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

31 Medium

Remembering +4 | -1

In Freundlich adsorption isotherm, the value of

1/n

is

View answer and explanation

Correct option: A

between 0 and 1 in all cases

Option A (between 0 and 1 in all cases) is correct because in the Freundlich adsorption isotherm, the value of 1/n lies between 0 and 1, indicating the favorability of adsorption.

Solution approach

Correct Answer + Core Reason: between 0 and 1 in all cases. In Freundlich adsorption isotherm, the value of $1/n$ lies between 0 and 1.

Step-by-Step Solution:

Step 1: Recall the Freundlich adsorption isotherm: $\frac{x}{m} = k{p^{1/n}}$

Step 2: Understand the constant $n$ : In the Freundlich isotherm, $n$ is an empirical constant related to the adsorption capacity and intensity. The value of $1/n$ indicates the degree of non-linearity in the adsorption process.

Step 3: Range of $1/n$ : The value of $1/n$ typically lies between 0 and 1. If $1/n = 0$ , adsorption is independent of pressure. If $1/n = 1$ , adsorption is directly proportional to pressure.

Why other options are wrong:
- Option B: between 2 and 4 in all cases is incorrect because $1/n$ is typically between 0 and 1.
- Option C: 1 in case of physical adsorption is incorrect because $1/n$ can be 1, but it's not specific to physical adsorption.
- Option D: 1 in case of chemisorption is incorrect because $1/n$ can be 1, but it's not specific to chemisorption.

Key Concepts:
- Freundlich Adsorption Isotherm
- Adsorption Isotherms

Memory Aids:
- Freundlich: 1/n is a fraction between 0 and 1.

Quick Tips:
- Remember the range of $1/n$ in the Freundlich isotherm.

32 Medium

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Understanding +4 | -1

Air can oxidize sodium sulphite in aqueous solution but cannot do so in the case of sodium arsenite. If however, air is passed through a solution containing both sodium sulphite and sodium arsenite then both are oxidized. This is an example of :

View answer and explanation

Correct option: C

Induced catalysis

Option C (Induced catalysis) is correct because the oxidation of sodium arsenite is induced by the oxidation of sodium sulphite.

Solution approach

Correct Answer + Core Reason: Induced catalysis. The reaction of one substance influences the reaction of another.

Step-by-Step Solution:

Step 1: Define induced catalysis.

Step 2: Induced catalysis occurs when one reaction influences the rate or occurrence of another reaction.

Step 3: In this case, the oxidation of sodium sulphite induces the oxidation of sodium arsenite.

Step 4: Therefore, it is an example of induced catalysis.

Why other options are wrong:
- Option A: Positive catalysis is incorrect because it refers to a catalyst increasing the rate of a reaction, but doesn't involve one reaction inducing another.
- Option B: Negative catalysis is incorrect because it refers to a catalyst decreasing the rate of a reaction.
- Option D: Autocatalysis is incorrect because it refers to a product of the reaction acting as a catalyst.

Key Concepts:
- Induced catalysis
- Catalysis

Memory Aids:
- Induced: "one reaction helps another".

Quick Tips:
- Look for a situation where one reaction enables another to occur.

33 Medium

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Application +4 | -1

For a gas-solid adsorbent system, the adsorption isotherm

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

is applicable, where

n=5

in the specified case. Point out the correct statement:

View answer and explanation

Correct option: C

\log \frac{x}{m}

Vs.

\log p

is a straight line with slope equal to 0.2

Option C is correct because taking the logarithm of the adsorption isotherm equation results in a linear equation where the slope is 1/n, which equals 0.2.

Solution approach

Correct Answer + Core Reason: $\log \frac{x}{m}$ Vs. $\log p$ is a straight line with slope equal to 0.2. This follows from taking the logarithm of the Freundlich isotherm.

Step-by-Step Solution:

Step 1: Write the adsorption isotherm equation: $\frac{x}{m} = K p^{1/n}$

Step 2: Take the logarithm of both sides: $\log(\frac{x}{m}) = \log(K) + \frac{1}{n} \log(p)$

Step 3: Identify the slope: The equation is now in the form of a straight line, $y = mx + c$ , where $y = \log(\frac{x}{m})$ , $x = \log(p)$ , $m = \frac{1}{n}$ , and $c = \log(K)$ .

Step 4: Calculate the slope: Given that $n = 5$ , the slope is $\frac{1}{n} = \frac{1}{5} = 0.2$ .

Step 5: Therefore, the plot of $\log(\frac{x}{m})$ vs. $\log(p)$ is a straight line with a slope of 0.2.

Why other options are wrong:
- Option A: The $\frac{x}{m}$ vs. $p$ plot is not linear.
- Option B: The slope of the $\log \frac{x}{m}$ vs. $\log p$ plot is 0.2, not 5.
- Option D: Option C is correct.

Key Concepts:
- Adsorption isotherm
- Logarithms
- Straight line equation

Memory Aids:
- Remember the logarithmic form of the Freundlich isotherm.

Quick Tips:
- Taking the logarithm can linearize exponential relationships.

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Application +4 | -1

3.6 gram of oxygen is adsorbed on

1.2 \mathrm{~g}

of metal powder. What volume of oxygen adsorbed per gram of the adsorbent at

1 \mathrm{~atm}

and

273 \mathrm{~K}

?

View answer and explanation

Correct option: C

2.1 \mathrm{L} \mathrm{g}^{-1}

Option C (2.1 L g^-1) is correct because it represents the volume of oxygen adsorbed per gram of metal powder, calculated using the ideal gas law.

Solution approach

Correct Answer + Core Reason: $2.1 \text{{ L g}}^{-1}$ . The volume of oxygen adsorbed per gram of adsorbent is calculated using the ideal gas law.

Step-by-Step Solution:

Step 1: Calculate the number of moles of oxygen adsorbed: $n = \frac{mass}{Molar \, mass} = \frac{3.6 \text{{ g}}}{32 \text{{ g/mol}}} = 0.1125 \text{{ mol}}$

Step 2: Calculate the volume of oxygen adsorbed using the ideal gas law: $V = \frac{nRT}{P} = \frac{0.1125 \text{{ mol}} \times 0.0821 \text{{ L atm mol}}^{-1} \text{{ K}}^{-1} \times 273 \text{{ K}}}{1 \text{{ atm}}} = 2.52 \text{{ L}}$

Step 3: Calculate the volume of oxygen adsorbed per gram of adsorbent: $\frac{V}{mass \, of \, adsorbent} = \frac{2.52 \text{{ L}}}{1.2 \text{{ g}}} = 2.1 \text{{ L g}}^{-1}$

Why other options are wrong:
- **Option A ( $0.19 \text{{ Lg}}^{-1}$ ): Incorrect calculation.
- Option B ( $1 \text{{ Lg}}^{-1}$ ): Incorrect calculation.
- Option D (None of these): Incorrect, as option C is correct.

