Surface Chemistry

NEET Chemistry · 98 questions · Page 8 of 10 · Click an option or "Show Solution" to reveal answer

Q71
A particular adsorption process has the following characteristics : (i) It arises due to van der Waals forces and (ii) it is reversible. Identify the correct statement that describes the above adsorption process :
A Enthalpy of adsorption is greater than 100 kJ mol−1.
B Energy of activation is low.
C Adsorption is monolayer.
D Adsorption increases with increase in temperature.
Correct Answer
Option B
Solution

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.

Q72
Which of the following statements about colloids is False ?
A Freezing point of colloidal solution is lower than true solution at same concentration of a solute.
B Colloidal particles can pass through ordinary filter paper.
C When silver nitrate solution is added to potassium iodide solution, a negatively charged colloidal solution is formed.
D When excess of electolyte is added to colloidal solution, colloidal particle will be precipitated.
Correct Answer
Option A
Solution

Freezing point of colloidal solution is same as true solution at same concentration of a solute.

Q73
The enthalpy change for the adsorption process and micelle formation respectively are
A ΔHads>0 and ΔHmic<0\mathrm{\Delta H_{ads} > 0~and~\Delta H_{mic} < 0}
B ΔHads>0 and ΔHmic>0\mathrm{\Delta H_{ads} > 0~and~\Delta H_{mic} > 0}
C ΔHads<0 and ΔHmic<0\mathrm{\Delta H_{ads} < 0~and~\Delta H_{mic} < 0}
D ΔHads0\mathrm{\Delta H_{ads} 0}
Correct Answer
Option D
Solution

Adsorption is the process of a substance (adsorbate) being attracted to a surface (adsorbent).

The enthalpy change for adsorption, ΔHads\mathrm{\Delta H_{ads}}, is negative because the adsorbent and adsorbate are brought closer together, which releases energy.

Micelle formation is the process of surfactant molecules clustering together in water to form micelles.

The micelles have a hydrophobic (water-repelling) core and a hydrophilic (water-loving) surface.

The enthalpy change for micelle formation, ΔHmic\mathrm{\Delta H_{mic}}, is positive because the surfactant molecules are moving from a state of high entropy (dispersed in water) to a state of low entropy (clustered together).

Here is a table summarizing the enthalpy changes for adsorption and micelle formation: .tg .tg Process Enthalpy change Adsorption ΔHads<0\mathrm{\Delta H_{ads} < 0} Micelle formation ΔHmic>0\mathrm{\Delta H_{mic} > 0}

Q74
Adsorption of a gas on a surface follows Freundlich adsorption isotherm. Plot of log xm{x \over m} versus log p gives a straight line with slope equal to 0.5, then: ( xm{x \over m} is the mass of the gas adsorbed per gram of adsorbent)
A Adsorption is independent of pressure.
B Adsorption is proportional to the pressure.
C Adsorption is proportional to the square root of pressure.
D Adsorption is proportional to the square of pressure.
Correct Answer
Option C
Solution

According to Freundlich adsorption isotherm, in the median range of pressure

xmP1n{x \over m} \propto {P^{{1 \over n}}}

\Rightarrow

xm{x \over m}

= kP

1n^{{1 \over n}}

taking log both sides, we get, log

xm{x \over m}

= logk +

1n{1 \over n}

logP Here in graph between log

xm{x \over m}

and logP, slope is

1n{1 \over n}

and intercepts = log k. Given that, slope = 0.5

\therefore\,\,\,
1n{1 \over n}

= 0.5 \Rightarrow n = 2

\therefore\,\,\,
xm{x \over m}

= kP

12^{{1 \over 2}}

= k

P\sqrt P

\therefore Adsorption is proportional to the square root of pressure.

