Dr. Sangeeta Khanna Ph.D 1. Sangeeta Khanna Ph.D Dr. Sangeeta Khanna Ph.D 6 CHEMISTRY COACHING CIRCLE G:\+2 Grand Test\Grand test-2 Moleconcept, Sol. & Surface Level-I.doc 29. The

$Page 1: Dr. Sangeeta Khanna Ph.D 1. Sangeeta Khanna Ph.D Dr. Sangeeta Khanna Ph.D 6 CHEMISTRY COACHING CIRCLE G:\+2 Grand Test\Grand test-2 Moleconcept, Sol. & Surface Level-I.doc 29. The$
Dr. Sangeeta Khanna Ph.D 1 CHEMISTRY COACHING CIRCLE G:\+2 Grand Test\Grand test-2 Moleconcept, Sol. & Surface Level-I.doc

Dr. Sangeeta Khanna Ph.D


Level – 1

READ INSTRUCTIONS CAREFULLY 1. The test is of 1 hour duration.

2. The maximum marks are 150.

3. This test consists of 50 questions.

4. For each question in Section - A, B & C you will be awarded 3 marks if you have darkened

only the bubble corresponding to the correct answer and zero mark if no bubbles are

darkened. Minus one (-1) mark will be awarded for wrong answer

5. Keep Your mobiles switched off during Test in the Halls.

Section – A (Single Correct Choice Type) Negative Marking

This Section contains 34 multiple choice questions. Each question has four choices A), B), C) and D) out

of which ONLY ONE is correct. Marks: 34 × 3 = 102

1. In which case should N2(g) be more soluble in water? a. The total pressure is 5 atm and the partial pressure of N2 is 1 atm b. The total pressure is 1 atm and the partial pressure of N2 is 0.03 atm. c. The total pressure is 1 atm and the partial pressure of N2 is 0.5 atm d. The total pressure is 3 atm and the partial pressure of N2 is 2 atm D 2. Which of the following does not affect the solubility of a solute in a given solvent? a. polarity of the solute b. polarity of the solvent c. rate of stirring d. temperature of the solvent and solute C 3. Intermolecular forces in liquid A are considerably large than intermolecular forces in liquid B. Which of

the following properties is NOT expected to be larger for A than B? a. The vapour pressure at 20 °C b. The temperature at which the vapour pressure is 100mm Hg c. The critical temperature d. The heat of vaporization (Hvap) A

4. What is the molality of the 870 g solution made by dissolving 120 g Br2 in CHC3.

[M.wt of Br2= 160] a. 1 b. 1/2 c. 3/4 d. 1/4 A 5. Gold number of a lyophilic sol is such property that:

a. the larger its value, the greater is the peptizing power b. the lower its value, the greater is the peptizing power c. the lower its value, the greater is the protecting power d. the larger its value, the greater is the protective power C

6. An oil-soluble dye is shaken with the given emulsion under study. We observe that whole background appears coloured. This indicates that emulsion is :

a. water-in-oil type b. oil-in-water type c. liquid under study is pure oil d. liquid under study is pure water A



7. Cloud bursts occur due to one of the following reasons:

a. The clouds are attracted towards the electrical charge on the earth. b. Large amount of water is present in the cloud c. Dense clouds are present in the upper atmosphere d. Mutual discharge of oppositely charged clouds resulting in the Coagulation of water droplets. D

8. When AgNO3 (excess) is added to Kl, charge on colloid is due to adsorption of:- a. K+ b. Ag+ c. I– d. NO3

– B

9. The colour of colloidal particles of gold obtained by different methods differ because of: a. Variable valency of gold b. Different concentration of gold particles c. Different types of impurities d. Different diameters of colloidal particles D

10. Purple of Cassius is: a. Silver sol b. Gold sol c. Platinum sol d. [As2S3]S

– sol B

11. Colloid of which of the following can be prepared by electrical dispersion method as well as reduction method: a. Sulphur b. Ferric hydroxide c. Arsenious sulphide d. Gold D

12. Alum help is purifying water by: a. Forming Si complex with clay particles b. Sulphate part which combines with the dirt & removes it c. Aluminium ion which coagulates the mud particles d. Making mud water soluble C

13. When ammonia gas is brought in contact with water surface, its pressure falls due to: a. Physical adsorption b. Chemical adsorption c. Absorption d. None of the above C

14. Y gm of a non-volatile solute of molar mass M is dissolved in 250 g of benzene. If Kb is molal elevation constant, the value of T is given by:

a. YK

N4

b

b. M

YK4 b c. 4M

Y Kb d.

