Chemistry Objective Questions and Answers: Surface Chemistry
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Multiple Choice Questions
(a) \( \ce{C17H35COONa} \)
(b) \( \ce{RC6H4SO3Na} \)
(c) \( \ce{C16H33N+(CH3)3Cl-} \)
(d) \( \ce{C16H33C6H4NHCl} \)
Show Answer & Explanation
Answer: (c) \( \ce{C16H33N+(CH3)3Cl-} \)
Explanation:
1. A cationic detergent has a long hydrocarbon tail attached to a positively charged head group.
2. Cetyltrimethylammonium chloride has a quaternary ammonium head; the others are anionic (soap, alkylbenzene sulphonate) or not a typical surfactant.
Teacher's Note:
1. Quaternary ammonium salts are the standard cationic detergents.
(a) Water glass
(b) Water gas
(c) Quartz silica
(d) Pearl glass
Show Answer & Explanation
Answer: (a) Water glass
Explanation:
1. Sodium silicate is soluble in water and is known as water glass.
2. Water gas is \( \ce{CO + H2} \); the other options are not names for sodium silicate.
Teacher's Note:
1. Water glass is used to make silica gel by acidification.
(a) Sols of metal sulphides are lyophilic
(b) According to the Schulze-Hardy law, the bigger the size of the ion, the greater its coagulating power
(c) One would expect charcoal to adsorb hydrogen sulphide more strongly than chlorine
(d) Brownian movement is more pronounced for smaller particles than for bigger ones
Show Answer & Explanation
Answer: (d) Brownian movement is more pronounced for smaller particles than for bigger ones
Explanation:
1. Smaller particles are knocked about more by solvent molecules, so Brownian motion is more pronounced.
2. Metal sulphide sols are lyophobic; coagulating power depends on the ionic charge, not size; and chlorine (critical temperature 417 K) is more easily liquefied than hydrogen sulphide (373 K), so charcoal adsorbs chlorine more strongly.
Teacher's Note:
1. Particle size is inversely related to Brownian movement.
(a) Finely divided nickel
(b) Colloidal solution of palladium
(c) Finely divided platinum
(d) Colloidal \( \ce{Fe(OH)3} \)
Show Answer & Explanation
Answer: (b) Colloidal solution of palladium
Explanation:
1. Palladium is well known for taking up very large volumes of hydrogen (hundreds of times its own volume).
2. In colloidal form its surface area is also enormous, so it adsorbs more than nickel or platinum.
Teacher's Note:
1. Pd is the classic hydrogen-absorbing metal.
(a) \( \ce{Cl-} \)
(b) \( \ce{[Fe(CN)6]^{4-}} \)
(c) \( \ce{SO4^{2-}} \)
(d) \( \ce{PO4^{3-}} \)
Show Answer & Explanation
Answer: (b) \( \ce{[Fe(CN)6]^{4-}} \)
Explanation:
1. For a positive sol, the coagulating ion is the anion; the higher its charge, the smaller the flocculation value (Hardy-Schulze rule).
2. \( \ce{[Fe(CN)6]^{4-}} \) has the highest negative charge (4-).
Teacher's Note:
1. Greater charge means lower flocculation value.
(a) Raise both the surface tension and viscosity of water
(b) Lower both the surface tension and the viscosity of water
(c) None of these
(d) Lower the surface tension and raise the viscosity of water
Show Answer & Explanation
Answer: (d) Lower the surface tension and raise the viscosity of water
Explanation:
1. Surface-active molecules gather at the surface and lower the surface tension.
2. Formation of micelles increases the viscosity of the solution.
Teacher's Note:
1. Surfactants reduce surface tension.
(a) The salt behaves as a normal, strong electrolyte
(b) The surfactant molecules or ions undergo association to form clusters
(c) The viscosity of the solution increases abruptly
(d) Substances like grease and fat dissolve colloidally
Show Answer & Explanation
Answer: (a) The salt behaves as a normal, strong electrolyte
Explanation:
1. Below the CMC the surfactant exists as free ions, behaving like an ordinary strong electrolyte.
2. Association, viscosity jump and solubilisation of grease happen only above the CMC.
Teacher's Note:
1. Micelle formation starts at the CMC.
(a) The two sols will stabilise each other
(b) The sol particles will acquire more electric charge
(c) The sols will coagulate each other mutually
(d) A true solution will be obtained
Show Answer & Explanation
Answer: (c) The sols will coagulate each other mutually
Explanation:
1. With excess KI, the AgI sol adsorbs \( \ce{I-} \) and is negative; with excess \( \ce{AgNO3} \), it adsorbs \( \ce{Ag+} \) and is positive.
