JEE Chemistry Surface Chemistry MCQs Set 12

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

Explanation:
1. Arsenic sulphide sol is negatively charged, so the coagulating ion is the cation.
2. By the Hardy-Schulze rule, coagulating power increases with the charge on the effective ion.
3. Hence Na\(^+\) < Ba\(^{2+}\) < Al\(^{3+}\).

Teacher's Note:
1. Match the sign of the ion to the opposite sign of the sol. Higher charge means greater coagulating power.

Answer: (d) All of the above are correct for different ranges of pressure

Explanation:
1. The Freundlich isotherm is \( \frac{x}{m} = kp^{1/n} \) with n greater than 1.
2. At low pressure the dependence is linear (\( p^{1} \)), at intermediate pressure it is \( p^{1/n} \), and at high pressure the adsorption saturates (\( p^{0} \)).
3. So all three forms apply in different pressure ranges.

Teacher's Note:
1. Remember the three regions of the adsorption curve: linear, fractional power and saturation.

Answer: (c) the mass of gas striking a given area of surface is proportional to the pressure of the gas

Explanation:
1. Langmuir assumes a monolayer, with one molecule per site, so (b) is wrong.
2. The rate of desorption is proportional to the fraction of surface covered, so (a) is wrong.
3. The rate of collision of gas molecules with the surface, and hence the mass striking a given area, is proportional to the gas pressure.

Teacher's Note:
1. Langmuir's model assumes a uniform surface, a monolayer and no interaction between adsorbed molecules.

Answer: (b) Mixing the sols has no effect

Explanation:
1. Oppositely charged sols neutralise each other's charge when mixed, so they mutually coagulate. Statement (b) is therefore incorrect.
2. Na\(^+\) coagulates the negative gold sol and \( \ce{SO4^{2-}} \) coagulates the positive iron(III) hydroxide sol, so (c) is correct.
3. Mg\(^{2+}\) (divalent, acting on gold) is more effective than Cl\(^-\) (monovalent, acting on iron(III) hydroxide), so (d) is correct.

Teacher's Note:
1. Mutual coagulation of oppositely charged sols is a standard result. Apply the Hardy-Schulze rule to each sol separately.

Answer: (a) \( \frac{V_C}{V_S} \approx 10^{3} \)

Explanation:
1. Colloidal particles are much larger than the particles of a true solution, so the ratio must be greater than 1.
2. A colloidal particle about 10 times wider than a molecule has a volume about \( 10^{3} \) times larger, whereas \(10^{23}\) is unrealistic. So the ratio is about \(10^{3}\).

Teacher's Note:
1. Colloidal particle sizes range from 1 nm to 1000 nm. Volume scales as the cube of the diameter.

Answer: (c) Zero

Explanation:
1. On tungsten the surface is saturated with adsorbed hydrogen, so the rate of decomposition does not depend on the gas pressure.
2. A rate independent of concentration means zero order.

Teacher's Note:
1. Surface reactions on a saturated catalyst surface are usually zero order.

Answer: (b) coagulation of negatively charged turbidity

Explanation:
1. Suspended clay particles in turbid water carry a negative charge.
2. Al\(^{3+}\) ions neutralise this charge and bring about coagulation, which makes the particles settle.

Teacher's Note:
1. Alum is used as a coagulant, not a peptising agent.

Answer: (b) Tyndall effect

Explanation:
1. Electro-osmosis, electrophoresis and coagulation all arise from the charge on the particles.
2. The Tyndall effect is the scattering of light by the particles, and is independent of their charge.

Teacher's Note:
1. Optical properties depend on particle size; electrical properties depend on charge.

Answer: (d) Enzymes are least reactive at the optimum temperature

Explanation:
1. Enzymes are mostly proteins, act specifically and are denatured by UV rays and heat, so (a), (b) and (c) are correct.
2. At the optimum temperature the enzyme activity is the highest, not the least, so (d) is incorrect.

Teacher's Note:
1. Enzyme activity rises with temperature up to the optimum and then falls because of denaturation.

Answer: (a) the adsorption sites are equivalent in their ability to adsorb the particles

Explanation:
1. Langmuir assumed a homogeneous surface with equivalent sites, a constant heat of adsorption, no interaction between adsorbed molecules and a single layer.
2. Only statement (a) matches.

Teacher's Note:
1. Multilayer adsorption is the BET model, not Langmuir.

Answer: (a) Dodecyl trimethyl ammonium chloride

Explanation:
1. Micelles form from molecules having a long hydrocarbon (hydrophobic) tail and an ionic or polar (hydrophilic) head.
2. Dodecyl trimethyl ammonium chloride is a cationic surfactant of this kind, whereas glucose, urea and pyridinium chloride have no long hydrocarbon chain.

Teacher's Note:
1. Look for a long non-polar chain with a polar head group when asked about micelles.

