Among the electrolytes ${\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_4}{\rm{,CaC}}{{\rm{l}}_2}{\rm{,A}}{{\rm{l}}_2}{\left( {{\rm{S}}{{\rm{O}}_4}} \right)_3}\;{\rm{and}}\;{\rm{N}}{{\rm{H}}_4}{\rm{Cl}}$, the most effective coagulation agent for ${\rm{S}}{{\rm{b}}_2}{{\rm{S}}_3}$ sol is

${\rm{A}}{{\rm{l}}_2}{\left( {{\rm{S}}{{\rm{O}}_4}} \right)_3}$

${\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_4}$

${\rm{N}}{{\rm{H}}_4}{\rm{Cl}}$

Have general formula ${M_{x/n}}\left[ {{{\left( {{\rm{Al}}{{\rm{O}}_2}} \right)}_x}{{\left( {{\rm{Si}}{{\rm{O}}_2}} \right)}_4}} \right] \cdot m{{\rm{H}}_2}{\rm{O}}$

Have pore sizes between 260 pm to 740 pm

Addition of ${\rm{FeC}}{{\rm{l}}_3}$ to ${{\rm{K}}_4}\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right]$ in dilute and cold solution gives :

${\rm{F}}{{\rm{e}}_4}{\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right]_3}$ sol

SURFACE CHEMISTRY MCQs for NEET UG CHEMISTRY — Easy

1 Easy

1 of 56Question permalink

Understanding +4 | -1

For adsorption of gas on solid surface, the plots of

\log \;x/m\mathrm{ } \;vs.

\log \;P

is linear with a slope equal to :

View answer and explanation

Correct option: D

\mathrm{l}/n

(

n

being integer)

Option D (

\mathrm{l}/n

(

n

being integer)) is correct because the Freundlich adsorption isotherm is represented by the equation

\frac{x}{m} = kP^{\frac{1}{n}}

, and the slope of the plot of

\log \;x/m\mathrm{ } \;vs.

\log \;P

is

\frac{1}{n}

.

Solution approach

Correct Answer + Core Reason: $\mathrm{l}/n$ ( $n$ being integer). The Freundlich adsorption isotherm is represented by the equation $\frac{x}{m} = kP^{\frac{1}{n}}$ , and the slope of the plot of $\log \;x/m\mathrm{ } \;vs.$ $\log \;P$ is $\frac{1}{n}$ .

Step-by-Step Solution:

Step 1: Recall the Freundlich adsorption isotherm:
$\frac{x}{m} = kP^{\frac{1}{n}}$
where:
- $x$ is the mass of the gas adsorbed
- $m$ is the mass of the adsorbent
- $P$ is the pressure
- $k$ and $n$ are constants

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

Step 3: Compare the equation with the equation of a straight line ( $y = mx + c$ ):
Here, $y = \log \frac{x}{m}$ , $x = \log P$ , $m = \frac{1}{n}$ , and $c = \log k$ .

Step 4: Identify the slope of the plot of $\log \frac{x}{m}$ vs $\log P$ :
The slope is $\frac{1}{n}$ .

Why other options are wrong:
- Option A: This is incorrect because $K$ is the Freundlich constant, not the slope.
- Option B: This is incorrect because $\log K$ is the intercept, not the slope.
- Option C: This is incorrect because $\frac{1}{n}K$ is not the slope of the plot.

Key Concepts:
- Adsorption
- Freundlich adsorption isotherm

Memory Aids:
- Remember the Freundlich adsorption isotherm: $\frac{x}{m} = kP^{\frac{1}{n}}$

Quick Tips:
- Understand the relationship between the Freundlich adsorption isotherm and the plot of $\log \frac{x}{m}$ vs $\log P$ .

2 Easy

2 of 56Question permalink

Application +4 | -1

Which of the following has maximum coagulation power with ferric hydroxide sol?

View answer and explanation

Correct option: D

{\mathrm{K}_4}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]

Option D (

{\mathrm{K}_4}[{\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}}])

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

[Fe(CN)_6]^{4-}

has the highest charge.

Solution approach

Correct Answer + Core Reason: ${\mathrm{K}_4}[{\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}}]$ . 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 ferric hydroxide sol.

Step 2: Ferric hydroxide 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:
- Cryolite: Complex fluoride, difficult to determine exact charge without knowing the specific formula, but likely less than -4.
- ${\mathrm{K}_2}{\mathrm{C}_2}{\mathrm{O}_4}$ : $C_2O_4^{2-}$ (charge -2)
- ${\mathrm{K}_3}[{\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}}]$ : $[Fe(CN)_6]^{3-}$ (charge -3)
- ${\mathrm{K}_4}[{\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}}]$ : $[Fe(CN)_6]^{4-}$ (charge -4)

Step 5: $[Fe(CN)_6]^{4-}$ has the highest charge (-4), so ${\mathrm{K}_4}[{\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}}]$ has the maximum coagulation power.

Why other options are wrong:
- Option A: Cryolite has a lower charge than $[Fe(CN)_6]^{4-}$ .
- Option B: ${\mathrm{K}_2}{\mathrm{C}_2}{\mathrm{O}_4}$ has a lower charge (-2) than $[Fe(CN)_6]^{4-}$ .
- Option C: ${\mathrm{K}_3}[{\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}}]$ has a lower charge (-3) than $[Fe(CN)_6]^{4-}$ .

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.

3 Easy

3 of 56Question permalink

Application +4 | -1

Among the electrolytes

\mathrm{N}{\mathrm{a}_2}\mathrm{S}{\mathrm{O}_4}\mathrm{,CaC}{\mathrm{l}_2}\mathrm{,A}{\mathrm{l}_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3}\;\mathrm{and}\;\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}

, the most effective coagulation agent for

\mathrm{S}{\mathrm{b}_2}{\mathrm{S}_3}

sol is

View answer and explanation

Correct option: C

\mathrm{A}{\mathrm{l}_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3}

Option C (

\mathrm{A}{\mathrm{l}_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3}

) is correct because, according to the Hardy-Schulze rule, the ion with the highest charge will be the most effective in coagulating a sol. Since

Sb_2S_3

is a negative sol, the cation with the highest charge (

Al^{3+}

) will be the most effective.

Solution approach

Correct Answer + Core Reason: $\mathrm{A}{\mathrm{l}_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3}$ . According to the Hardy-Schulze rule, the ion with the highest charge has the greatest coagulating power.

