SURFACE CHEMISTRY MCQs for NEET UG CHEMISTRY — Medium

1 Medium

Application +4 | -1

Graph between

\log \frac{{x}}{{m}}

and

\log {P}

is a straight line inclined at an angle

\theta=45^{\circ}

. When pressure is

0.5

atm and

\log {k}= 0.699,

the amount of solute adsorbed per

\mathrm{g}

of adsorbent will be:

View answer and explanation

Correct option: C

2.5 \mathrm{~g} / \mathrm{g}

adsorbent

Option C (2.5 g / g adsorbent) is correct because the amount of solute adsorbed per gram of adsorbent is calculated using the Freundlich isotherm equation.

Solution approach

Correct Answer + Core Reason: $2.5 \text{{ g / g adsorbent}}$ . The amount 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: Determine the value of $\frac{1}{n}$ : The angle is $45^\circ$ , so the slope of the line is $\tan(45^\circ) = 1$ . Therefore, $\frac{1}{n} = 1$ .

Step 4: Identify the given values: $\log(k) = 0.699$ and $p = 0.5 \text{{ atm}}$ .

Step 5: Calculate k: $k = 10^{0.699} = 10^{\log 5} = 5$

Step 6: Substitute the values into the Freundlich isotherm equation: $\frac{x}{m} = 5 \cdot (0.5)^1 = 2.5 \text{{ g/g adsorbent}}$

Why other options are wrong:
- **Option A ( $0.397 \text{{ g / g adsorbent}}$ ): Incorrect calculation.
- Option B ( $1.5 \text{{ g / g adsorbent}}$ ): Incorrect calculation.
- Option D ( $0.25 \text{{ g / g adsorbent}}$ ): Incorrect calculation.

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 $\log(k)$ .

2 Medium

2 of 59Question permalink

Understanding +4 | -1

The magnitude of colligative properties in all colloidal dispersions is …..than solution :

View answer and explanation

Correct option: B

Lower

Option B (Lower) is correct because colloidal dispersions have fewer particles than true solutions for the same mass concentration, leading to lower colligative properties.

Solution approach

Correct Answer + Core Reason: Lower. Colligative properties depend on the number of solute particles. Colloidal dispersions have fewer particles than true solutions.

Step-by-Step Solution:

Step 1: Define colligative properties: Colligative properties are properties of solutions that depend on the number of solute particles in a solution and not on the nature of the solute.

Step 2: Consider particle concentration in colloidal dispersions: In colloidal dispersions, the dispersed phase consists of aggregates of molecules or ions. Therefore, for the same mass concentration, the number of particles is much lower in a colloidal dispersion compared to a true solution.

Step 3: Relate particle concentration to colligative properties: Since colligative properties depend on the number of particles, a lower particle concentration in colloidal dispersions results in lower colligative properties compared to true solutions.

Step 4: Conclusion: The magnitude of colligative properties in colloidal dispersions is lower than in solutions.

Why other options are wrong:
- Option A: This option is incorrect because colloidal dispersions have lower particle concentrations than true solutions, leading to lower colligative properties.
- Option C: This option is incorrect because colligative properties are lower in colloidal dispersions.
- Option D: This option is incorrect because the colligative properties are definitely different.

Key Concepts:
- Colligative Properties
- Colloidal Dispersions
- Solutions

Memory Aids:
- Colligative properties depend on the *number* of particles.

Quick Tips:
- Remember that colloids have larger particle sizes, so fewer particles for the same mass.

3 Medium

3 of 59Question permalink

Remembering +4 | -1

Which of the following is a property of physisorption?

View answer and explanation

Correct option: C

Non-specificity

Option C (Non-specificity) is correct because physisorption involves weak van der Waals forces and does not require specific interactions between the adsorbate and adsorbent.

Solution approach

Correct Answer + Core Reason: Non-specificity. Physisorption is a non-specific process due to weak van der Waals forces.

Step-by-Step Solution:
Step 1: Physisorption involves weak intermolecular forces (van der Waals forces) between the adsorbate and adsorbent.
Step 2: These forces are non-specific, meaning they can occur between any two molecules.
Step 3: Therefore, physisorption does not require specific chemical interactions or bonding.

Why other options are wrong:
- Option A (High specificity): High specificity is a characteristic of chemisorption, where chemical bonds are formed.
- Option B (Irreversibility): Irreversibility is a characteristic of chemisorption, while physisorption is generally reversible.
- Option D (None of these): Non-specificity is a property of physisorption.

Key Concepts:
- Physisorption
- Specificity
- Van der Waals forces
- Chemisorption

Memory Aids:
- Physisorption = Physical = Non-specific

Quick Tips:
- Remember the difference between physisorption (non-specific) and chemisorption (specific).

4 Medium

4 of 59Question permalink

Remembering +4 | -1

The stabilization of a dispersed phase in a lyophobic colloid is due to

View answer and explanation

Correct option: C

The formation of an electrical layer between two phases

Option C (The formation of an electrical layer between two phases) is correct because the stabilization of lyophobic colloids is due to the electrical double layer formed at the interface.

Solution approach

Correct Answer + Core Reason: The formation of an electrical layer between two phases. This layer prevents aggregation of the dispersed phase.

Step-by-Step Solution:
Step 1: Lyophobic colloids are inherently unstable due to the lack of attraction between the dispersed phase and the dispersion medium.
Step 2: Stabilization is achieved by the formation of an electrical double layer at the interface between the dispersed phase and the dispersion medium.
Step 3: This electrical layer consists of a layer of adsorbed ions and a diffuse layer of counterions, creating a repulsive force that prevents aggregation.

Why other options are wrong:
- Option A (The absorption of charged substances on dispersed phase): While adsorption of charged substances contributes to the electrical layer, it's not the complete picture. The double layer formation is key.
- Option B (The large electro-kinetic potential developed in the colloid): The electro-kinetic potential (zeta potential) is a result of the electrical double layer, not the cause of stabilization itself.
- Option D (The viscosity of the medium): Viscosity can influence the stability, but the primary stabilization mechanism is the electrical double layer.

Key Concepts:
- Lyophobic colloids
- Electrical double layer
- Zeta potential
- Colloid stability

Memory Aids:
- Lyophobic = Liquid hating = Needs electrical layer to stabilize

Quick Tips:
- Remember that lyophobic colloids require external stabilization mechanisms, such as the electrical double layer.