Key Concepts:
- Adsorption
- Ideal gas law
- Moles

Memory Aids:**
- Ideal gas law: $PV = nRT$

Quick Tips:
- Ensure consistent units when using the ideal gas law.

35 Medium

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Analyzing +4 | -1

Assertion (A): When

{SnO}_2

is reacted with

{NaOH}

, then its sol. particles are attracted towards cathode.

Reason (R): When

{SnO}_2

is reacted with

{NaOH}

, then it gives

{SnO}_3^{2-}

which is absorbed by

{SnO}_2

, so it is negatively charged sol.

View answer and explanation

Correct option: D

(A) is false but (R) is true

Option D is correct because the assertion is false (particles move towards anode), but the reason is true (SnO2 absorbs SnO3^2- making it negatively charged).

Solution approach

Correct Answer + Core Reason: (A) is false but (R) is true. SnO2 sol particles are negatively charged and move towards the anode, not the cathode.

Step-by-Step Solution:

Step 1: Analyze the assertion. The assertion states that when SnO2 is reacted with NaOH, its sol particles are attracted towards the cathode. This is incorrect. When SnO2 reacts with NaOH, it forms a negatively charged sol.

Step 2: Analyze the reason. The reason states that when SnO2 is reacted with NaOH, it gives SnO3^2- which is absorbed by SnO2, so it is a negatively charged sol. This is correct. The reaction of SnO2 with NaOH leads to the formation of SnO3^2- ions, which are adsorbed onto the SnO2 particles, making the sol negatively charged.

Step 3: Determine the direction of movement of charged particles in an electric field. Negatively charged particles move towards the anode (positive electrode), and positively charged particles move towards the cathode (negative electrode).

Step 4: Conclusion. Since the SnO2 sol is negatively charged, its particles will be attracted towards the anode, not the cathode. Therefore, the assertion is false, and the reason is true.

Why other options are wrong:
- Option A: Both (A) and (R) are true and (R) is the correct explanation of (A). This is incorrect because the assertion is false.
- Option B: Both (A) and (R) are true and (R) is not the correct explanation of (A). This is incorrect because the assertion is false.
- Option C: (A) is true but (R) is false. This is incorrect because the reason is true.

Key Concepts:
- Colloidal solutions
- Charge of colloidal particles
- Electrophoresis
- Adsorption

Memory Aids:
- Anode attracts negative ions (anions).
- Cathode attracts positive ions (cations).

Quick Tips:
- Remember that negatively charged particles move towards the anode and positively charged particles move towards the cathode in an electric field.

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Remembering +4 | -1

The size of colloidal particle is

View answer and explanation

Correct option: C

10^{-6}

to

10^{-9} {~m}

Option C (

10^{-6}

to

10^{-9} {~m}

) is correct because it represents the generally accepted size range for colloidal particles.

Solution approach

Correct Answer + Core Reason: $10^{-6}$ to $10^{-9} {~m}$ . This is the standard size range for colloidal particles.

Step-by-Step Solution:

Step 1: Recall the definition of colloidal particles.

Step 2: Colloidal particles are larger than particles in a true solution but smaller than particles in a suspension.

Step 3: The size range for colloidal particles is typically between 1 nm and 1000 nm, which is equivalent to $10^{-9}$ m to $10^{-6}$ m.

Why other options are wrong:
- Option A: $10^{-3}$ to $10^{-9} {~m}$ - This range is too broad and includes sizes larger than typical colloids.
- Option B: $10^{-3}$ to $10^{-12} {~m}$ - This range is also too broad and includes sizes smaller than typical colloids.
- Option D: $10^{-12}$ to $10^{-19} {~m}$ - This range is far too small; these are atomic or molecular dimensions, not colloidal.

Key Concepts:
- Colloidal solutions
- Particle size
- Suspensions
- True solutions

Memory Aids:
- Remember the size order: True solution < Colloid < Suspension

Quick Tips:
- Know the general size ranges for different types of mixtures.

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Analyzing +4 | -1

Assertion (A):

{SO}_2

gas is easily liquified while

{H}_2

is difficult to liquify.

Reason (R):

{SO}_2

has low critical temperature while

{H}_2

has high critical temperature.

View answer and explanation

Correct option: C

(A) is true but (R) is false

The correct answer is '(A) is true but (R) is false' because SO2 is easily liquefied due to stronger intermolecular forces, but it has a higher critical temperature than H2.

Solution approach

Correct Answer + Core Reason: (A) is true but (R) is false. SO2 is easily liquefied due to stronger van der Waals forces, and it has a higher critical temperature than H2.

Step-by-Step Solution:

Step 1: Analyze the assertion: SO2 is easily liquefied because it has a higher molecular weight and stronger intermolecular forces (van der Waals forces) compared to H2.

Step 2: Analyze the reason: SO2 has a higher critical temperature than H2 because of its stronger intermolecular forces. The statement says SO2 has a low critical temperature, which is false.

Step 3: Therefore, the assertion is true, but the reason is false.

Why other options are wrong:
- Option A: The reason is false.
- Option B: The reason is false.
- Option D: The assertion is true.

Key Concepts:
- Liquefaction of gases
- Critical temperature
- Intermolecular forces

Memory Aids:
- Higher molecular weight and stronger intermolecular forces lead to easier liquefaction and higher critical temperatures.

Quick Tips:
- Relate intermolecular forces to physical properties like boiling point and critical temperature.

38 Medium

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Application +4 | -1

In which of the following reactions colloids are not prepared by the double decomposition method?

View answer and explanation

Correct option: D

\mathrm{Cu} + \mathrm{HgC}{\mathrm{l}_2} \to \mathrm{CuC}{\mathrm{l}_2} + \mathrm{Hg}

Option D (

\mathrm{Cu} + \mathrm{HgC}{\mathrm{l}_2} \to \mathrm{CuC}{\mathrm{l}_2} + \mathrm{Hg}

) is correct because it represents a single displacement redox reaction, not a double decomposition reaction, which is typically used to prepare colloids.

Solution approach

Correct Answer + Core Reason: $\mathrm{Cu} + \mathrm{HgC}{\mathrm{l}_2} \to \mathrm{CuC}{\mathrm{l}_2} + \mathrm{Hg}$ . This reaction is a single displacement reaction, not a double decomposition.

Step-by-Step Solution:

Step 1: Understand the double decomposition method for preparing colloids. It involves the reaction between two soluble salts to form an insoluble precipitate, which then forms the colloidal solution.

