JEE Chemistry Surface Chemistry MCQs Set 09

Answer: (a) It is usually monolayer

Explanation:
1. Physical adsorption is held by weak van der Waals forces, is reversible and has a low heat of adsorption (20-40 kJ/mol).
2. It can build up many layers at low temperature and high pressure, so it is multilayer, not monolayer.
3. Statement (a) is therefore the incorrect one.

Teacher's Note:
1. Physisorption is multilayer; chemisorption is monolayer.
2. Remember: weak forces, reversible, low enthalpy.

Answer: (d) It does not require activation energy

Explanation:
1. Chemisorption forms chemical bonds, is irreversible and has a high enthalpy (80-240 kJ/mol).
2. It usually needs activation energy, so it becomes appreciable at higher temperature.
3. Hence the statement that it needs no activation energy is wrong.

Teacher's Note:
1. Chemisorption behaves like a chemical reaction.
2. Rate rises with temperature up to a point.

Answer: (b) Extent of adsorption is independent of temperature

Explanation:
1. Adsorption rises with pressure and depends on the adsorbent's nature and surface area.
2. Chemisorption stops once the surface is covered by a monolayer, so it is limited.
3. Adsorption is exothermic and depends strongly on temperature, so (b) is incorrect.

Teacher's Note:
1. Adsorption always varies with temperature.
2. Use Le Chatelier's principle for the exothermic process.

Answer: (a) accumulates on the surface of another substance

Explanation:
1. Adsorption is the higher concentration of molecules on the surface of a solid or liquid.
2. Penetration into the body of the substance is absorption.

Teacher's Note:
1. Adsorption is a surface effect; absorption is a bulk effect.
2. Sponge in water: absorption. Gas on charcoal surface: adsorption.

Answer: (a) temperature is increased

Explanation:
1. Physical adsorption is exothermic.
2. Raising the temperature shifts the equilibrium toward desorption.
3. Increasing pressure or concentration would favour more adsorption.

Teacher's Note:
1. Apply Le Chatelier's principle to the exothermic process.
2. Heating charcoal releases adsorbed gas.

Answer: (b) increases with increase of pressure

Explanation:
1. Higher gas pressure brings more molecules to the surface per second, so the rate of chemisorption rises.
2. Chemisorption also depends on temperature because it needs activation energy.

Teacher's Note:
1. Rate increases with pressure for adsorption of a gas on a surface.
2. Do not confuse rate with extent.

Answer: (c) it is a multilayer phenomenon

Explanation:
1. Chemisorption is irreversible, highly specific and has a high heat of adsorption.
2. It stops after one layer, so it is monolayer, not multilayer.

Teacher's Note:
1. Chemisorption = one layer. Physisorption = many layers.
2. Specific means only certain gases are chemisorbed by a given solid.

Answer: (a) adsorption of Ca²⁺ and Mg²⁺ ions of hard water replacing Na⁺ ions

Explanation:
1. Zeolite (permutit) is a sodium aluminium silicate.
2. Ca²⁺ and Mg²⁺ from hard water exchange with Na⁺ in the zeolite.
3. Aluminium stays in the framework, so only (a) is right.

Teacher's Note:
1. Zeolite works by ion exchange of Na⁺ with Ca²⁺ and Mg²⁺.
2. Hardness ions are removed, Na⁺ goes into the water.

Answer: (d) Ni is used as a catalyst in the manufacture of ammonia

Explanation:
1. Catalysts are specific, work in small amounts and are more active when finely divided (more surface vacancies).
2. Ammonia is made by the Haber process using iron with molybdenum as a promoter; nickel is used for hydrogenation, not ammonia.

Teacher's Note:
1. Haber process: Fe catalyst, Mo promoter.
2. Nickel: hydrogenation of oils.

Answer: (c) decreasing activation energy

Explanation:
1. A catalyst offers an alternate path with lower activation energy.
2. It changes neither ΔH nor the internal energy of reactants and products.

Teacher's Note:
1. Catalysts lower Ea but do not change ΔH or the equilibrium position.
2. Faster rate comes from more molecules crossing the lower barrier.

Answer: (a) different diameters of colloidal gold particles

Explanation:
1. Colour of a sol depends on how light is scattered.
2. Scattering depends on particle size, so different sizes give different colours.

Teacher's Note:
1. Particle size decides the colour of a metal sol.
2. Gold sols can be red, violet or blue.

Answer: (d) both (a) and (b)

Explanation:
1. Extrinsic colloids (lyophobic) are metal sulphides, hydroxides and metals; they need a stabiliser and are easily coagulated.
2. As₂S₃ and Fe(OH)₃ sols are both lyophobic; egg albumin is lyophilic.

Teacher's Note:
1. Lyophobic: sulphides, hydroxides, metals.
2. Lyophilic: gelatin, starch, albumin, gum.

Answer: (a) Arsenious sulphide

Explanation:
1. Metal hydroxides (Al(OH)₃, Fe(OH)₃) are positive.
2. AgI in excess AgNO₃ adsorbs Ag⁺ and is positive.
3. As₂S₃ is a negative sol.

