

Step 5: Conclusion. Therefore, the substance that gets adsorbed on the surface of solid is called adsorbate.

Why other options are wrong:
- Option B: Adsorbent is the material doing the adsorbing, not the substance being adsorbed.
- Option C: Micelle is an aggregate of surfactant molecules dispersed in a liquid colloid. It is not related to adsorption in this context.
- Option D: Absorbent is a substance that absorbs another substance, meaning it permeates the bulk of the absorbent, rather than just adhering to the surface.

Key Concepts:
- Adsorption
- Adsorbate
- Adsorbent

Memory Aids:
- Adsorbate is \"what is AT the surface.\" Adsorbent is \"what DOES the adsorbing.\"

Quick Tips:
- Remember the difference between adsorption and absorption. Adsorption is a surface phenomenon, while absorption involves the entire volume of the material.

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Analyzing +4 | -1

Assertion: Whipped cream is a colloid.

Reason: Whipped cream is a foam, which is a liquid dispersed in a gas.

View answer and explanation

Correct option: C

If Assertion is correct but Reason is incorrect.

Option C is correct because whipped cream is a colloid, but it's a gas dispersed in a liquid, not a liquid dispersed in a gas.

Solution approach

Correct Answer + Core Reason: If Assertion is correct but Reason is incorrect. Whipped cream is a colloid, but its dispersed phase and dispersion medium are incorrectly stated.

Step-by-Step Solution:

Step 1: Analyze the assertion. The assertion states that whipped cream is a colloid. This is correct because whipped cream consists of air bubbles (gas) dispersed in a liquid (cream).

Step 2: Analyze the reason. The reason states that whipped cream is a foam, which is a liquid dispersed in a gas. This is incorrect. A foam is a gas dispersed in a liquid. In whipped cream, air (gas) is dispersed in cream (liquid).

Step 3: Determine the relationship between the assertion and the reason. The assertion is correct, but the reason is incorrect. Therefore, the correct answer is that the assertion is correct, but the reason is incorrect.

Why other options are wrong:
- Option A: If both Assertion and Reason are correct and Reason is the correct explanation of Assertion. This is incorrect because the reason is incorrect.
- Option B: If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion. This is incorrect because the reason is incorrect.
- Option D: If both the Assertion and Reason are incorrect. This is incorrect because the assertion is correct.

Key Concepts:
- Colloids
- Foams
- Dispersed phase and dispersion medium

Memory Aids:
- Foam: Gas in liquid
- Emulsion: Liquid in liquid

Quick Tips:
- Remember the definitions of different types of colloids based on their dispersed phase and dispersion medium.

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Which one of the following equation represents Freundlich adsorption isotherm?

View answer and explanation

Correct option: B

\frac{x}{m} = k{p^n}

Option B is correct because the Freundlich adsorption isotherm is represented by the equation x/m = k*p^n, where n is between 0 and 1.

Solution approach

Correct Answer + Core Reason: $\frac{x}{m} = k{p^n}$ . This is the correct Freundlich adsorption isotherm equation.

Step-by-Step Solution:

Step 1: Recall the Freundlich adsorption isotherm. The Freundlich adsorption isotherm is an empirical equation that describes the adsorption of a gas on a solid surface.

Step 2: State the Freundlich adsorption isotherm equation. The equation is given by:
$\frac{x}{m} = k{p^n}$ \nwhere:
- x is the mass of the gas adsorbed
- m is the mass of the adsorbent
- p is the pressure of the gas
- k and n are constants that depend on the nature of the adsorbent and the gas at a particular temperature. The value of n is between 0 and 1.

Why other options are wrong:
- Option A: $\frac{x}{m} = kp$ . This is a simplified version of the Freundlich isotherm where n=1, which is not the general form.
- Option C: $\log \frac{x}{m} = k{p^n}$ . This is not the correct form of the Freundlich isotherm.
- Option D: $\log \frac{x}{m} = kn\log p$ . This is the logarithmic form of the Freundlich isotherm, but the question asks for the equation itself, not its logarithmic form. The correct logarithmic form is $\log \frac{x}{m} = \log k + n \log p$ .

Key Concepts:
- Adsorption
- Adsorption isotherm
- Freundlich adsorption isotherm

Memory Aids:
- Remember the equation: x/m = k*p^n

Quick Tips:
- Know the different types of adsorption isotherms and their equations.