Step 2: Analyze each reaction:
- A. $2{\mathrm{H}_3}\mathrm{As}{\mathrm{O}_4} + 3{\mathrm{H}_2}\mathrm{S} \to \mathrm{A}{\mathrm{s}_2}{\mathrm{S}_3} + 6{\mathrm{H}_2}\mathrm{O}$ : This is a double decomposition reaction where $As_2S_3$ is formed as a colloid.
- B. $3{\mathrm{K}_4}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right] + 4\mathrm{FeC}{\mathrm{l}_3} \to \mathrm{F}{\mathrm{e}_4}{\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]_3} + 12\mathrm{KCl}$ : This is a double decomposition reaction where $Fe_4[Fe(CN)_6]_3$ is formed as a colloid.
- C. $\mathrm{Mg}{\left( {\mathrm{CN}} \right)_2} + {\mathrm{H}_2}\mathrm{S} \to \mathrm{HgS} + 2\mathrm{HCN}$ : This appears to be a typo, should be $\mathrm{Hg}{\left( {\mathrm{CN}} \right)_2} + {\mathrm{H}_2}\mathrm{S} \to \mathrm{HgS} + 2\mathrm{HCN}$ . This is a double decomposition reaction where $HgS$ is formed as a colloid.
- D. $\mathrm{Cu} + \mathrm{HgC}{\mathrm{l}_2} \to \mathrm{CuC}{\mathrm{l}_2} + \mathrm{Hg}$ : This is a single displacement redox reaction where copper displaces mercury from mercury chloride. No colloid is formed by double decomposition.

Step 3: Identify the reaction that is not a double decomposition method.

Why other options are wrong:
- Option A: $2{\mathrm{H}_3}\mathrm{As}{\mathrm{O}_4} + 3{\mathrm{H}_2}\mathrm{S} \to \mathrm{A}{\mathrm{s}_2}{\mathrm{S}_3} + 6{\mathrm{H}_2}\mathrm{O}$ : This is a double decomposition reaction.
- Option B: $3{\mathrm{K}_4}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right] + 4\mathrm{FeC}{\mathrm{l}_3} \to \mathrm{F}{\mathrm{e}_4}{\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]_3} + 12\mathrm{KCl}$ : This is a double decomposition reaction.
- Option C: $\mathrm{Mg}{\left( {\mathrm{CN}} \right)_2} + {\mathrm{H}_2}\mathrm{S} \to \mathrm{HgS} + 2\mathrm{HCN}$ : Assuming this is a typo and should be Hg(CN)2, this is a double decomposition reaction.

Key Concepts:
- Double decomposition reaction
- Colloid preparation methods
- Redox reaction

Memory Aids:
- Double decomposition: Exchange of ions between two reactants.

Quick Tips:
- Identify the type of reaction involved in each option.
- Double decomposition reactions typically involve the formation of a precipitate.

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Understanding +4 | -1

In the formation of

\mathrm{S}{\mathrm{O}_3}

by

\mathrm{S}{\mathrm{O}_2}

and

{\mathrm{O}_2}

using NO as catalyst, the catalytic action of

\mathrm{N}{\mathrm{O}_2}

is evidenced by :

View answer and explanation

Correct option: C

Brown vapours

Option C (Brown vapours) is correct because

NO_2

is a brown gas and its presence indicates the catalytic action.

Solution approach

Correct Answer + Core Reason: Brown vapours. $NO_2$ is a brown gas, indicating its presence as an intermediate in the catalytic cycle.

Step-by-Step Solution:

Step 1: Understand the role of NO as a catalyst in the formation of $\mathrm{S}{\mathrm{O}_3}$ .

Step 2: NO reacts with ${\mathrm{O}_2}$ to form $NO_2$ , which then reacts with $\mathrm{S}{\mathrm{O}_2}$ to form $\mathrm{S}{\mathrm{O}_3}$ and regenerates NO.

Step 3: The presence of $NO_2$ is evidenced by its brown color.

Why other options are wrong:
- Option A: Green vapours are not associated with nitrogen oxides in this reaction.
- Option B: Violet vapours are not associated with nitrogen oxides in this reaction.
- Option D: None of these is incorrect because brown vapours are observed due to the presence of $NO_2$ .

Key Concepts:
- Catalysis
- Nitrogen oxides

Memory Aids:
- $NO_2$ : "brown gas".

Quick Tips:
- Remember the colors of common gases.

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Understanding +4 | -1

Alumns purify muddy water by

View answer and explanation

Correct option: D

Coagulation

Option D (Coagulation) is correct because alum acts as a coagulant, neutralizing the charge on the mud particles and causing them to clump together and settle out.

Solution approach

Correct Answer + Core Reason: Coagulation. Alum neutralizes charges, causing particles to clump and settle.

Step-by-Step Solution:

Step 1: Understand the role of alum in water purification.

Step 2: Alum (aluminum sulfate) is used as a coagulant in water treatment.

Step 3: Alum neutralizes the negative charge on the suspended clay particles in muddy water.

Step 4: This neutralization causes the particles to clump together (coagulate) and form larger aggregates that settle out of the water.

Why other options are wrong:
- Option A: Dialysis is a separation technique based on the size of particles, not directly related to alum's function.
- Option B: Absorption involves the penetration of one substance into the bulk of another, which is not how alum works.
- Option C: Adsorption involves the adhesion of molecules to a surface, but coagulation is the primary mechanism of alum.

Key Concepts:
- Coagulation
- Colloids
- Water purification

Memory Aids:
- Alum = Aggregation

Quick Tips:
- Remember that alum is a *coagulant*.

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Understanding +4 | -1

Kraft temperature is the temperature

View answer and explanation

Correct option: D

Above which the formation of micelles takes place

Option D (Above which the formation of micelles takes place) is correct because the Kraft temperature is the minimum temperature at which surfactants can form micelles in solution.

Solution approach

Correct Answer + Core Reason: Above which the formation of micelles takes place. This is the definition of Kraft temperature.

Step-by-Step Solution:

Step 1: Define Kraft temperature.

Step 2: The Kraft temperature (Tk) is the minimum temperature at which surfactants begin to form micelles in solution.

Step 3: Below the Kraft temperature, the solubility of the surfactant is too low to form micelles.

Why other options are wrong:
- Option A: The freezing point of the solution is not directly related to the Kraft temperature.
- Option B: Micelle formation does not take place *below* the Kraft temperature.
- Option C: The boiling point of the solution is not directly related to the Kraft temperature.

Key Concepts:
- Micelles
- Surfactants
- Kraft temperature

Memory Aids:
- Kraft = Micelle Formation

Quick Tips:
- Remember that the Kraft temperature is a *minimum* temperature.