Teacher's Note:
1. Sulphide sols and metal sols are generally negative.
2. Hydroxide sols are positive.

Answer: (a) dialysis

Explanation:
1. Dialysis removes dissolved ions and molecules through a semi-permeable membrane while colloidal particles stay inside.
2. Peptization makes a sol; ordinary filtration does not separate colloids from solutes.

Teacher's Note:
1. Dialysis = removal of electrolytes.
2. Ultra-filtration separates colloids from small molecules.

Answer: (b) conversion of a precipitate into a colloidal sol

Explanation:
1. Peptisation is the dispersal of a freshly formed precipitate into a sol by adding a small amount of electrolyte (the peptising agent).
2. Option (c) describes Bredig's arc method.

Teacher's Note:
1. Peptisation is the reverse of coagulation.
2. Example: Fe(OH)₃ precipitate + a little FeCl₃ gives a sol.

Answer: (c) the blood is coagulated and thus the blood vessel is sealed

Explanation:
1. Blood is a negatively charged sol.
2. Fe³⁺ ions neutralise the charge and coagulate the blood, sealing the wound.

Teacher's Note:
1. Trivalent ions coagulate negative sols most effectively.
2. Alum is used in the same way.

Answer: (c) the lower its value, the greater is the protecting power

Explanation:
1. Gold number is the minimum mass (mg) of a protective colloid that prevents coagulation of 10 mL of a standard gold sol on adding 1 mL of 10% NaCl.
2. A smaller mass needed means a better protector, so a lower gold number means greater protecting power.

Teacher's Note:
1. Gold number is inversely related to protective power.
2. Gelatin has a very low gold number.

Answer: (a) lyophilic

Explanation:
1. A protective colloid is a lyophilic sol that surrounds the lyophobic particles and prevents their coagulation.

Teacher's Note:
1. Protective colloids: gelatin, gum arabic, starch.
2. They are solvent-loving colloids.

Answer: (c) Al³⁺

Explanation:
1. As₂S₃ is a negative sol, so the coagulating ion is the cation.
2. By the Hardy-Schulze rule the greater the charge, the greater the coagulating power: Al³⁺ > Mg²⁺ > K⁺.

Teacher's Note:
1. The ion with charge opposite to the sol coagulates it.
2. Higher valence gives higher coagulating power.

Answer: (d) SO₄²⁻

Explanation:
1. Fe(OH)₃ is a positive sol, so the anion coagulates it.
2. SO₄²⁻ has the highest charge, hence the greatest coagulating power.

Teacher's Note:
1. Anions coagulate positive sols.
2. Count the charge on each anion.

Answer: (b) Emulsion

Explanation:
1. A liquid dispersed in a liquid is an emulsion (e.g. milk, oil in water).

Teacher's Note:
1. Know the dispersed phase and medium for each colloid type.
2. Emulsion = liquid in liquid.

Answer: (c) associate

Explanation:
1. Above CMC the surfactant molecules associate to form micelles.

Teacher's Note:
1. Micelles are associated colloids.
2. Soap is a common example.

Answer: (a) Syneresis

Explanation:
1. Slow exudation of liquid from a gel on standing is syneresis.
2. Thixotropy is the gel-to-sol change on shaking.

Teacher's Note:
1. Imbibition is swelling by absorbing liquid.
2. Syneresis = squeezing out.

Answer: (a) It can be easily solvated

Explanation:
1. Lyophobic (solvent-hating) sols are not solvated, carry charge, are irreversible on coagulation and are less stable.
2. Hence the statement that they are easily solvated is wrong.

Teacher's Note:
1. Lyophobic = little attraction to the medium.
2. Lyophilic sols are solvated.

Answer: (a) liquid in a solid

Explanation:
1. A gel is a colloid with a liquid dispersed in a solid medium.
2. Option (b) is a solid aerosol, (c) an emulsion and (d) a solid foam.

Teacher's Note:
1. Gel = liquid in solid.
2. Use the dispersed phase and medium.

Answer: (b) chemisorption of hexanethiol onto the gold surface renders it hydrophobic

Explanation:
1. The thiol forms a strong Au-S bond, which is chemisorption; excess water cannot wash it away.
2. The exposed hydrocarbon chains make the surface hydrophobic, so water does not spread.

Teacher's Note:
1. Washing with excess water does not remove a chemisorbed layer.
2. Hydrocarbon tails repel water.

Answer: (d) increases with p at low pressure and remains the same at high pressure

Explanation:
1. At low p, bp is much smaller than 1, so x ≈ ap and x increases with p.
2. At high p, bp is much larger than 1, so x ≈ a/b, a constant (saturation).

Teacher's Note:
1. Check the limits p → 0 and p → ∞.
2. This is the Langmuir isotherm.

Answer: (a) 90 J

Explanation:
1. Total area = 2.4 × 10¹⁸ × 12.5 × 10⁻¹⁶ = 3000 m².
2. Energy = surface tension × area = 0.03 × 3000 = 90 J.

Teacher's Note:
1. Surface energy = γ × increase in area.
2. Multiply the number of droplets by the area of one.