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Which adsorption takes place at low temperature?

View answer and explanation

Correct option: A

Physical

Option A is correct because physical adsorption (physisorption) is favored at low temperatures due to weak van der Waals forces.

Solution approach

Correct Answer + Core Reason: Physical. Physical adsorption is favored at low temperatures.

Step-by-Step Solution:

Step 1: Define physical adsorption (physisorption). Physical adsorption is a type of adsorption in which the adsorbate is held to the surface of the adsorbent by weak van der Waals forces.

Step 2: Define chemical adsorption (chemisorption). Chemical adsorption is a type of adsorption in which the adsorbate is held to the surface of the adsorbent by chemical bonds.

Step 3: Explain the effect of temperature on physical adsorption. Physical adsorption is an exothermic process, meaning it releases heat. According to Le Chatelier's principle, decreasing the temperature will favor the exothermic process, thus increasing the extent of physical adsorption.

Step 4: Explain the effect of temperature on chemical adsorption. Chemical adsorption typically requires an activation energy to break and form chemical bonds. Increasing the temperature can provide the necessary activation energy, thus increasing the extent of chemical adsorption.

Step 5: Conclusion. Physical adsorption takes place at low temperatures because it is an exothermic process favored by lower temperatures.

Why other options are wrong:
- Option B: Chemical. Chemical adsorption is generally favored at higher temperatures because it requires activation energy.
- Option C: Both (1) and (2). This is incorrect because physical adsorption is favored at low temperatures, while chemical adsorption is favored at high temperatures.
- Option D: None of these. This is incorrect because physical adsorption takes place at low temperatures.

Key Concepts:
- Adsorption
- Physical adsorption (physisorption)
- Chemical adsorption (chemisorption)
- Le Chatelier's principle

Memory Aids:
- Physisorption: \"Physical\" = weak forces, low temperature.
- Chemisorption: \"Chemical\" = strong bonds, high temperature.

Quick Tips:
- Remember the difference between physical and chemical adsorption and how temperature affects each.

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Application +4 | -1

Which of the following is the most effective in the coagulation of gold sol?

View answer and explanation

Correct option: B

\mathrm{MgC}{\mathrm{l}_2}

Option B (

\mathrm{MgC}{\mathrm{l}_2}

) is correct because gold sol is negatively charged, and according to the Hardy-Schulze rule, the cation with the highest charge will be most effective for coagulation.

Mg^{2+}

has a higher charge than

Na^+

.

Solution approach

Correct Answer + Core Reason: $\mathrm{MgC}{\mathrm{l}_2}$ . Higher charge on the cation leads to better coagulation of negatively charged gold sol.

Step-by-Step Solution:

Step 1: Recall the Hardy-Schulze rule.

Step 2: Gold sol is negatively charged.

Step 3: According to the Hardy-Schulze rule, the coagulating power of an electrolyte is directly proportional to the valency of the ion causing coagulation.

Step 4: Compare the charges of the cations in each option:
- $\mathrm{NaN}{\mathrm{O}_3}$ : $Na^+$ (charge +1)
- $\mathrm{MgC}{\mathrm{l}_2}$ : $Mg^{2+}$ (charge +2)
- $\mathrm{N}{\mathrm{a}_3}\mathrm{P}{\mathrm{O}_4}$ : $Na^+$ (charge +1)
- ${\mathrm{K}_4}[{\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}}]$ : This is an anion, so it won't coagulate a negative sol.

Step 5: $Mg^{2+}$ has the highest charge, so $\mathrm{MgC}{\mathrm{l}_2}$ will be the most effective in coagulating gold sol.

Why other options are wrong:
- Option A: $\mathrm{NaN}{\mathrm{O}_3}$ contains $Na^+$ , which has a +1 charge, less effective than $Mg^{2+}$ .
- Option C: $\mathrm{N}{\mathrm{a}_3}\mathrm{P}{\mathrm{O}_4}$ contains $Na^+$ , which has a +1 charge, less effective than $Mg^{2+}$ .
- Option D: ${\mathrm{K}_4}[{\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}}]$ is an anionic complex, which would stabilize the negatively charged gold sol rather than coagulate it.

Key Concepts:
- Hardy-Schulze rule
- Coagulation
- Gold sol

Memory Aids:
- Higher charge = higher coagulating power.