2. Oppositely charged sols neutralise each other and coagulate.
Teacher's Note:
1. Mixing oppositely charged sols causes mutual coagulation.
(a) Oil and grease can be absorbed into the hydrophilic centres of soap micelles and washed away
(b) Oil and grease can be absorbed into both hydrophilic and hydrophobic centres but not washed away
(c) Oil and grease can be absorbed into the hydrophobic centres of soap micelles and washed away
(d) None of these
Show Answer & Explanation
Answer: (c) Oil and grease can be absorbed into the hydrophobic centres of soap micelles and washed away
Explanation:
1. The hydrocarbon tails form the non-polar interior of the micelle, which dissolves oil and grease.
2. The polar heads face the water, so the micelle carries the grease away.
Teacher's Note:
1. Like dissolves like: oil goes into the non-polar core.
(a) \( \ce{SO2} > \ce{CO2} > \ce{H2} \)
(b) \( \ce{H2} > \ce{CO2} > \ce{SO2} \)
(c) \( \ce{CO2} > \ce{H2} > \ce{SO2} \)
(d) \( \ce{H2} = \ce{CO2} > \ce{SO2} \)
Show Answer & Explanation
Answer: (a) \( \ce{SO2} > \ce{CO2} > \ce{H2} \)
Explanation:
1. Easily liquefiable gases (higher critical temperature, stronger van der Waals forces) are adsorbed more.
2. Critical temperatures: SO2 430 K, CO2 304 K, H2 33 K.
Teacher's Note:
1. Adsorption increases with ease of liquefaction.
(a) They have less iron number
(b) They have any metal number
(c) None of the above
(d) They stabilise lyophobic sols
Show Answer & Explanation
Answer: (d) They stabilise lyophobic sols
Explanation:
1. A lyophilic colloid forms a protective layer around lyophobic particles.
2. This prevents coagulation when electrolytes are added.
Teacher's Note:
1. Protective power is measured by the gold number.
(a) 10 mL of Au sol and 10 mL of 10% NaCl solution
(b) 10 mL of Au sol and 1 mL of 10% NaCl
(c) 10 mL of NaCl sol and 1 mL of 10% Au solution
(d) 10 mL of Au sol and x mg of NaCl
Show Answer & Explanation
Answer: (b) 10 mL of Au sol and 1 mL of 10% NaCl
Explanation:
1. The gold number is the minimum mass in mg of protective colloid that just prevents the coagulation of 10 mL of gold sol on adding 1 mL of 10% NaCl solution.
Teacher's Note:
1. Gold number definition uses 10 mL gold sol and 1 mL 10% NaCl.
(a) \( \ce{SO4^{2-}} \)
(b) \( \ce{PO4^{3-}} \)
(c) \( \ce{Ca^{2+}} \)
(d) \( \ce{Al^{3+}} \)
Show Answer & Explanation
Answer: (b) \( \ce{PO4^{3-}} \)
Explanation:
1. \( \ce{Fe(OH)3} \) sol is positively charged, so anions coagulate it.
2. The anion with the highest charge, \( \ce{PO4^{3-}} \), is the most effective.
Teacher's Note:
1. Opposite charge and higher magnitude coagulate best.
(a) Absorption of light by dust particles
(b) Reflection of light by dust particles
(c) None of these
(d) Scattering of light by dust particles
Show Answer & Explanation
Answer: (d) Scattering of light by dust particles
Explanation:
1. Short-wavelength (blue) light is scattered most strongly by the colloidal particles in the atmosphere (Tyndall effect).
Teacher's Note:
1. Scattering depends strongly on wavelength.
(a) Neutralisation of charge on colloids
(b) Centrifugation
(c) Absorption
(d) Addition of electrolyte
Show Answer & Explanation
Answer: (a) Neutralisation of charge on colloids
Explanation:
1. Smoke particles are charged colloids; they are neutralised and deposited on oppositely charged electrodes.
Teacher's Note:
1. The Cottrell precipitator uses high voltage electrodes.
(a) A > B > C
(b) B > C > A
(c) C > B > A
(d) A = B = C
Show Answer & Explanation
Answer: (c) C > B > A
Explanation:
1. The path with the lowest activation energy is the easiest.
2. Hence C is the easiest, then B, then A.
Teacher's Note:
1. Lower barrier means faster path.
(a) Dative bonding
(b) Ionic bonding
(c) Van der Waals forces
(d) Polar covalent bonding
Show Answer & Explanation
Answer: (c) Van der Waals forces
Explanation:
1. Multimolecular colloids consist of aggregates of small molecules (for example sulphur sols) bound by weak van der Waals forces.
Teacher's Note:
1. Macromolecular colloids contain covalent bonds instead.
(a) Heating charcoal with steam so that it becomes more porous
(b) Adding \( \ce{Ca3(PO4)2} \) to charcoal
(c) Adding impurity to charcoal
(d) Reacting with conc. \( \ce{HNO3} \)
Show Answer & Explanation
Answer: (a) Heating charcoal with steam so that it becomes more porous
Explanation:
1. Activation involves heating charcoal in steam (or air) to remove adsorbed hydrocarbons and create pores, increasing its surface area.
