Surface Chemistry

Tyndall effect is observed only when

(i) The diameter of the dispersed particle is not much smaller than the wavelength of the light used.

(ii) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.

So, (b) and (d) are correct.

Gold number of a protective colloid is a minimum weight of it in milligrams which must be added to 10 ml of a standard red gold sol. so that no coagulation of the gold sol. (i.e. change of colour from red to blue) takes place when 1 ml of 10% sodium chloride solution is rapidly added to it. Obviously, smaller the gold number of a protective colloid, the greater is the protective actioin.

It is given that the gelatin has lowest gold number thus, it would serve as a better protective colloid than colloids in the rest of the options.

From the given characteristics, the adsorption is an example of physical adsorption.

(i) Physical adsorption is multimolecular layer.

(ii) It decreases with increase in temperature.

(iii) Enthalpy of adsorption is low 20–40 kJ mol⁻¹ for physical adsorption.

(iv) Energy of activation is low because in this type of adsorption no chemical reaction takes place.

Only the transparent part of an egg has albumin.

According to Freundlich adsorption isotherm

$\log {x \over m} = \log K + {1 \over n}\log \,P$

Thus if a graph is plotted between log $(x/m)$

and log $P,$ a straight line will be obtained



The slope of the line is equal to $1/n$ and the intercept on log $x/m$ axis is will correspond to $log$ $K. $

Let the weight of acetic acid initially be

${w_1}$ in $50$ $ml$ of $0.060$ $N$ solution.

Let the $N = {{{w_1} \times 1000} \over {M.wt. \times 50}}\,\,\,$

(Normality $=0.06$ $N$)

$0.06 = {{{w_1} \times 1000} \over {60 \times 50}}$

$ \Rightarrow \,\,\,\,\,{w_1} = {{0.06 \times 60 \times 50} \over {1000}}$

$ = 0.18\,g = 180\,mg.$

After an hour, the strength of acetic acid

$=0.042$ $N$

so, let the weight of acetic acid be ${w_2}$

$N = {{{w_2} \times 1000} \over {60 \times 50}};0.042 = {{{w_2} \times 1000} \over {3000}}$

$ \Rightarrow \,\,\,\,\,\,\,\,\,\,{w_2} = 0.126g = 126mg$

So amount of acetic acid adsorbed per $3g$

$ = 180 - 126\,mg = 54\,mg$

Amount of acetic acid adsorbed per $g$

$ = {{54} \over 3} = 18\,mg$

According to Hardly Schulze rule, greater the charge or cation, greater is its coagulating power for negatively charged solution. Hence the correct order of coagulating power : $N{a^ + } < B{a^{2 + }} < A{l^{3 + }}$

The Freundlich adsorption isotherm is mathematically represented as

${x \over m} = k{P^{1/n}}$

at high pressure ${1 \over n}$ = 0 Hence, $x\,/\,m \propto {p^ \circ }$

at low pressure ${1 \over n}$ = 1 Hence, $x/m\,\, \propto \,\,p'$

In adsorption when two molecule comes closer then bond is created between the molecules and we know when bond is created between molecules the energy gets released so, Adsorption is an exothermic process, hence $\Delta H$ will always be negative .

For a protective colloid $\mu $ lesser the value of gold number better is the protective power.

Thus the correct order of protective power of $A,B,C$ and $D$ is

$ \Rightarrow $ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $\left( A \right)\,\, < \,\,\left( C \right)\,\, < \,\,\left( B \right)\,\, < \,\,\left( D \right)$

Gold number $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.50\,\,\,\,\,\,0.10\,\,\,\,\,\,0.01\,\,\,\,\,\,0.005$

Hence $(c)$ is the correct answer

According to Langmuir's Model of adsorption of gas on a solid surface the mass of gas adsorbed(x) per gram of the adsorbent (m) is directly proportional to the pressure of the gas $(p)$ at constant temperature i.e. $${x \over m} \propto p$$

Particle size of colloidal particle

$ = 1\,m\mu \,\,$ to $100\,m\mu $ (suppose $10\,m\mu $ )

${V_C} = {4 \over 3}\pi {r^3} = {4 \over 3}\pi {\left( {10} \right)^3}$

Particle size of true solution particle $ = 1\,m\mu $

$Vs = {4 \over 3}\pi {\left( 1 \right)^3}$

hence now ${{Vc} \over {Vs}} = {10^3}$

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

As adsorption is an exothermic process.

$\therefore$ Rise in temperature will decrease adsorption (according to Le-chatelier principle).

It is zero order reaction

[NOTE : Adsorption of gas on metal surface is of zero order]

Alum furnishes $A{l^{3 + }}$ ions which bring about coagulation of negatively charged clay particles, bacteria etc.