Quick Tips:
- Apply the Hardy-Schulze rule to determine the most effective coagulating agent.

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Which of the following statements is correct about Langmuir’s adsorption isotherm?

View answer and explanation

Correct option: A

It forms monolayer

Option A (It forms monolayer) is correct because Langmuir's adsorption isotherm assumes the formation of a monolayer of adsorbate on the adsorbent surface.

Solution approach

Correct Answer + Core Reason: It forms monolayer. Langmuir's adsorption isotherm is based on the assumption of monolayer adsorption.

Step-by-Step Solution:

Step 1: Langmuir's adsorption isotherm is a model that describes the adsorption of gas molecules onto a solid surface.

Step 2: The key assumption of the Langmuir isotherm is that adsorption occurs only at specific localized sites on the surface, and each site can hold at most one molecule (monolayer adsorption).

Step 3: The Langmuir isotherm is represented by the equation: $\theta = \frac{{Kp}}{{1 + Kp}}$ , where $\theta$ is the fractional surface coverage, $p$ is the pressure, and $K$ is the equilibrium constant for adsorption.

Step 4: The model assumes that once a site is occupied, no further adsorption can take place at that site, leading to the formation of a monolayer.

Why other options are wrong:
- Option B: Langmuir's adsorption isotherm can be considered reversible, as it involves an equilibrium between adsorption and desorption. However, the primary defining characteristic is monolayer formation, making option A more directly relevant.
- Option C: While temperature can influence adsorption, Langmuir's isotherm itself doesn't inherently dictate that it only occurs at low temperatures. The isotherm can be applied at various temperatures, although the equilibrium constant K is temperature-dependent.
- Option D: Option A is correct, so "None of the above" is incorrect.

Key Concepts:
- Langmuir Adsorption Isotherm
- Monolayer Adsorption
- Adsorption Equilibrium

Memory Aids:
- Langmuir = Layer (Monolayer)

Quick Tips:
- Remember the key assumptions of the Langmuir isotherm: monolayer adsorption, homogeneous surface, and no interaction between adsorbed molecules.

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Analyzing +4 | -1

Flocculation value of

\mathrm{BaCl}_2

is much less than that of

\mathrm{KCl}

for sol

\mathrm{A}

and flocculation value of

\mathrm{Na}_2 \mathrm{SO}_4

is much less than that of

\mathrm{NaBr}

for sol

\mathrm{B}

. The correct statement among the following is

View answer and explanation

Correct option: B

Sol A is positively charged and Sol B is negatively charged.

Option B is correct because the flocculation value is inversely proportional to the charge on the sol. Lower flocculation value implies higher charge, hence A is positive and B is negative.

Solution approach

Correct Answer + Core Reason: Sol A is positively charged and Sol B is negatively charged. Flocculation value is inversely proportional to the charge of the ion causing flocculation.

Step-by-Step Solution:

Step 1: Understand the concept of flocculation value. Flocculation value is the minimum concentration of an electrolyte required to cause coagulation or precipitation of a sol.

Step 2: Relate flocculation value to the charge of the ion. According to the Hardy-Schulze rule, the flocculation value is inversely proportional to the valency of the ion causing coagulation. This means that ions with higher charges are more effective in causing coagulation and thus have lower flocculation values.

Step 3: Analyze the given information for sol A. BaCl2 has a much lower flocculation value than KCl. This indicates that the Ba2+ ion is more effective in causing coagulation than the K+ ion. Therefore, sol A must be negatively charged, as positive ions cause the coagulation of negative sols.

Step 4: Analyze the given information for sol B. Na2SO4 has a much lower flocculation value than NaBr. This indicates that the SO42- ion is more effective in causing coagulation than the Br- ion. Therefore, sol B must be positively charged, as negative ions cause the coagulation of positive sols.

Step 5: Conclusion. Sol A is negatively charged, and sol B is positively charged. Therefore, the correct statement is that sol A is positively charged and sol B is negatively charged.

Why other options are wrong:
- Option A: Both the sols A and B are negatively charged. This is incorrect because the flocculation data indicates that sol B is positively charged.
- Option C: Both the sols A and B are positively charged. This is incorrect because the flocculation data indicates that sol A is negatively charged.
- Option D: Sol A is negatively charged and sol B is positively charged. This is incorrect because the analysis in Step 3 and 4 is reversed.