5 Medium

5 of 59Question permalink

Analyzing +4 | -1

Assertion (A): Ethyl Alcohol vapour is passed over anhydrous alumina as catalyst at

350^\circ C

acetaldehyde is obtained.
Reason (R): Emulsions contain liquid dispersed in liquid.

View answer and explanation

Correct option: D

(A) is false but (R) is true

Option D ((A) is false but (R) is true) is correct because the oxidation of ethyl alcohol to acetaldehyde requires a catalyst like copper at 573K, not alumina at 350°C, and emulsions are indeed liquid dispersed in liquid.

Solution approach

Correct Answer + Core Reason: (A) is false but (R) is true. The reaction of ethyl alcohol to acetaldehyde requires a different catalyst and temperature, and emulsions are liquid dispersed in liquid.

Step-by-Step Solution:

Step 1: Analyze Assertion (A): Ethyl Alcohol vapor is passed over anhydrous alumina as a catalyst at $350^\circ C$ , and acetaldehyde is obtained. This is false. Alumina at that temperature would likely cause dehydration to ethene. The oxidation of ethanol to acetaldehyde typically uses a copper catalyst at a higher temperature (around 573 K).

Step 2: Analyze Reason (R): Emulsions contain liquid dispersed in liquid. This is true. Emulsions are a type of colloid where both the dispersed phase and the dispersion medium are liquids.

Step 3: Determine the relationship between (A) and (R). (R) is a correct statement, but it does not explain (A).

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

Key Concepts:
- Catalysis
- Oxidation of alcohols
- Emulsions

Memory Aids:
- Oxidation of alcohols: Primary alcohols can be oxidized to aldehydes or carboxylic acids depending on the oxidizing agent and conditions.
- Emulsion: Liquid in liquid.

Quick Tips:
- Remember common catalysts and reaction conditions for organic reactions.

6 Medium

6 of 59Question permalink

Application +4 | -1

A cube of

1 \mathrm{~m}

edge length is crushed in order to get small cubes of edge length

1 \mathrm{~mm}

. By which factor does the surface area increase?

View answer and explanation

Correct option: B

10^3

Option B (

10^3

) is correct because the surface area increases by a factor of 1000 when the cube is crushed into smaller cubes with 1/1000th the edge length.

Solution approach

Correct Answer + Core Reason: $10^3$ . The surface area increases by a factor of 1000.

Step-by-Step Solution:

Step 1: Calculate the surface area of the original cube.\nSurface area of original cube = $6 \times (1 \text{ m})^2 = 6 \text{ m}^2$

Step 2: Calculate the volume of the original cube.\nVolume of original cube = $(1 \text{ m})^3 = 1 \text{ m}^3$

Step 3: Calculate the volume of each small cube.\nEdge length of small cube = $1 \text{ mm} = 10^{-3} \text{ m}$ \nVolume of small cube = $(10^{-3} \text{ m})^3 = 10^{-9} \text{ m}^3$

Step 4: Calculate the number of small cubes.\nNumber of small cubes = $\frac{\text{Volume of original cube}}{\text{Volume of small cube}} = \frac{1 \text{ m}^3}{10^{-9} \text{ m}^3} = 10^9$

Step 5: Calculate the surface area of each small cube.\nSurface area of small cube = $6 \times (10^{-3} \text{ m})^2 = 6 \times 10^{-6} \text{ m}^2$

Step 6: Calculate the total surface area of all small cubes.\nTotal surface area = Number of small cubes $\times$ Surface area of small cube = $10^9 \times 6 \times 10^{-6} \text{ m}^2 = 6 \times 10^3 \text{ m}^2$

Step 7: Calculate the factor by which the surface area increases.\nFactor = $\frac{\text{Total surface area}}{\text{Surface area of original cube}} = \frac{6 \times 10^3 \text{ m}^2}{6 \text{ m}^2} = 10^3$

Why other options are wrong:
- Option A: $10^4$ is incorrect. It likely arises from miscalculating the surface area or the number of smaller cubes.
- Option C: $10^5$ is incorrect. This could be due to an error in calculating the ratio of volumes or surface areas.
- Option D: $10^2$ is incorrect. This might result from only considering the change in edge length without accounting for the number of cubes.

Key Concepts:
- Surface area
- Volume
- Ratios

Memory Aids:
- Surface area: "area of all faces".
- Volume: "space occupied".

Quick Tips:
- Pay attention to units and conversions.

7 Medium

7 of 59Question permalink

Remembering +4 | -1

Sulphur colloid is prepared by

View answer and explanation

Correct option: B

Oxidation

Option B (Oxidation) is correct because sulphur colloid is prepared by oxidizing hydrogen sulfide or other sulfur-containing compounds.

Solution approach

Correct Answer + Core Reason: Oxidation. Sulphur colloid is typically prepared by oxidizing a sulphur compound.

Step-by-Step Solution:

Step 1: Understand the formation of sulphur colloid.

Step 2: Sulphur colloid can be formed by the oxidation of hydrogen sulfide ( $H_2S$ ) or by the reaction of sodium thiosulfate with an acid.

Step 3: The oxidation process involves the loss of electrons, leading to the formation of elemental sulphur in colloidal form.
$H_2S + O_2 \rightarrow S + H_2O$

Why other options are wrong:
- Option A: Mechanical dispersion involves breaking down larger particles into smaller ones, which is not the primary method for sulphur colloid preparation.
- Option C: Electrical dispersion (Bredig's arc method) is used for metals, not typically for sulphur.
- Option D: Reduction involves gaining electrons, which is the opposite of what happens in the formation of sulphur colloid from $H_2S$ .

Key Concepts:
- Colloid preparation
- Oxidation
- Sulphur chemistry

Memory Aids:
- Remember that sulphur often gains oxygen in its colloidal preparation.

Quick Tips:
- Consider the chemical reactions involved in colloid formation to determine the method.

8 Medium

8 of 59Question permalink

Remembering +4 | -1

An example of homogeneous catalysis is

View answer and explanation

Correct option: C

Lead chamber process

Option C is correct because the lead chamber process involves gaseous reactants and catalysts in the same phase, making it a homogeneous catalysis.

Solution approach

Correct Answer + Core Reason: Lead chamber process. In the lead chamber process, the reactants and catalyst are in the same phase (gas phase), making it a homogeneous catalysis.

Step-by-Step Solution:

Step 1: Define homogeneous catalysis. Homogeneous catalysis is a type of catalysis in which the catalyst and the reactants are in the same phase (i.e., all are gases or all are liquids).