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Remembering +4 | -1

Reducing fog is mixture of

View answer and explanation

Correct option: B

Smoke, fog and

\mathrm{S}{\mathrm{O}_\mathrm{2}}

Option B (Smoke, fog and

\mathrm{S}{\mathrm{O}_\mathrm{2}}

) is correct because reducing smog, also known as classical smog, is a mixture of smoke, fog, and sulfur dioxide.

Solution approach

Correct Answer + Core Reason: Smoke, fog and $\mathrm{S}{\mathrm{O}_\mathrm{2}}$ . Reducing fog (classical smog) is a mixture of smoke, fog, and sulfur dioxide.

Step-by-Step Solution:

Step 1: Define reducing smog: Reducing smog, also known as classical smog or London smog, is a type of air pollution that is characterized by high concentrations of smoke, fog, and sulfur dioxide.

Step 2: Identify the components: The main components of reducing smog are:
- Smoke (particulate matter)
- Fog (water droplets)
- Sulfur dioxide ( $\mathrm{S}{\mathrm{O}_\mathrm{2}}$ )

Step 3: Relate to the question: The question asks for the mixture that constitutes reducing fog, which is smoke, fog, and $\mathrm{S}{\mathrm{O}_\mathrm{2}}$ .

Why other options are wrong:
- Option A: Smoke, fog and ${\mathrm{O}_\mathrm{3}}$ is incorrect because ozone is a component of photochemical smog, not reducing smog.
- Option C: Smoke, fog and $C{H_2} = CH - CHO$ is incorrect because acrolein is a component of photochemical smog, not reducing smog.
- Option D: Smoke, fog and ${\mathrm{N}_\mathrm{2}}{\mathrm{O}_\mathrm{3}}$ is incorrect because nitrogen oxides are more characteristic of photochemical smog.

Key Concepts:
- Reducing Smog (Classical Smog)
- Air Pollution

Memory Aids:
- Reducing Smog: Smoke + Fog + SO2 (Sulfur).

Quick Tips:
- Distinguish between reducing smog and photochemical smog.

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Understanding +4 | -1

A catalyst is most effective in finely divided state because

View answer and explanation

Correct option: B

greater number of active sites

Option B (greater number of active sites) is correct because a finely divided catalyst has a larger surface area, thus more active sites.

Solution approach

Correct Answer + Core Reason: greater number of active sites. A finely divided catalyst has a larger surface area, leading to more active sites for the reaction to occur.

Step-by-Step Solution:

Step 1: Understand the concept of active sites on a catalyst.

Step 2: Active sites are the locations on the catalyst surface where the reaction takes place.

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

Step 4: A larger surface area means more active sites are available.

Why other options are wrong:
- Option A: low surface area is incorrect because a finely divided catalyst has a high surface area.
- Option C: It dissolves easily in water is incorrect because solubility is not the primary reason for using a finely divided catalyst.
- Option D: It reduces the activation energy to the larger extent is incorrect because the extent of activation energy reduction is a property of the catalyst itself, not its state of division.

Key Concepts:
- Active sites
- Surface area
- Catalysis

Memory Aids:
- More surface: "more active sites".

Quick Tips:
- Remember that the number of active sites is directly related to the surface area of the catalyst.

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Application +4 | -1

1 \mathrm{~g}

of charcoal adsorbs

100 \mathrm{~mL} \,0.5 \,\mathrm{MCH}_3 \mathrm{COOH}

to form a monolayer, and thereby the molarity of

\mathrm{CH}_3 \mathrm{COOH}

reduces to 0.49. Calculate the surface area of the charcoal covered by each molecule of acetic acid. Surface area of charcoal

=3.01 \times 10^2 \mathrm{~m}^2 / \mathrm{g}

.

View answer and explanation

Correct option: D

5 \times 10^{-19} \mathrm{~m}^2

Option D (5 × 10^-19 m²) is correct because it represents the surface area covered by each acetic acid molecule, calculated from the amount adsorbed and the total surface area.

Solution approach

Correct Answer + Core Reason: $5 \times 10^{-19} \text{{ m}}^2$ . The surface area per molecule is calculated from the amount adsorbed and the total surface area.

Step-by-Step Solution:

Step 1: Calculate the change in molarity of acetic acid: $\Delta M = 0.5 \text{{ M}} - 0.49 \text{{ M}} = 0.01 \text{{ M}}$

Step 2: Calculate the moles of acetic acid adsorbed: $n = \Delta M \times V = 0.01 \text{{ M}} \times 0.1 \text{{ L}} = 0.001 \text{{ mol}}$

Step 3: Calculate the number of molecules of acetic acid adsorbed: $N = n \times N_A = 0.001 \text{{ mol}} \times 6.022 \times 10^{23} \text{{ molecules/mol}} = 6.022 \times 10^{20} \text{{ molecules}}$

Step 4: Calculate the surface area of charcoal: $A = 3.01 \times 10^2 \text{{ m}}^2/\text{{g}} \times 1 \text{{ g}} = 3.01 \times 10^2 \text{{ m}}^2$

Step 5: Calculate the surface area covered by each molecule: $Area \, per \, molecule = \frac{Total \, surface \, area}{Number \, of \, molecules} = \frac{3.01 \times 10^2 \text{{ m}}^2}{6.022 \times 10^{20} \text{{ molecules}}} = 0.5 \times 10^{-18} \text{{ m}}^2 = 5 \times 10^{-19} \text{{ m}}^2$

Why other options are wrong:
- **Option A ( $0.5 \times 10^{-19} \text{{ m}}^2$ ): Incorrect calculation.
- Option B ( $4 \times 10^{-20} \text{{ m}}^2$ ): Incorrect calculation.
- Option C ( $0.5 \times 10^{-10} \text{{ m}}^2$ ): Incorrect calculation.

Key Concepts:
- Adsorption
- Molarity
- Avogadro's number
- Surface area

Memory Aids:**
- Avogadro's number: $N_A = 6.022 \times 10^{23}$

Quick Tips:
- Pay attention to units and conversions.

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Remembering +4 | -1

The outcome of internal liquid of gels on shear is called :

View answer and explanation

Correct option: B

Thixotropy

Option B is correct because thixotropy is the property of certain gels to become fluid when stirred or shaken and then to return to the semi-solid state upon standing.

Solution approach

Correct Answer + Core Reason: Thixotropy. Thixotropy refers to the phenomenon where gels become fluid upon shear and revert to a gel-like state upon standing.

Step-by-Step Solution:

Step 1: Define gels. Gels are semi-solid systems consisting of a solid network that is swollen by a liquid.