Teacher's Note:
1. Activation means increasing the surface area.
(a) \( \ce{CH3(CH2)11OSO3^-Na+} \)
(b) \( \ce{CH3(CH2)6COO^-Na+} \)
(c) \( \ce{CH3(CH2)11N+(CH3)3Br-} \)
(d) \( \ce{CH3(CH2)15N+(CH3)3Br-} \)
Show Answer & Explanation
Answer: (d) \( \ce{CH3(CH2)15N+(CH3)3Br-} \)
Explanation:
1. CMC decreases as the hydrocarbon chain gets longer.
2. The \( \ce{C16} \) trimethylammonium bromide has the longest chain (16 carbons) among the ionic surfactants listed.
Teacher's Note:
1. Longer hydrophobic tail gives lower CMC.
(a) Colloidal solution formation
(b) Adsorption
(c) Catalytic properties
(d) Absorption
Show Answer & Explanation
Answer: (b) Adsorption
Explanation:
1. Silica gel has a very large surface area and adsorbs water vapour on its surface.
Teacher's Note:
1. Silica gel is a common adsorbent.
(a) Ziegler-Natta catalyst, heterogeneous catalyst
(b) Fe/Mo, heterogeneous catalyst / inhibitor
(c) Pd/BaSO4 catalyst, catalytic poison is BaSO4
(d) Shape selective catalyst, ZSM-5
Show Answer & Explanation
Answer: (b) Fe/Mo, heterogeneous catalyst / inhibitor
Explanation:
1. In the Haber process Fe is the catalyst and Mo is a promoter (it increases the catalyst's activity), not an inhibitor.
2. The other three matches are correct: Ziegler-Natta is heterogeneous, BaSO4 partly poisons Pd in the Rosenmund reduction, and ZSM-5 is a shape-selective zeolite.
Teacher's Note:
1. Mo in the Haber process is a promoter.
(a) True solution
(b) Hypertonic solution
(c) Hypotonic solution
(d) Colloidal solution
Show Answer & Explanation
Answer: (d) Colloidal solution
Explanation:
1. Colloidal particles are large enough to scatter light (Tyndall effect).
2. Because there are few particles per unit mass, colligative properties are small.
Teacher's Note:
1. Colloids scatter light but have very small colligative effects.
(a) \( \ce{CH3COO^-K+} \)
(b) \( \ce{CH3(CH2)16COO^-Na+} \)
(c) \( \ce{CH3(CH2)11SO4^-Na+} \)
(d) \( \ce{CH3(CH2)16N+(CH3)3Br-} \)
Show Answer & Explanation
Answer: (a) \( \ce{CH3COO^-K+} \)
Explanation:
1. A surfactant needs a long hydrocarbon tail (about 10 or more carbons) and a polar head.
2. Potassium acetate has only a two-carbon chain, too short to give a hydrophobic core.
Teacher's Note:
1. No long tail, no micelles.
(a) Peptisation
(b) Electrophoresis
(c) Electrodialysis
(d) Electro-osmosis
Show Answer & Explanation
Answer: (c) Electrodialysis
Explanation:
1. Electrodialysis uses an electric field to speed the removal of ionic impurities through a membrane.
2. Peptisation prepares a sol; electrophoresis and electro-osmosis demonstrate charge.
Teacher's Note:
1. Dialysis and electrodialysis purify sols.
(a) Critical temperature
(b) Boyle's temperature
(c) Kraft temperature
(d) Inversion temperature
Show Answer & Explanation
Answer: (c) Kraft temperature
Explanation:
1. Micelles form only above the Kraft temperature, at which the solubility of the surfactant equals its CMC.
Teacher's Note:
1. Kraft temperature is the minimum temperature for micelle formation.
(a) Water-in-oil emulsions are formed when the emulsifying agent at the interface is chiefly in the water phase
(b) Emulsions are made by shaking two immiscible liquids, say oil and water, with an emulsifying agent
(c) Water-in-oil emulsions are formed when the emulsifying agent at the interface is chiefly in the oil phase
(d) Milk is an oil-in-water emulsion stabilised by casein
Show Answer & Explanation
Answer: (a) Water-in-oil emulsions are formed when the emulsifying agent at the interface is chiefly in the water phase
Explanation:
1. By Bancroft's rule, the phase in which the emulsifier is more soluble becomes the continuous phase.
2. For a water-in-oil emulsion the continuous phase is oil, so the emulsifier must be chiefly in the oil phase; the statement about the water phase is wrong.
Teacher's Note:
1. The emulsifier lives in the continuous phase.
(a) (c) and (d)
(b) (a) and (b)
(c) (a), (b) and (c)
(d) (a) and (d)
Show Answer & Explanation
Answer: (d) (a) and (d)
Explanation:
1. Lyophobic sols are stabilised by the like charge acquired through preferential adsorption of ions (a) and by the zeta potential across the double layer (d).
2. Solvent adsorption is the stabilising factor in lyophilic sols, and opposite-charge attraction would cause coagulation.
Teacher's Note:
1. Charge repulsion keeps lyophobic sols apart.
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