Step 2: Define heterogeneous catalysis. Heterogeneous catalysis is a type of catalysis in which the catalyst and the reactants are in different phases (e.g., a solid catalyst and gaseous reactants).

Step 3: Analyze the given options.
- Contact process: The contact process involves the oxidation of sulfur dioxide to sulfur trioxide using a solid catalyst (V2O5), so it is heterogeneous.
- Haber’s process: The Haber process involves the synthesis of ammonia from nitrogen and hydrogen using a solid catalyst (iron), so it is heterogeneous.
- Lead chamber process: The lead chamber process involves the oxidation of sulfur dioxide to sulfur trioxide using nitrogen oxides (NO) as catalysts in the gas phase, so it is homogeneous.
- Hydrogenation of oils: Hydrogenation of oils typically uses a solid catalyst (e.g., nickel), so it is heterogeneous.

Why other options are wrong:
- Option A: Contact process. The contact process is heterogeneous catalysis because it uses a solid catalyst (V2O5) and gaseous reactants.
- Option B: Haber’s process. The Haber process is heterogeneous catalysis because it uses a solid catalyst (iron) and gaseous reactants.
- Option D: Hydrogenation of oils. Hydrogenation of oils is typically heterogeneous catalysis because it uses a solid catalyst (e.g., nickel) and liquid reactants.

Key Concepts:
- Catalysis
- Homogeneous catalysis
- Heterogeneous catalysis

Memory Aids:
- Homogeneous: \"Same phase\"
- Heterogeneous: \"Different phases\"

Quick Tips:
- Remember the definitions of homogeneous and heterogeneous catalysis and be able to identify examples of each.

9 Medium

9 of 59Question permalink

Remembering +4 | -1

The spontaneous outcome of internal liquid from gels is called :

View answer and explanation

Correct option: A

Synerisis

Option A is correct because syneresis is the spontaneous expulsion of liquid from a gel due to the contraction of the gel network.

Solution approach

Correct Answer + Core Reason: Synerisis. Syneresis is the spontaneous expulsion of liquid from a gel.

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 syneresis. Syneresis is the spontaneous contraction of a gel, resulting in the expulsion of a liquid. This occurs when the gel network shrinks, forcing the liquid out.

Step 3: Define thixotropy. 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.

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

Step 5: Conclusion: The spontaneous outcome of internal liquid from gels is called syneresis.

Why other options are wrong:
- Option B: Thixotropy is the change in viscosity upon shear, not the spontaneous expulsion of liquid.
- Option C: Swelling is the absorption of liquid, not the expulsion of liquid.
- Option D: Syneresis is the correct answer.

Key Concepts:
- Gels
- Syneresis
- Thixotropy
- Swelling

Memory Aids:
- Think of syneresis as "squeeze out".

Quick Tips:
- Distinguish between syneresis (spontaneous liquid expulsion) and swelling (liquid absorption).

10 Medium

10 of 59Question permalink

Understanding +4 | -1

Which of the following is true in respect of adsorption?

View answer and explanation

Correct option: B

\Delta G < 0;\Delta S < 0;\Delta H < 0

Option B (

\Delta G < 0;\Delta S < 0;\Delta H < 0

) is correct because adsorption is a spontaneous process (

\Delta G < 0

) that decreases the randomness of the system (

\Delta S < 0

) and releases heat (

\Delta H < 0

).

Solution approach

Correct Answer + Core Reason: $\Delta G < 0;\Delta S < 0;\Delta H < 0$ . Adsorption is spontaneous and exothermic.

Step-by-Step Solution:

Step 1: Recall the thermodynamic criteria for spontaneity: $\Delta G < 0$ .

Step 2: Adsorption is a spontaneous process, so $\Delta G$ must be negative.

Step 3: During adsorption, gas molecules are held on the surface of the adsorbent, which decreases their freedom of movement. This leads to a decrease in entropy, so $\Delta S < 0$ .

Step 4: Adsorption is generally an exothermic process, meaning heat is released during the process. Therefore, $\Delta H < 0$ .

Why other options are wrong:
- Option A: $\Delta S > 0$ is incorrect because adsorption decreases the randomness of the system.
- Option C: $\Delta G > 0$ is incorrect because adsorption is a spontaneous process.
- Option D: $\Delta H > 0$ is incorrect because adsorption is an exothermic process.

Key Concepts:
- Gibbs free energy
- Entropy
- Enthalpy
- Spontaneity

Memory Aids:
- Spontaneous processes have $\Delta G < 0$ .
- Adsorption decreases randomness: $\Delta S < 0$ .
- Adsorption releases heat: $\Delta H < 0$ .

Quick Tips:
- Remember the relationship between $\Delta G$ , $\Delta H$ , and $\Delta S$ : $\Delta G = \Delta H - T\Delta S$ .

11 Medium

11 of 59Question permalink

Understanding +4 | -1

If

(x/m)

is the mass of adsorbate adsorbed per unit mass of adsorbent.

p

is the pressure of the adsorbate gas and

a

and

b

are constants, which of the following represents “Langmuir adsorption isotherm”?

View answer and explanation

Correct option: D

\frac{1}{{\left( {x/m} \right)}} = \frac{b}{a} + \frac{1}{{ap}}

Option D (

\frac{1}{{\left( {x/m} \right)}} = \frac{b}{a} + \frac{1}{{ap}}

) is correct because it represents the reciprocal form of the Langmuir adsorption isotherm equation.

Solution approach

Correct Answer + Core Reason: $\frac{1}{{\left( {x/m} \right)}} = \frac{b}{a} + \frac{1}{{ap}}$ . This is the reciprocal form of the Langmuir adsorption isotherm.