Q75
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 36 mg
B 42 mg
C 54 mg
D 18 mg
Correct Answer
Option D
Solution

Let the weight of acetic acid initially be

w1{w_1}

in

5050
mlml

of

0.0600.060
NN

solution. Let the

N=w1×1000M.wt.×50N = {{{w_1} \times 1000} \over {M.wt. \times 50}}\,\,\,

(Normality

=0.06=0.06
NN

)

0.06=w1×100060×500.06 = {{{w_1} \times 1000} \over {60 \times 50}}
w1=0.06×60×501000\Rightarrow \,\,\,\,\,{w_1} = {{0.06 \times 60 \times 50} \over {1000}}
=0.18g=180mg.= 0.18\,g = 180\,mg.

After an hour, the strength of acetic acid

=0.042=0.042
NN

so, let the weight of acetic acid be

w2{w_2}
N=w2×100060×50;0.042=w2×10003000N = {{{w_2} \times 1000} \over {60 \times 50}};0.042 = {{{w_2} \times 1000} \over {3000}}
w2=0.126g=126mg\Rightarrow \,\,\,\,\,\,\,\,\,\,{w_2} = 0.126g = 126mg

So amount of acetic acid adsorbed per

3g3g
=180126mg=54mg= 180 - 126\,mg = 54\,mg

Amount of acetic acid adsorbed per

gg
=543=18mg= {{54} \over 3} = 18\,mg
Q76
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R Assertion A: Amongst He,Ne,Ar\mathrm{He}, \mathrm{Ne}, \mathrm{Ar} and Kr\mathrm{Kr}; 1g\mathrm{g} of activated charcoal adsorbs more of Kr\mathrm{Kr}. Reason R: The critical volume Vc(cm3 mol1)\mathrm{V}_{\mathrm{c}}\left(\mathrm{cm}^{3} \mathrm{~mol}^{-1}\right) and critical pressure Pc\mathrm{P}_{\mathrm{c}} (atm) is highest for Krypton but the compressibility factor at critical point Zc\mathrm{Z}_{\mathrm{c}} is lowest for Krypton. In the light of the above statements, choose the correct answer from the options given below
A Both A and R are true and R is the correct explanation of A
B A is false but R is true
C Both A and R are true but R is NOT the correct explanation of A
D A is true but R is false
Correct Answer
Option D
Solution

MW order, Kr>Ar>Ne>He\mathrm{Kr}>\mathrm{Ar}>\mathrm{Ne}>\mathrm{He} ZZ (at critical point) =38=\dfrac{3}{8}

Q77
The coagulating power of electrolytes having ions Na+, Al3+ and Ba2+ for arsenic sulphide sol increases in the order:
A Na+ < Ba2+ < Al3+
B Al3+ < Ba2+ < Na+
C Ba2+ < Na+ < Al3+
D Al3+ < Na+ < Ba2+
Correct Answer
Option A
Solution

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 :

Na+<Ba2+<Al3+N{a^ + } < B{a^{2 + }} < A{l^{3 + }}
Q78
For coagulation of arsenious sulphate sol, which one of the following salt solution will be most effective ?
A BaCl2
B AlCl3
C NaCl
D Na3PO4
Correct Answer
Option B
Solution

Sulphide is -ve charged colloid.

As coagulation is directly proportional to the effective charge, so cation with maximum charge will be most effective for coagulation.

Al3+ > Ba2+ > Na+ coagulating power.

Q79
Among the colloids cheese (C), milk (M), and smoke (S), the correct combination of the dispersed phase and dispersion medium, respectively is :
A C : liquid in solid ; M : liquid in liquid ; S : solid in gas
B C : liquid in solid ; M : liquid in solid ; S : solid in gas
C C : solid in liquid ; M : liquid in liquid ; S : gas in solid
D C : solid in liqid ; M : solid in liquid ; S : solid in gas
Correct Answer
Option A
Solution

Cheese (C) is liquid dispersed in solid. Milk (M) is liquid dispersed in liquid. Smoke (S) is solid dispersed in gas.

Q80
The Tyndall effect is observed only when following conditions are satisfied: (a) The diameter of the dispersed particles is much smaller than the wavelength of the light used. (b) The diameter of the dispersed particle is not much smaller than the wavelength of the light used. (c) The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude. (d) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.
A (b) and (d)
B (a) and (c)
C (b) and (c)
D (a) and (d)
Correct Answer
Option A
Solution

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.

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