Step-by-Step Solution:

Step 1: Determine the charge of the $\mathrm{S}{\mathrm{b}_2}{\mathrm{S}_3}$ sol. Since it's a sulphide sol, it's negatively charged.

Step 2: Identify the ions provided by each electrolyte:
- $\mathrm{N}{\mathrm{a}_2}\mathrm{S}{\mathrm{O}_4}$ provides $Na^{+}$ ions.
- $\mathrm{CaC}{\mathrm{l}_2}$ provides $Ca^{2+}$ ions.
- $\mathrm{A}{\mathrm{l}_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3}$ provides $Al^{3+}$ ions.
- $\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}$ provides $NH_4^{+}$ ions.

Step 3: According to the Hardy-Schulze rule, the ion with the highest charge will be the most effective in coagulating the sol. Since the sol is negatively charged, we consider the cations.

Step 4: Compare the charges of the cations: $Al^{3+}$ has the highest charge (+3), followed by $Ca^{2+}$ (+2), $Na^{+}$ (+1), and $NH_4^{+}$ (+1).

Step 5: Therefore, $\mathrm{A}{\mathrm{l}_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3}$ is the most effective coagulation agent.

Why other options are wrong:
- Option A: $\mathrm{N}{\mathrm{a}_2}\mathrm{S}{\mathrm{O}_4}$ : $Na^{+}$ ions have a lower charge than $Al^{3+}$ ions, so its coagulating power is less.
- Option B: $\mathrm{CaC}{\mathrm{l}_2}$ : $Ca^{2+}$ ions have a lower charge than $Al^{3+}$ ions, so its coagulating power is less.
- Option D: $\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}$ : $NH_4^{+}$ ions have a lower charge than $Al^{3+}$ ions, so its coagulating power is less.

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

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

Quick Tips:
- Identify the charge of the sol and the ions provided by the electrolytes.
- Apply the Hardy-Schulze rule to determine the coagulating power.

4 Easy

4 of 56Question permalink

Remembering +4 | -1

Zeolites :

View answer and explanation

Correct option: D

All of the above

The correct answer is D (All of the above) because zeolites are microporous aluminosilicates with a specific formula and pore sizes.

Solution approach

Correct Answer + Core Reason: All of the above. Zeolites are microporous aluminosilicates with a specific formula and pore sizes within a certain range.

Step-by-Step Solution:

Step 1: Recall the properties of zeolites.

Step 2: Zeolites are microporous aluminosilicates.

Step 3: Zeolites have the general formula ${M_{x/n}}\left[ {\left( {\mathrm{Al}{\mathrm{O}_2}} \right)_x{{\left( {\mathrm{Si}{\mathrm{O}_2}} \right)}_4}} \right] \cdot m{\mathrm{H}_2}\mathrm{O}$

Step 4: Zeolites have pore sizes between 260 pm to 740 pm.

Why other options are wrong:
- Option A: Zeolites are microporous aluminosilicates, but this is not the only correct statement.
- Option B: Zeolites have the given formula, but this is not the only correct statement.
- Option C: Zeolites have the given pore sizes, but this is not the only correct statement.

Key Concepts:
- Zeolites
- Aluminosilicates
- Microporous materials

Memory Aids:
- Zeolites are aluminosilicates with pores.

Quick Tips:
- Know the general properties and characteristics of zeolites.

5 Easy

5 of 56Question permalink

Remembering +4 | -1

Milk can be preserved by adding a few drops of :

View answer and explanation

Correct option: B

Formaldehyde solution

Option B (Formaldehyde solution) is correct because formaldehyde acts as a preservative by killing bacteria and preventing spoilage in milk.

Solution approach

Correct Answer + Core Reason: Formaldehyde solution. Formaldehyde acts as a preservative due to its antibacterial properties.

Step-by-Step Solution:

Step 1: Understand the need for preservation. Milk spoils due to bacterial growth.

Step 2: Identify suitable preservatives. Preservatives inhibit or kill bacteria.

Step 3: Consider the given options. Formaldehyde is known for its antibacterial properties and can be used (though not recommended for consumption) as a preservative.

Why other options are wrong:
- Option A: Formic acid solution is an acid and may curdle the milk rather than preserve it.
- Option C: Acetic acid solution (vinegar) is an acid and will curdle the milk.
- Option D: Acetaldehyde solution is an aldehyde, but formaldehyde is a more effective preservative.

Key Concepts:
- Preservation: Preventing spoilage by inhibiting bacterial growth.
- Antibacterial properties: The ability to kill or inhibit bacteria.

Memory Aids:
- Formaldehyde is a known preservative.

Quick Tips:
- Consider the properties of each chemical and its effect on milk.

6 Easy

6 of 56Question permalink

Understanding +4 | -1

The rate of a certain biochemical reaction catalysed by an enzyme in human body is

{10^4}

times faster than when it carried out in the laboratory. The activation energy of this reaction :

View answer and explanation

Correct option: B

Is different in two cases

Option B (Is different in two cases) is correct because enzymes lower the activation energy significantly compared to the uncatalyzed reaction.

Solution approach

Correct Answer + Core Reason: Is different in two cases. Enzymes lower the activation energy, leading to a faster reaction rate compared to the uncatalyzed reaction.

Step-by-Step Solution:

Step 1: Understand the role of enzymes as catalysts.

Step 2: Enzymes lower the activation energy of biochemical reactions.

Step 3: A lower activation energy results in a faster reaction rate.

Step 4: Since the reaction is ${10^4}$ times faster with the enzyme, the activation energy must be different.

Why other options are wrong:
- Option A: Is zero is incorrect because activation energy cannot be zero for a reaction to occur, even with a catalyst.
- Option C: Is the same in both the cases is incorrect because the enzyme changes the activation energy.
- Option D: None of the above is incorrect because option B is correct.

Key Concepts:
- Enzymes
- Activation energy
- Catalysis

Memory Aids:
- Enzymes: "lower activation energy, faster reaction".

Quick Tips:
- Remember that catalysts, including enzymes, lower the activation energy of a reaction.

7 Easy

7 of 56Question permalink

Application +4 | -1

Which is more powerful to coagulate the negative colloid?

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because, according to the Hardy-Schulze rule, the ion with the highest charge will be the most effective in coagulating a negative colloid.

Al^{3+}

has the highest positive charge among the options.

Solution approach

Correct Answer + Core Reason: $\mathrm{AlC}{\mathrm{l}_3}$ . According to the Hardy-Schulze rule, the ion with the highest charge has the greatest coagulating power for a negative colloid.

Step-by-Step Solution:

Step 1: Identify that the colloid is negative.