Step-by-Step Solution:

Step 1: Recall the Langmuir adsorption isotherm equation: $\frac{x}{m} = \frac{{ap}}{{1 + bp}}$

Step 2: Take the reciprocal of both sides: $\frac{1}{{\left( {x/m} \right)}} = \frac{{1 + bp}}{{ap}}$

Step 3: Separate the terms on the right side: $\frac{1}{{\left( {x/m} \right)}} = \frac{1}{{ap}} + \frac{{bp}}{{ap}}$

Step 4: Simplify the second term: $\frac{1}{{\left( {x/m} \right)}} = \frac{1}{{ap}} + \frac{b}{a}$

Step 5: Rearrange the terms: $\frac{1}{{\left( {x/m} \right)}} = \frac{b}{a} + \frac{1}{{ap}}$

Why other options are wrong:
- Option A: $\log \left( {\frac{x}{m}} \right) = \log \left( {\frac{a}{b}} \right) + \frac{1}{a}\log p$ is incorrect because it is not related to the Langmuir adsorption isotherm.
- Option B: $\frac{x}{m} = \frac{b}{a} + \frac{1}{{ap}}$ is incorrect because it is not the correct form of the Langmuir adsorption isotherm.
- Option C: $\frac{x}{m} = \frac{{1 + bp}}{{ap}}$ is incorrect because it is the inverse of the Langmuir adsorption isotherm.

Key Concepts:
- Langmuir adsorption isotherm
- Adsorption
- Isotherms

Memory Aids:
- Langmuir: "adsorption, monolayer".

Quick Tips:
- Remember the Langmuir adsorption isotherm equation and its reciprocal form.

12 Medium

12 of 59Question permalink

Analyzing +4 | -1

The disperse phase in colloidal iron (III) hydroxide and colloidal gold is positively and negatively charged, respectively. Which of the following statements is not correct?

View answer and explanation

Correct option: C

Sodium sulphate solution causes coagulation in both sols

Option C is correct because sodium sulfate will coagulate the positively charged iron(III) hydroxide sol, but it will not coagulate the negatively charged gold sol.

Solution approach

Correct Answer + Core Reason: Sodium sulphate solution causes coagulation in both sols. This statement is incorrect because sodium sulphate will coagulate the positively charged iron(III) hydroxide sol, but it will not coagulate the negatively charged gold sol.

Step-by-Step Solution:

Step 1: Identify the charges of the sols. Colloidal iron (III) hydroxide is positively charged, and colloidal gold is negatively charged.

Step 2: Understand the principle of coagulation. Coagulation occurs when the charge on the colloidal particles is neutralized, causing them to aggregate and precipitate out of the solution. This is typically achieved by adding an electrolyte with ions of opposite charge to the sol.

Step 3: Analyze the effect of sodium sulfate (Na2SO4). Sodium sulfate dissociates into 2Na+ and SO42- ions. The sulfate ions (SO42-) are negatively charged and will effectively coagulate the positively charged iron (III) hydroxide sol. However, the sodium ions (Na+) are positively charged and will not effectively coagulate the negatively charged gold sol.

Step 4: Analyze the other options.
- A. Coagulation in both sols can be brought about by electrophoresis: Electrophoresis can cause coagulation by moving charged particles to an electrode where they lose their charge and aggregate.
- B. Mixing the sols has no effect: Mixing oppositely charged sols can lead to mutual coagulation.
- D. Magnesium chloride solution coagulates the gold sol more readily than the iron (III) hydroxide sol: MgCl2 provides Mg2+ ions, which are more effective at coagulating the negatively charged gold sol than the Cl- ions are at coagulating the positively charged iron (III) hydroxide sol.

Step 5: Conclusion. Sodium sulfate solution will coagulate the iron (III) hydroxide sol but not the gold sol. Therefore, the statement that sodium sulfate solution causes coagulation in both sols is incorrect.

Why other options are wrong:
- Option A: Coagulation in both sols can be brought about by electrophoresis. This is a correct statement.
- Option B: Mixing the sols has no effect. This is incorrect, as mixing oppositely charged sols can lead to mutual coagulation.
- Option D: Magnesium chloride solution coagulates the gold sol more readily than the iron (III) hydroxide sol. This is a correct statement.

Key Concepts:
- Colloidal solutions
- Charge of colloidal particles
- Coagulation
- Hardy-Schulze rule
- Electrophoresis

Memory Aids:
- Opposites attract: Oppositely charged ions cause coagulation of sols.

Quick Tips:
- Remember that the effectiveness of an ion in causing coagulation depends on its charge and the charge of the sol.

13 Medium

13 of 59Question permalink

Understanding +4 | -1

For which of the given colloidal substances the dispersed phase and dispersion medium are in liquid state?

View answer and explanation

Correct option: A

Butter

Option A (Butter) is correct because butter is an emulsion where both the dispersed phase (fat) and the dispersion medium (water) are in the liquid state at room temperature.

Solution approach

Correct Answer + Core Reason: Butter. Both dispersed phase (fat) and dispersion medium (water) are liquid.

Step-by-Step Solution:

Step 1: Analyze the composition of each option.

Step 2: Butter is an emulsion of water in fat. Both fat and water are liquid at room temperature.

Why other options are wrong:
- Option B: Jellies - Jellies are solid or semi-solid gels, where the dispersed phase is solid and the dispersion medium is liquid.
- Option C: Paints - Paints are suspensions of solid pigments in a liquid medium.
- Option D: Insecticide sprays - These are often solutions or suspensions of solid or liquid insecticides in a liquid medium.

Key Concepts:
- Colloidal solutions
- Emulsions
- Dispersed phase
- Dispersion medium

Memory Aids:
- Emulsion: Liquid in liquid.

Quick Tips:
- Consider the physical state of each component at the given temperature.

14 Medium

14 of 59Question permalink

Understanding +4 | -1

Which statement is wrong?

View answer and explanation

Correct option: B

Reaction with higher activation energy has higher rate constant

Option B (Reaction with higher activation energy has higher rate constant) is wrong because reactions with higher activation energy have *lower* rate constants, as they require more energy to proceed.

Solution approach

Correct Answer + Core Reason: Reaction with higher activation energy has higher rate constant. Higher activation energy implies a slower reaction rate.

Step-by-Step Solution:

Step 1: Recall the Arrhenius equation: $k = A e^{-E_a/RT}$ , where k is the rate constant, A is the pre-exponential factor, $E_a$ is the activation energy, R is the gas constant, and T is the temperature.

Step 2: Analyze the relationship between the rate constant (k) and the activation energy ( $E_a$ ).

Step 3: From the Arrhenius equation, it is clear that the rate constant (k) is inversely proportional to the exponential of the activation energy ( $E_a$ ). Therefore, a higher activation energy leads to a smaller rate constant, and vice versa.

Why other options are wrong:
- Option A: Catalysts do not alter the equilibrium of a reaction; they only speed up the rate at which equilibrium is reached.
- Option C: In an endothermic reaction, the activation energy is higher than the heat of reaction because energy is required to break bonds before new bonds can form.
- Option D: The half-life period of a first-order reaction is independent of the initial concentration, as it only depends on the rate constant.

