Surface chemistry practice · Mixed difficulty

Option C (Coagulation) is correct because alum purifies muddy water through the process of coagulation, where it neutralizes the charge on the colloidal particles, causing them to clump together and settle down.

Option C (A gel) is correct because a gel is defined as a colloidal solution where the dispersed phase is a liquid and the dispersion medium is a solid.

Option C (To act as a catalyst promoter and increase activity of catalyst) is correct because molybdenum acts as a promoter in Haber's process, enhancing the iron catalyst's efficiency.

Option B (1) is correct because 1 mole of lead nitrate ($Pb^{2+}$) is required to coagulate 2 moles of colloidal $[AgI]I^-$.

Option A (Petrochemical industries during cracking) is correct because zeolites are commonly used as catalysts in cracking processes within the petrochemical industry.

Option D (Cetyltrimethylammonium bromide) is correct because it is a cationic surfactant that forms cationic micelles in solution.

Option B (${\mathrm{V}_2}{\mathrm{O}_5}$) is correct because vanadium(V) oxide (${\mathrm{V}_2}{\mathrm{O}_5}$) is the catalyst used in the contact process for the manufacture of sulphuric acid.

Option A (Multilayers are formed) is correct because after a certain limit, the adsorbent surface is covered, and further adsorption leads to multilayer formation.

Option D (Water vapours) is correct because activated charcoal has a high affinity for water vapor due to its polar nature and ability to form hydrogen bonds.

Option C (Vanadium pentoxide or Pt (asbestos)) is correct because these are the catalysts used in the contact process for sulfuric acid production.

Option B ($100\,ml\,\,0.2\,M\,AgN{O_3} + 100\,ml\,\,0.1\,M\,KI$) is correct because an excess of $AgNO_3$ leads to the adsorption of $Ag^+$ ions, resulting in a positively charged colloid.

Option D (Adsorption) is correct. Activated charcoal removes coloring matter through adsorption, where the colored substances adhere to the surface of the charcoal.

Option A (Dispersion of a solid or liquid in a gas) is correct because an aerosol is defined as a colloidal dispersion of solid particles or liquid droplets in a gas.

Option C (Increases the rate of forward and backward reaction equally) is correct because a catalyst lowers the activation energy for both the forward and reverse reactions to the same extent, thus increasing the rate of both equally.

Option A (Linseed oil) is correct because linseed oil is a common and relatively inexpensive oil used in soap making.

Option D (25) is correct because the gold number is defined as the amount of protective colloid in milligrams required to prevent coagulation of 10 cm³ of gold sol.

Option B (Smoke, fog and $\mathrm{S}{\mathrm{O}_{\mathrm{2}}}$) is correct because reducing smog is a mixture of smoke, fog, and sulfur dioxide.

Option A (Milk is an emulsion of fat in water) is correct because milk consists of fat droplets dispersed in a water-based solution.

Option D (Positively and negatively charged sols, respectively) is correct because hemoglobin is a positively charged sol, while gold sol is a negatively charged sol.

Option D is correct because in physical adsorption (physisorption), gas molecules are held on the solid surface by weak Van der Waals' forces.

Option B (Adsorption) is correct because charcoal removes the color of litmus solution through adsorption, where litmus molecules adhere to the charcoal's surface.

Option B is correct because colloidal solutions are heterogeneous mixtures, not homogeneous.

Option A (Solid sol) is correct because pumice stone is a solid with gas dispersed within its solid matrix, which defines a solid sol.

Option D (An inhibitor) is correct because a catalytic poison acts as an inhibitor, reducing or eliminating the activity of a catalyst.

Option A (Urease) is correct because urease is the enzyme that catalyzes the hydrolysis of urea into ammonia and carbon dioxide.

Option C (Acid catalyst) is correct because AlCl3 acts as a Lewis acid to generate the electrophile in Friedel-Crafts reactions.

Option D ($\mathrm{M}{\mathrm{g}_3}{\left( {\mathrm{P}{\mathrm{O}_4}} \right)_2}$) is correct because according to the Hardy-Schulze rule, ions with higher charges are more effective in coagulation, and $PO_4^{3-}$ has the highest negative charge.

Option A (Peptization) is correct because peptization is the process of converting a precipitate into a colloidal solution by adding a suitable electrolyte.

Option C (Both (1) and (2)) is correct because negative catalysts can either remove active intermediates or terminate chain reactions.

Option A (Synerisis) is correct because synerisis is the process where a gel contracts and exudes liquid.