Step 2: Identify the ions responsible for coagulation in each option:
- $\mathrm{ZnS}{\mathrm{O}_4}$ provides $Zn^{2+}$ ions.
- $\mathrm{N}{\mathrm{a}_3}\mathrm{P}{\mathrm{O}_4}$ provides $Na^{+}$ ions.
- $\mathrm{AlC}{\mathrm{l}_3}$ provides $Al^{3+}$ ions.
- ${\mathrm{K}_4}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]$ provides $K^{+}$ ions (coagulation is by positive ions for a negative colloid).

Step 3: Compare the charges of the positive ions. $Al^{3+}$ has the highest charge (+3), followed by $Zn^{2+}$ (+2), and $Na^{+}$ and $K^{+}$ (+1).

Step 4: Based on the Hardy-Schulze rule, the electrolyte with $Al^{3+}$ will be the most powerful to coagulate the negative colloid.

Why other options are wrong:
- Option A: $\mathrm{ZnS}{\mathrm{O}_4}$ : $Zn^{2+}$ ions have a lower charge than $Al^{3+}$ ions, so its coagulating power is less.
- Option B: $\mathrm{N}{\mathrm{a}_3}\mathrm{P}{\mathrm{O}_4}$ : $Na^{+}$ ions have the lowest charge among the options, resulting in the least coagulating power.
- Option D: ${\mathrm{K}_4}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]$ : $K^{+}$ ions have a lower charge than $Al^{3+}$ ions, so its coagulating power is less.

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

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

Quick Tips:
- Identify the charge of the sol and the ions provided by the electrolytes.
- Apply the Hardy-Schulze rule to determine the coagulating power.

8 Easy

8 of 56Question permalink

Understanding +4 | -1

Which of the following is an example for heterogeneous catalysis reaction?

View answer and explanation

Correct option: D

Hydrolysis of liquid in the presence of aqueous mineral acid

Option C (

2{\mathrm{H}_2}{\mathrm{O}_2}\left( l \right)\xrightarrow{{pt(s)}}2{\mathrm{H}_2}\mathrm{O}\left( l \right) + {\mathrm{O}_2}\left( g \right)

) is correct because the catalyst (Pt) and reactants are in different phases.

Solution approach

Correct Answer + Core Reason: $2{\mathrm{H}_2}{\mathrm{O}_2}\left( l \right)\xrightarrow{{pt(s)}}2{\mathrm{H}_2}\mathrm{O}\left( l \right) + {\mathrm{O}_2}\left( g \right)$ . Heterogeneous catalysis involves catalyst and reactants in different phases.

Step-by-Step Solution:

Step 1: Define heterogeneous catalysis.

Step 2: Heterogeneous catalysis occurs when the catalyst and reactants are in different phases.

Step 3: Analyze each option to determine the phases of the catalyst and reactants.

Step 4: In option C, the catalyst is solid (Pt(s)) and the reactant is liquid ( ${\mathrm{H}_2}{\mathrm{O}_2}\left( l \right)$ ), hence it is heterogeneous.

Why other options are wrong:
- Option A: $2\mathrm{S}{\mathrm{O}_2}\left( g \right) + {\mathrm{O}_2}\left( g \right)\xrightarrow{{NO\;(g)}}2\mathrm{S}{\mathrm{O}_3}\left( g \right)$ is incorrect because all reactants and the catalyst are in the same phase (gas), making it homogeneous.
- Option B: Hydrolysis of aqueous sucrose solution in the presence of aqueous mineral acid is incorrect because all reactants and the catalyst are in the same phase (aqueous), making it homogeneous.
- Option D: Hydrolysis of liquid in the presence of aqueous mineral acid is incorrect because all reactants and the catalyst are in the same phase (liquid), making it homogeneous.

Key Concepts:
- Heterogeneous catalysis
- Homogeneous catalysis
- Phase of matter

Memory Aids:
- Heterogeneous: "different phases".
- Homogeneous: "same phase".

Quick Tips:
- Identify the phases of the reactants and catalyst to determine if the catalysis is homogeneous or heterogeneous.

9 Easy

9 of 56Question permalink

Remembering +4 | -1

In a chemical reaction, catalyst

View answer and explanation

Correct option: A

Decrease the energy of activation

The correct answer is A (Decrease the energy of activation) because catalysts lower the activation energy of a reaction.

Solution approach

Correct Answer + Core Reason: Decrease the energy of activation. Catalysts provide an alternative reaction pathway with a lower activation energy.

Step-by-Step Solution:

Step 1: Recall the effect of a catalyst on the activation energy of a reaction.

Step 2: Catalysts lower the activation energy, making the reaction proceed faster.

Why other options are wrong:
- Option B: Catalysts decrease, not increase, the activation energy.
- Option C: Catalysts change the activation energy.
- Option D: Catalysts do affect the activation energy.

Key Concepts:
- Catalysis
- Activation energy

Memory Aids:
- Catalysts lower activation energy.

Quick Tips:
- Catalysts speed up reactions by lowering the activation energy.

10 Easy

10 of 56Question permalink

Understanding +4 | -1

An emulsifier is a substance which

View answer and explanation

Correct option: A

Stabilises the emulsion

Option A (Stabilises the emulsion) is correct because an emulsifier's primary function is to prevent the separation of the emulsion's components, thus stabilizing it.

Solution approach

Correct Answer + Core Reason: Stabilises the emulsion. Emulsifiers prevent the separation of immiscible liquids in an emulsion.

Step-by-Step Solution:

Step 1: Define an emulsion. An emulsion is a mixture of two or more liquids that are normally immiscible (unmixable or unblendable).

Step 2: Define an emulsifier. An emulsifier is a substance that stabilizes an emulsion by increasing its kinetic stability.

Step 3: Explain the mechanism. Emulsifiers typically have a hydrophobic (water-repelling) part and a hydrophilic (water-attracting) part. This allows them to reside at the interface between the two liquids, reducing surface tension and preventing coalescence.

Step 4: Conclusion. Therefore, an emulsifier stabilizes the emulsion.

Why other options are wrong:
- Option B: Homogenisation is a process that reduces the particle size of the dispersed phase, but it doesn't necessarily stabilize the emulsion in the long term without an emulsifier.
- Option C: Coagulation is the process of particles clumping together, which is the opposite of what an emulsifier does.
- Option D: While emulsifiers aid in the initial dispersion, their primary role is to maintain the stability of the dispersion over time, not just accelerate it initially.

