JEE Chemistry Surface Chemistry MCQs Set 07

Answer: (b) The adsorption is accompanied by a decrease in enthalpy

Explanation:
1. Adsorption of a dye on charcoal is physical adsorption (physisorption), held by van der Waals forces.
2. Physisorption is exothermic, so the enthalpy decreases, needs no activation energy, decreases with a rise in temperature and is reversible.

Teacher's Note:
1. Adsorption always has a negative \(\Delta H\) because the freedom of the adsorbate falls.
2. Chemisorption is the type that needs activation and is irreversible.

Answer: (a) Preferential adsorption of ions on their surface from the solution

Explanation:
1. Lyophobic particles adsorb ions common to the dispersion medium preferentially and so acquire a like electric charge.
2. Particles with the same charge repel one another and do not aggregate.
3. This repulsion is the cause of the stability.

Teacher's Note:
1. The charge on the sol particles is the main reason for their stability.
2. Removing the charge with oppositely charged ions leads to coagulation.

Answer: (a) \(\ce{CH3(CH2)12N+(CH3)3Br-}\)

Explanation:
1. The critical micelle concentration falls as the length of the hydrophobic hydrocarbon chain increases.
2. Option (a) has a chain of 13 carbons plus the head group, the longest of the four; (b) and (d) have 12 carbons and (c) has only 7.
3. The surfactant with the longest chain therefore forms micelles at the lowest concentration.

Teacher's Note:
1. Longer hydrocarbon tails are more strongly driven out of water, so micelles form earlier.
2. Compare the chain lengths when asked about CMC.

Answer: (c) The enthalpy of adsorption (\(\Delta H_{adsorption}\)) is low and positive

Explanation:
1. Easily liquefiable gases have strong van der Waals forces and are adsorbed more.
2. At high pressure physisorption builds up multimolecular layers.
3. The enthalpy of physisorption is low (20 to 40 kJ/mol) but it is negative, since adsorption is exothermic, so statement (c) is incorrect.

Teacher's Note:
1. Adsorption is exothermic, so \(\Delta H\) is always negative.
2. Physisorption has a smaller negative \(\Delta H\) than chemisorption.

Answer: (d) \(\ce{Na+} < \ce{Ba^2+} < \ce{Al^3+}\)

Explanation:
1. Arsenic sulphide is a negatively charged sol, so the effective coagulating ions are cations.
2. By the Hardy-Schulze rule, the higher the charge on the cation, the greater the coagulating power.
3. The order of increasing power is \(\ce{Na+} < \ce{Ba^2+} < \ce{Al^3+}\).

Teacher's Note:
1. Use the charge of the ion opposite to that of the sol.
2. The Hardy-Schulze rule is about the valency of the coagulating ion.

Answer: (b) Only \(1/n\) appears as the slope

Explanation:
1. The Freundlich isotherm is \(\dfrac xm=kp^{1/n}\).
2. Taking logarithms, \(\log\dfrac xm=\log k+\dfrac1n\log p\).
3. This is a straight line of slope \(1/n\) and intercept \(\log k\), so only \(1/n\) appears as the slope.

Teacher's Note:
1. Compare with \(y = mx + c\).
2. The constant k appears only in the intercept.

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

Explanation:
1. Scattering of light needs particles that are not much smaller than the wavelength of light, so condition (b) is required.
2. The dispersed phase must also have a refractive index very different from that of the medium, otherwise there is no scattering, so condition (d) is required.

Teacher's Note:
1. Particles in true solutions are too small to scatter visible light.
2. Colloidal particles have sizes of about 1 to 1000 nm.

Answer: (d) \(V_C/V_S \approx 10^{3}\)

Explanation:
1. A solute particle in a true solution has a size of about 1 nm, while colloidal particles range from 1 to 1000 nm in diameter.
2. A typical colloidal particle is thus about 10 times larger in each dimension, making its volume about \(10^3\) times larger.
3. Hence \(V_C/V_S\approx10^3\).

Teacher's Note:
1. Volume scales as the cube of the diameter.
2. A factor of 10 in diameter is a factor of 1000 in volume.

Answer: (c) A colloidal sol

Explanation:
1. Scattering of light (the Tyndall effect) shows that the particles are of colloidal size.
2. Colloidal particles pass through ordinary filter paper, so no residue is left.
3. A suspension would be left on the paper and a true solution would not scatter light.

Teacher's Note:
1. Scattering plus passing through filter paper indicates a colloid.
2. Only ultrafiltration or a membrane stops colloidal particles.

Answer: (a) \(\ce{Fe^3+}\) ions coagulate blood, which is a negatively charged sol

Explanation:
1. Blood is a negatively charged colloidal sol.
2. The highly charged \(\ce{Fe^3+}\) cations neutralise this charge and coagulate the blood, forming a clot that seals the wound.
3. The chloride ion has the same charge as the sol and is not the coagulating ion.

Teacher's Note:
1. This is an application of the Hardy-Schulze rule.
2. Alum is used in the same way.