Surface chemistry practice · Easy

Option D (Chorcoal) is correct because charcoal is a good adsorbent and can adsorb methylene blue dye, causing it to become colorless.

Option A (Oxidation of ammonia to form nitric acid) is correct because Platinum acts as a catalyst in the Ostwald process for the oxidation of ammonia to nitric oxide, which is a precursor to nitric acid.

Option A (Glucose) is correct because the hydrolysis of maltose by maltase yields two molecules of glucose.

Option A (Gel formation) is correct because cementation involves the formation of a gel-like structure as water molecules are held within the cement particles.

Option C is correct because in an aerosol, the dispersion medium is a gas, and the dispersed phase can be either solid or liquid.

Option A (Saliva) is correct because ptyalin (salivary amylase) is an enzyme present in saliva used for the digestion of food.

Option B (Formaldehyde solution) is correct because formaldehyde acts as a preservative for milk by inhibiting bacterial growth.

Option D (All of the above) is correct because all the given statements accurately describe the characteristics of zeolites.

Option A ($\frac{x}{m} = K{p^{0.3}}$) is correct because the Freundlich adsorption isotherm is represented by $\frac{x}{m} = K{p^{1/n}}$, where 0 < 1/n < 1. 0.3 falls within this range.

Option B (Pt) is correct because platinum (Pt) is used as a catalyst in the Ostwald process for the manufacture of nitric acid.

Option B (Sol A is positively charged and Sol B is negatively charged) is correct because a lower flocculation value for BaCl2 compared to KCl indicates that sol A is negatively charged and a lower flocculation value for Na2SO4 compared to NaBr indicates that sol B is positively charged.

The correct answer is D (Negatively as well as positively charged particles) because blood contains both positively and negatively charged ions, but its overall charge is considered negative due to the presence of negatively charged proteins.

Option D is correct because, according to the Freundlich adsorption isotherm, the slope of the plot of $\log \frac{x}{m}$ vs $\log p$ is equal to $\frac{1}{n}$.

Option D (Clouds) is correct because clouds are aerosols, not emulsions. Emulsions have both dispersed phase and dispersion medium in liquid state.

Option A (Cellulose nitrate in a alcohol-ether) is correct because collodion is a solution of cellulose nitrate in a mixture of alcohol and ether, used for various applications including preparing ultrafilters.

Option B ($\Delta H$ is of the order of 40 KJ) is correct because the enthalpy change for chemisorption is typically much larger than 40 kJ/mol.

Option B (${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}$) is correct because ethanol (${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}$) acts as a negative catalyst to retard the oxidation of chloroform, preventing the formation of phosgene.

Option A (May be defined as the milligram of the dry material of which the hydrophilic sol is prepared and which when added to 10 mL of red gold sol, will prevent it from coagulation on the addition of 1 mL of 10 per cent sodium chloride solution) is correct because it accurately describes the definition of gold number.

Option B ($\log \frac{X}{m} = \log k + \frac{1}{n}\log p$) is correct because the Freundlich adsorption isotherm is expressed as $\frac{X}{m} = k{p^{1/n}}$, which can be written in logarithmic form as $\log \frac{X}{m} = \log k + \frac{1}{n}\log p$.

Option C (Synthetic gems) is correct because synthetic gems are a solid dispersed in another solid, forming a solid-solid system.

Option C (Bredig’s Arc method) is correct because Bredig's arc method is a common method for preparing colloidal solutions of metals like gold.

Option B is correct. Peptisation is the process of converting a precipitate into a colloidal solution by shaking it with a dispersion medium in the presence of a small amount of electrolyte.

Option D (Zero activation energy) is correct because chemical adsorption involves the formation of chemical bonds, which requires a significant activation energy.

Option C (Solid dispersed in gas) is correct because smoke consists of solid particles (like carbon) dispersed within a gaseous medium (like air).

Option C (Size of particles) is correct because the efficiency of a catalyst is highly dependent on its surface area, which is related to the size of its particles.

Option C (Viscosity is of the order of that of water) is the false statement because hydrophilic sols have a much higher viscosity than water.

Option B is correct. A catalyst in finely divided form has a larger surface area, leading to more active sites where reactions can occur.

Option C (Hydrogenation of oils requires iron as catalyst) is correct because hydrogenation of oils typically uses nickel (Ni) as a catalyst, not iron.

Option C ($\mathrm{SO}_2$) is correct because readily liquefiable gases like sulfur dioxide are more easily adsorbed by activated charcoal due to stronger van der Waals forces.

Option A (Heterogeneous catalyst) is correct because the catalyst (platinized asbestos) is in a different phase (solid) than the reactants (gases).