Key Concepts:
- Arrhenius equation
- Activation energy
- Rate constant
- Catalysis
- First-order reactions

Memory Aids:
- Higher activation energy = slower reaction rate.
- Arrhenius equation: $k = A e^{-E_a/RT}$

Quick Tips:
- Remember the inverse relationship between activation energy and the rate constant.

15 Medium

15 of 59Question permalink

Application +4 | -1

Which of the following mixture will lead to the formation of negatively charged colloid

[AgI]{I^ - }

View answer and explanation

Correct option: B

50\,\mathrm{ml}\,\mathrm{of}\,0.1M\,AgN{O_3}\, + \,50\,\mathrm{ml}\,\mathrm{of}\,0.2M\,KI

Option B is correct because the excess of

I^-

ions from KI in the mixture leads to the adsorption of

I^-

on the AgI colloid, resulting in a negatively charged colloid.

Solution approach

Correct Answer + Core Reason: $50\,\mathrm{ml}\,\mathrm{of}\,0.1M\,AgN{O_3}\, + \,50\,\mathrm{ml}\,\mathrm{of}\,0.2M\,KI$ . Excess iodide ions ( $I^-$ ) are adsorbed onto the AgI precipitate, forming a negatively charged colloid.

Step-by-Step Solution:

Step 1: Identify the reaction: $AgNO_3 + KI \rightarrow AgI + KNO_3$

Step 2: Calculate the moles of $AgNO_3$ and $KI$ in each option.

Step 3: Determine which mixture has an excess of $KI$ .

Step 4: In option B, moles of $AgNO_3 = 50 \times 0.1 = 5$ mmol and moles of $KI = 50 \times 0.2 = 10$ mmol. Since $KI$ is in excess, $I^-$ ions will be adsorbed, forming $[AgI]I^-$ .

Why other options are wrong:
- Option A: Moles of $AgNO_3$ and $KI$ are equal, so no preferential adsorption occurs.
- Option C: $AgNO_3$ is in excess, so $Ag^+$ ions would be adsorbed, forming a positively charged colloid.
- Option D: $AgNO_3$ and $KI$ are in equal amounts, so no preferential adsorption occurs.

Key Concepts:
- Colloid formation
- Preferential adsorption
- Charge of colloidal particles

Memory Aids:
- Excess of an ion leads to its adsorption on the colloid.

Quick Tips:
- Calculate moles to determine which reactant is in excess.

16 Medium

16 of 59Question permalink

Remembering +4 | -1

The correct option representing a Freundlich adsorption isotherm is

View answer and explanation

Correct option: A

\frac{x}{m}\, = \,K{p^{0.3}}

Option A (

\frac{x}{m}\, = \,K{p^{0.3}}

) is correct because the exponent of pressure in Freundlich isotherm lies between 0 and 1.

Solution approach

Correct Answer + Core Reason: $\frac{x}{m}\, = \,K{p^{0.3}}$ . The Freundlich adsorption isotherm has the pressure exponent between 0 and 1.

Step-by-Step Solution:

Step 1: Recall the Freundlich adsorption isotherm equation: $\frac{x}{m} = Kp^{\frac{1}{n}}$ , where $x$ is the mass of the gas adsorbed, $m$ is the mass of the adsorbent, $p$ is the pressure, $K$ is a constant, and $n > 1$ .

Step 2: Recognize that $\frac{1}{n}$ must be between 0 and 1 (i.e., $0 < \frac{1}{n} < 1$ ).

Step 3: Examine the given options and identify the one where the exponent of $p$ falls within the range of 0 to 1.

Step 4: In option A, the exponent is 0.3, which satisfies the condition $0 < 0.3 < 1$ .

Why other options are wrong:
- Option B: The exponent 2.5 is greater than 1, which is not consistent with the Freundlich isotherm.
- Option C: The exponent -0.5 is negative, which is not consistent with the Freundlich isotherm.
- Option D: The exponent -1 is negative, which is not consistent with the Freundlich isotherm.

Key Concepts:
- Freundlich Adsorption Isotherm
- Adsorption

Memory Aids:
- Remember that the exponent in the Freundlich isotherm is a fraction between 0 and 1.

Quick Tips:
- The Freundlich isotherm is an empirical equation, so knowing the general form is crucial. Pay attention to the range of the exponent.

17 Medium

17 of 59Question permalink

Remembering +4 | -1

For which colloidal substance the dispersed phase and dispersion medium are liquid and solid respectively?

View answer and explanation

Correct option: D

Jellies

Option D is correct because jellies have a liquid dispersed phase and a solid dispersion medium.

Solution approach

Correct Answer + Core Reason: Jellies. Jellies have a liquid dispersed phase and a solid dispersion medium.

Step-by-Step Solution:

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

Step 2: Identify the dispersed phase and dispersion medium in jellies. In jellies, a liquid is dispersed within a solid network. Therefore, the dispersed phase is liquid, and the dispersion medium is solid.

Step 3: Consider the other options.
- Butter: Butter is a solid dispersed in a liquid (fat in water).
- Cell fluids: Cell fluids are complex mixtures, but generally liquid in liquid.
- Whipped cream: Whipped cream is a gas dispersed in a liquid (air in cream).

Why other options are wrong:
- Option A: Butter. Butter is a solid dispersed in a liquid, not a liquid dispersed in a solid.
- Option B: Cell fluids. Cell fluids are complex mixtures, but generally liquid in liquid.
- Option C: Whipped cream. Whipped cream is a gas dispersed in a liquid, not a liquid dispersed in a solid.

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

Memory Aids:
- Jellies: Think of \"jelly\" as a solid structure with liquid inside.

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

18 Medium

18 of 59Question permalink

Remembering +4 | -1

In petrochemical industries, alcohols are directly converted to gasoline by passing over heated

View answer and explanation

Correct option: B

ZSM-5

The correct answer is B (ZSM-5) because ZSM-5 is a zeolite catalyst used to convert alcohols directly to gasoline.

Solution approach

Correct Answer + Core Reason: ZSM-5. ZSM-5 is a zeolite catalyst used in the petrochemical industry to convert alcohols directly into gasoline.

Step-by-Step Solution:

Step 1: Recall the catalysts used in petrochemical industries for converting alcohols to gasoline.

