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

SURFACE CHEMISTRY MCQs for NEET UG CHEMISTRY — Mixed difficulty

1 Hard

1 of 136Question permalink

Application +4 | -1

A cationic colloidal electrolyte forms micelle at

10^{-3} \mathrm{M}

concentration in water. Its

1 \mathrm{~mm}^3

solution contains

2 \times 10^{13}

micelle structures. What is the number of cations contained in one micelle? Take

\mathrm{N}_{\mathrm{A}}=6 \times 10^{23}

.

View answer and explanation

Correct option: A

30

Option A (30) is correct because it involves calculating the number of micelles and then dividing the total number of cations by the number of micelles.

Solution approach

Correct Answer + Core Reason: 30. The number of cations in one micelle is calculated using the concentration of the electrolyte and the number of micelle structures.

Step-by-Step Solution:

Step 1: Convert the concentration to number of cations per $mm^3$ :
- Concentration = $10^{-3} M = 10^{-3} mol/L$
- Number of moles in $1 mm^3 = 10^{-3} mol/L \times 10^{-6} L = 10^{-9} mol$
- Number of cations in $1 mm^3 = 10^{-9} mol \times 6 \times 10^{23} cations/mol = 6 \times 10^{14}$ cations

Step 2: Calculate the number of cations per micelle:
- Number of micelle structures in $1 mm^3 = 2 \times 10^{13}$
- Number of cations per micelle = $\frac{Total number of cations}{Number of micelles} = \frac{6 \times 10^{14}}{2 \times 10^{13}} = 30$

Why other options are wrong:
- Option B: 0 is incorrect. The calculation leads to a value of 30.
- Option C: 0 is incorrect. The calculation leads to a value of 30.
- Option D: 0 is incorrect. The calculation leads to a value of 30.

Key Concepts:
- Molarity
- Avogadro's number
- Micelle formation

Memory Aids:
- Remember Avogadro's number: $6 \times 10^{23}$

Quick Tips:
- Pay attention to units and convert them appropriately.
- Understand the relationship between concentration, volume, and number of particles.

2 Medium

2 of 136Question permalink

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 $\log(k)$ .

3 Easy

3 of 136Question 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 $\log P$ .

4 Hard

4 of 136Question permalink

Understanding +4 | -1

Most suitable salt which can be used for efficient clotting of blood will be

View answer and explanation

Correct option: D

FeC{l_3}

Option D (

FeC{l_3}

) is correct because, according to the Hardy-Schulze rule, ions with higher charges are more effective in causing coagulation.

Fe^{3+}

has a higher charge than

Fe^{2+}

,

Mg^{2+}

, or

Na^+

. Blood is a negatively charged sol, so a cation is needed.

Solution approach

Correct Answer + Core Reason: $FeC{l_3}$ . Higher charge on the cation leads to better coagulation of negatively charged blood.

Step-by-Step Solution:

Step 1: Recall the Hardy-Schulze rule.

Step 2: The Hardy-Schulze rule states that the coagulating power of an electrolyte is directly proportional to the valency of the ion causing coagulation.

Step 3: Blood is a negatively charged sol, so the cation will cause coagulation.

Step 4: Compare the charges of the cations in each option: $Na^+$ , $Fe^{2+}$ , $Mg^{2+}$ , $Fe^{3+}$ .

Step 5: $Fe^{3+}$ has the highest charge, so $FeC{l_3}$ will be the most effective in coagulating blood.

Why other options are wrong:
- Option A: $NaHC{O_3}$ contains $Na^+$ , which has a +1 charge, less effective than $Fe^{3+}$ .
- Option B: $FeS{O_4}$ contains $Fe^{2+}$ , which has a +2 charge, less effective than $Fe^{3+}$ .
- Option C: $Mg{(HC{O_3})_2}$ contains $Mg^{2+}$ , which has a +2 charge, less effective than $Fe^{3+}$ .

Key Concepts:
- Hardy-Schulze rule
- Coagulation
- Electrolytes

Memory Aids:
- Higher charge = higher coagulating power.

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

5 Hard

5 of 136Question permalink

Application +4 | -1

The density of gold is

19\;\mathrm{g/c}{\mathrm{m}^3}

. If

{1.9 \times {{10}^{ - 4}}\;\mathrm{g}}

of gold is dispersed in 1 L of water to give a sol having spherical gold particles of radius

10\;\mathrm{ }nm

, then the number of gold particles per

m{m^3}

of the sol will be

View answer and explanation

Correct option: D

2.4 \times {10^6}

Option D (

2.4 \times {10^6}

) is correct because it is calculated by finding the volume of a single gold particle, then the total number of particles, and finally the number of particles per unit volume.

Solution approach

Correct Answer + Core Reason: $2.4 \times {10^6}$ . The number of gold particles per $mm^3$ is calculated using the density and volume of the gold particles.

