Surface Chemistry

According to Freundlich adsorption isotherm,

$ \begin{aligned} \frac{x}{m} & =k p^{1 / n} \\ \frac{10}{500} & =k(10)^{1 / n} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \frac{14}{500} & =k(20)^{1 / n} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ \frac{10}{14} & =\left(\frac{10}{20}\right)^{1 / n} \\ 0.71 & =(0.5)^{1 / n} \\ \log (0.71) & =\log (0.5)^{1 / n} \\ -0.14 & =1 / n \log (0.5) \\ & =1 / n(-0.3) \\ \frac{1}{n} & =\frac{0.14}{0.3}=0.46 .=0.5 \end{aligned} $

Moles of HCl added to coagulate 200 mL of colloid

$ \begin{aligned} & =\frac{\text { mass of } \mathrm{HCl} \text { taken }}{\text { molecular mass of } \mathrm{HCl}} \\ & =\frac{0.73}{(1+35.5)}=0.02 \mathrm{~mol} \\ & =20 \mathrm{~m} \mathrm{~mol} \end{aligned} $

Now, we know

Flocculation value is the minimum number of moles of electrolyte per litre required to coagulate a colloid.

200 mL of sol required 20 mmol of HCl

So, 1000 mL of sol required

$ =\frac{20}{200} \times 1000=100 \mathrm{~m} \mathrm{~mol} $

Thus, the flocculation value of HCl for the colloid is 100 .

Tyndall effect is light scattering by particles in colloid. The particles size is smaller than that of suspension but greater than solution that means Tyndall effect is shown only by colloidal solutions. Out of given options, clouds, milk and suspension are colloidal solutions. So, they can show Tyndall effect. But sugar solution is a pure solution so, it does not show Tyndall effect.

$1 \mathrm{~cm}=10^7 \mathrm{~nm}$

Edge length of original cube $=10^7 \mathrm{~nm}$

Volume of original cube $(V)=\left(10^7\right)^3 \mathrm{~nm}$

Edge length of new cube $=1 \mathrm{~nm}$

Volume of new cube $=(1)^3 \mathrm{~nm}$

If the number of new cubes formed is $n$, then volume of total new cubes will be $\left(V_1\right)=n \times(1)^3 \mathrm{~nm}$

But $V=V_1$

$ \therefore\left(10^7\right)^3=n \times(1)^3 \therefore n=10^{21} $

The surface area of the original cube

$ (S)=6 \times\left(10^7\right)^2 \mathrm{~nm}^2=6 \times a^2 $

The surface area of the total cubes formed after crushing the original cube

$ \begin{aligned} & =\left(S_1\right)=6 \times 10^{21} \times(1)^2 \mathrm{~nm}^2 \\ \frac{S_1}{S} & =\frac{6 \times 10^{21} \times(1)^2}{6 \times\left(10^7\right)^2}=10^7 \end{aligned} $

I. Esterification reaction is catalysed by mineral acid such as HCl (l).

II. Conversion of $\mathrm{SO}_2$ into $\mathrm{SO}_3$ is an important reaction in manufacture of sulphuric acid which is catalysed by $\mathrm{Pt}(s)$.

III. In lead chamber process conversion of $\mathrm{SO}_2$ into $\mathrm{SO}_3$ by reaction with $\mathrm{O}_2$ is catalysed by $\mathrm{NO}(\mathrm{g})$.

IV. In Haber's process ammonia is manufactured by reacting $\mathrm{N}_2$ and $\mathrm{H}_2$ in presence of catalysed $\mathrm{Fe}(\mathrm{s})$.

According to Hardy Schulze rule, greater the valency of the coagulating ion, greater is its coagulating power. $\mathrm{As}_2 \mathrm{~S}_3$ is a negatively charged sol. For coagulation, oppositely charged sol is needed. Thus, out of the given, $\mathrm{FeCl}_3\left(\mathrm{Fe}^{3+}\right)$ is most effective for causing coagulation of $\mathrm{As}_2 \mathrm{~S}_3$ sol.

