JEE Chemistry Surface Chemistry MCQs Set 13

Answer: (c) 18 mg

Explanation:
1. Fall in normality = 0.06 - 0.042 = 0.018 N; milli-equivalents lost = \( 0.018\times 50 = 0.9 \).
2. Mass = \( 0.9\times 60 = 54 \) mg for 3 g, i.e. 18 mg per gram.

Teacher's Note:
1. Equivalent mass of acetic acid = 60.
2. Divide by the mass of charcoal to get the amount per gram.

Answer: (d) 8

Explanation:
1. Millimoles = \( 2\times 1 = 2 \) mmol in 250 mL.
2. Per litre: \( 2\times 1000/250 = 8 \) mmol/L.

Teacher's Note:
1. Coagulating value is mmol per litre of sol.

Answer: (b) 8 mg

Explanation:
1. Millimoles in 20 mL = \( 3.33\times 20/1000 = 0.0666 \) mmol.
2. Mass = \( 0.0666\times 120.4 = 8.0 \) mg.

Teacher's Note:
1. Scale the flocculation value to the volume, then multiply by the molar mass.

Answer: (c) \( 3.3\times10^{16} \)

Explanation:
1. Mass adsorbed = \( 1\text{ ppm}\times 20\text{ g} = 2\times10^{-5} \) g (20 mL of dilute solution is about 20 g).
2. Molecules = \( 2\times10^{-5}\times 6.022\times10^{23}/180 = 6.7\times10^{16} \).
3. Per cm\( ^2 \) of the 2 cm\( ^2 \) total area: \( 3.3\times10^{16} \).

Teacher's Note:
1. 1 ppm = 1 mg per kg of solution = \( 10^{-6} \) g per g.
2. Count both faces of the paper.

Answer: (d) 10

Explanation:
1. Total volume = 10 mL; \( [\ce{Ba^2+}] = 0.1\times 1/10 = 0.01 \) M = 10 mmol/L.

Teacher's Note:
1. Dilute on mixing: 1 mL in 10 mL.
2. The effective ion is the cation (sol is negative).

Answer: (a) 2

Explanation:
1. Molecules needed = \( 500\times10^{-4}/(0.2\times10^{-18}) = 2.5\times10^{17} \), i.e. \( 4.15\times10^{-7} \) mol.
2. Concentration = 5.12/256 = 0.02 mol/L, so \( V = 4.15\times10^{-7}/0.02 = 2.08\times10^{-5} \) L.
3. \( 10^5 V = 2.08 \approx 2 \).

Teacher's Note:
1. Area of a monolayer = number of molecules x area per molecule.

Answer: (c) 60

Explanation:
1. 1 mm\( ^3 \) = \( 10^{-6} \) L holds \( 10^{-9} \) mol = \( 6.022\times10^{14} \) units.
2. Per particle: \( 6.022\times10^{14}/10^{13} \approx 60 \).

Teacher's Note:
1. Convert 1 mm3 to litres: 1e-6 L.

Answer: (a) \( -30 \) kJ/mol; physical adsorption

Explanation:
1. Moles adsorbed = \( 10\times10^{-3}/100 = 10^{-4} \) mol.
2. \( \Delta H = -3\text{ J}/10^{-4}\text{ mol} = -3\times10^{4} \) J/mol = -30 kJ/mol.
3. A value of 20-40 kJ/mol is typical of van der Waals (physical) adsorption; chemisorption is 80-240 kJ/mol.

Teacher's Note:
1. Heat released means \( \Delta H \) is negative.
2. Physisorption about 20-40 kJ/mol.

Answer: (c) \( 5\times10^{-19} \) m\( ^2 \)

Explanation:
1. Moles adsorbed = \( (0.50-0.49)\times 0.100 = 10^{-3} \) mol, i.e. \( 6.022\times10^{20} \) molecules.
2. Area per molecule = \( 3.01\times10^{2}/6.022\times10^{20} = 5.0\times10^{-19} \) m\( ^2 \).

Teacher's Note:
1. Monolayer: total area divided by the number of molecules.

Answer: (d) NaCl > \( \ce{Na2SO4} \) > \( \ce{Na3PO4} \)

Explanation:
1. Migration towards the cathode means the sol is positive, so the anion coagulates it.
2. Higher anionic charge means higher coagulating power and so a lower coagulating value: \( \ce{PO4^3-} \lt \ce{SO4^2-} \lt \ce{Cl-} \) in coagulating value.

