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111.
Which of the following statements is not correct for chemisorption and physisorption ?
A
Physical adsorption occurs at a low temperature and chemisorption occurs at all temperatures.
B
Magnitude of chemisorption decreases with rise in temperature while physisorption increases with rise in temperature.
C
Chemisorption is irreversible and physisorption is reversible.
D
In physisorption activation energy is low while in chemisorption it is high.
Answer :
Magnitude of chemisorption decreases with rise in temperature while physisorption increases with rise in temperature.
Chemisorption increases and physisorption decreases with rise in temperature.
112.
Which property of colloidal solution is independent of charge on the colloidal particles?
A
Coagulation
B
Electrophoresis
C
Electroosmosis
D
Tyndall effect
Answer :
Tyndall effect
Coagulation is generally brought about by the addition of electrolytes. When an electrolyte is added to a colloidal solution, the particles of the sol take up the ions which are oppositely charged. As a result their charge gets neutralised. Electrophoresis The movement of colloidal patticles under an applied electric potential is called electrophoresis. Electroosmosis may be defined as a phenomenon in which the molecules of the dispersion medium are allowed to more under the influence of an electric field whereas colloidal particles are not allowed to more. Tyndall effect is the scattering of light by sol particles, which cannot be affected by charge on them.
113.
Method by which lyophobic sol can be protected
A
by addition of oppositely charged sol
B
by addition of an electrolyte
C
by addition of lyophilic sol
D
by boiling
Answer :
by addition of lyophilic sol
Lyophilic colloids have a unique property of protecting lyophobic colloids. When a lyophilic sol is added to the lyophobic sol, the lyophilic particles form a layer around lyophobic particles and thus protect the latter from electrolytes. Lyophilic colloids used for this purpose are called protective colloids
114.
Colloidal gold is prepared by
A
Mechanical dispersion
B
Peptisation
C
Bredig’s Arc method
D
Hydrolysis
Answer :
Bredig’s Arc method
Colloidal gold is prepared by Bredig's arc method.
115.
Position of non-polar and polar parts in micelle is
A
polar at outer surface but non-polar at inner surface
B
polar at inner surface but non-polar at outer surface
C
distributed all over the surface
D
present in the surface only
Answer :
polar at outer surface but non-polar at inner surface
Micelles are the clusters formed by the association of colloids. They are formed by lyophilic and lyophobic groups. As the concentration increases, the lyophobic parts receding away from the solvent approach each ( non-polar part ) other and form a cluster, the lyophobic ends are in the interior, lyophilic groups ( polar part ) projecting outward in contact with the solvent.
116.
What happens when a lyophilic sol is added to a lyophobic sol?
A
Lyophilic sol is protected.
B
Lyophobic sol is protected.
C
Both the sols are coagulated.
D
Electrophoresis takes place.
Answer :
Lyophobic sol is protected.
Lyophobic sol is protected since a film of lyophilic sol is formed over lyophobic sol.
117.
Select the correct statements.
(i) Physical adsorption is weak, multilayer, non-directional and non-specific.
(ii) Chemical adsorption is unilayer, directional and strong.
(iii) Chemical adsorption decreases with temperature.
(iv) Chemical adsorption is more stronger than physical adsorption.
A
Only (i) and (iii)
B
Only (i), (ii) and (iv)
C
Only (iii)
D
All of these
Answer :
Only (i), (ii) and (iv)
Chemisorption first increases with temperature as it requires energy of activation.
118.
The separation of an emulsion into its constituent liquids is known as
A
emulsification
B
protection of colloid
C
coagulation
D
demulsification
Answer :
demulsification
The separation of an emulsion into its constituents is called demulsification. Various methods of demulsification are freezing, boiling, centrifugation or chemical methods.
119.
$$3 g$$ of activated charcoal was added to $$50 mL$$ of acetic acid solution $$(0.06N)$$ in a flask. After an hour it was filtered and the strength of the filtrate was found to be $$0.042 N.$$ The amount of acetic acid adsorbed ( per gram of charcoal ) is :
A
42 $$mg$$
B
54 $$mg$$
C
18 $$mg$$
D
36 $$mg$$
Answer :
18 $$mg$$
Let the weight of acetic acid initially be w1 in 50 ml of 0.060 N solution.
$$\eqalign{
& {\text{Let}}\,{\text{the}}\,N = \frac{{{w_1} \times 1000}}{{M.wt. \times 50}}\,\,\,\,\,\,\,\,\left( {{\text{Normality}} = 0.06\,N} \right) \cr
& 0.06 = \frac{{{w_1} \times 1000}}{{60 \times 50}} \cr
& \Rightarrow \,\,\,{w_1} = \frac{{0.06 \times 60 \times 50}}{{1000}} \cr
& = 0.18\,g \cr
& = 180\,mg. \cr} $$
After an hour, the strength of acetic acid = 0.042 N
so, let the weight of acetic acid be w2
$$\eqalign{
& N = \frac{{{w_2} \times 1000}}{{60 \times 50}} \cr
& 0.042 = \frac{{{w_2} \times 1000}}{{3000}} \cr
& \Rightarrow {w_2} = 0.126\,g \cr
& = 126\,mg \cr} $$
So amount of acetic acid adsorbed per 3g
= 180 -126 $$mg$$ = 54 $$mg$$
Amount of acetic acid adsorbed per g
$$ = \frac{{54}}{3} = 18mg$$
120.
Under ambient conditions, which among the following surfactants will form micelles in aqueous solution at lowest molar concentration ?
Greater the surface area, greater the van der waal forces of attraction and therefore at
lesser concentration micelle formation will take place. In case of $$C{H_3}{\left( {C{H_2}} \right)_{15}}\mathop N\limits^ \oplus {\left( {C{H_3}} \right)_3}B{r^ - }$$ due to greater chain length, greater will be van der waal forces.