What Do You Know About Surface Chemistry? Trivia Quiz
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- 1.
Catalyst is required in very small non-stoichiometric quantities because:
- A.
They are consumed during the reaction
- B.
They are not consumed during the reaction
- C.
They increase the temperature of the reaction
- D.
They slow down the reaction process
Correct Answer
B. They are not consumed during the reactionExplanation
Catalysts are substances that speed up chemical reactions by providing an alternative pathway with lower activation energy. They are not consumed during the reaction, meaning they are not permanently changed or used up in the process. Instead, they facilitate the reaction by lowering the energy barrier required for the reaction to occur, allowing it to proceed at a faster rate. This property makes catalysts highly efficient and effective in small quantities, as they can be reused multiple times without being depleted.Rate this question:
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- 2.
Rate of physical adsorption increase with:
- A.
Increase in temperature
- B.
Increase in pressure
- C.
Decrease in temperature
- D.
Decrease in pressure
Correct Answer
C. Decrease in temperatureExplanation
The rate of physical adsorption refers to the speed at which a substance is adsorbed onto a surface. When the temperature is decreased, the kinetic energy of the adsorbate molecules decreases, causing them to move more slowly. This slower movement allows for a greater chance of interaction and adsorption onto the surface. Therefore, a decrease in temperature increases the rate of physical adsorption.Rate this question:
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- 3.
The basic principle of Cottrell's precipitator is:
- A.
Le Chatelier’s Principle
- B.
Peptisation
- C.
Neutralisation of charge on colloidal particles
- D.
Scattering of light
Correct Answer
C. Neutralisation of charge on colloidal particlesExplanation
The basic principle of Cottrell's precipitator is the neutralization of charge on colloidal particles. This principle states that when an electric field is applied to a mixture of charged particles, the particles will migrate towards the oppositely charged electrode. In the case of Cottrell's precipitator, the colloidal particles are charged and the electric field causes them to move towards the oppositely charged plates, where they are neutralized and precipitated out of the solution. This process is used to remove particulate matter from air or gas streams.Rate this question:
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- 4.
In Freundlich Adsorption isotherm predicts a limiting adsorption at:
- A.
Low pressure
- B.
Intermediate pressure
- C.
High pressure
- D.
None of these
Correct Answer
C. High pressureExplanation
The Freundlich Adsorption isotherm predicts a limiting adsorption at high pressure. This means that as the pressure increases, the amount of adsorbate molecules that can be adsorbed onto the surface of the adsorbent reaches a maximum value. At high pressures, the adsorbate molecules are more likely to collide with and stick to the surface of the adsorbent, resulting in a higher adsorption capacity. Therefore, the correct answer is high pressure.Rate this question:
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- 5.
Alum helps in purifying water by:
- A.
Coagulating the mud particles
- B.
Making mud water soluble
- C.
Forming Si complex with clay particles
- D.
Sulphate part which combines with dirt and removes it
Correct Answer
A. Coagulating the mud particlesExplanation
Alum helps in purifying water by coagulating the mud particles. When alum is added to water, it forms a gel-like substance called floc. This floc attracts and binds together the suspended particles in the water, including mud particles. As a result, the coagulated particles become heavier and settle down to the bottom, making it easier to remove them through filtration or sedimentation. This process helps in removing impurities and making the water clearer and safer to drink.Rate this question:
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- 6.
During the adsorption of Krypton on activated charcoal at low temperature:
- A.
ΔH < 0 and ΔS < 0
- B.
ΔH > 0 and ΔS < 0
- C.
ΔH > 0 and ΔS > 0
- D.
ΔH < 0 and ΔS > 0
Correct Answer
A. ΔH < 0 and ΔS < 0Explanation
When Krypton is adsorbed on activated charcoal at low temperature, the enthalpy change (ΔH) is negative, indicating that the process is exothermic and releases heat. Additionally, the entropy change (ΔS) is negative, indicating a decrease in disorder or randomness. This suggests that the adsorption of Krypton on activated charcoal at low temperature is an exothermic process that reduces the randomness of the system.Rate this question:
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- 7.
Chromatography is based on the principle of:
- A.
Chemical adsorption
- B.
Physical adsorption
- C.
Hydrogen bonding
- D.
Sedimentation
Correct Answer
B. Physical adsorptionExplanation
Chromatography is a technique used to separate and analyze different components of a mixture. It is based on the principle of physical adsorption, where the components of the mixture interact differently with the stationary phase (adsorbent) and mobile phase (solvent). The stationary phase has a high surface area and the components of the mixture can adsorb onto it to different extents. As the mobile phase moves through the stationary phase, the components are desorbed at different rates, leading to their separation. This separation is based on the differences in physical properties such as polarity, size, and charge of the components.Rate this question:
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- 8.
The most effective coagulating agent for Sb2S3 sol is:
- A.
Na2SO4
- B.
CaCl2
- C.
Al2 (SO4)3
- D.