Teacher's Note:
1. Coagulating value is the reverse of coagulating power.

Answer: (b) the sols coagulate each other (mutual coagulation)

Explanation:
1. With excess iodide the \( \ce{AgI} \) particles adsorb \( \ce{I-} \) and are negative; with excess silver ion they adsorb \( \ce{Ag+} \) and are positive.
2. Oppositely charged sols neutralise each other on mixing and precipitate.

Teacher's Note:
1. Mutual coagulation needs oppositely charged sols in suitable proportion.

Answer: (d) \( \ce{S^2-} \)

Explanation:
1. The sol is made with excess \( \ce{H2S} \); \( \ce{S^2-} \) is the ion common to \( \ce{H2S} \) and \( \ce{As2S3} \) and is preferentially adsorbed, giving a negative sol.

Teacher's Note:
1. Preferential adsorption of the common ion decides the charge of the sol.

Answer: (c) raises the activation energy of the reaction

Explanation:
1. A negative catalyst (inhibitor) provides a path of higher activation energy, so the reaction slows.
2. It does not change the equilibrium constant.

Teacher's Note:
1. Inhibitor example: glycerol or acetanilide for \( \ce{H2O2} \) decomposition.

Answer: (b) osmosis

Explanation:
1. Water moves out of the skin cells through a semipermeable membrane into the more concentrated salt water.

Teacher's Note:
1. Osmosis is the flow of solvent through a semipermeable membrane towards the higher solute concentration.

Answer: (a) emulsification

Explanation:
1. Soap molecules form micelles with hydrocarbon tails towards grease and polar heads towards water, dispersing the grease as a stable emulsion.

Teacher's Note:
1. Soaps and detergents: cleaning action is based on micelle formation and emulsification.

Answer: (c) poison the catalyst partially so that the aldehyde is not reduced further

Explanation:
1. \( \ce{BaSO4} \) deactivates (poisons) palladium, preventing over-reduction of the aldehyde to the alcohol.

Teacher's Note:
1. A catalytic poison decreases catalytic activity by preferential adsorption.
2. Promoters increase activity.

Answer: (a) promoter

Explanation:
1. Mo increases the activity of iron by increasing the number of active sites on its surface.

Teacher's Note:
1. Promoter: raises activity. Poison: lowers it (CO, \( \ce{H2S} \) for iron).

Answer: (b) At very high pressure \( x/m \) becomes constant, equal to \( a/b \)

Explanation:
1. At high pressure \( bP \gg 1 \) so \( x/m \to a/b \) (monolayer saturated).
2. At low pressure \( bP \ll 1 \) and \( x/m \approx aP \).

Teacher's Note:
1. Langmuir: monolayer saturation at high pressure.

Answer: (b) \( 1/a \) and \( b/a \)

Explanation:
1. Inverting \( x/m = aP/(1+bP) \): \( m/x = \dfrac{1}{aP} + \dfrac{b}{a} \).
2. So the slope against \( 1/P \) is \( 1/a \) and the intercept is \( b/a \).

Teacher's Note:
1. Invert the isotherm to get a straight line.

Answer: (a) zeta (electrokinetic) potential

Explanation:
1. The electrical double layer gives a potential difference called zeta potential, which keeps lyophobic sols stable by mutual repulsion.

Teacher's Note:
1. Larger zeta potential means a more stable lyophobic sol.

Answer: (d) The freezing point of a colloidal solution is lower than that of a true solution of the same mass concentration

Explanation:
1. A colloid contains far fewer particles than a true solution of the same mass concentration, so its colligative effects are much smaller.
2. The freezing-point depression is therefore smaller, and the freezing point is higher, not lower.

Teacher's Note:
1. Colligative properties of colloids are very small because the number of particles is small.

Answer: (b) Combustion of coal

Explanation:
1. Ostwald's process uses Pt-Rh, Haber's iron and hydrogenation nickel, all solid catalysts with gaseous or liquid reactants.
2. Combustion of coal is an uncatalysed reaction.

Teacher's Note:
1. Solid catalyst with gas or liquid reactants = heterogeneous.

Answer: (a) Latex is a colloidal solution of positively charged rubber particles

Explanation:
1. Rubber latex is a negatively charged sol.
2. The other statements are standard facts about colloids.

Teacher's Note:
1. Latex is negatively charged and coagulates with acids and electrolytes.