Step 2: Define thixotropy. Thixotropy is the property of certain gels to become fluid when stirred or shaken (shear) and then to return to the semi-solid state upon standing. This is a time-dependent shear thinning property.

Step 3: Define syneresis. Syneresis is the spontaneous contraction of a gel, resulting in the expulsion of a liquid.

Step 4: Define swelling. Swelling is the increase in volume of a substance when it absorbs a liquid.

Step 5: Conclusion: The outcome of internal liquid of gels on shear is called thixotropy.

Why other options are wrong:
- Option A: Syneresis is the spontaneous expulsion of liquid from a gel, not the outcome on shear.
- Option C: Swelling is the absorption of liquid by a substance, not the outcome on shear.
- Option D: Thixotropy is the correct answer.

Key Concepts:
- Gels
- Thixotropy
- Syneresis
- Swelling

Memory Aids:
- Think of thixotropy as "shake and thin".

Quick Tips:
- Distinguish between thixotropy (shear-induced thinning) and syneresis (spontaneous liquid expulsion).

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Understanding +4 | -1

According to Langmuir adsorption isotherm the amount of gas adsorbed at very high pressure :

View answer and explanation

Correct option: A

Reaches a constant limiting value

Option A (Reaches a constant limiting value) is correct because, at very high pressure, the surface of the adsorbent is fully covered, leading to a maximum adsorption.

Solution approach

Correct Answer + Core Reason: Reaches a constant limiting value. According to Langmuir adsorption isotherm, at very high pressure, the surface is fully covered, and adsorption reaches a maximum.

Step-by-Step Solution:

Step 1: State Langmuir Adsorption Isotherm: The Langmuir adsorption isotherm describes the relationship between the pressure of a gas and the amount of gas adsorbed on a solid surface at a constant temperature.

Step 2: Langmuir Adsorption Isotherm Equation: The Langmuir isotherm is given by the equation: $\frac{x}{m} = \frac{aP}{1 + bP}$ , where x is the mass of gas adsorbed, m is the mass of adsorbent, P is the pressure, and a and b are constants.

Step 3: Analyze at very high pressure: At very high pressure (P -> infinity), the term '1' in the denominator becomes negligible compared to 'bP'. Therefore, the equation becomes: $\frac{x}{m} = \frac{aP}{bP} = \frac{a}{b}$ .

Step 4: Interpret the result: The equation $\frac{x}{m} = \frac{a}{b}$ shows that at very high pressure, the amount of gas adsorbed per unit mass of adsorbent (x/m) becomes constant and equal to a/b. This means the adsorption reaches a limiting value.

Step 5: Conclusion: At very high pressure, the amount of gas adsorbed reaches a constant limiting value.

Why other options are wrong:
- Option B: This option is incorrect because, at very high pressure, the surface becomes saturated, and adsorption cannot increase indefinitely.
- Option C: This option is incorrect because adsorption does not decrease with increasing pressure; it approaches a constant value.
- Option D: This option is incorrect because adsorption increases and then reaches a constant value, not decreases.

Key Concepts:
- Langmuir Adsorption Isotherm
- Adsorption
- Surface Chemistry

Memory Aids:
- Langmuir: Think \

47 Medium

47 of 59Question permalink

Analyzing +4 | -1

Freundlich adsorption isotherm is given by the expression

\frac{x}{m} = KP{\,^{1/n}}

, which of the following conclusions can be drawn from this expression.
a) When,

\frac{1}{n} = 0

, the adsorption is independent of pressure
b) When,

\frac{1}{n} = 0

, the adsorption is directly proportional to pressure
c) When,

n = 0,{\mkern 1mu} \frac{x}{m}\,\,VS

P graph is a line parallel to x-axis
d) When, n = 0, plot of

\frac{x}{m}VS\,\,\,P

is a curve

View answer and explanation

Correct option: B

a,c

Option B (a,c) is correct because when 1/n = 0, adsorption is independent of pressure, and when n = 0, the x/m vs P graph is a line parallel to the x-axis.

Solution approach

Correct Answer + Core Reason: a,c. The Freundlich isotherm equation simplifies to pressure-independent adsorption when 1/n = 0, and results in a horizontal line when n = 0.

Step-by-Step Solution:

Step 1: Analyze the Freundlich adsorption isotherm: $\frac{x}{m} = KP^{\frac{1}{n}}$

Step 2: Consider the case when $\frac{1}{n} = 0$ : $\frac{x}{m} = KP^0 = K$ . This means the amount adsorbed (x/m) is constant and independent of pressure.

Step 3: Consider the case when $n = 0$ : The equation becomes undefined. However, if we consider the limit as n approaches 0, the adsorption becomes independent of pressure, resulting in a horizontal line on the graph of x/m vs P.

Step 4: Therefore, statements a and c are correct.

Why other options are wrong:
- Option A: Statement b is incorrect because when 1/n = 0, adsorption is independent of pressure, not directly proportional to it.
- Option C: Statement b is incorrect because when 1/n = 0, adsorption is independent of pressure, not directly proportional to it. Statement d is incorrect because when n=0, the graph is a line parallel to x-axis, not a curve.
- Option D: Statement d is incorrect because when n=0, the graph is a line parallel to x-axis, not a curve.

Key Concepts:
- Freundlich adsorption isotherm
- Pressure dependence of adsorption

Memory Aids:
- Remember the Freundlich isotherm equation and how it changes with different values of n.

Quick Tips:
- Analyze the equation for extreme values of the parameters to understand the behavior of the system.

48 Medium

48 of 59Question permalink

Understanding +4 | -1

For the coagulation of methylene blue sol, the flocculating power of which of the following ion is maximum ?

View answer and explanation

Correct option: A

PO_4^{3 - }

Option A (

PO_4^{3 - }

) is correct because, according to the Hardy-Schulze rule, ions with higher charges have greater flocculating power for oppositely charged sols. Methylene blue sol is positively charged, so a highly negative ion will have the greatest power.

Solution approach

Correct Answer + Core Reason: $PO_4^{3 - }$ . According to Hardy-Schulze rule, higher the charge, greater the flocculating power for oppositely charged sols.

Step-by-Step Solution:

Step 1: State the Hardy-Schulze rule.

Step 2: The Hardy-Schulze rule states that the flocculating power of an ion is directly proportional to its charge.

Step 3: Methylene blue sol is a positively charged sol.

Step 4: Therefore, the ion with the highest negative charge will have the maximum flocculating power.

Step 5: Comparing the charges: $PO_4^{3 - }$ has a charge of -3, $A{l^{3 + }}$ has a charge of +3, $C{l^ - }$ has a charge of -1, and $B{a^{2 + }}$ has a charge of +2.