Step 2: ZSM-5 is a zeolite catalyst used for this conversion.

Why other options are wrong:
- Option A: Platinum is used in other catalytic processes, but not directly for converting alcohols to gasoline.
- Option C: Iron is used in the Haber process and other industrial applications, but not for this conversion.
- Option D: Nickel is used in hydrogenation reactions, but not directly for converting alcohols to gasoline.

Key Concepts:
- Catalysis
- Zeolites
- ZSM-5

Memory Aids:
- ZSM-5 converts alcohols to gasoline.

Quick Tips:
- Know the specific catalysts used in important industrial processes.

19 Medium

19 of 59Question permalink

Remembering +4 | -1

Which of the following is an anionic detergent?

View answer and explanation

Correct option: D

Sodium stearate

Option D is correct because sodium stearate is an anionic detergent, containing a negatively charged head group.

Solution approach

Correct Answer + Core Reason: Sodium stearate. Sodium stearate is an anionic detergent because it has a negatively charged head group.

Step-by-Step Solution:

Step 1: Define anionic detergent. Anionic detergents are detergents that have a negatively charged head group. When dissolved in water, the head group carries a negative charge.

Step 2: Analyze the given options.
- Sodium lauryl sulphate & sodium phosphate: Sodium lauryl sulphate is an anionic detergent, but sodium phosphate is not a detergent. It is a builder that enhances the effectiveness of detergents.
- Cetyltrimethyl ammonium bromide: Cetyltrimethyl ammonium bromide is a cationic detergent because it has a positively charged head group.
- Glyceryl oleate: Glyceryl oleate is a non-ionic surfactant.
- Sodium stearate: Sodium stearate is an anionic detergent. It is the sodium salt of stearic acid (a fatty acid). In water, it dissociates into Na+ and stearate ions. The stearate ion has a long hydrocarbon tail and a negatively charged carboxylate head group (CH3(CH2)16COO-).

Why other options are wrong:
- Option A: Sodium lauryl sulphate & sodium phosphate. While sodium lauryl sulphate is anionic, sodium phosphate is not a detergent.
- Option B: Cetyltrimethyl ammonium bromide. This is a cationic detergent, not an anionic detergent.
- Option C: Glyceryl oleate. This is a non-ionic surfactant, not an anionic detergent.

Key Concepts:
- Detergents
- Anionic detergents
- Cationic detergents
- Non-ionic surfactants

Memory Aids:
- Anionic: \"Anion\" = negative charge.
- Cationic: \"Cation\" = positive charge.

Quick Tips:
- Remember the different types of detergents and their characteristic charge.

20 Medium

20 of 59Question permalink

Remembering +4 | -1

Bredig’s arc method cannot be used for the preparation of colloidal sol of

View answer and explanation

Correct option: D

Sodium

Option D (Sodium) is correct because Bredig's arc method is unsuitable for preparing colloidal sols of highly reactive metals like sodium, as they react rapidly with water.

Solution approach

Correct Answer + Core Reason: Sodium. Bredig’s arc method cannot be used for sodium due to its high reactivity with water.

Step-by-Step Solution:

Step 1: Understand Bredig's arc method. Bredig's arc method involves striking an electric arc between two metal electrodes immersed in a dispersion medium (usually water) to vaporize the metal and then condense it into colloidal particles.

Step 2: Consider the reactivity of metals. Highly reactive metals like sodium react violently with water, making it impossible to prepare a stable colloidal sol using this method.

Why other options are wrong:
- Option A: Copper is a less reactive metal, and its colloidal sol can be prepared by Bredig's arc method.
- Option B: Gold is a noble metal and can be easily prepared as a colloidal sol by Bredig's arc method.
- Option C: Silver is also a relatively less reactive metal, and its colloidal sol can be prepared by Bredig's arc method.

Key Concepts:
- Bredig's arc method: Electrical dispersion method for preparing colloidal sols of metals.
- Reactivity of metals: The tendency of a metal to react with other substances.

Memory Aids:
- Bredig's Arc: Not for Reactive Metals like Sodium.

Quick Tips:
- Remember that Bredig's arc method is suitable for less reactive metals.

21 Medium

21 of 59Question permalink

Remembering +4 | -1

Which among the following is used to purify colloidal solution?

View answer and explanation

Correct option: B

Dialysis

Option B (Dialysis) is correct because dialysis is a process used to purify colloidal solutions by removing impurities.

Solution approach

Correct Answer + Core Reason: Dialysis. Dialysis is a method of separating colloidal particles from crystalloids by using a semipermeable membrane.

Step-by-Step Solution:

Step 1: Understand the concept of colloidal solutions and their purification.

Step 2: Define dialysis: Dialysis is a process involving the use of a semipermeable membrane to separate smaller solute molecules (crystalloids) from larger colloidal particles.

Step 3: Explain the process: The colloidal solution is placed in a bag made of semipermeable membrane, which is then immersed in a suitable solvent. The smaller ions and molecules diffuse out through the membrane, leaving behind the purified colloidal solution.

Why other options are wrong:
- Option A: Tyndall effect is the scattering of light by colloidal particles and is used to detect the presence of colloids, not to purify them.
- Option C: Brownian movement is the random motion of colloidal particles and doesn't purify the solution.
- Option D: Electrophoresis is the movement of charged colloidal particles under the influence of an electric field and is used for separation based on charge, not general purification.

Key Concepts:
- Colloidal solutions
- Dialysis
- Semipermeable membrane

Memory Aids:
- Dialysis: Think of kidney dialysis, which purifies blood in a similar way.

Quick Tips:
- Remember that purification methods separate the desired substance from impurities.

22 Medium

22 of 59Question permalink

Analyzing +4 | -1

Assertion : Physical absorption of molecules takes place on surface only.

Reason : In this process, the bonds of the absorbed molecules are broken.

View answer and explanation

Correct option: D

If both the Assertion and Reason are incorrect.

The correct answer is 'If both the Assertion and Reason are incorrect' because physical adsorption occurs on the surface, but the reason is incorrect as physical adsorption doesn't break bonds.

Solution approach

Correct Answer + Core Reason: If both the Assertion and Reason are incorrect. Physical adsorption can occur in multilayers, and it does not involve breaking bonds.

Step-by-Step Solution:

Step 1: Analyze the assertion: Physical adsorption can occur in multilayers, not just on the surface.

Step 2: Analyze the reason: Physical adsorption involves weak van der Waals forces, not the breaking of bonds.