M

Y Kb

B

Sol. T = Kb × M

YK4

250M

1000YK

wm

1000w bb

AB

B

15. What is the correct sequence of osmotic pressure of 0.01 M aqueous solution of:

(1) Al2(SO4)3 (2) K3PO4 (3) BaCl2 (4) Urea

a. 4 > 3 > 2 > 1 b. 1 > 2 > 3 > 4 c. 1 = 2 > 3 = 4 d. 2 > 4 > 1 > 3

B

Sol. At same molar concentration, the osmotic pressure will depend on van’t Hoff factor.

= iCRT

Thus, osmotic pressure will lie in the sequence.

1(i = 5) > 2(i = 4) > 3(i = 3) > 4 (i = 1)



16. Which of the following solutions has maximum freezing point depression at equimolal concentration?

a. [Co(H2O)6]Cl3 b. [Co(H2O)5Cl]Cl2H2O

c. [Co(H2O)4Cl2]Cl2H2O d. [Co(H2O)3Cl3]3H2O

A

Sol. [Co(H2O)6]Cl3 [Co(H2O)6]3+ + 3Cl–

This complex gives maximum number of ions, hence its depression in freezing point will also be

maximum .

17. A solute forms a pentamer when dissolved in a solvent. The van’t Hoff factor ‘I’ for the solute will be:

a. 0.2 b. 0.8 c. 0.5 d. 0.6

A

Sol. n/11

i1

1 = 5/11

i1

i = 0.2

18. The van’t Hoff factor for a 0.1 M Al2(SO4)3 solution is 4.20. The degree of dissociation is

a. 80% b. 90% c. 78% d. 83%

A

19. Negative deviations from Raoult’s law are exhibited by binary liquid mixtures

a. in which the molecules tend to attract each other and hence their escape into the vapour phase is

retarded

b. in which the molecules tend to repel each other and hence their escape into the vapour phase is

retarded

c. in which the molecules tend to attract each other and hence their escape into the vapour phase is

speeded up

d. in which the molecules tend to repel each other and hence their escape into the vapour phase is

speeded up

A

20. A substance will be deliquescent if its vapour pressure is

a. equal to the atmospheric pressure

b. equal to the vapour pressure of water vapour in the air

c. greater than the vapour pressure of water vapour in the air d. less than the vapour pressure of water vapour in the air D

21. In homogeneous catalytic reaction, there are three

alternative paths A, B and C (shown in the figure). Which one of the following indicates the relative ease with which the reaction can take place?

a. A > B > C

b. C > B > A

c. B > C > A

d. A = B = C B

A

B

C

Reaction coordinate

Potential energy



22. Lake test of aluminium ion is based on adsorption of blue litmus on:

a. solid surface of Al b. solid surface of colloidal Al(OH)3

c. solid surface of Al2O3 d. solid surface of AlCl3 B 23. Which is the correct for graph of physical adsorption? a. T1 > T2 > T3 b. T1 > T2 < T3 c. T1 < T2 < T3 d. T1 < T2 > T3 C Sol. As temperature increases the physical adsorption decreases. Hence less the temperature more the adsorption Hence choice (c) is correct while all others are wrong. 24. What is the freezing point of a 0.50 m solution of Cs2SO4 in water? Kf for water is 1.860C/m. a. – 0.930C b. -1.90C c. -6.50C d. -2.80C D 25. Given below are a few electrolytes, indicate which one among them will bring about the coagulation of

a gold sol (negative) quickest and in the least of concentration? a. NaCl b. MgSO4 c. Al2(SO4)3 d. K4[Fe(CN)6] C 26. The dispersed phase in colloidal iron (III) hydroxide and colloidal gold is positively and negatively

charged respectively which of the following statement is not correct? a. Magnesium chloride solution coagulated the gold sol more readily that the iron (III) hydroxide sol b. Sodium sulphate solution causes coagulation in both sols c. Mixing the sols has no effect d. Coagulation in both sol can be brought about by electrophoresis. C

27. Which of the following false regarding physical adsorption?

a. It is mathematically represented by Freundlich equation

b. log m

x versus log P is a straight line with slope = 1/n

c. log m

x versus log C is a straight line with slope = n

d. m

x = k C1/n

C 28. Barium ions, CN– & Co+2 form an ionic complex. If this complex is 75% ionised in aqueous solution with

van’t Hoff’s factor equal to four. What will be the molecular formula of complex.

a. Ba2[Co(CN)5] b. Ba3[Co(CN)5]2 c. Ba[Co(CN)5] d. Ba3[Co(CN)5] B

Sol. i = 1 + (n – 1) 4 = 1 + (n – 1).75 n = 5

T1

T2

T3

Pressure

m

x



29. The freezing point of a solution containing 50 cm3 of ethylene glycol in 50 g of water is found to be -34°C. Assuming ideal behaviour, calculate the density of ethylene glycol. (Kf for water is 1.86K kg mol–1)

a. 1.133 g cm–3 b. 2.133 g cm–3 c. 11.33 g cm–3 d. 113.3 g cm–3

A

Sol. Kf for water = 1.86 K Kg mol–1

Volume of ethylene glycol = 50 cm3. Wt. of ethylene glycol = 50 × d; Mol. mass of glycol

(CH2OH.CH2OH) = [12 + (2 × 1) + 16 + 1 + 12 + (2 × 1) + 16 + 1] = 62 g mol– ; d = density of glycol.