Step 6: Since we need a negative ion to coagulate a positive sol, $PO_4^{3 - }$ is the best option.

Why other options are wrong:
- Option B: $A{l^{3 + }}$ - This is a positive ion and will not effectively coagulate a positive sol.
- Option C: $C{l^ - }$ - This has a lower negative charge than $PO_4^{3 - }$ .
- Option D: $B{a^{2 + }}$ - This is a positive ion and will not effectively coagulate a positive sol.

Key Concepts:
- Hardy-Schulze rule
- Flocculation
- Coagulation
- Colloidal solutions
- Charge of ions

Memory Aids:
- Hardy-Schulze: Higher charge, higher coagulation.

Quick Tips:
- Remember that oppositely charged ions are required for coagulation.

49 Medium

49 of 59Question permalink

Understanding +4 | -1

Which is an example of auto-catalyst?

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because in all listed reactions, a product of the reaction acts as a catalyst.

Solution approach

Correct Answer + Core Reason: All of the above. In each case, a product of the reaction acts as a catalyst.

Step-by-Step Solution:

Step 1: Define autocatalysis.

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

Step 3: Analyze each option to determine if a product acts as a catalyst.

Step 4: Hydrolysis of methyl acetate is autocatalyzed by the produced acid.

Step 5: Decomposition of TNG is autocatalyzed by the produced gases.

Step 6: Oxidation of oxalic acid by $\mathrm{KMn}{\mathrm{O}_4}$ is autocatalyzed by $Mn^{2+}$ ions.

Step 7: Therefore, all options are examples of autocatalysis.

Why other options are wrong:
- Option A: Hydrolysis of methyl acetate is an example of autocatalysis, but not the only one.
- Option B: Decomposition of TNG is an example of autocatalysis, but not the only one.
- Option C: Oxidation of oxalic acid by $\mathrm{KMn}{\mathrm{O}_4}$ is an example of autocatalysis, but not the only one.

Key Concepts:
- Autocatalysis
- Catalyst

Memory Aids:
- Autocatalysis: "self-catalyzing".

Quick Tips:
- Look for a product of the reaction that also speeds up the reaction.

50 Medium

50 of 59Question permalink

Application +4 | -1

On adding

\left| m \right|

solution of 10 % NaCl to 10 ml gold Sol in presence of 0.25 g starch so that coagulation is just prevented, what will be gold number of starch

View answer and explanation

Correct option: C

250

Option C (250) is correct because the gold number is defined as the mass in milligrams of a protective colloid that prevents the coagulation of 10 mL of a gold sol when 1 mL of a 10% NaCl solution is added.

Solution approach

Correct Answer + Core Reason: 250. Gold number is the mass of protective colloid in mg that prevents coagulation of 10 mL gold sol by 1 mL of 10% NaCl.

Step-by-Step Solution:

Step 1: Define Gold Number: The gold number is the minimum amount of protective colloid (in milligrams) that will prevent the coagulation of 10 mL of a standard gold sol on the addition of 1 mL of a 10% NaCl solution.

Step 2: Given: 250 mg of starch prevents coagulation of 10 mL gold sol when 1 mL of 10% NaCl is added.

Step 3: Calculate the gold number: Since 250 mg of starch is required, the gold number of starch is 250.

Why other options are wrong:
- Option A: 0.25 is incorrect because the gold number is directly the amount in mg, not a fraction of it.
- Option B: 2.5 is incorrect because the gold number is directly the amount in mg, not a fraction of it.
- Option D: 0.025 is incorrect because the gold number is directly the amount in mg, not a fraction of it.

Key Concepts:
- Gold number
- Protective colloids
- Coagulation

Memory Aids:
- Gold number is in milligrams.

Quick Tips:
- Directly relate the given mass in mg to the definition of the gold number.

51 Medium

51 of 59Question permalink

Remembering +4 | -1

A curve showing the variation of pressure with temperature for a given amount of adsorption is called

View answer and explanation

Correct option: C

Adsorption isoster

Option C (Adsorption isoster) is correct because an adsorption isoster is a graph that shows the variation of pressure with temperature for a given amount of adsorption (constant amount adsorbed).

Solution approach

Correct Answer + Core Reason: Adsorption isoster. An isoster shows the pressure-temperature relationship at constant adsorption.

Step-by-Step Solution:

Step 1: Define adsorption isotherms. Adsorption isotherms are graphs that show the amount of adsorbate adsorbed on an adsorbent as a function of pressure at a constant temperature.

Step 2: Define adsorption isobars. Adsorption isobars are graphs that show the amount of adsorbate adsorbed on an adsorbent as a function of temperature at a constant pressure.

Step 3: Define adsorption isosteres. Adsorption isosteres are graphs that show the variation of pressure with temperature for a constant amount of adsorption.

Step 4: Define adsorption isochore. Adsorption isochore is a graph plotting the relationship between the amount adsorbed and temperature at constant volume.

Step 5: Conclusion. A curve showing the variation of pressure with temperature for a given amount of adsorption is called an adsorption isoster.

Why other options are wrong:
- Option A: Adsorption isobar shows the variation of adsorption with temperature at constant pressure.
- Option B: Adsorption isotherm shows the variation of adsorption with pressure at constant temperature.
- Option D: Adsorption isochore is not a standard term in adsorption studies in the context of pressure vs temperature.

Key Concepts:
- Adsorption Isotherm
- Adsorption Isobar
- Adsorption Isoster

Memory Aids:
- Isotherm: constant TEMPERATURE
- Isobar: constant PRESSURE
- Isoster: constant amount adsorbed (\"STER\" = amount)

Quick Tips:
- Remember the definitions of isotherm, isobar, and isoster to identify the correct graph.

52 Medium

52 of 59Question permalink

Remembering +4 | -1

The Langmuir adsorption isotherm is deduced using the assumption

View answer and explanation

Correct option: A

The adsorption sites are equivalent in their ability to adsorb the particles

The correct answer is A (The adsorption sites are equivalent in their ability to adsorb the particles) because this is a key assumption in the Langmuir adsorption isotherm.

Solution approach

Correct Answer + Core Reason: The adsorption sites are equivalent in their ability to adsorb the particles. This is a fundamental assumption of the Langmuir adsorption isotherm.

Step-by-Step Solution:

Step 1: Recall the assumptions of the Langmuir adsorption isotherm.

Step 2: One of the key assumptions is that all adsorption sites are equivalent.

Why other options are wrong:
- Option B: The Langmuir isotherm assumes a constant heat of adsorption.
- Option C: The Langmuir isotherm assumes no interaction between adsorbed molecules.
- Option D: The Langmuir isotherm assumes monolayer adsorption.

