Ex.1 The charge on the colloidal particle of soap in its solution developed -(A) By prefrential adsorption of ions
(B) The effective ion of soap micelle carries negative charge (C) The effective ion of soap micelle carries positive charge (D) None of the above
Sol. B
RCOONa RCOO¯ + Na+
The active species of soap is negatively charged.
Ex.2 In order to coagulate the impurities present in water. Which is the most effective ion present in the
‘alum’
-(A) K+ (B) SO42– (C) Al3+ (D) None of these
Sol. C
The impurities carry negative charge hence Al3+ will exhibit maximum coagulating power.
Ex.3 Which is not the property of the hydrophilic
sols-(A) High concentration of dispered phase can be easily attained (B) Coagulation is reversible
(C) The charge of the particle depends upon the pH values of the dispersion medium. It may be positive or negative
(D) Viscosity and that surface tension are about the same as of dispersion medium.
Sol. D
Viscosity of hydrophilic sol is much higher than dispersion medium while surface tension of the sol is much lesser than water. This is attributed to the higher concentration of the sol itself.
Ex.4 If water is chosen as dispersion medium, then the sol formed will be called -(A) Alcohosol (B) Benzosol (C) Hydrosol (D) Aerosol Sol. C
If water is chosen as dispersion medium then the colloidal system will be called hydrosol.
Ex.5 The number of moles of lead nitrate needed to coagulate 2 mole of colloidal [AgI]I–
is-(A) 2 (B) 1 (C)
2
1 (D)
3 2
Sol. 2[Ag I] I– + Pb2+ PbI2 + 2AgI
Thus, one mole of Pb(NO3)2 is required to coagulate 2 mole of [AgI] I–.
Ex.6 Maximum hydrophobic character will be shown by
(A) Glycine (B) Stearic acid (C) Glucose (D) Adenine Sol. D
Adenine exhibits maximum hydrophobic character.
Ex.7 Fe(OH)3 sol obtained by peptization when subjected to ‘electrophoretic effect’ -(A) Colloidal particles will migrate towards anode
(B) Colloidal particles will migrate towards cathode (C) Colloidal particles remain stationary
(D) Colloidal particles migrate towards both the electrode Sol. B
Ferric hydroxide is a positive sol Fe(OH)3/Fe3+, it will migrate towards cathode.
Ex.8 Which of the following solution is positively charged
-(A) As2S3 (B) Fe(OH)3 (C) Au (D) Starch
Sol. B
Fe(OH)3/Fe3+ is a positively sol particle.
Ex.9 Dispersion of AgCl in water is called
-(A) Hydrophilic sol (B) Emulsion (C) Benzosol (D)Hydrophobic colloid Sol. D
Colloidal system AgCl is called, hydrophobic colloid.
Ex.10 The stability of lyophilic colloid is due to
-(A) Charge on the solution particles (B) Layer of dispersion medium over the solution paricles (C) Smaller size of solution particles (D) Bigger size of solution particles
Sol. B
Lyophilic colloids are of bigger size and this accounts for their stability.
Ex.11 The volume of nitrogen gas at 0ºC and 1.013 bar required to cover a sample of silica gel with unimolecular layer is 129 cm3 g–1 of gel. Calculate the surface area per gram of the gel if each nitrogen molecule occupies 16.2 × 10–20 m2.
Ex.12 The rate of decomposition of acetaldehyde into methane and CO in the presence of I2 at 800 K follows the rate law
Rate = k[CH3CHO][I2]
The decomposition is belived to go by the two steps mechanism CH3CHO + I2 CH3I + HI + CO
CH3I + HI CH4 + I2
What is the catalyst for the reaction ? Which of the two steps is a slower one ? Sol. I2 is catalyst ; first step is slow.
Ex.13 A catalyst lowered the activation energy by 25 kJ/mol at 25ºC. By how many times will the rate slow ?
Ex.14 Discuss the effect of temperature on the degree of adsorption of N2 on the surface of iron.
Sol. At room temperature, practically there is no adsorption of N2 gas on the surface of iron. At 83 K, nitrogen is physically adsorbed on the surface of iron. Its degree of adsorption decreases with rise in temperature and becomes almost nil at room temperature. At 773 K and above, there is chemical adsorption of nitrogen on the surface of iron.
Ex.15 One gram of charcoal adsorbs 100 mL of 0.5 M CH3COOH to form a mono-layer and thereby the molarity of acetic acid is reduced to 0.49 M. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal = 3.01 × 102 m2/gm.
Sol. Number of moles of acetic acid initially present = 1000
Number of moles of acetic acid left = 1000 Number of moles of acetic acid adsorbed = 0.05 – 0.049 = 0.01 mol
Number of molecules of acid adsorbed = 0.001 × 6.023 × 1023 = 6.023 × 1020 Area occupied by single molecule of acetic acid = Numberof moleculesadsorbed
area molecule is 0.16 nm2. What is the surface area per gram of silica gel ?
(NA = 6.023 × 1023)
(A) 1.6 m2 g–1 (B) 5.568 m2 g–1 (C) 3.48 m2 g–1 (D) 4.42 m2 g–1 Sol. B
Number of N2 molecules adsorbed per gram of silica gel
= 22400
Area of cross-section of molecule = 0.16 nm2
= 1.6 × 10–19 m2
Surface area covered per gram of silica gel
= 1.6 × 10–19 × 3.48 × 1019 = 5.568 m2
x against log P is a straight line inclined at an angle of 45º. When the pressure is 0.5 atm and Freundlich parameter, k is 10, the amount of solute adsorbed per gram of adsorbent will be : (log 5 = 0.6990)
Ex.18 In homogeneous catalytic reactions, there are three alternative paths A, B and C (shown in the figure).
Which one of the following indicates the relative case with which the reaction can take place ?
A B C
Reaction coordinate
Potential energy (A) A > B > C (B) C > B > A
(C) B > C > A (D) A = B = C
Sol. Activation energy in the different paths lies in the following sequence : C < B < A.
Lesser is the activation energy, greater is the ease with which the reaction can take place.
C > B > A (Deceasing case with which the reaction can take place.)
Ex.19 The coagulation of 100 mL of a colloidal solution of gold is completely prevented by adding 0.25 g of starch to it before adding 10 mL of 10% NaCl solution. Find out the gold number of starch.
Sol. 10 mL of 10% NaCl solution is added to 100 mL of solution of gold.
Thus, 1 mL of 10% NaCl solution has been added to 10 mL solution of gold.
Since, 100 mL gold solution required = 0.25 g starch
= 0.25 × 103 mg starch
So, 10 mL gold solution required
= 100 10 0.25 3
× 10
= 25 mg starch
Thus, by definition, the gold number of starch is 25.
Ex.20 The coagulation of 100 mL of colloidal solution of gold is completely prevented by addition of 0.25 g of a substance “X” to it before addition of 1 mL of 10 % NaCl solution. The gold number of “X”
is-(A) 0.25 (B) 25 (C) 250 (D) 2.5
Sol. Number of milligrams of protective colloid added in 10 mL of colloidal gold to prevent its coagulation on addition of 1 mL of 10% NaCl solution is called its gold number.
Gold number of present colloid = 25