Answer: (d) adsorption lowers the activation energy of the reaction

Explanation:
1. A catalyst provides an alternative path of lower activation energy.
2. In the adsorption theory, reactant molecules adsorbed on the surface react through this lower-energy path, so the rate increases.

Teacher's Note:
1. Catalysts increase the rate by lowering the activation energy, not by supplying heat.

Answer: (a) polar at the outer surface but non-polar at the inner surface

Explanation:
1. In water the hydrophobic hydrocarbon tails cluster in the interior away from water.
2. The hydrophilic polar heads face the water at the outer surface.

Teacher's Note:
1. A micelle has a hydrocarbon core and an ionic or polar shell.

Answer: (a) On increasing temperature, adsorption increases continuously

Explanation:
1. Adsorption is exothermic, so by Le Chatelier's principle it decreases as the temperature rises (physisorption), although chemisorption first increases and then decreases. It does not increase continuously.
2. Enthalpy and entropy changes are both negative, adsorption is specific and it is reversible, so (b), (c) and (d) are correct.

Teacher's Note:
1. Adsorption is exothermic and lowers the freedom of the gas, so both \( \Delta H \) and \( \Delta S \) are negative.

Answer: (a) associate

Explanation:
1. Below the CMC the surfactant exists as individual ions.
2. At the CMC, the molecules begin to associate to form micelles.

Teacher's Note:
1. CMC is the concentration above which micelles form.

Answer: (c) 4

Explanation:
1. 2 mL of 1 M NaCl contains 2 mmol of NaCl.
2. This coagulates 500 mL of sol, so 1000 mL of sol needs \( 2 \times 2 = 4 \) mmol.
3. The flocculation value is 4.

Teacher's Note:
1. Flocculation value is the minimum number of millimoles of electrolyte per litre of sol needed to coagulate it.

Answer: (b) 1

Explanation:
1. 0.01 mg of gelatin protects 10 mL of gold sol.
2. For 1000 mL we need \( 0.01 \times \frac{1000}{10} = 1 \) mg.

Teacher's Note:
1. Scale the gold number in proportion to the volume of the sol.

Answer: (b) 3.84 g

Explanation:
1. Moles of HCl adsorbed \( = 0.5263 \times (0.5 - 0.4) = 0.05263 \) mol.
2. Mass adsorbed \( = 0.05263 \times 36.5 = 1.92 \) g.
3. \( x/m = 1.92 / 0.5 = 3.84 \) g per gram of charcoal.

Teacher's Note:
1. Find the fall in concentration, convert to mass, and divide by the mass of adsorbent.

Answer: (c) 3

Explanation:
1. Colloidal particles have diameters between 1 nm and 1000 nm (1 micrometre).
2. \( 1000 = 10^{3} \) nm, so x = 3.

Teacher's Note:
1. 1 micrometre equals 1000 nm.

Answer: (b) 2 and 3 only

Explanation:
1. Micelles form only above the critical micelle concentration and above the Kraft temperature, so statement 1 is false.
2. Statement 2 is true. On dilution below the CMC micelles break into ions, so 3 is true.
3. Soaps behave as normal electrolytes only below the CMC, so 4 is false.

Teacher's Note:
1. CMC and Kraft temperature are the two conditions for micelle formation.

Answer: (b) 2 and 3

Explanation:
1. Physical adsorption increases with the ease of liquefaction, i.e. with a high critical temperature and strong van der Waals forces.
2. Hydrogen has very weak van der Waals forces and a very low critical temperature, so it is adsorbed only slightly (2 and 3).

Teacher's Note:
1. Easily liquefiable gases such as ammonia and carbon dioxide are adsorbed much more than hydrogen.

Answer: (b) 2 and 4

Explanation:
1. Oppositely charged sols precipitate each other, so 1 is false.
2. Similar charges cause repulsion and so stabilise a colloid (2 true).
3. Too much dispersed phase breaks an emulsion (3 false).
4. Brownian motion opposes settling under gravity and stabilises the sol (4 true).

Teacher's Note:
1. Colloidal stability comes from like charges and Brownian motion.

Answer: (a) 1 and 4

Explanation:
1. A negatively charged emulsion is coagulated by cations of an electrolyte.
2. KCl and NaCl are electrolytes, whereas glucose and urea are non-electrolytes and have no effect.

Teacher's Note:
1. Only electrolytes can coagulate charged colloids.

Answer: (c) 3 and 4

Explanation:
1. Lyophilic (reversible) colloids are stabilised by a solvent layer around the particles as well as charge, so they are very stable.
2. Lyophobic (irreversible) colloids are readily coagulated by small amounts of electrolyte.

Teacher's Note:
1. Reversible and lyophilic colloids describe the same stable class.

Answer: (a) 1 and 4

Explanation:
1. The porous chalk absorbs the solvent of the ink and rises up by capillary action.
2. The coloured substance is adsorbed on the surface of the chalk and so stays near the bottom (1 and 4).

Teacher's Note:
1. The solvent is absorbed whereas the dye is adsorbed.