Key Concepts:
- Emulsions
- Emulsifiers
- Surface Tension
- Stability

Memory Aids:
- Emulsifiers = Emulsion Stabilizers

Quick Tips:
- Think of emulsifiers as \"glue\" that holds the emulsion together.

11 Easy

11 of 56Question permalink

Remembering +4 | -1

Decomposition of urea into

\mathrm{N}{\mathrm{H}_3}\;\mathrm{and}\;\mathrm{C}{\mathrm{O}_2}

is followed by the action of enzyme :

View answer and explanation

Correct option: A

Urease

The correct answer is A (Urease) because urease is the enzyme that catalyzes the decomposition of urea into ammonia and carbon dioxide.

Solution approach

Correct Answer + Core Reason: Urease. Urease is the specific enzyme that catalyzes the hydrolysis of urea into ammonia and carbon dioxide.

Step-by-Step Solution:

Step 1: Recall the function of enzymes in biological reactions.

Step 2: Identify the enzyme that specifically catalyzes the decomposition of urea.

Step 3: Urease is the enzyme known to catalyze the decomposition of urea into ammonia and carbon dioxide.

Why other options are wrong:
- Option B: Pepsin is involved in protein digestion in the stomach.
- Option C: Trypsin is involved in protein digestion in the small intestine.
- Option D: Urease is the correct enzyme.

Key Concepts:
- Enzymes
- Catalysis
- Urease

Memory Aids:
- Urease breaks down urea.

Quick Tips:
- Know the specific enzymes involved in common biochemical reactions.

12 Easy

12 of 56Question permalink

Understanding +4 | -1

Which of the following statement is not correct for the catalyst?

View answer and explanation

Correct option: B

It alters

\Delta G

of the reaction

Option B is correct because a catalyst does not alter the Gibbs free energy change (

\Delta G

) of a reaction; it only affects the rate at which equilibrium is reached.

Solution approach

Correct Answer + Core Reason: It alters $\Delta G$ of the reaction. Catalysts do not affect the thermodynamics of a reaction, only the kinetics.

Step-by-Step Solution:

Step 1: Understand the role of a catalyst. A catalyst speeds up a reaction by lowering the activation energy.

Step 2: Recall the effect of a catalyst on equilibrium. A catalyst speeds up both the forward and reverse reactions equally, thus not affecting the equilibrium constant.

Step 3: Consider the effect on Gibbs Free Energy. The Gibbs Free Energy change ( $\Delta G$ ) is a thermodynamic property and is not affected by the catalyst.

Step 4: Conclude that a catalyst does not alter $\Delta G$ .

Why other options are wrong:
- Option A: It catalyses the forward and backward reactions to the same extent. This is correct; catalysts affect both reactions equally.
- Option C: It is a substance that does not change the equilibrium constant of a reaction. This is correct; catalysts do not shift the equilibrium.
- Option D: It provides an alternate mechanism by reducing activation energy between reactants and products. This is correct; catalysts lower the activation energy.

Key Concepts:
- Catalysis
- Gibbs Free Energy
- Equilibrium Constant
- Activation Energy

Memory Aids:
- Catalysts speed up reactions but don't change equilibrium or thermodynamics.

Quick Tips:
- Remember that catalysts only affect the rate of a reaction, not its equilibrium position or thermodynamic properties.

13 Easy

13 of 56Question permalink

Understanding +4 | -1

Which forms micelles in aqueous solution above certain concentration?

View answer and explanation

Correct option: B

Dodecyl trimethyl ammonium chloride

Option B (Dodecyl trimethyl ammonium chloride) is correct because it is a surfactant that can form micelles in aqueous solution above a certain concentration.

Solution approach

Correct Answer + Core Reason: Dodecyl trimethyl ammonium chloride. It is a surfactant that forms micelles above CMC.

Step-by-Step Solution:

Step 1: Understand the requirements for micelle formation.

Step 2: Micelles are formed by amphiphilic molecules (surfactants) that have both hydrophobic and hydrophilic parts.

Step 3: Dodecyl trimethyl ammonium chloride is a surfactant with a long hydrophobic alkyl chain and a hydrophilic ammonium head group.

Why other options are wrong:
- Option A: Glucose - Glucose is a sugar and does not have a long hydrophobic chain necessary for micelle formation.
- Option C: Urea - Urea is a small, polar molecule and does not form micelles.
- Option D: Pyridinium chloride - While it has a charged group, the hydrophobic portion is not long enough to effectively form micelles.

Key Concepts:
- Micelles
- Surfactants
- Amphiphilic molecules
- Critical micelle concentration (CMC)

Memory Aids:
- Micelle formation requires a molecule with both hydrophobic and hydrophilic parts.

Quick Tips:
- Look for molecules with long alkyl chains and charged head groups.

14 Easy

14 of 56Question permalink

Understanding +4 | -1

In autocatalysis

View answer and explanation

Correct option: B

One of the product acts as catalyst

Option B (One of the product acts as catalyst) is correct because autocatalysis is defined as a reaction where a product catalyzes the reaction itself.

Solution approach

Correct Answer + Core Reason: One of the product acts as catalyst. This is the definition of autocatalysis.

Step-by-Step Solution:

Step 1: Define autocatalysis.

Step 2: In autocatalysis, one of the products of the reaction acts as a catalyst for the same reaction.

Step 3: Consider the given options:
- Reactant acting as a catalyst is not autocatalysis.
- Vessel acting as a catalyst is heterogeneous catalysis, not autocatalysis.

Why other options are wrong:
- Option A: A reactant acting as a catalyst is simply catalysis, not autocatalysis.
- Option C: The vessel acting as a catalyst is an example of heterogeneous catalysis, not autocatalysis.
- Option D: Option B is correct, so this is incorrect.

Key Concepts:
- Catalysis
- Autocatalysis

Memory Aids:
- Auto = Self; Autocatalysis = Self-catalyzed

Quick Tips:
- Remember that the *product* must act as the catalyst in autocatalysis.

15 Easy

15 of 56Question permalink

Understanding +4 | -1

The cotterells precipitator is used to :

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because the Cottrell precipitator works by neutralizing the charge on particles, causing them to coagulate and be removed from the air, which involves cataphoresis.

Solution approach

Correct Answer + Core Reason: All of the above. Cottrell precipitator neutralizes charge, coagulates particles, and involves cataphoresis.

Step-by-Step Solution:

Step 1: Understand the Cottrell precipitator. The Cottrell precipitator is a device used to remove particulate matter from industrial emissions.