Key Concepts:
- Flocculation value
- Hardy-Schulze rule
- Colloidal solutions

Memory Aids:
- Hardy-Schulze rule: Higher the charge, greater the coagulating power.

Quick Tips:
- Remember that oppositely charged ions cause coagulation of sols.

50 Easy

50 of 56Question permalink

Remembering +4 | -1

In physical adsorption gas molecules are bound on the solid surface by

View answer and explanation

Correct option: D

Van der Waals’ forces

Option D (Van der Waals’ forces) is correct because physical adsorption involves weak intermolecular forces, specifically Van der Waals forces, between the gas molecules and the solid surface.

Solution approach

Correct Answer + Core Reason: Van der Waals’ forces. Physical adsorption is characterized by weak Van der Waals interactions.

Step-by-Step Solution:

Step 1: Define physical adsorption (physisorption). Physisorption is a type of adsorption in which the adsorbate adheres to the surface through weak intermolecular forces (Van der Waals forces).

Step 2: Describe Van der Waals forces. Van der Waals forces include dipole-dipole interactions, London dispersion forces, and dipole-induced dipole interactions.

Step 3: Contrast with chemical adsorption (chemisorption). Chemisorption involves stronger chemical bonds between the adsorbate and the adsorbent.

Step 4: Conclusion. In physical adsorption, gas molecules are bound on the solid surface by Van der Waals’ forces.

Why other options are wrong:
- Option A: Chemical forces are involved in chemisorption, not physisorption.
- Option B: Electrostatic forces can play a role in some types of adsorption, but Van der Waals forces are the primary forces in physical adsorption.
- Option C: Graphical forces is not a scientific term related to adsorption.

Key Concepts:
- Physical Adsorption (Physisorption)
- Van der Waals forces
- Chemical Adsorption (Chemisorption)

Memory Aids:
- Physisorption: \"PHYSICAL forces (Van der Waals)\"
- Chemisorption: \"CHEMICAL bonds\"

Quick Tips:
- Remember the difference between physisorption and chemisorption based on the strength of the interaction.

51 Easy

51 of 56Question permalink

Analyzing +4 | -1

Select wrong statement.

View answer and explanation

Correct option: A

If a very small amount of

\mathrm{AlC}{\mathrm{l}_3}

is added to gold sol, coagulation occurs, but if a large quantity of

\mathrm{AlC}{\mathrm{l}_3}

is added, there is no coagulation.

Option A is the wrong statement because adding a large quantity of electrolyte to a sol can lead to peptization, not necessarily preventing coagulation.

Solution approach

Correct Answer + Core Reason: If a very small amount of $\mathrm{AlC}{\mathrm{l}_3}$ is added to gold sol, coagulation occurs, but if a large quantity of $\mathrm{AlC}{\mathrm{l}_3}$ is added, there is no coagulation. This statement is incorrect because a large amount of electrolyte can cause peptization, but it doesn't necessarily prevent coagulation.

Step-by-Step Solution:

Step 1: Analyze each statement:
- A: If a very small amount of $\mathrm{AlC}{\mathrm{l}_3}$ is added to gold sol, coagulation occurs, but if a large quantity of $\mathrm{AlC}{\mathrm{l}_3}$ is added, there is no coagulation. This is incorrect. Adding a large quantity of electrolyte generally leads to coagulation due to charge neutralization. However, in some cases, it can lead to peptization, but it doesn't guarantee no coagulation.
- B: Organic ions are more strongly adsorbed on charged surfaces in comparison to inorganic ions. This is generally true due to the larger size and polarizability of organic ions.
- C: Both emulsifier and peptising agents stabilise colloids but their actions are different. This is true. Emulsifiers reduce interfacial tension, while peptizing agents introduce a common ion.
- D: Colloidal solutions are thermodynamically stable. This is incorrect. Colloidal solutions are thermodynamically unstable and require stabilizers.

Step 2: Identify the incorrect statement: Statement A is the most likely to be incorrect because while peptization can occur with excess electrolyte, it doesn't guarantee the prevention of coagulation.

