JEE Chemistry Surface Chemistry MCQs Set 08

Answer: (a) 2

Explanation:
1. (A) false: physisorption is not specific. (B) false: enthalpy is low (20–40 kJ mol⁻¹).
2. (C) true: adsorption is exothermic and falls as temperature rises. (D) false: multilayers form.
3. (E) true: no activation energy is needed. So 2 statements are correct.

Teacher's Note:
1. Physisorption is due to weak van der Waals forces; chemisorption involves chemical bonds.
2. Chemisorption is specific, has high enthalpy and forms a monolayer.

Answer: (d) starts at the origin, rises with increasing substrate concentration and then levels off to a constant maximum rate

Explanation:
1. At low substrate concentration the rate rises almost in proportion to the concentration (first order).
2. When all active sites of the enzyme are occupied, the rate reaches a maximum and becomes independent of substrate concentration (zero order).
3. Hence the curve rises and then saturates.

Teacher's Note:
1. Enzyme reactions show saturation kinetics (Michaelis–Menten).
2. Enzymes have a limited number of active sites.

Answer: (c) 0.12 mL

Explanation:
1. Molarity of the acid = 1210 g L⁻¹ / 24.2 g mol⁻¹ = 50 M.
2. Millimoles of NaOH = 25 × 0.24 = 6; the acid is monobasic, so millimoles of acid = 6.
3. Volume = 6/50 = 0.12 mL (12 × 10⁻² mL).

Teacher's Note:
1. Use M₁V₁ = M₂V₂ for a monobasic acid and a monoacidic base.
2. The acid is the pure liquid itself, so its molarity comes from density/molar mass.

Answer: (b) 2

Explanation:
1. (A) is false: a bulk molecule is attracted equally in all directions, giving zero net force, which is not the cause of surface tension.
2. (B) is true: surface molecules feel a net inward pull. (D) is true: molecules escaping from the surface give the vapour pressure.
3. (C) is false: molecules continually exchange between bulk and surface. So 2 statements are correct.

Teacher's Note:
1. Surface molecules have fewer neighbours, so they feel a net inward attraction.
2. Molecules in a liquid keep moving between the bulk and the surface.

Answer: (a) 2

Explanation:
1. (A) is incorrect: anhydrous CaCl₂ takes up water by absorption (it forms hydrates throughout its bulk).
2. (B) and (D) are correct: surface energy and entropy both decrease on adsorption.
3. (C) is incorrect: as the surface gets covered, the heat released per mole falls, so ΔH becomes less negative. Incorrect statements: A and C = 2.

Teacher's Note:
1. Absorption takes place throughout the bulk; adsorption only on the surface.
2. Adsorption sites are used up as coverage increases, so the heat evolved decreases.

Answer: (a) 10 mL of 0.2 mol dm⁻³ AlCl₃

Explanation:
1. Excess NaOH makes the Fe(OH)₃ sol negatively charged (it adsorbs OH⁻ ions).
2. A negative sol is coagulated by cations; by the Hardy–Schulze rule the cation with the highest charge is most effective.
3. Al³⁺ (trivalent, 0.2 M) is more effective than Ca²⁺ or Na⁺, so AlCl₃ coagulates fastest.

Teacher's Note:
1. Hardy–Schulze rule: coagulating power increases with the charge of the oppositely charged ion.
2. First decide the sign of the sol, then look at the oppositely charged ions.

Answer: (b) A-III, B-I, C-IV, D-II

Explanation:
1. Osmosis: solvent moves into the solution (III). Reverse osmosis: solvent moves towards the pure solvent side under external pressure (I).
2. Electro-osmosis: the dispersion medium moves (IV). Electrophoresis: colloidal particles move (II).
3. So A-III, B-I, C-IV, D-II.

Teacher's Note:
1. In electrophoresis the particles move; in electro-osmosis the medium moves.
2. Reverse osmosis needs a pressure larger than the osmotic pressure.

Answer: (d) 1.6

Explanation:
1. Freundlich equation: log(x/m) = log k + (1/n) log p. Slope 1/n = tan 45° = 1; log k = 0.6020 = 2 log 2, so k = 4.
2. x/m = k p^1/n = 4 × 0.4 = 1.6 (that is 16 × 10⁻¹).

Teacher's Note:
1. Slope of the Freundlich line is 1/n and the intercept is log k.
2. tan 45° = 1.

Answer: (c) the polar head groups are in the interior and the hydrocarbon tails are directed towards the non-polar solvent

Explanation:
1. Like dissolves like: the non-polar tails prefer the non-polar solvent, the polar heads avoid it.
2. So the heads cluster in the core and the tails point outward. In water the arrangement is the opposite (normal micelle).

Teacher's Note:
1. Micelles in water: tails inside, heads outside. In non-polar solvents: heads inside, tails outside.
2. The core always hides the part that is not compatible with the solvent.

Answer: (d) A-II, B-I, C-IV, D-III

Explanation:
1. Physisorption has low enthalpy of adsorption, 20–40 kJ mol⁻¹ (A-II). Chemisorption forms a monolayer (B-I).
2. The Haber process on iron is heterogeneous catalysis (C-IV). Chromatography is an analytical use of adsorption (D-III).

Teacher's Note:
1. Chemisorption forms a single layer; physisorption can form multilayers.
2. Solid catalysts act through chemisorption of reactants.

Answer: (c) 3 and 2.505

Explanation:
1. Freundlich isotherm: log(x/m) = log K + (1/n) log P, which has the form y = mx + c.
2. Comparing with y = 3x + 2.505: slope 1/n = 3 and intercept log K = 2.505.

Teacher's Note:
1. The slope of the plot is 1/n and the intercept is log K.
2. Compare the given line with y = mx + c.