Step 2: Explain the working principle. It works by ionizing the gas stream, charging the particles, and then collecting them on charged plates. This involves neutralizing the charge on the particles, causing them to coagulate and be removed.

Step 3: Relate to the given options. The process involves neutralizing the charge on carbon particles, causing them to coagulate, and the movement of charged particles in an electric field is called cataphoresis.

Why other options are wrong:
- Option A: Neutralize charge on carbon particles in air in smoke is a correct description of the function.
- Option B: Coagulate carbon atoms of smoke is a correct description of the outcome.
- Option C: Bring in cataphoresis in carbon particles is correct because the charged particles move under the influence of an electric field.

Key Concepts:
- Cottrell precipitator: A device for removing particulate matter from gas streams.
- Cataphoresis: The movement of charged particles in an electric field.
- Coagulation: The process of particles clumping together.

Memory Aids:
- Cottrell: Neutralize, Coagulate, Cataphoresis.

Quick Tips:
- Remember the three main functions of the Cottrell precipitator.

16 Easy

16 of 56Question permalink

Understanding +4 | -1

Which of the following is not a characteristic of chemisorption?

View answer and explanation

Correct option: D

Adsorption is specific

Option C (Adsorption may be multimolecular layer) is not a characteristic of chemisorption. Chemisorption typically forms a unimolecular layer, while physisorption can form multimolecular layers.

Solution approach

Correct Answer + Core Reason: Adsorption may be multimolecular layer. Chemisorption is generally unimolecular.

Step-by-Step Solution:

Step 1: Recall the characteristics of chemisorption.

Step 2: Chemisorption involves the formation of chemical bonds between the adsorbate and the adsorbent. This process is specific and typically results in the formation of a monolayer.

Step 3: Physisorption, on the other hand, involves weak van der Waals forces and can lead to the formation of multiple layers of adsorbate on the adsorbent surface.

Why other options are wrong:
- Option A: $\Delta H$ for chemisorption is typically in the order of 400 kJ/mol due to the formation of chemical bonds.
- Option B: Chemisorption is generally irreversible because it involves the formation of strong chemical bonds.
- Option D: Chemisorption is specific because it involves the formation of chemical bonds between specific adsorbate and adsorbent pairs.

Key Concepts:
- Chemisorption
- Physisorption
- Monolayer vs. Multilayer adsorption

Memory Aids:
- Chemisorption: chemical bonds, monolayer; Physisorption: weak forces, multilayer.

Quick Tips:
- Remember that chemisorption is specific and forms a monolayer, while physisorption is non-specific and can form multilayers.

17 Easy

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Understanding +4 | -1

Which of the following is not a characteristic of physisorption of a gas on solid surface?

View answer and explanation

Correct option: B

Uni-molecular layer formation

Option B (Uni-molecular layer formation) is correct because physisorption results in multi-molecular layers, not uni-molecular.

Solution approach

Correct Answer + Core Reason: Uni-molecular layer formation. Physisorption involves weak van der Waals forces leading to multi-layer adsorption, not a single layer.

Step-by-Step Solution:

Step 1: Understand the characteristics of physisorption.

Step 2: Physisorption involves weak van der Waals forces.

Step 3: Weak forces allow for the formation of multiple layers of adsorbate on the adsorbent surface.

Step 4: Therefore, uni-molecular layer formation is not a characteristic of physisorption.

Why other options are wrong:
- Option A: Reversible in nature is a characteristic of physisorption because the weak forces allow easy desorption.
- Option C: Involves van der Waals forces is a characteristic of physisorption as these are the primary forces involved.
- Option D: Low enthalpy of adsorption is a characteristic of physisorption because the forces involved are weak.

Key Concepts:
- Physisorption
- Van der Waals forces
- Adsorption isotherms

Memory Aids:
- Physisorption: Think "physical" - weak, reversible.

Quick Tips:
- Remember the difference between physisorption (weak forces, multi-layer) and chemisorption (strong forces, single-layer).

18 Easy

18 of 56Question permalink

Application +4 | -1

In an experiment,

200 \mathrm{~mL}

of

0.5 \mathrm{M}

oxalic acid is shaken with

10 \mathrm{~g}

of activated charcoal and filtered. The concentration of the filtrate is reduced to

0.4 \mathrm{M}

. The amount of adsorption

\left(\frac{x}{m}\right)

is

View answer and explanation

Correct option: C

0.18

Option C (0.18) is correct because it represents the amount of oxalic acid adsorbed per gram of charcoal, calculated from the change in concentration.

Solution approach

Correct Answer + Core Reason: 0.18. The amount adsorbed is calculated from the change in concentration of the oxalic acid solution.

Step-by-Step Solution:

Step 1: Calculate the initial moles of oxalic acid: $n_i = M_i \times V = 0.5 \text{{ M}} \times 0.2 \text{{ L}} = 0.1 \text{{ mol}}$

Step 2: Calculate the final moles of oxalic acid: $n_f = M_f \times V = 0.4 \text{{ M}} \times 0.2 \text{{ L}} = 0.08 \text{{ mol}}$

Step 3: Calculate the moles of oxalic acid adsorbed: $n_{adsorbed} = n_i - n_f = 0.1 \text{{ mol}} - 0.08 \text{{ mol}} = 0.02 \text{{ mol}}$

Step 4: Calculate the mass of oxalic acid adsorbed: $m_{adsorbed} = n_{adsorbed} \times Molar \, mass \, of \, oxalic \, acid$ . The molar mass of oxalic acid ( $H_2C_2O_4$ ) is approximately 90 g/mol. So, $m_{adsorbed} = 0.02 \text{{ mol}} \times 90 \text{{ g/mol}} = 1.8 \text{{ g}}$

Step 5: Calculate the amount of adsorption (x/m): $\frac{x}{m} = \frac{m_{adsorbed}}{mass \, of \, charcoal} = \frac{1.8 \text{{ g}}}{10 \text{{ g}}} = 0.18$

Why other options are wrong:
- Option A (0.9): This is incorrect. It might be obtained by incorrectly dividing the moles adsorbed by the mass of charcoal without converting to grams.
- Option B (1.8): This is the mass of oxalic acid adsorbed, not the amount adsorbed per gram of charcoal.
- Option D (0.09): This is incorrect and likely a calculation error.

Key Concepts:
- Adsorption
- Molarity
- Moles

Memory Aids:
- Molarity = Moles / Volume

Quick Tips:
- Always convert volumes to liters before calculations.
- Pay attention to units.