NH4Cl
Correct Answer
C. Al2 (SO4)3Explanation
Al2(SO4)3 is the most effective coagulating agent for Sb2S3 sol. This is because Al2(SO4)3 is a trivalent cation, meaning it has a higher charge density compared to the other options. This higher charge density allows Al2(SO4)3 to form stronger bonds with the negatively charged particles in the sol, causing them to coagulate and form larger, more easily separable particles. Na2SO4, CaCl2, and NH4Cl are not as effective because they have lower charge densities and therefore weaker bonding interactions with the sol particles.Rate this question:
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- 9.
The addition of I mL solution of 10% NaCl to 10m gold sol in the presence of 0.25 G. The gold number of starch is:
- A.
0.25
- B.
2.5
- C.
25.0
- D.
250
Correct Answer
D. 250Explanation
The gold number is a measure of the amount of gold sol that can be stabilized by a given amount of protective colloid. In this case, 0.25 g of starch is added to 10 mL of gold sol. The gold number is calculated by dividing the weight of gold sol required for coagulation by the weight of protective colloid used. Since 0.25 g of starch is used to stabilize 10 mL of gold sol, the gold number of starch is 250.Rate this question:
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- 10.
Identify the gas which is readily adsorbed by activated charcoal:
- A.
N2
- B.
O2
- C.
H2
- D.
SO2
Correct Answer
D. SO2Explanation
Activated charcoal is known for its high adsorption capacity, especially for gases and volatile compounds. It works by trapping molecules on its porous surface, and the adsorption process is influenced by factors such as surface area, temperature, and the nature of the gas. Among the given options, sulfur dioxide (SO2) is readily adsorbed by activated charcoal due to its polar nature and ability to form weak chemical bonds with the charcoal surface. Nitrogen (N2), oxygen (O2), and hydrogen (H2) are nonpolar gases and are not as easily adsorbed by activated charcoal.Rate this question:
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- 11.
Which of the following is correct for lyophilic sols:
- A.
They are irreversible
- B.
They are self stabilised
- C.
They are readily coagulated by addition of electrolytes
- D.
They are formed by inorganic substances
Correct Answer
B. They are self stabilisedExplanation
Lyophilic sols are self-stabilized because the particles in these sols have a strong affinity for the dispersing medium, which prevents them from coagulating or settling. This stability is due to the presence of a protective layer of solvent molecules adsorbed on the surface of the particles, creating a repulsive force that keeps them dispersed. Unlike lyophobic sols, lyophilic sols do not require the addition of electrolytes to coagulate, as their stability is inherent. These sols can be formed by both organic and inorganic substances.Rate this question:
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- 12.
Faster coagulation of ions with high valency is explained by:
- A.
Tyndall Effect
- B.
Hardy-Schulze Law
- C.
Freundlich law
- D.
Bredig’s Arc Method
Correct Answer
B. Hardy-Schulze LawExplanation
The Hardy-Schulze Law explains the faster coagulation of ions with high valency. According to this law, ions with higher valency have a greater charge density, which results in stronger attractions between the ions and the oppositely charged particles in the dispersion medium. This stronger attraction leads to faster coagulation, as the particles come closer together and form larger aggregates.Rate this question:
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- 13.
- A.
Irreversible reaction
- B.
Heterogeneous catalysis
- C.
Homogeneous catalysis
- D.
Neutralisation reaction
Correct Answer
B. Heterogeneous catalysis -
- 14.
Emulsifying agent is a substance which:
- A.
Makes the emulsion homogeneous
- B.
Helps in coagulating the emulsion
- C.
Accelerates the dispersion
- D.
Stabilises the emulsion
Correct Answer
D. Stabilises the emulsionExplanation
An emulsifying agent is a substance that stabilizes the emulsion. It prevents the separation of immiscible liquids by reducing the surface tension between them. This allows the emulsion to remain homogeneous and prevents the dispersed particles from coagulating or settling. The emulsifying agent acts as a barrier between the two phases, preventing them from separating and ensuring the stability of the emulsion.Rate this question:
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- 15.
The best coagulant for the precipitation of ferric hydroxide sol is:
- A.
Na3PO4
- B.
NaNO3
- C.
MgSO4
- D.
KCl
Correct Answer
A. Na3PO4Explanation
Na3PO4 is the best coagulant for the precipitation of ferric hydroxide sol because it contains phosphate ions (PO4^3-) which can react with the ferric ions (Fe^3+) in the sol to form insoluble ferric phosphate (FePO4). This reaction helps in the coagulation of the sol particles, causing them to come together and form larger, settleable particles. NaNO3, MgSO4, and KCl do not contain ions that can effectively react with the ferric ions to form insoluble precipitates, hence they are not suitable coagulants for this purpose.Rate this question:
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Current Version
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Mar 22, 2023Quiz Edited by
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Dec 04, 2013Quiz Created by
Tanmay Shankar
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