Step-by-Step Solution:

Step 1: Calculate the volume of a single gold particle:
- Radius, $r = 10 nm = 10 \times 10^{-7} cm = 10^{-6} cm$
- Volume, $V_{particle} = \frac{4}{3} \pi r^3 = \frac{4}{3} \times \pi \times (10^{-6} cm)^3 = \frac{4}{3} \pi \times 10^{-18} cm^3$

Step 2: Calculate the total volume of gold:
- Mass of gold, $m = 1.9 \times 10^{-4} g$
- Density of gold, $\rho = 19 g/cm^3$
- Total volume, $V_{total} = \frac{m}{\rho} = \frac{1.9 \times 10^{-4} g}{19 g/cm^3} = 10^{-5} cm^3$

Step 3: Calculate the number of gold particles:
- Number of particles, $N = \frac{V_{total}}{V_{particle}} = \frac{10^{-5} cm^3}{\frac{4}{3} \pi \times 10^{-18} cm^3} = \frac{3 \times 10^{13}}{4 \pi} \approx \frac{3 \times 10^{13}}{12.56} \approx 0.238 \times 10^{13} \approx 2.38 \times 10^{12}$

Step 4: Calculate the number of gold particles per $mm^3$ :
- Volume of sol = 1 L = $1000 cm^3 = 1000 \times (10 mm)^3 = 10^6 mm^3$
- Number of particles per $mm^3 = \frac{N}{Volume} = \frac{2.38 \times 10^{12}}{10^6 mm^3} = 2.38 \times 10^6 \approx 2.4 \times 10^6$

Why other options are wrong:
- Option A: $1.9 \times {10^{12}}$ is incorrect. It's likely a miscalculation in the number of particles.
- Option B: $6.3 \times {10^{14}}$ is incorrect. It's likely a miscalculation in the number of particles.
- Option C: $6.3 \times {10^{10}}$ is incorrect. It's likely a miscalculation in the number of particles per $mm^3$ .

Key Concepts:
- Density
- Volume of a sphere
- Colloidal solutions

Memory Aids:
- Remember the formula for the volume of a sphere: $V = \frac{4}{3} \pi r^3$ .

Quick Tips:
- Pay attention to units and convert them appropriately.
- Use the density formula to relate mass and volume.

6 Medium

6 of 136Question 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.

7 Medium

7 of 136Question 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).

8 Medium

8 of 136Question 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.

9 Medium

9 of 136Question 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.

10 Easy

10 of 136Question 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.

11 Medium

11 of 136Question 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.

12 Easy

12 of 136Question 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.

13 Medium

13 of 136Question 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.

14 Medium

14 of 136Question 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.

15 Medium

15 of 136Question 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).

16 Easy

16 of 136Question 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.

17 Easy

17 of 136Question 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.

18 Hard

18 of 136Question permalink

Application +4 | -1

The coagulation of

100 \mathrm{~mL}

of a colloidal sol of gold is completely prevented by addition of 250

\mathrm{mg}

of a substance to it before adding

10 \mathrm{~mL}

of

1 \% \mathrm{NaCl}

solution. What is the gold number of this substance?

View answer and explanation

Correct option: A

25

Option A (25) is correct because the gold number is defined as the mass in milligrams of a protective colloid that prevents the coagulation of 10 mL of a gold sol when 1 mL of a 10% NaCl solution is added.

Solution approach

Correct Answer + Core Reason: 25. Gold number is the mass of protective colloid in mg that prevents coagulation of 10 mL gold sol by 1 mL of 10% NaCl.

Step-by-Step Solution:

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

Step 2: Given: 250 mg of a substance prevents coagulation of 100 mL of gold sol when 10 mL of 1% NaCl is added.

Step 3: Adjust for the standard conditions (10 mL gold sol and 1 mL of 10% NaCl):
- We have 100 mL gold sol and 10 mL of 1% NaCl. To get to the standard conditions, we need to consider the ratio.
- The ratio of gold sol volume is $\frac{10 mL}{100 mL} = \frac{1}{10}$ .
- The ratio of NaCl concentration is $\frac{10%}{1%} = 10$ .
- Therefore, the equivalent mass of the protective colloid for standard conditions is $\frac{250 mg}{10} = 25 mg$ .

Step 4: Calculate the gold number: Since 25 mg of the substance is required, the gold number of the substance is 25.

Why other options are wrong:
- Option B: 0 is incorrect. The calculation leads to a value of 25.
- Option C: 0 is incorrect. The calculation leads to a value of 25.
- Option D: 0 is incorrect. The calculation leads to a value of 25.

Key Concepts:
- Gold number
- Protective colloids
- Coagulation

Memory Aids:
- Gold number is in milligrams.

Quick Tips:
- Adjust the given conditions to match the standard conditions for gold number determination.

19 Medium

19 of 136Question 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 $\Delta S$ : $\Delta G = \Delta H - T\Delta S$ .

20 Medium

20 of 136Question 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.

21 Easy

21 of 136Question 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.

22 Easy

22 of 136Question 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.

23 Medium

23 of 136Question 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.

24 Easy

24 of 136Question 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.

25 Easy

25 of 136Question 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.

26 Medium

26 of 136Question 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.

27 Medium

27 of 136Question 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.

28 Medium

28 of 136Question 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.

29 Medium

29 of 136Question 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.

30 Easy

30 of 136Question 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.

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Surface chemistry practice · Mixed difficulty

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