Surface Chemistry MCQ Questions & Answers in Physical Chemistry | Chemistry

With excess of $$KI,$$  colloidal particles will be $$\left[ {AgI} \right]{I^ - }$$
\[\underset{1\,\,mol}{\mathop{\left[ AgI \right]{{I}^{-}}}}\,+\underset{1\,\,mol}{\mathop{AgN{{O}_{3}}}}\,\to AgI\downarrow +\]     \[AgI\downarrow +NO_{3}^{-}\]
\[\underset{\begin{smallmatrix} 2\,\,mol \\ 1\,\,mol \end{smallmatrix}}{\mathop{2\left[ AgI \right]{{I}^{-}}}}\,+\underset{\begin{smallmatrix} 1\,\,mol \\ \frac{1}{2}\,mol \end{smallmatrix}}{\mathop{Pb{{\left( N{{O}_{3}} \right)}_{2}}}}\,\to \]     \[2AgI\downarrow +Pb{{I}_{2}}\downarrow +2NO_{3}^{-}\]
\[\underset{\begin{smallmatrix} 3\,\,mol \\ 1\,\,mol \end{smallmatrix}}{\mathop{3\left[ AgI \right]{{I}^{-}}}}\,+\underset{\begin{smallmatrix} 1\,\,mol \\ \frac{1}{3}\,mol \end{smallmatrix}}{\mathop{Fe{{\left( N{{O}_{3}} \right)}_{3}}}}\,\to \]     \[3AgI\downarrow +Fe{{I}_{3}}\downarrow +3NO_{3}^{-}\]
∴ Molar ratio required for coagulation of same amount of $$\left[ {AgI} \right]{I^ - }\,\,{\text{is}} = 1:\frac{1}{2}:\frac{1}{3} = 6:3:2$$

Critical temp. $$ \uparrow $$ Attractive force $$ \uparrow $$ Extent of adsorption $$ \uparrow .$$

Gases, liquids or solids can be adsorbed on the solid surfaces.

Hydrolysis of urea can be represented as follows

Since it involves biological catalyst (enzyme), so it is an example of biochemical catalysis.

According to Le-Chatelier's principle, exothermic reactions are favoured at low temperature.

$$\eqalign{ & {\text{Particle size of colloidal particle = 1}}m\mu {\text{ to 100 }}m\mu \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{(suppose 10 }}m\mu ) \cr & {{\text{V}}_C} = \frac{4}{3}\pi {r^3} = \frac{4}{3}\pi {\left( {10} \right)^3} \cr & {\text{Particle size of true solution particle = }}1m\mu \cr & {\text{Vs}}\,{\text{ = }}\frac{4}{3}\pi {\left( 1 \right)^3} \cr & {\text{hence}}\,{\text{now}}\,\,\frac{{Vc}}{{Vs}} = {10^3} \cr} $$

When oppositely charged sols are mixed their charges are neutralised. Both sols may be partially or completely precipitated.

Desorption removes products from the surface to create space for other reactants.

Aqueous solution of slaked lime is called lime water whereas suspension solution of slaked lime is called milk of lime.

$$\eqalign{ & \frac{x}{m} = k{p^{\frac{1}{n}}} \cr & \ln \frac{x}{m} = \ln k + \frac{1}{n}\ln \,p\,\,\,\,\left[ {\because \,{\text{slope}} = \frac{1}{n} = 0} \right] \cr & {\text{Thus,}}\,\frac{x}{m} = k{p^0} \Rightarrow \frac{x}{m} = k \cr} $$