Wt. of water = 50 g = 1000

50 = 0.05 kg

Tf = 34°C = 34 K

we know that; Tf = kg 05.0mol g 62

dcm 50mol kgK 1.86K 34 ;

kg in WM

g in WK

-

3-

12

2f

d = 3-

-

cm 50 mol kgK 86.1

kg 05.0mol g 62K 34

= 1.133 g cm–3

30. The formula of an acid is HXO2. The mass of 0.0242 moles of the acid is 1.657 g. What is the atomic weight of X?

a. 35.5 b. 28.1 c. 128 d. 19.0 A

Sol. 0.0242 mole have mass = 1.657 gm

1 mole will have = gm5.680242.0

657.1 .

At. mass of X = 68.5 – 1 – 32 = 35.5 31. Under ambient conditions, which among the following surfactants will form micelles in aqueous solution

at lowest molar concentration?

a. NaOSO)CH(CH 31323 b.

Br)CH(N)CH(CH 331123

c. NaCOO)CH(CH Θ823 d.

Θ

331523 Br)CH(N)CH(CH

D 32. Determination of the molar mass of acetic acid in benzene using freezing point depression is affected

by:

a. dissociation b. partial ionization c. complex formation d. association D 33. A solution at 20°C is composed of 1.5 mol of benzene and 3.5 mol of toluene. If the vapour pressure of

pure benzene and pure toluene at this temperature are 74.7 torr and 22.3 torr, respectively, then the total vapour pressure of the solution and the benzene mole fraction in equilibrium with it will be, respectively :

a. 35.8 torr and 0.280 b. 35.0 torr and 0.480 c. 38.0 torr and 0.589 d. 30.5 torr and 0.389 C

34. 3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in 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 D



SECTION – B (Assertion and Reason) Negative Marking

This Section contains 6 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONLY ONE is correct. Marks: 6 × 3 = 18

(A) Mark a if both A and R are correct and R is the correct reason of A.

(b) Mark b if both A and R are correct and R is not the correct reason of A.

(c) Mark c if A is correct and R is wrong.

(d) Mark d if A is wrong and R is correct. 1. Assertion: A colloidal sol of Al(OH)3 prepared by adding H2O in AlCl3 is more readily coagulated by

0.1 M NaCl than by 0.1 M Na2SO4. Reason: The coagulating power of an electrolyte is related to the valency of the active ions. a. (a) b. (b) c. (c) d. (d) D 2. Assertion (A): The activity of a catalyst is enhanced in the presence of a promoter. Reason (R): The promoter make the surface of the catalyst more uneven and thereby increases the

number of active centres. a. (a) b. (b) c. (c) d. (d) A 3. Assertion. Both soaps and detergents adsorb on the surface of dust and are called associated colloids.

Reason. They contain hydrophilic and hydrophobic moieties in their molecule and thus show adsorption and association (micellization). a. (a) b. (b) c. (c) d. (d) A

4. Assertion: Associated colloids are formed at a particular concentration, called critical micellisation concentration. Reason: Soap and synthetic detergent have CMC level 10–4 to 10–3 mole 1–1.

a. (a) b. (b) c. (c) d. (d) B

5. Assertion: 1 Molar aqueous solution is less concentrated than 1 molal aqueous solution. Reason: Molarity is temperature depend term. a. (a) b. (b) c. (c) d. (d) D

6. Assertion: Solubility of Na2SO410H2O increases with temperature upto 32° then decreases with further increase in temperature. Reason: Dissolution of glaubersalt is exothermic a. (a) b. (b) c. (c) d. (d) C



SECTION – C (Paragraph Type) This Section contain 1 Comprehension. Each of these questions has four choices A), B), C) and D) out of which ONLY ONE is correct. 10 × 3 = 30

Comprehension - 1 Compounds isolated from natural sources and prepared in the laboratory are not pure. Therefore

purification process are required for these. There are large number of methods (like filtration, crystallisation, sublimation, distillation, chromatography) for purification of substances. The selection of method depends upon the nature of the substance and type of impurities present in it.