19 Easy

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Understanding +4 | -1

Isoelectric point is the pH at which colloidal particles

View answer and explanation

Correct option: C

Coagulate

Option C (Coagulate) is correct because at the isoelectric point, the net charge of the colloidal particles is zero, leading to reduced electrostatic repulsion and subsequent coagulation.

Solution approach

Correct Answer + Core Reason: Coagulate. At the isoelectric point, colloidal particles lose their charge and tend to aggregate and coagulate.

Step-by-Step Solution:

Step 1: Define Isoelectric Point (IEP). The isoelectric point (pI) is the pH at which a molecule or surface carries no net electrical charge.

Step 2: Explain Colloidal Stability. Colloidal particles are typically stabilized by electrostatic repulsion due to their surface charge. This prevents them from aggregating.

Step 3: Explain what happens at IEP. At the isoelectric point, the surface charge of the colloidal particles becomes zero. This eliminates the electrostatic repulsion between the particles.

Step 4: Explain Coagulation. Without electrostatic repulsion, van der Waals forces become dominant, causing the particles to aggregate and coagulate (clump together).

Step 5: Conclusion. Therefore, colloidal particles coagulate at the isoelectric point.

Why other options are wrong:
- Option A: At the isoelectric point, the particles have minimal or no net charge, not become electrically charged.
- Option B: The movement towards electrodes (electrophoresis) requires a net charge on the particles, which is absent at the isoelectric point.
- Option D: Option C is the correct answer.

Key Concepts:
- Isoelectric Point
- Colloids
- Electrostatic Repulsion
- Coagulation
- Zeta Potential

Memory Aids:
- IEP = No Charge = Coagulation

Quick Tips:
- Remember that the isoelectric point is a point of instability for colloids.

20 Easy

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Understanding +4 | -1

Addition of

\mathrm{FeC}{\mathrm{l}_3}

to

{\mathrm{K}_4}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]

in dilute and cold solution gives :

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because the addition of

\mathrm{FeC}{\mathrm{l}_3}

to

{\mathrm{K}_4} {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}}

can lead to the formation of Prussian blue sol,

\mathrm{F}{\mathrm{e}_4} {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} _3}

sol, and a positive sol.

Solution approach

Correct Answer + Core Reason: All of these. The reaction can lead to the formation of Prussian blue sol, $\mathrm{F}{\mathrm{e}_4} {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} _3}$ sol, and a positive sol.

Step-by-Step Solution:

Step 1: Identify the reactants and potential products.\nReactants: $\mathrm{FeC}{\mathrm{l}_3}$ and ${\mathrm{K}_4} {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}}$ \nPotential Products: Prussian blue, $\mathrm{F}{\mathrm{e}_4} {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} _3}$ , and a positive sol.

Step 2: Consider the possible reactions.
$\mathrm{FeC}{\mathrm{l}_3}$ reacts with ${\mathrm{K}_4} {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}}$ to form Prussian blue, which is a colloidal solution.
$\mathrm{FeC}{\mathrm{l}_3}$ can also react to form $\mathrm{F}{\mathrm{e}_4} {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} _3}$ sol.\nThe resulting sol can be positively charged due to the presence of $\mathrm{Fe}^{3+}$ ions.

Step 3: Conclusion\nSince all three products are possible, the correct answer is "All of these".

Why other options are wrong:
- Option A: Prussian blue sol is a possible product, but it's not the only one.
- Option B: $\mathrm{F}{\mathrm{e}_4} {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} _3}$ sol is a possible product, but it's not the only one.
- Option C: Positive sol is a possible product, but it's not the only one.

Key Concepts:
- Colloidal solutions
- Complex ion reactions
- Formation of Prussian blue

Memory Aids:
- Prussian blue is formed by the reaction of ferric ions with ferrocyanide ions.

Quick Tips:
- Remember that multiple products can form in a chemical reaction, especially in complex ion chemistry.

21 Easy

21 of 56Question permalink

Understanding +4 | -1

Which of the following is not a characteristic off chemical adsorption of a gas on solid surface ?

View answer and explanation

Correct option: D

Zero activation energy

The correct answer is D (Zero activation energy) because chemisorption requires activation energy to form chemical bonds.

Solution approach

Correct Answer + Core Reason: Zero activation energy. Chemisorption involves the formation of chemical bonds and therefore requires activation energy.

Step-by-Step Solution:

Step 1: Recall the characteristics of chemisorption.

Step 2: Chemisorption involves the formation of chemical bonds.

Step 3: The formation of chemical bonds requires activation energy.

Why other options are wrong:
- Option A: Chemisorption is irreversible.
- Option B: Chemisorption is specific in nature.
- Option C: Chemisorption forms a monomolecular layer.

Key Concepts:
- Chemisorption
- Activation energy

Memory Aids:
- Chemisorption involves chemical bonds and activation energy.

Quick Tips:
- Distinguish between physisorption and chemisorption based on the type of forces and activation energy.

22 Easy

22 of 56Question permalink

Understanding +4 | -1

Clouds, mist, fog and aerosols are colloidal solutions of :

View answer and explanation

Correct option: C

Liquid in a gas

Option C (Liquid in a gas) is correct because clouds, mist, fog, and aerosols consist of liquid droplets dispersed in a gaseous medium (air).

Solution approach

Correct Answer + Core Reason: Liquid in a gas. These are examples of liquid droplets dispersed in a gas.

Step-by-Step Solution:

Step 1: Analyze the composition of clouds, mist, fog, and aerosols.

Step 2: These are all examples of liquid droplets (water or other liquids) dispersed in air (a gas).

Step 3: Consider the definitions:
- Clouds: Water droplets or ice crystals suspended in the atmosphere.
- Mist: Fine droplets of water suspended in the air.
- Fog: A cloud at ground level.
- Aerosols: Suspension of fine solid particles or liquid droplets in a gas.

Why other options are wrong:
- Option A: Solid in a gas would be smoke or dust in the air.
- Option B: Gas in a solid is not a common type of colloid.
- Option D: Gas in a liquid would be a foam, like whipped cream.

Key Concepts:
- Colloids
- Aerosols
- Dispersed phase
- Dispersion medium

Memory Aids:
- Think of rain (liquid) falling from the sky (gas).

Quick Tips:
- Identify the dispersed phase and the dispersion medium.

23 Easy

23 of 56Question permalink

Remembering +4 | -1

The temperature above which micelle formation occurs is :

View answer and explanation

Correct option: D

Kraft’s temperature

Option D (Kraft’s temperature) is correct because it is the minimum temperature required for micelle formation.