Key Concepts:
- Langmuir adsorption isotherm
- Adsorption

Memory Aids:
- Langmuir assumes equivalent sites, monolayer adsorption, and no interaction.

Quick Tips:
- Know the assumptions and limitations of the Langmuir isotherm.

53 Medium

53 of 59Question permalink

Understanding +4 | -1

Which of the following reactions lead to the formation of colloidal solution?

View answer and explanation

Correct option: B

2\mathrm{HN}{\mathrm{O}_3} + 3{\mathrm{H}_2}\mathrm{S} \to 3\mathrm{S} + {\mathrm{H}_2}\mathrm{O} + 2\mathrm{NO}

The correct answer is B (

2\mathrm{HN}{\mathrm{O}_3} + 3{\mathrm{H}_2}\mathrm{S} \to 3\mathrm{S} + {\mathrm{H}_2}\mathrm{O} + 2\mathrm{NO}

) because it involves the formation of sulfur (S) as a precipitate in the colloidal range.

Solution approach

Correct Answer + Core Reason: $2\mathrm{HN}{\mathrm{O}_3} + 3{\mathrm{H}_2}\mathrm{S} \to 3\mathrm{S} + {\mathrm{H}_2}\mathrm{O} + 2\mathrm{NO}$ . Formation of sulfur sol.

Step-by-Step Solution:

Step 1: Analyze each reaction to determine if a colloidal solution is formed.

Step 2: In option B, sulfur (S) is formed as a precipitate. If the particle size of sulfur is in the colloidal range (1-1000 nm), a colloidal solution is formed.

Step 3: The reaction $2\mathrm{HN}{\mathrm{O}_3} + 3{\mathrm{H}_2}\mathrm{S} \to 3\mathrm{S} + {\mathrm{H}_2}\mathrm{O} + 2\mathrm{NO}$ produces sulfur sol.

Why other options are wrong:
- Option A: $\mathrm{Cu} + \mathrm{HgC}{\mathrm{l}_2} \to \mathrm{CuC}{\mathrm{l}_2} + \mathrm{Hg}$ forms mercury, which is a liquid, not necessarily a colloidal solution.
- Option C: $2\mathrm{Mg} + \mathrm{C}{\mathrm{O}_2} \to 2\mathrm{MgO} + \mathrm{C}$ forms magnesium oxide and carbon, but not necessarily in colloidal form.
- Option D: $\mathrm{Cu} + \mathrm{CuC}{\mathrm{l}_2} \to 2\mathrm{CuCl}$ forms copper(I) chloride, which is a soluble salt, not a colloidal solution.

Key Concepts:
- Colloidal solution
- Precipitation
- Particle size

Memory Aids:
- Colloidal solutions have particle sizes between 1 and 1000 nm.

Quick Tips:
- Look for reactions that produce precipitates with particle sizes in the colloidal range.

54 Medium

54 of 59Question permalink

Understanding +4 | -1

If

x

is amount of adsorbate and

m

is amount of adsorbent, which of the following relations is related to adsorption process?

View answer and explanation

Correct option: A

\frac{x}{m} = P \times T

Option B (

\frac{x}{m} = f\left( P \right)

at constant T) is correct because it represents the relationship between the amount of adsorbate adsorbed per unit mass of adsorbent as a function of pressure at a constant temperature, which is the basis of adsorption isotherms.

Solution approach

Correct Answer + Core Reason: $\frac{x}{m} = f\left( P \right)$ at constant T. Adsorption isotherms relate x/m to pressure at constant temperature.

Step-by-Step Solution:

Step 1: Define the terms: x = amount of adsorbate, m = amount of adsorbent.

Step 2: Understand that adsorption is a process where a substance (adsorbate) adheres to the surface of another substance (adsorbent).

Step 3: Recognize that the extent of adsorption ( $\frac{x}{m}$ ) depends on various factors, including pressure (P) and temperature (T).

Step 4: At a constant temperature, the relationship between $\frac{x}{m}$ and P is described by an adsorption isotherm, which is a function of pressure: $\frac{x}{m} = f(P)$ at constant T.

Why other options are wrong:
- Option A: $\frac{x}{m} = P \times T$ is incorrect because it suggests a direct proportionality between $\frac{x}{m}$ and both P and T, which is not generally true for adsorption.
- Option C: $\frac{x}{m} = f\left( T \right)$ at constant P describes the relationship between $\frac{x}{m}$ and temperature at a constant pressure, which is not the standard representation of an adsorption isotherm.
- Option D: $P = f\left( T \right)$ at constant $(x/m)$ describes the relationship between pressure and temperature at a constant extent of adsorption, which is not the standard representation of an adsorption isotherm.

Key Concepts:
- Adsorption
- Adsorbate
- Adsorbent
- Adsorption isotherm

Memory Aids:
- Adsorption isotherm: $\frac{x}{m}$ vs. P at constant T.

Quick Tips:
- Remember that adsorption isotherms are typically studied at constant temperature to understand the effect of pressure on the extent of adsorption.

55 Medium

55 of 59Question permalink

Remembering +4 | -1

The heat evolved in physisorption lies in the range (kJ/mol) of

View answer and explanation

Correct option: A

20-40

Option A (20-40) is correct because the heat evolved in physisorption is generally in the range of 20-40 kJ/mol due to weak van der Waals forces.

Solution approach

Correct Answer + Core Reason: 20-40 kJ/mol. Physisorption involves weak van der Waals forces, resulting in low heat of adsorption.

Step-by-Step Solution:
Step 1: Physisorption is a surface phenomenon where adsorbate molecules are held by weak van der Waals forces.
Step 2: The heat evolved during physisorption is relatively low because the interactions are weak.
Step 3: The typical range for the heat of physisorption is 20-40 kJ/mol.

Why other options are wrong:
- Option B (40-100): This range is more typical of chemisorption, which involves stronger chemical bonds.
- Option C (100-200): This range is significantly higher than what is observed in physisorption.
- Option D (200-400): This range is characteristic of strong chemical bonding interactions, not physisorption.

Key Concepts:
- Physisorption
- Heat of Adsorption
- Van der Waals forces

Memory Aids:
- Physisorption = Physical = Weak forces = Low heat

Quick Tips:
- Remember that physisorption is driven by weak intermolecular forces, leading to lower energy changes.

56 Medium

56 of 59Question permalink

Application +4 | -1

The graph of

\log \frac{x}{m}

vs

\log p

for an adsorption process is a straight line inclined at an angle of

45^{\circ}

with intercept equal to 0.6020 . The mass of gas adsorbed per unit mass of adsorbent at the pressure of _____

0.4 \mathrm{~atm}

is

\times 10^{-1}

(Nearest integer).