Coagulating power $\propto$ Charge

Charge on $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4^{-}}>\mathrm{SO}_4^{2-}>\mathrm{Cl}^{-}$

$\therefore$ Correct order of coagulating power is

$\text { I }>\text { III }>\text { II }$

Macromolecular colloids are the colloids in which macromolecules form a solution. The size of the particles lies in the range of colloidal particles size. Starch solution is a naturally occurring macromolecular colloid. Synthetic rubber is a synthetic or artificial macromolecular colloid,

Animal skin is colloidal in nature and has positively charged particles. When animal skin is soaked in tannin, which contains negatively charged colloidal particles, mutual coagulation takes place. This results in the hardening of leather. Both Assertion and Reason are correct but Reason is not a correct explanation of Assertion.

A colloid is a mixture in which one of the substances split into very minute particles which are dispersed throughout a second substance. The size of colloid particles generally ranges between 1 nm to 1000 nm .

When AgNO₃ is added to KI,

Iodide ion gets adsorbed on the surface of AgI. This will result in formation of negatively charged colloids.

When KI is added to AgNO₃ following reaction takes place.

Silver cation gets adsorbed on surface of AgI, resulting in formation of positively charged colloids.

According to Hardy-Schulze rule, for negatively charged sol, most (+ve) charged ion is needed for efficient coagulation.

Blood is a negatively charged sol. Hence FeCl₃ can be used for blood clotting and it from Fe³⁺ ion.

Statements (b) and (c) are correct whereas statements (a) and (d) are incorrect.

(a) Process of precipitating colloidal sol is known as coagulation.

(b) Concentration of colloidal solutions is small due to large molar mass and thus, their colligative properties are very small compared to true solutions. Hence, these solution freezes at higher temperature.

(c) Micelles are formed at the critical micelle concentration (CMC) which depends to temperature.

(d) Micelles and macromolecular colloids are two different types of colloid.

Statements (1), (2) and (3) are correct.

(1) Adsorption isotherm is a curve that expresses the variation in the amount of gas adsorbed by adsorbent with temperature at constant pressure. It fails at high pressure.

(2) During adsorption, enthalpy and entropy of system are negative and $\Delta$G must be negative, so, process is spontaneous.

(3) Physical adsorption is non-specific in nature as there are more significantly strong attraction between solute and solvent. Any gas can be adsorbed on surface of solid to some extant.

(4) Due to presence of strong attraction between solute and solvent, chemical adsorption is irreversible while physisorption is reversible.

$\begin{aligned} & \frac{x}{m}=k p^{1 / n} \text { [adsorption isotherm] } \\ & \log \frac{x}{m}=\log k+\frac{1}{n} \log p \end{aligned}$

$\therefore$ Plot of $\log \frac{x}{m} v s \log p$ is linear with slope $=\frac{1}{n}$ and intercept $=\log k$

Hence, $l / n=\tan \theta=\tan 45 \Upsilon=1$ i.e. $n=1$

$\log k=0.3010$

or $k=\operatorname{antilog}(0.3010)=2$

At $p=0.5 \mathrm{~atm}$,

$x / m=k p^{1 / n} \Rightarrow x / m=2 \times(0.5)^1=1.0$

Partition or ascending paper chromatography is used to separate coloured chemicals or substance. It is an analytical technique in which the mobile phase moves upward through the stationary phase.

Mobile phase $\to$ Liquid (solvent)

Stationary phase $\to$ Liquid (Whatmann, paper contain cellulose).

While in gas- liquid chromatography, gas is stationary phase and liquid is mobile phase, TLC contain silica which is solid (stationary phase) and solvent (mobile phase).

Lyophilic colloid or sol are unstable, so they are stabilised by adding some protective which protect them from precipitation. Thus, lyophilic colloids are protecting colloids. Their protecting power is denoted in terns of gold number.

It defined as the minimum amount of lyophilic colloid in mg which present the flocculation of 10 mL gold sol by the addition of 1mL of 10% NaCl solution. Lesser the gold number, higher the protecting power.

The empirical relation $x / m=k p^{1 / n}$ put forward by Freundlich is known as Freundlich adsorption isotherm.

Taking logarithm, $\log \frac{x}{m}=\log k+\frac{1}{n} \log p$

Compare with, $y=m x+c$

Slope $(m)=1 / n$

$y \text {-axis }=\log \frac{x}{m} ; x \text {-axis }=\log p$