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SURFACE CHEMISTRY

MULTIPLE CHOICE QUESTIONS

1. In the coagulation of a positive sol, the flocculation powers

of Cl–, 2 3

4 4SO , PO and [Fe(CN)6]4– are in the order of

(a) 2 4 3

4 6 4Cl SO [Fe(CN) ] PO (b) 3 2 4

4 4 6Cl PO SO [Fe(CN) ]

(c) 4 3 2

6 4 4[Fe(CN) ] PO SO Cl (d) 2 3 4

4 4 6Cl SO PO [Fe(CN) ]

Sol. (c)

2. 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 10 ml of 10% NaCl solution.

The gold number of "X" is

(a) 0.25 (b) 25

(c) 250 (d) 2.5

Sol. (b)

3. Which one of the following electrolytes will be most

effective in coagulation of arsenoussulphide (As2S3) sol?

(a) KCl (b) NaCl

(c) MgCl2 (d) AlCl3

Sol. (d)

As2S3 is –vely charged sol. So it can be coagulated by

cations. More the charge of cations, more effective will be

the coagulation.

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4. The stability of the dispersed phase in a lyophobic colloid

is due to

(a) high viscosity of the medium.

(b) the formation of electrical layer between two phases.

(c) high surface tension of solution.

(d) none of the answer is correct.

Sol. (b)

5. Which equation represents Freundlisch adsorption isotherm

(physical adsorption is basis of

this theory):

(a) 1/nk(P)m

x wherex is amount of gas adsorbed on mass ‘m’

at pressure P

(b) 1log log k log P

m n

x

(c) kPm

x at low pressure and k

m

x at high pressure

(d) All of these

Sol. (d)

These are different forms of Freundlich equation.

6. Lyophilic sols have …… between disperse phase and

dispersion medium.

(a) strong attractive interaction.

(b) little attractive interaction.

(c) repulsive interaction.

(d) none of these

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Sol. (a)

To form colloidal solution there should be strong forces of

attraction between dispersed phase and dispersion medium.

7. Which of the following will have highest coagulating

power for As2S3 colloid?

(a) 3

4PO (b) 2

4SO

(c) Al3+ (d) Na+

Sol. (c)

As2S3 sol is negatively charged and hence the ion with

maximum positive charge shall be required in minimum

concentration to cause its precipitation.

8. For a gas-solid system, classical adsorption isotherm is

applicable. If m g of solid absorbs x g of gas at pressure p,

then which of the following plots will be a straight line?

(a) Vs. pm

x (b) pVs. p

/ mx

(c) log Vs. log pm

x (d) xlog Vs. p

m

Sol. (c)

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9. Examine the following table:

Colloid Gelatin Gum

Arabic

Sodium

Oleate

Starch

Gold

No.

0.01 0.25 1.0 25

Which of the following is better protective colloid?

(a) Gelatin (b) Gum Arabic

(c) Sodium Oleate (d) Starch

Sol. (a)

Lower is the gold number, less is the lyophilic colloid

required to prevent the precipitation and more is the

protective power of the colloid.

10. The stabilization of the dispersed phase in a lyophobic sol

is due to

(a) the viscosity of the medium.

(b) the surface tension of the medium.

(c) affinity for the medium.

(d) the formation of an electrical layer between the two

phase.

Sol. (d)

The stabilization of the dispersed phase is due to adsorption

of the ions.

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11. A catalyst is a substance which

(a) Increases the equilibrium concentration of the product.

(b) Changes the equilibrium constant of the reaction.

(c) Shortens the time to reach equilibrium.

(d) Supplies energy to the reaction.

Sol. (c)

A catalyst is a substance which alters the rate of reaction

and shortens the time to reach equilibrium.

12. The ability of an ion to bring about coagulation of a given

colloid depends upon

(a) Its size

(b) The magnitude of its charge only

(c) The sign of its charge

(d) Both the magnitude and the sign of its charge

Sol. (d) The ability of an ion to bring about coagulation of a

given calloiddepend upon both the magnitude and sign of

its charge.