Step 3: Since both the assertion and the reason are incorrect, the correct answer is that both are incorrect.

Why other options are wrong:
- Option A: The reason is not a correct explanation of the assertion, and both are incorrect.
- Option B: Both are incorrect.
- Option C: The assertion is incorrect.

Key Concepts:
- Physical adsorption
- Van der Waals forces

Memory Aids:
- Physical adsorption involves weak intermolecular forces.

Quick Tips:
- Distinguish between physical and chemical adsorption based on the type of forces involved.

23 Medium

23 of 59Question permalink

Understanding +4 | -1

Which might have lowest value of critical micelle concentration in water

View answer and explanation

Correct option: D

Sodium dodecanoate

Option D (Sodium dodecanoate) is correct because longer hydrocarbon chains in surfactants generally lead to lower critical micelle concentrations (CMC).

Solution approach

Correct Answer + Core Reason: Sodium dodecanoate. Longer hydrocarbon chains lead to lower CMC.

Step-by-Step Solution:

Step 1: Define critical micelle concentration (CMC).

Step 2: CMC is the concentration above which micelles start to form.

Step 3: Surfactants with longer hydrophobic chains have a greater tendency to aggregate and form micelles, thus requiring a lower concentration to reach the CMC.

Step 4: Comparing the hydrocarbon chain lengths: Sodium octanoate (8 carbons), Sodium decanoate (10 carbons), Sodium hexanoate (6 carbons), Sodium dodecanoate (12 carbons).

Step 5: Sodium dodecanoate has the longest hydrocarbon chain and therefore the lowest CMC.

Why other options are wrong:
- Option A: Sodium octanoate - Shorter chain length, higher CMC.
- Option B: Sodium decanotate - Shorter chain length than dodecanoate, higher CMC.
- Option C: Sodium hexanoate - Shortest chain length, highest CMC.

Key Concepts:
- Critical micelle concentration (CMC)
- Surfactants
- Hydrophobic interactions
- Micelle formation

Memory Aids:
- Longer chain, lower CMC.

Quick Tips:
- CMC decreases with increasing hydrophobic chain length.

24 Medium

24 of 59Question permalink

Understanding +4 | -1

Which equation represents Freundlich adsorption isotherm (physical adsorption is basis of this theory)?

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because all the given equations represent different forms or aspects of the Freundlich adsorption isotherm.

Solution approach

Correct Answer + Core Reason: All of the above. All options are valid representations of the Freundlich adsorption isotherm.

Step-by-Step Solution:

Step 1: Recall the Freundlich adsorption isotherm equation: $\frac{x}{m} = K P^{1/n}$ , where x is the mass of gas adsorbed, m is the mass of adsorbent, P is the pressure, K is a constant, and n > 1.

Step 2: Verify that option A is the Freundlich adsorption isotherm equation.

Step 3: Take the logarithm of both sides of the equation in Step 1: $\log(\frac{x}{m}) = \log(K P^{1/n}) = \log(K) + \frac{1}{n} \log(P)$ . This is option B.

Step 4: At low pressure, $\frac{x}{m}$ is approximately proportional to P, so $\frac{x}{m} = KP$ . At high pressure, $\frac{x}{m}$ approaches a constant value, so $\frac{x}{m} = K$ . This is option C.

Step 5: Since all options are valid representations of the Freundlich adsorption isotherm, the correct answer is "All of the above".

Why other options are wrong:
- Option A: This is a correct representation of the Freundlich isotherm, but it's not the only one.
- Option B: This is a correct logarithmic form of the Freundlich isotherm, but it's not the only one.
- Option C: This describes the behavior of the Freundlich isotherm at low and high pressures, but it's not the only representation.

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

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

Quick Tips:
- Be familiar with different forms of the Freundlich adsorption isotherm equation.

25 Medium

25 of 59Question permalink

Application +4 | -1

50 \mathrm{ml}

of

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

sol is completty coagulated by

50 \mathrm{ml}

of

0.2 \mathrm{M} \mathrm{KBr}

. Coagulation value is

View answer and explanation

Correct option: A

200

Option A (200) is correct because the coagulation value is defined as the number of milligrams of electrolyte required to coagulate 1 liter of the colloidal solution. We convert the molarity of KBr to mg/L and then adjust for the volume of the sol.

Solution approach

Correct Answer + Core Reason: 200. The coagulation value is the millimoles of electrolyte needed to coagulate 1 L of sol.

Step-by-Step Solution:

Step 1: Calculate the moles of KBr used: $50 \text{ ml} \times 0.2 \text{ M} = 50 \times 0.2 \text{ mmol} = 10 \text{ mmol}$

Step 2: Since 10 mmol of KBr coagulates 50 ml of $Fe(OH)_3$ sol, we need to find the amount needed for 1000 ml (1 L) of sol.

Step 3: Use the ratio: $\frac{10 \text{ mmol}}{50 \text{ ml}} = \frac{x \text{ mmol}}{1000 \text{ ml}}$

Step 4: Solve for x: $x = \frac{10 \text{ mmol} \times 1000 \text{ ml}}{50 \text{ ml}} = 200 \text{ mmol}$

Step 5: Therefore, the coagulation value is 200.

Why other options are wrong:
- Option B: 300 is incorrect as it doesn't match the correct calculation based on the given data.
- Option C: 500 is incorrect as it doesn't match the correct calculation based on the given data.
- Option D: 100 is incorrect as it doesn't match the correct calculation based on the given data.

Key Concepts:
- Coagulation value
- Molarity

Memory Aids:
- Coagulation value is expressed in millimoles per liter.

Quick Tips:
- Always convert volumes to liters or milliliters consistently.

26 Medium

26 of 59Question permalink

Remembering +4 | -1

Blue colour of the sky is due to

View answer and explanation

Correct option: A

Scattering of light by dust particles

Option A (Scattering of light by dust particles) is correct because the blue color of the sky is due to Rayleigh scattering, where shorter wavelengths (blue) are scattered more effectively by small particles in the atmosphere.

Solution approach

Correct Answer + Core Reason: Scattering of light by dust particles. The blue color of the sky is due to Rayleigh scattering.

Step-by-Step Solution:

Step 1: Understand Rayleigh scattering. Rayleigh scattering is the scattering of electromagnetic radiation (including light) by particles of a wavelength much larger than the particles.