Why other options are wrong:
- Option B: Organic ions are generally more strongly adsorbed than inorganic ions due to their larger size and polarizability. This statement is correct.
- Option C: Emulsifiers and peptizing agents do stabilize colloids through different mechanisms. This statement is correct.
- Option D: Colloidal solutions are thermodynamically unstable, requiring stabilizers. This statement is correct.

Key Concepts:
- Coagulation
- Peptization
- Adsorption
- Colloidal stability

Memory Aids:
- Coagulation is the aggregation of colloidal particles.
- Peptization is the dispersion of a precipitate into a colloidal solution by adding an electrolyte.

Quick Tips:
- Understand the difference between coagulation and peptization.
- Remember that colloidal solutions are generally thermodynamically unstable.

52 Easy

52 of 56Question permalink

Application +4 | -1

Which of the following will be the most effective in the coagulation of

\mathrm{Fe}{\left( {\mathrm{OH}} \right)_3}

sol?

View answer and explanation

Correct option: D

\mathrm{M}{\mathrm{g}_3}{\left( {\mathrm{P}{\mathrm{O}_4}} \right)_2}

Option D (

\mathrm{M}{\mathrm{g}_3}{\left( {\mathrm{P}{\mathrm{O}_4}} \right)_2}

) is correct because, according to the Hardy-Schulze rule, the ion with the highest charge will have the maximum coagulation power. Since

\mathrm{Fe}{\left( {\mathrm{OH}} \right)_3}

is a positive sol, we need to look for the anion with the highest charge, which is

PO_4^{3-}

.

Solution approach

Correct Answer + Core Reason: $\mathrm{M}{\mathrm{g}_3}{\left( {\mathrm{P}{\mathrm{O}_4}} \right)_2}$ . The anion with the highest charge has the maximum coagulation power for a positive sol.

Step-by-Step Solution:

Step 1: Identify the charge of the $\mathrm{Fe}{\left( {\mathrm{OH}} \right)_3}$ sol.

Step 2: $\mathrm{Fe}{\left( {\mathrm{OH}} \right)_3}$ sol is positively charged.

Step 3: According to the Hardy-Schulze rule, the anion with the highest charge will have the maximum coagulation power.

Step 4: Determine the charge of the anions in each option:
- $KCN$ : $CN^-$ (charge -1)
- $\mathrm{BaC}{\mathrm{l}_2}$ : $Cl^-$ (charge -1)
- $NaCl$ : $Cl^-$ (charge -1)
- $\mathrm{M}{\mathrm{g}_3}{\left( {\mathrm{P}{\mathrm{O}_4}} \right)_2}$ : $PO_4^{3-}$ (charge -3)

Step 5: $PO_4^{3-}$ has the highest charge (-3), so $\mathrm{M}{\mathrm{g}_3}{\left( {\mathrm{P}{\mathrm{O}_4}} \right)_2}$ will be the most effective in coagulating $\mathrm{Fe}{\left( {\mathrm{OH}} \right)_3}$ sol.

Why other options are wrong:
- Option A: $KCN$ contains $CN^-$ , which has a -1 charge, less effective than $PO_4^{3-}$ .
- Option B: $\mathrm{BaC}{\mathrm{l}_2}$ contains $Cl^-$ , which has a -1 charge, less effective than $PO_4^{3-}$ .
- Option C: $NaCl$ contains $Cl^-$ , which has a -1 charge, less effective than $PO_4^{3-}$ .

Key Concepts:
- Hardy-Schulze rule
- Coagulation
- Charge of ions

Memory Aids:
- Higher charge = higher coagulation power.

Quick Tips:
- Apply the Hardy-Schulze rule to determine the most effective coagulating agent.

53 Easy

53 of 56Question permalink

Remembering +4 | -1

The process which is catalysed by one of the products formed during the reaction is known as :

View answer and explanation

Correct option: A

Autocatalysis

The correct answer is A (Autocatalysis) because autocatalysis is when a product of the reaction acts as a catalyst.

Solution approach

Correct Answer + Core Reason: Autocatalysis. Autocatalysis is the process where a product of the reaction itself acts as a catalyst.

Step-by-Step Solution:

Step 1: Define autocatalysis.

Step 2: Autocatalysis is when a product of the reaction acts as a catalyst.

Why other options are wrong:
- Option B: Anticatalysis is not a standard term.
- Option C: Negative catalysis slows down the reaction.
- Option D: Acid catalysis involves acids as catalysts.