Given :

\log 2=0.3010

View answer and explanation

Correct option: A

16

Option A (16) is correct because the mass of gas adsorbed per unit mass of adsorbent is calculated using the Freundlich isotherm, resulting in 1.6, which is 16 × 10^-1.

Solution approach

Correct Answer + Core Reason: 16. The mass adsorbed is calculated using the Freundlich isotherm equation.

Step-by-Step Solution:

Step 1: Write the Freundlich isotherm equation: $\frac{x}{m} = k \cdot p^{1/n}$

Step 2: Take the logarithm of both sides: $\log(\frac{x}{m}) = \log(k) + \frac{1}{n} \log(p)$

Step 3: Identify the given values: The slope of the line is $\frac{1}{n} = \tan(45^\circ) = 1$ . The intercept is $\log(k) = 0.6020$ .

Step 4: Calculate k: $k = 10^{0.6020} = 10^{\log 4} = 4$

Step 5: Substitute the values into the Freundlich isotherm equation: $\frac{x}{m} = 4 \cdot (0.4)^1 = 1.6$

Step 6: Express the answer in the required format: $1.6 = 16 \times 10^{-1}$

Why other options are wrong:
- Option B (0): This is incorrect. It indicates no adsorption, which is not the case.
- Option C (0): This is incorrect for the same reason as option B.
- Option D (0): This is incorrect for the same reason as option B.

Key Concepts:
- Freundlich Adsorption Isotherm
- Logarithms

Memory Aids:
- Remember the Freundlich isotherm equation: $\frac{x}{m} = k \cdot p^{1/n}$

Quick Tips:
- The slope of the $\log(\frac{x}{m})$ vs. $\log(p)$ plot is equal to $\frac{1}{n}$ .
- The intercept of the $\log(\frac{x}{m})$ vs. $\log(p)$ plot is equal to $\log(k)$ .

57 Medium

57 of 59Question permalink

Remembering +4 | -1

Which of the following is an example of associated colloid

View answer and explanation

Correct option: A

Soap in water

Option A (Soap in water) is correct because soap molecules in water aggregate to form micelles, which are characteristic of associated colloids.

Solution approach

Correct Answer + Core Reason: Soap in water. Soap molecules aggregate to form micelles in water, which is a characteristic of associated colloids.

Step-by-Step Solution:

Step 1: Define associated colloids. Associated colloids are substances that behave as normal electrolytes at low concentrations but exhibit colloidal properties at higher concentrations due to the formation of aggregates (micelles).

Step 2: Consider the given options. Soap in water forms micelles above the critical micelle concentration (CMC), making it an associated colloid.

Why other options are wrong:
- Option B: Protein in water can form colloidal solutions, but it's not a typical example of an associated colloid. Proteins are usually classified as lyophilic colloids.
- Option C: Rubber in Benzene is a lyophilic colloid, not an associated colloid.
- Option D: $AgNO_3$ in water forms a true solution, not a colloidal solution.

Key Concepts:
- Associated colloids: Substances forming micelles at higher concentrations.
- Micelles: Aggregates of surfactant molecules.

Memory Aids:
- Soap = Associated Colloid.

Quick Tips:
- Remember that soaps and detergents are common examples of associated colloids.

58 Medium

58 of 59Question permalink

Understanding +4 | -1

Mention the type of reaction to obtain

\mathrm{Au}\left( {\mathrm{sol}} \right)

.
Reaction,

2\mathrm{AuC}{\mathrm{l}_3} + 3\mathrm{HCHO} + 3{\mathrm{H}_2}\mathrm{O} \to 2\mathrm{Au}\left( {\mathrm{sol}} \right) + 3\mathrm{HCOOH} + 6\mathrm{HCl}

View answer and explanation

Correct option: C

Reduction

Option C (Reduction) is correct because the gold ions (

\mathrm{AuC}{\mathrm{l}_3}

) are reduced to gold atoms (

\mathrm{Au}

) in the reaction.

Solution approach

Correct Answer + Core Reason: Reduction. The oxidation state of Au decreases from +3 to 0.

Step-by-Step Solution:

Step 1: Identify the oxidation states of Au in the reactant and product.

Step 2: In $2\mathrm{AuC}{\mathrm{l}_3}$ , the oxidation state of Au is +3.

Step 3: In $2\mathrm{Au}(\mathrm{sol})$ , the oxidation state of Au is 0.

Step 4: Since the oxidation state of Au decreases from +3 to 0, it undergoes reduction.

Why other options are wrong:
- Option A: Hydrolysis involves the reaction of a substance with water. This reaction involves reduction of gold ions, not hydrolysis.
- Option B: Oxidation involves an increase in oxidation state. Here, the oxidation state of Au decreases, so it's not oxidation.
- Option D: Double decomposition involves the exchange of ions between two reactants. This reaction is a redox reaction, not a double decomposition.

Key Concepts:
- Oxidation state
- Reduction
- Redox reaction

Memory Aids:
- OIL RIG (Oxidation Is Loss, Reduction Is Gain of electrons)

Quick Tips:
- Determine oxidation states to identify redox reactions.

59 Medium

59 of 59Question permalink

Analyzing +4 | -1

Assertion : In chemisorption, adsorption first increases with temperature and then decreases.

Reason : Heat keeps on providing more and more activation energy.

View answer and explanation

Correct option: C

If Assertion is correct but Reason is incorrect.

The correct answer is 'If Assertion is correct but Reason is incorrect' because chemisorption initially increases with temperature, but the reason is incorrect as heat eventually decreases adsorption.

Solution approach

Correct Answer + Core Reason: If Assertion is correct but Reason is incorrect. Chemisorption initially increases with temperature due to activation energy, but decreases at higher temperatures due to being exothermic.

Step-by-Step Solution:

Step 1: Analyze the assertion: Chemisorption is an activated process, so initially, increasing the temperature increases the rate of adsorption. However, chemisorption is exothermic, so at higher temperatures, the equilibrium shifts to favor desorption, and adsorption decreases.

Step 2: Analyze the reason: While heat initially provides activation energy, it doesn't keep providing more and more activation energy indefinitely. At higher temperatures, the reverse process (desorption) becomes more favorable.

Step 3: Therefore, the assertion is correct, but the reason is incorrect.

Why other options are wrong:
- Option A: The reason is not the correct explanation of the assertion.
- Option B: The reason is incorrect.
- Option D: The assertion is correct.

Key Concepts:
- Chemisorption
- Activation energy
- Exothermic process

Memory Aids:
- Chemisorption involves chemical bonds and is exothermic.

Quick Tips:
- Consider the temperature dependence of both the forward and reverse processes in adsorption.

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