13. Which one of the following is not a surfactant?

(a) |

3 2 15 3|

CH3

CH (CH ) N CH Br

CH3

(b) 3 2 14 2 2CH (CH ) CH NH

(c) 3 2 16 2 2CH (CH ) CH OSO Na

(d) 2 14 2OHC (CH ) CH COO Na

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Sol. (b)

Surfactant are those which have charge on their tail e.g.,

cetyltrimethyl ammonium bromide.

3

|

3 2 15 3|

3

CH

CH (CH ) N CH Br

CH

Surfactants are those, which dissociate in water to yield

positively charged ion.

14. A catalyst is used in a reaction to

(a) Change the nature of reaction products

(b) Increase the reaction yield

(c) Decrease the need for reactants

(d) Decrease the time required for the reaction

Sol. (d)

A catalyst is used to decrease the time required for the

reaction hence it can decease or increase the rate of

reaction.

15. Gold number gives

(a) The amount of gold present in the colloid.

(b) The amount of gold required to break the colloid.

(c) The amount of gold required to protect the colloid.

(d) None of these

Sol. (d)

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Gold no. is a measure of protective power of a lyophillic

colloid.

16. Identify the gas which is readily adsorbed by activated

charcol

(a) 2N (b) 2SO

(c) 2H (d) 2O

Sol. (b)

Easily liquefiable gases like 2 3 2SO , NH ,CO are adsorbed to a

greater extent than the elemental gases like 2 2 2N ,O ,H .

17. At the equilibrium position in the process of adsorption

___________.

(a) ΔH > 0

(b) ΔH = TΔS

(c) ΔH > TΔS

(d) ΔH < TΔS

Ans. (b)

At equilibrium ΔG=0

ΔG=ΔH-TΔS

Therefore, ΔH=TΔS

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18. Which of the following forms cationic miscelles above

certain concentration

(a) Urea

(b) Cetyltrimethylammonium bromide

(c) Sodium dodecyl sulphate

(d) Sodium acetate

Sol. (d)

Sodium acetate forms cationic micelles in the molecule of

soap and detergent the negative ions aggregate to form a

micelle of colloidal size. The negative ion has a long

hydrocarbon chain and a polar group ( COO ) at one end.

19. Which of the following colloid can be prepared by

electrical dispersion as well as reduction method?

(a) Sulphur (b) Ferric hydroxide

(c) Arsenioussulphide (d) Gold

Sol. (d)

Of the given four options, only gold sol can be prepared by

electrical dispersion method i.e., Bredig’s arc method as

well as reduction method when AuCl3 is reduced by SnCl2

2AuCl3 + 3SnCl2 2Au + 3SnCl4

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20. Silver iodide is used for producing artificial rains because

silver iodide

(a) is easy to spray at high altitude

(b) is insoluble in water

(c) has crystal structure similar to ice

(d) is easy to synthesize

Sol. (c)

Silver iodide is used for producing artificial rains because it

has crystal structure similar to ice. When AgI is sprayed

over clouds, it causes water droplets of clouds to coalesce

to form bigger water drops and induce rains.

21. Existence of colloidal particles is due to

(a) The colloidal particles are neutral in presence of

dispersion medium.

(b) They carry the charge only of one type.

(c) They carry oppositely charged particles.

(d) Colloidal particles are not reactive.

Sol. (b)

22. When an electric current is passed through the colloidal

solution and the experiment so arranged that only

dispersion medium moves, the phenomenon is called:

(a) Electrophoresis (b) Electrodisintegration

(c) Electroosmosis (d) Electrolysis

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Sol. (c)

23. The property of colloidal solutions that their particles are

charged is used

(I) Sewage disposal

(II) Smoke screens

(III) Delta formation

(a) (I) (b) (I) and (III)

(c) (I), (II) and (III) (d) All of the above

Sol. (d)

INTEGER TYPE QUESTIONS

24. 50 ml of 1 M oxalic acid is shaken with 0.5 gm of wood

charcoal. The final concentration of the solution after

adsorption is 0.5 M. Amount of oxalic acid absorbed per

gm of charcoal is

Sol. 126 1 50W

1000

6.3

(Molecular weight of oxalic acid 163)

6.30.5 gm

2

6.31 gm 1

2 0.5

6.3 gm.

