JEE Chemistry Surface Chemistry MCQs Set 18

Answer: (a) \( p^{2/3} \)

Explanation:
1. \( \log(x/m) = \log K + (1/n) \log p \), so the slope equals \( 1/n \).
2. With \( 1/n = 2/3, x/m = K p^{2/3} \).

Teacher's Note:
1. The slope of the \( \log-\log \) plot is the exponent of pressure. The intercept gives \( \log K \).

Answer: (c) A-iii, B-i, C-ii, D-iv

Explanation:
1. \( V_{2}O_{5} \) is the catalyst of the contact process for H₂SO₄.
2. TiCl₄ with Al(CH₃)₃ is the Ziegler-Natta catalyst for polyethylene.
3. PdCl₂ is used in the Wacker process to make ethanal, and iron (iron oxide based) catalyses ammonia synthesis.

Teacher's Note:
1. Learn each industrial catalyst together with its product. Ziegler-Natta, contact, Haber and Wacker processes are frequent in JEE.

Answer: (b) increase in p and decrease in T

Explanation:
1. \( x/m \) is proportional to \( p^{0.5} \), so it rises with pressure.
2. Physical adsorption is exothermic, so by Le Chatelier's principle it falls as temperature rises.

Teacher's Note:
1. Physisorption increases with pressure and decreases with temperature. Chemisorption first rises with temperature and then falls.

Answer: (b) the process of converting a precipitate into a colloidal solution

Explanation:
1. In peptisation a freshly formed precipitate is shaken with a small amount of electrolyte, the peptising agent.
2. The precipitate adsorbs ions, becomes charged and breaks up into colloidal particles.

Teacher's Note:
1. Peptisation is the reverse of coagulation. Fresh Fe(OH)₃ precipitate turns into a sol on adding a little FeCl₃.

Answer: (c) The osmotic pressure of a colloidal solution is higher than that of a true solution of the same concentration.

Explanation:
1. Colligative properties depend on the number of particles.
2. Colloidal particles are large and few in number, so the osmotic pressure of a colloid is lower than that of a true solution of equal concentration.

Teacher's Note:
1. Colloids show very small colligative effects. That is why osmotic pressure is used to find molar masses of macromolecules only with sensitive methods.

Answer: (b) AlCl₃

Explanation:
1. As₂S₃ sol is negatively charged, so it is coagulated by cations.
2. By the Hardy-Schulze rule the cation of highest charge is most effective: Al³⁺ > Ba²⁺ > Na⁺.

Teacher's Note:
1. Coagulating power depends on the charge of the ion opposite in sign to that of the sol.

Answer: (a) A-R, B-Q, C-P

Explanation:
1. Benzaldehyde is held on the surface of the solid, so it is the adsorbate.
2. Alumina is the solid adsorbent, the stationary phase.
3. Acetonitrile moves over the adsorbent, so it is the dynamic phase.

Teacher's Note:
1. In chromatography the stationary phase is the adsorbent, and the substance being carried along is in the mobile phase.

Answer: (d) positively and negatively charged sols, respectively

Explanation:
1. Haemoglobin is a positively charged sol.
2. Metal sols such as gold are negatively charged.

Teacher's Note:
1. Positively charged sols include Fe(OH)₃ and haemoglobin. Negatively charged sols include metal sols, As₂S₃ and starch.

Answer: (b) H₂ + I₂ → 2HI

Explanation:
1. Reactivity of halogens decreases down the group: \( F_{2} \) > Cl₂ > Br₂ > \( I_{2} \).
2. Iodine is the least reactive, so its reaction with hydrogen needs a catalyst.

Teacher's Note:
1. The trend of halogen reactivity decides which reaction is slow.

Answer: (a) C: liquid in solid; M: liquid in liquid; S: solid in gas

Explanation:
1. Cheese is a gel, a liquid dispersed in a solid.
2. Milk is an emulsion, a liquid in a liquid, and smoke is a solid in a gas.

Teacher's Note:
1. Gel = liquid in solid, emulsion = liquid in liquid, aerosol = solid or liquid in gas.

Answer: (d) Latex is a colloidal solution of rubber particles which are positively charged.

Explanation:
1. Latex is a stable dispersion of rubber particles that carry a negative charge.
2. The other three statements are correct (\( a \) lyophilic sol needs a large amount of electrolyte).

Teacher's Note:
1. Rubber latex is negatively charged and is coagulated by adding acid or electrolytes.

Answer: (b) II and IV

Explanation:
1. Scattering needs particles comparable with the wavelength of light.
2. It also needs a large difference between the refractive indices of the dispersed phase and the medium.

Teacher's Note:
1. True solutions do not scatter light because their particles are too small.

Answer: (b) A-ii, B-i, C-iv, D-iii

Explanation:
1. TiCl₃ is used in Ziegler-Natta polymerisation of ethene.
2. PdCl₂ is used in the Wacker process, CuCl₂ in Deacon's process (chlorine from HCl), and \( V_{2}O_{5} \) in the contact process.

Teacher's Note:
1. Each of these industrial processes is identified by its catalyst.

Answer: (b) Na⁺ < Ba²⁺ < Al³⁺

Explanation:
1. As₂S₃ is a negative sol, so the coagulating ions are the cations.
2. By the Hardy-Schulze rule, power rises with charge: Na⁺ < Ba²⁺ < Al³⁺.

Teacher's Note:
1. Higher charge on the oppositely charged ion means a lower coagulation value.

Answer: (d) Enthalpy of adsorption is slow and positive.

Explanation:
1. Adsorption is exothermic, so ΔH(adsorption) is always negative.
2. Physisorption is fast and reversible; the other statements are correct.

Teacher's Note:
1. Physisorption is fast, reversible, multilayer, with small negative \( \DeltaH \) of about 20 to 40 kJ per mole.

Answer: (c) A < C < B < D

Explanation:
1. The lower the gold number, the greater the protective power.
2. Gold numbers: \( A 0.50 \gt C 0.10 \gt B 0.01 \gt D 0.005 \), so protective power \( A \lt C \lt B \lt D \).

Teacher's Note:
1. Protective power varies inversely as the gold number.

Answer: (b) Mixing the sols has no effect.

Explanation:
1. Oppositely charged sols neutralise each other on mixing, so they coagulate.
2. Hence the statement that mixing has no effect is incorrect.

Teacher's Note:
1. Mutual coagulation of oppositely charged sols is a standard way to precipitate colloids.

Answer: (a) V_C/V_S ≈ 10³

Explanation:
1. Colloidal particles are about 10 nm across and solute particles about 1 nm.
2. The volume ratio is \( (10/1)^{3} = 10^{3} \).

Teacher's Note:
1. Volume scales as the cube of the diameter.

Answer: (c) zero

Explanation:
1. Reactions on a solid catalyst surface at saturation coverage do not depend on gas concentration.
2. Hence the decomposition on tungsten follows a zero order rate law.

Teacher's Note:
1. Surface reactions become zero order when the surface is saturated with reactant.