Key Concepts:
- Autocatalysis
- Catalysis

Memory Aids:
- Auto means self, so autocatalysis is self-catalyzed.

Quick Tips:
- Autocatalysis involves a product acting as a catalyst.

54 Easy

54 of 56Question permalink

Remembering +4 | -1

The enzyme which converts starch to maltose is

View answer and explanation

Correct option: A

Diastase

Option A is correct because diastase is the enzyme that specifically catalyzes the hydrolysis of starch into maltose.

Solution approach

Correct Answer + Core Reason: Diastase. Diastase is the enzyme that converts starch to maltose.

Step-by-Step Solution:

Step 1: Understand the function of enzymes. Enzymes are biological catalysts that speed up chemical reactions in living organisms.

Step 2: Identify the enzyme that converts starch to maltose. Diastase is the enzyme that specifically catalyzes the hydrolysis of starch into maltose.

Step 3: Consider the other options.
- Zymase: Zymase is a complex of enzymes that converts glucose to ethanol and carbon dioxide during fermentation.
- Maltase: Maltase is an enzyme that converts maltose to glucose.
- Invertase: Invertase is an enzyme that converts sucrose to glucose and fructose.

Why other options are wrong:
- Option B: Zymase. Zymase converts glucose to ethanol and carbon dioxide, not starch to maltose.
- Option C: Maltase. Maltase converts maltose to glucose, not starch to maltose.
- Option D: Invertase. Invertase converts sucrose to glucose and fructose, not starch to maltose.

Key Concepts:
- Enzymes
- Catalysis
- Starch
- Maltose

Memory Aids:
- Diastase digests starch.

Quick Tips:
- Remember the specific enzymes that catalyze different biochemical reactions.

55 Easy

55 of 56Question permalink

Remembering +4 | -1

The critical micelle concentration (CMC) is

View answer and explanation

Correct option: A

The concentration at which micellisation starts

Option A (The concentration at which micellisation starts) is correct because CMC is defined as the minimum concentration at which micelles begin to form.

Solution approach

Correct Answer + Core Reason: The concentration at which micellisation starts. CMC is the minimum concentration required for micelle formation.

Step-by-Step Solution:

Step 1: Define micellisation. Micellisation is the process of forming micelles, which are aggregates of surfactant molecules.

Step 2: Define CMC. The critical micelle concentration (CMC) is the concentration above which micelles spontaneously form.

Why other options are wrong:
- Option B: The concentration at which the true solution is formed is incorrect because micelle formation is different from forming a true solution.
- Option C: The concentration at which one molar electrolyte is present per 1000 g of the solution is unrelated to CMC.
- Option D: The concentration at which $\Delta H = 0$ is incorrect because CMC is related to concentration, not enthalpy change.

Key Concepts:
- Micellisation: The formation of micelles.
- Critical Micelle Concentration (CMC): The concentration at which micellisation begins.

Memory Aids:
- CMC = Concentration at which Micelles Commence.

Quick Tips:
- Remember the definition of CMC as the starting point of micelle formation.

56 Easy

56 of 56Question permalink

Understanding +4 | -1

Which is not correct for catalyst? It :

View answer and explanation

Correct option: B

Changes enthalpy of reaction

Option B (Changes enthalpy of reaction) is correct because a catalyst does not change the enthalpy of the reaction; it only affects the rate.

Solution approach

Correct Answer + Core Reason: Changes enthalpy of reaction. A catalyst only affects the rate of reaction, not the enthalpy change.

Step-by-Step Solution:

Step 1: Understand the role of a catalyst in a chemical reaction.

Step 2: A catalyst speeds up the reaction by lowering the activation energy.

Step 3: A catalyst does not affect the equilibrium position or the enthalpy change of the reaction.

Why other options are wrong:
- Option A: Enhances the rate of reaction in both directions is correct because a catalyst speeds up both forward and reverse reactions equally.
- Option C: Reduces activation energy of reaction is correct because this is how a catalyst speeds up the reaction.
- Option D: Specific in nature is correct because catalysts are usually specific to certain reactions.

Key Concepts:
- Catalyst
- Activation energy
- Enthalpy change

Memory Aids:
- Catalyst: Think "speed up, no change in enthalpy".

Quick Tips:
- Remember that a catalyst only affects the kinetics of a reaction, not the thermodynamics.

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