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25. On addition of one ml solution of 10%NaCl to 10 ml gold sol

in the presence of 0.25 gm of starch, the coagulation is just

prevented. Starch has the following gold number

Sol. By definition gold number of starch is the amount of starch

in mg added to 10 ml standard gold sol which prevents the

coagulation of gold on adding 1 ml of 10% NaCl solution.

So the amount of starch is 0.25g = 250 mg. Hence gold

number is 250.

26. The coagulation of 100 cm3 of gold solution is completely

prevented by addition of 0.25 g of starch to it before adding

10 ml of 10% NaCl solution. The gold number of starch is

Sol. Gold number of starch is the milligrams of starch added to

10 ml of standard gold sol just to prevent coagulation of Au

sol when 1 mL of 10% NaCl is added to it.

10 mL of 10% NaCl is added to 100 mL of gold sol. In

other words 1 mL of 10% aCl is added to 10 mL of gold

sol.

Quantity of starch added to 10 mL of gold sol = 25 mg

Gold number of starch = 25.

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27. For the coagulation of 115 mL of Sb2S3 sol, 10 mL of 1 M

NaCl is required. The flocculation value of NaCl is

Sol. Millimoles of NaCl added = 10

Volume of the sol containing NaCl solution = 115 + 10 =

125 mL.

Flocculation value is the amount of electrolyte (in

millimoles) that must be added to 1 L of the colloidal

solution containing NaCl solution so as to bring about

complete coagulation.

Flocculation value of NaCl = 10 1000

125

= 80 m mol L1.

28. A sample of charcoal weighing 6 g was brought into

contact with a gas contained in a vessel of one litre capacity

at 27°C. The pressure of the gas was found to fall from 700

to 400 mm of Hg. Calculate the volume of the gas (reduced

to STP) that is adsorbed per gram of the adsorbent under

the condition of the experiment (density of charcoal sample

is 1.5 g cm3).

Sol. The adsorption is taking place in a closed vessel, thus when

pressure falls there is correspondingly increase in volume

and to keep volume constant, excess of the volume of the

gas would be adsorbed.

UsingP1V1 = P2V2

V2 = 1 1

2

P V

P= 700 1000

400

= 1750 ml

Actual volume of the flask = 1000 volume of charcoal

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= 1000 6

1.50 = 996 ml

Volume of the gas adsorbed = 1750 996 = 754 ml

Volume of the gas adsorbed per gram of charcoal = 754

6 =

125.67 ml g1

Volume of the gas adsorbed per gram at STP 1 1 2 2

1 2

P V P Vu sin g

T T

= 125.67 400 273

300 760

= 60.19 ml g1.

29. A solution of palmitic acid (M = 256 g mol1) in benzene

contains 4.24 g of acid per dm3. When this solution is

dropped on a water surface the benzene evaporates and the

palmitic acid forms a monomolecular film of the solid type.

If we wish to cover an area of 500 cm2 with a monolayer,

what volume of solution should be used? The area covered

by one palmitic acid molecule may be taken to be 0.21 nm2.

Sol. Let the volume of palmitic acid solution required to cover

the desired area of 500 cm2 be V ml.

Number of molecules of palmitic acid in V ml = 234.24 V 6.023 10

1000 256

= 9.976 1018 V

The area covered by these molecules = 9.976 1018 V

0.21 1014 cm2

= 2.095 104 V cm2

But this area is equal to 500 cm2

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2.095 104 V = 500

or V= 0.024 ml.

30. The density of gold is 319 g / cm . If 41.9 10 g of gold is dispersed

in one litre of water to give a sol having spherical gold

particles of radius 10 nm, then the number of gold particles

per 3mm of the sol will be

Sol. Volume of the gold dispersed in one litre water Mass

Density

4

3

1.9 10 gm

19 gm cm

5 31 10 cm

Radius of gold sol particle 10 nm910 10 m

7 610 10 cm 10 cm

Volume of the gold sol particle 34r

3 6 34 22

(10 )3 7

18 34.19 10 cm

No. of gold sol particle in 5 31 10 cm5

18

1 10

4.19 10

122.38 10

No. of gold sol particle in one 3mm12

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6

2.38 102.38 10 2380000

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