Chromatography is one of the most modern and versatile method used for separation and purification

of organic compounds. Chromatography was discovered in 1906 by a Russian botanist, Tswett through his experiment. If a petroleum ethereal solution of chlorophyll is filtered through a column of calcium carbonate adsorbent filled in a narrow glass tube, then the pigments are resolved from top to bottom into various coloured zones. These zones are combinely called chromatogram. The different components of pigment mixture are resolved in CaCO3 column according to their adsorbing power. There are various types of chromatographic methods such as; column chromatography, high performance chromatography, thin layer chromatography, gas liquid chromatography and paper chromatography. Chromatography is based on two phases:

One phase is mobile phase, consists of mixture of components, which are to be separated. Second phase is fixed phase which is filled in column as in column chromatography. A solvent (Eluent) which is used to extract components of chromatogram. Eluent is passed through

column, it dissolves the different compounds in zones and collects the component of mixture in separate flasks. This process is called elution. In TLC, a thin layer of an adsorbent like silica is spread over a glass plate. A solution of mixture to be separated is applied as a small spot with the help of a fine capillary. Plate is placed in a closed jar containing suitable solvent. As the solvent moves up, the components of the mixture also moves up depending upon their extent of adsorption. The various components on the developed TLC plate are identified through their retention factor (Rf values).

1. The phenomenon involve in the separation of mixture of pigments is based on a. surface phenomenon b. Chemical phenomenon c. solubility phenomenon d. Both A & C D 2. The substance present in a zone towards bottom of column in column chromatography is a. most coloured b. least adsorbed c. most adsorbed d. least coloured B 3. The Rf value of A and B in a mixture determined by TLC in a solvent mixture are 0.65 and 0.42

respectively. If the mixture is separated by column chromatography using the same solvent mixture on a mobile phase, which of the two components, A or B will elute first ?

a. Component A b. Component B c. both A and B components are obtained together d. Elution is not related to R f values A Comprehension – 2 Gold sol is metallic sol and it is negatively charged. It can be prepared by reduction method and

Bredig’s arc method. When NaCl solution is added to a gold sol, it results in coagulation. But there is no coagulation of gold sol, when NaCl solution is added to a gold sol after adding gelatine. Protective action is more at low temperature. Gold sol has very little or no affinity with its dispersion medium.



4. The stability of gelatine is a. more than that of gold sol b. less than that of gold sol c. equal to that of gold sol d. it can’t be compared A 5. If the temperature of gold sol containing sodium chloride and gelatin increases to 70°C, then a. protective action of gelatine will increase b. protective action of gelatine will decrease c. protective action do not get affected by increasing temperature d. NaCl undergo more ionisation and adsorption of Na+ ions take place on surface of sol particle to

create zeta potential B 6. When excess quantity of NaCl electrolyte is added to a pure gold colloid a. it result in unstability of sol b. it result in stabilization of sol c. protection of gold sol takes place by NaCl electrolyte d. it is not related to stability of sol A

Comprehension – 3

The colligative properties of electrolytes require a slightly different approach than the one used for the colligative properties of non-electrolytes. The electrolytes dissociate into ions in a solution. It is the number of solute particles that determines the colligative properties of a solution. The electrolyte solutions, therefore, show abnormal colligative properties. To account for this effect we define a quantity called the van’t Hoff factor, given by

solution in dissolvednitially i units formula of Number

ondissociati after solution in particles of number Actuali

i = 1(for non-electrolytes); i > 1 (for electrolytes, undergoing dissociation) i < 1 (for solutes, undergoing association).

7. Benzoic acid undergoes dimerisation in benzene solution. The van’t Hoff factor ‘i' is related to the

degree of association ‘’ of the acid as:

a. i = 1 - b. i = 1 + c. i = 1 - 2

d.

21i

C 8. A substance trimerises partially when dissolved in a solvent A. The van’t Hoff factor ‘i’ for the solution

is:

a. 1 b. 1/3 c. 3 d. unpredictable D

9. For a solution of a non-electrolyte in water, the van’t Hoff factor is:

a. always equal to 0 b. 1 c. always equal to 2 d. > 1 but < 2 B

10. 0.1 M K4 [Fe(CN)6] is 60% ionized. What will be its van’t Hoff factor?

a. 1.4 b. 2.4 c. 3.4 d. 4.4 C

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Dr. Sangeeta Khanna Ph.D 1. Sangeeta Khanna Ph.D Dr. Sangeeta Khanna Ph.D 6 CHEMISTRY COACHING CIRCLE G:\+2 Grand Test\Grand test-2 Moleconcept, Sol. & Surface Level-I.doc 29. The