Solution approach

Correct Answer + Core Reason: Kraft’s temperature. It is the minimum temperature above which micelle formation occurs.

Step-by-Step Solution:

Step 1: Define micelle formation.

Step 2: Micelles are aggregates of surfactant molecules that form in solution above a certain concentration and temperature.

Step 3: The Kraft temperature is the minimum temperature required for micelle formation to occur.

Why other options are wrong:
- Option A: Critical temperature is a term used in thermodynamics and phase transitions, not directly related to micelle formation.
- Option B: Charles’ temperature is not a standard term. Charles's Law relates volume and temperature of a gas.
- Option C: Inversion temperature is the temperature at which the Joule-Thomson effect changes sign, not related to micelle formation.

Key Concepts:
- Micelle formation
- Kraft temperature
- Surfactants

Memory Aids:
- Kraft: "temperature for micelles".

Quick Tips:
- Remember the definition of Kraft temperature in the context of micelle formation.

24 Easy

24 of 56Question permalink

Remembering +4 | -1

Which one is hydrophobic in nature?

View answer and explanation

Correct option: B

Sulphur

Option B (Sulphur) is correct because it is water-repelling, unlike the other options which are water-attracting.

Solution approach

Correct Answer + Core Reason: Sulphur. Sulphur is hydrophobic, meaning it repels water.

Step-by-Step Solution:

Step 1: Define hydrophobic and hydrophilic substances.

Step 2: Hydrophobic substances repel water, while hydrophilic substances attract water.

Step 3: Consider the chemical nature of each option.

Step 4: Gelatin, starch, and protein are all polymers with polar groups that can form hydrogen bonds with water, making them hydrophilic.

Step 5: Sulphur is a nonpolar element and does not interact favorably with water, making it hydrophobic.

Why other options are wrong:
- Option A: Gelatin is hydrophilic due to its proteinaceous nature and polar groups.
- Option C: Starch is hydrophilic due to the presence of hydroxyl groups.
- Option D: Protein is hydrophilic due to the presence of polar amino acid residues.

Key Concepts:
- Hydrophobic
- Hydrophilic
- Polarity

Memory Aids:
- Hydrophobic: "water-fearing".
- Hydrophilic: "water-loving".

Quick Tips:
- Remember that nonpolar substances are generally hydrophobic, while polar substances are hydrophilic.

25 Easy

25 of 56Question permalink

Remembering +4 | -1

Platinum is used as a catalyst in :

View answer and explanation

Correct option: A

Oxidation of ammonia to form nitric acid

The correct answer is A (Oxidation of ammonia to form nitric acid) because platinum is a catalyst in the Ostwald process for nitric acid production.

Solution approach

Correct Answer + Core Reason: Oxidation of ammonia to form nitric acid. Platinum is used as a catalyst in the Ostwald process for the oxidation of ammonia to nitric oxide, which is then converted to nitric acid.

Step-by-Step Solution:

Step 1: Recall the industrial processes that use platinum as a catalyst.

Step 2: The Ostwald process uses platinum as a catalyst for the oxidation of ammonia.

Why other options are wrong:
- Option B: Hardening of oils typically uses nickel as a catalyst.
- Option C: Production of synthetic rubber involves different catalysts, not platinum.
- Option D: Synthesis of methanol typically uses copper-based catalysts.

Key Concepts:
- Catalysis
- Platinum catalyst
- Ostwald process

Memory Aids:
- Platinum in the Ostwald process for nitric acid.

Quick Tips:
- Know the common catalysts used in major industrial processes.

26 Easy

26 of 56Question permalink

Application +4 | -1

The arsenioussulphide sol has negative charge. The maximum coagulating power for precipitating it is of :

View answer and explanation

Correct option: D

0.1\;N\;\mathrm{AlC}{\mathrm{l}_3}

Option D (

0.1\;N\;\mathrm{AlC}{\mathrm{l}_3}

) is correct because, according to the Hardy-Schulze rule, ions with higher charges have greater coagulating power, and

Al^{3+}

has the highest positive charge among the given options for a negatively charged sol.

Solution approach

Correct Answer + Core Reason: $0.1\;N\;\mathrm{AlC}{\mathrm{l}_3}$ . The coagulating power is highest for the ion with the highest charge, according to the Hardy-Schulze rule.

Step-by-Step Solution:

Step 1: Identify the charge of the arsenious sulphide sol. It is given as negatively charged.

Step 2: Recall the Hardy-Schulze rule, which states that the coagulating power of an electrolyte is directly proportional to the valency of the ion having the charge opposite to that of the colloidal solution.

Step 3: Determine the ions responsible for coagulation in each option:
- A. $\mathrm{Zn}{\left( {\mathrm{N}{\mathrm{O}_3}} \right)_2}$ provides $Zn^{2+}$ ions.
- B. $\mathrm{N}{\mathrm{a}_3}\mathrm{P}{\mathrm{O}_4}$ provides $Na^{+}$ ions.
- C. $\mathrm{ZnS}{\mathrm{O}_4}$ provides $Zn^{2+}$ ions.
- D. $\mathrm{AlC}{\mathrm{l}_3}$ provides $Al^{3+}$ ions.

Step 4: Compare the charges of the positive ions. $Al^{3+}$ has the highest charge (+3), followed by $Zn^{2+}$ (+2), and $Na^{+}$ (+1).

Step 5: Based on the Hardy-Schulze rule, the electrolyte with $Al^{3+}$ will have the maximum coagulating power.

Step 6: Therefore, $0.1\;N\;\mathrm{AlC}{\mathrm{l}_3}$ has the maximum coagulating power.

Why other options are wrong:
- Option A: $\mathrm{0.1}\,N\;\mathrm{Zn}{\left( {\mathrm{N}{\mathrm{O}_3}} \right)_2}$ : $Zn^{2+}$ ions have a lower charge than $Al^{3+}$ ions, so its coagulating power is less.
- Option B: $\mathrm{0.1}\,N\;\mathrm{N}{\mathrm{a}_3}\mathrm{P}{\mathrm{O}_4}$ : $Na^{+}$ ions have the lowest charge among the options, resulting in the least coagulating power.
- Option C: $\mathrm{0.1}\,N\;\mathrm{ZnS}{\mathrm{O}_4}$ : $Zn^{2+}$ ions have a lower charge than $Al^{3+}$ ions, so its coagulating power is less.

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

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

Quick Tips:
- Identify the charge of the sol and the ions provided by the electrolytes.
- Apply the Hardy-Schulze rule to determine the coagulating power.