Step 2: Apply Rayleigh scattering to the atmosphere. In the Earth's atmosphere, smaller particles like air molecules and dust particles scatter shorter wavelengths (blue and violet) more effectively than longer wavelengths (red and orange).

Step 3: Explain the blue color. Since blue light is scattered more, it appears to come from all directions, making the sky appear blue.

Why other options are wrong:
- Option B: Absorption of light by dust particles would lead to a darker sky, not a blue one.
- Option C: Reflection of light by dust particles is not the primary reason for the blue color of the sky. Scattering is the dominant phenomenon.
- Option D: The presence of clouds can affect the color of the sky, but the blue color is primarily due to scattering by smaller particles, not the clouds themselves.

Key Concepts:
- Rayleigh scattering: Scattering of light by particles smaller than the wavelength of light.
- Wavelength of light: Different colors have different wavelengths.

Memory Aids:
- Blue Sky = Rayleigh Scattering.

Quick Tips:
- Remember that scattering is the key phenomenon behind the blue color of the sky.

27 Medium

27 of 59Question permalink

Remembering +4 | -1

Which of the following can reduce decomposition of

\mathrm{H}_{2} \mathrm{O}_{2}

on exposure to light

View answer and explanation

Correct option: A

Urea

The correct answer is A (Urea) because urea can act as a stabilizer to reduce the decomposition of H2O2.

Solution approach

Correct Answer + Core Reason: Urea. Urea acts as a stabilizer to inhibit the decomposition of H2O2 upon exposure to light.

Step-by-Step Solution:

Step 1: Recall the substances that can stabilize hydrogen peroxide.

Step 2: Urea is known to reduce the decomposition of H2O2.

Why other options are wrong:
- Option B: Alkali promotes the decomposition of H2O2.
- Option C: Glass containers do not prevent decomposition, although dark containers can help.
- Option D: Dust can catalyze the decomposition of H2O2.

Key Concepts:
- Hydrogen peroxide decomposition
- Stabilizers

Memory Aids:
- Urea stabilizes H2O2.

Quick Tips:
- Know the factors that affect the stability of hydrogen peroxide.

28 Medium

28 of 59Question permalink

Understanding +4 | -1

Using very little soap while washing clothes, does not serve the purpose of cleaning of clothes, because:

View answer and explanation

Correct option: C

The micelles are not formed due to concentration of soap, below its CMC value.

Option C (The micelles are not formed due to concentration of soap, below its CMC value) is correct because soap needs to be above its critical micelle concentration (CMC) to form micelles and effectively clean clothes.

Solution approach

Correct Answer + Core Reason: The micelles are not formed due to concentration of soap, below its CMC value. Micelles are essential for cleaning action of soap.

Step-by-Step Solution:

Step 1: Understand the cleaning action of soap.

Step 2: Soap cleans by forming micelles around grease and dirt, allowing them to be washed away with water.

Step 3: Micelles only form when the concentration of soap is above its critical micelle concentration (CMC).

Step 4: If the soap concentration is below the CMC, micelles will not form, and the soap will not effectively clean clothes.

Why other options are wrong:
- Option A: Soap particles remain floating in water as ions - While soap does ionize, this doesn't prevent cleaning if micelles are formed.
- Option B: The hydrophobic part of soap is not able to take away grease - The hydrophobic part can interact with grease, but micelles are needed to emulsify and remove it.
- Option D: Colloidal structure of soap in water is completely distributed - This is not the primary reason; the lack of micelle formation is the key issue.

Key Concepts:
- Micelles
- Critical micelle concentration (CMC)
- Soap
- Cleaning action

Memory Aids:
- CMC: Concentration needed to clean.

Quick Tips:
- Soap must be above its CMC to form micelles and clean effectively.

29 Medium

29 of 59Question permalink

Application +4 | -1

Plot of

\log \frac{x}{m}

against

\log p

is a straight line inclined at an angle of

45^\circ

. When the pressure is 0.5 atm and Freundlich parameter,

k

is 10, the amount of solute adsorbed per gram of adsorbent will be (log 5=0.6990)

View answer and explanation

Correct option: D

5\mathrm{ }\;g

Option D (5 g) is correct because, with a 45-degree angle, 1/n = 1, and the equation simplifies to x/m = kp, resulting in 5 g when the values are substituted.

Solution approach

Correct Answer + Core Reason: $5 \text{{ g}}$ . The amount adsorbed is calculated using the Freundlich isotherm with the given parameters.

Step-by-Step Solution:

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

Step 2: Determine the value of $\frac{1}{n}$ from the angle of the straight line: The angle is $45^\circ$ , so the slope of the line is $\tan(45^\circ) = 1$ . Therefore, $\frac{1}{n} = 1$ .

Step 3: Substitute the given values into the Freundlich isotherm equation: $\frac{x}{m} = 10 \cdot (0.5)^1 = 5$

Step 4: The amount of solute adsorbed per gram of adsorbent is 5 g.

Why other options are wrong:
- Option A (1 g): This is incorrect. It may be a result of misinterpreting the Freundlich isotherm or incorrect substitution.
- Option B (2 g): This is incorrect and likely a calculation error.
- Option C (3 g): This is incorrect and likely a calculation error.

Key Concepts:
- Freundlich Adsorption Isotherm
- Slope of a line

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 $\frac{1}{n}$ .

30 Medium

30 of 59Question permalink

Understanding +4 | -1

The velocity of oxidation of oxalic acid by acidified

\mathrm{KMn}{\mathrm{O}_4}

increase as the reaction progress. It is an example of

View answer and explanation

Correct option: C

Autocatalysis

Option C (Autocatalysis) is correct because one of the products of the reaction acts as a catalyst.

Solution approach

Correct Answer + Core Reason: Autocatalysis. The reaction is catalyzed by one of its products.

Step-by-Step Solution:

Step 1: Define autocatalysis.

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

Step 3: In the oxidation of oxalic acid by acidified $\mathrm{KMn}{\mathrm{O}_4}$ , $Mn^{2+}$ ions formed during the reaction act as a catalyst.

Step 4: Therefore, it is an example of autocatalysis.

Why other options are wrong:
- Option A: Promoters are substances that enhance the activity of a catalyst, but are not products of the reaction.
- Option B: Catalytic poisons are substances that decrease the activity of a catalyst.
- Option D: Inhibitors are substances that slow down or stop a reaction.

Key Concepts:
- Autocatalysis
- Catalyst
- Promoters
- Inhibitors

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

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

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