27 Easy

27 of 56Question permalink

Remembering +4 | -1

The catalyst iron, employed in the Haber’s process, contains molybdenum, the function of which is :

View answer and explanation

Correct option: C

To act as a catalyst promoter and increase activity of catalyst

The correct answer is C (To act as a catalyst promoter and increase activity of catalyst) because molybdenum acts as a promoter in the Haber process.

Solution approach

Correct Answer + Core Reason: To act as a catalyst promoter and increase activity of catalyst. Molybdenum acts as a promoter, enhancing the activity of the iron catalyst in the Haber process.

Step-by-Step Solution:

Step 1: Recall the role of promoters in catalysis.

Step 2: Identify the function of molybdenum in the Haber process.

Step 3: Molybdenum acts as a promoter, increasing the activity of the iron catalyst.

Why other options are wrong:
- Option A: Molybdenum does not directly increase the rate of combination of gases but enhances the catalyst's activity.
- Option B: Molybdenum's primary role is not to counterbalance impurities.
- Option D: Molybdenum's primary role is not to compensate for adverse conditions.

Key Concepts:
- Catalysis
- Catalyst promoter
- Haber process

Memory Aids:
- Molybdenum promotes iron in the Haber process.

Quick Tips:
- Know the roles of promoters and inhibitors in catalysis.

28 Easy

28 of 56Question permalink

Remembering +4 | -1

Identify the gas which is readily adsorbed by activated charcoal

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because sulfur dioxide is readily adsorbed by activated charcoal due to its higher critical temperature and stronger intermolecular forces compared to the other gases listed.

Solution approach

Correct Answer + Core Reason: $\mathrm{S}{\mathrm{O}_2}$ . Sulfur dioxide is readily adsorbed by activated charcoal due to its higher critical temperature and stronger intermolecular forces.

Step-by-Step Solution:

Step 1: Understand adsorption and activated charcoal: Activated charcoal is a highly porous material with a large surface area, making it an effective adsorbent.

Step 2: Factors affecting adsorption: The extent of adsorption depends on factors like:
- Nature of the gas (critical temperature, polarity, molecular weight)
- Nature of the adsorbent (surface area, pore size)
- Temperature
- Pressure

Step 3: Compare the gases: Among the given gases, $\mathrm{S}{\mathrm{O}_2}$ has a higher critical temperature and stronger intermolecular forces compared to ${\mathrm{H}_2}$ , ${\mathrm{N}_2}$ , and ${\mathrm{O}_2}$ . This makes it more readily adsorbed.

Why other options are wrong:
- Option A: ${\mathrm{H}_2}$ is a small, non-polar molecule with weak intermolecular forces, making it less readily adsorbed.
- Option B: ${\mathrm{N}_2}$ is a non-polar molecule with relatively weak intermolecular forces, making it less readily adsorbed compared to $\mathrm{S}{\mathrm{O}_2}$ .
- Option D: ${\mathrm{O}_2}$ is also a non-polar molecule with relatively weak intermolecular forces, making it less readily adsorbed compared to $\mathrm{S}{\mathrm{O}_2}$ .

Key Concepts:
- Adsorption
- Activated Charcoal
- Critical Temperature
- Intermolecular Forces

Memory Aids:
- Higher critical temperature = easier adsorption.

Quick Tips:
- Consider the intermolecular forces and critical temperatures of gases when predicting adsorption.

29 Easy

29 of 56Question permalink

Understanding +4 | -1

Which of the following is not correct?

View answer and explanation

Correct option: C

Physical adsorption increases with the increase in temperature

Option C (Finely divide platinum) is correct because platinum has a high surface area when finely divided, which allows for greater adsorption of hydrogen gas compared to the other options.

Solution approach

Correct Answer + Core Reason: Finely divide platinum. Platinum has a high surface area and good adsorption properties.

Step-by-Step Solution:
Step 1: Adsorption is a surface phenomenon, so the amount of gas adsorbed depends on the surface area of the adsorbent.
Step 2: Finely divided metals have a larger surface area compared to colloidal solutions or bulk materials.
Step 3: Platinum is known for its excellent catalytic and adsorption properties, especially for hydrogen gas.
Step 4: Therefore, finely divided platinum will adsorb a larger volume of hydrogen gas compared to the other options.

Why other options are wrong:
- Option A (Colloidal solution of palladium): While palladium can adsorb hydrogen, a colloidal solution has a lower effective surface area than finely divided platinum.
- Option B (Finely divided nickel): Nickel is also a good adsorbent, but platinum generally has better adsorption properties for hydrogen.
- **Option D (Colloidal $\mathrm{Fe}{\left( {\mathrm{OH}} \right)_3}$ ): Iron(III) hydroxide is not as effective at adsorbing hydrogen gas as platinum or nickel.

Key Concepts:
- Adsorption
- Surface Area
- Adsorbent Properties
- Catalysis

Memory Aids:
- Platinum (Pt) = Premium Adsorbent

Quick Tips:**
- Remember that surface area is crucial for adsorption, and some materials are better adsorbents than others.

30 Easy

30 of 56Question permalink

Remembering +4 | -1

Washing soap can be prepared by saponifying alkali with oil of :

View answer and explanation

Correct option: C

Ground nut oil

Option C (Ground nut oil) is correct because washing soaps are typically prepared by saponifying alkali with vegetable oils like groundnut oil.

Solution approach

Correct Answer + Core Reason: Ground nut oil. Washing soap is prepared by saponifying alkali with vegetable oils like groundnut oil.

Step-by-Step Solution:

Step 1: Understand the process of saponification. Saponification is the process of making soap by reacting a fat or oil with an alkali (base).

Step 2: Identify suitable oils for saponification. Vegetable oils like groundnut oil, coconut oil, and olive oil are commonly used in soap making.

Step 3: Consider the given options. Among the options, only groundnut oil is a suitable vegetable oil for saponification.

Why other options are wrong:
- Option A: Rose oil is an essential oil used for fragrance and is not suitable for saponification.
- Option B: Paraffin oil is a mineral oil and does not undergo saponification.
- Option D: Kerosene is a petroleum-derived fuel and is not suitable for saponification.

Key Concepts:
- Saponification: The process of making soap from fats or oils and an alkali.
- Vegetable oils: Oils derived from plants, suitable for saponification.

Memory Aids:
- Remember that soaps are made from fats/oils and alkali.

Quick Tips:
- Focus on identifying the correct type of oil (vegetable oil) used in soap making.

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