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Dr. Sangeeta Khanna Ph.D 1 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Physical\Test\Grand Test\Grand test-2\+2 Grand test-2.doc Test Venue: Lajpat Bhawan, Madhya Marg, Sector 15-B, Chandigarh

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Page 1: Test Venue: Lajpat Bhawan, Madhya Marg, Sector 15-B, Chandigarh · 2016-05-30 · Test Venue: Lajpat Bhawan, Madhya Marg, Sector 15-B, Chandigarh. ... which among the following surfactants

Dr. Sangeeta Khanna Ph.D 1 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Physical\Test\Grand Test\Grand test-2\+2 Grand test-2.doc

Test Venue:

Lajpat Bhawan, Madhya Marg, Sector 15-B, Chandigarh

Page 2: Test Venue: Lajpat Bhawan, Madhya Marg, Sector 15-B, Chandigarh · 2016-05-30 · Test Venue: Lajpat Bhawan, Madhya Marg, Sector 15-B, Chandigarh. ... which among the following surfactants

Dr. Sangeeta Khanna Ph.D

Dr. Sangeeta Khanna Ph.D 2 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Physical\Test\Grand Test\Grand test-2\+2 Grand test-2.doc

Test Dated 29.5.2016

READ INSTRUCTIONS CAREFULLY 1. The test is of 2 hours 15 minutes duration.

2. The maximum marks are 402.

3. This test consists of 94 questions.

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

Section – A (Single Correct Choice Type) Negative Marking

This Section contains 50 multiple choice questions. Each question has four choices A), B), C) and D) out

of which ONLY ONE is correct. Marks: 50 × 4 = 200

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

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Dr. Sangeeta Khanna Ph.D

Dr. Sangeeta Khanna Ph.D 3 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Physical\Test\Grand Test\Grand test-2\+2 Grand test-2.doc

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 in 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. 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 .

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Dr. Sangeeta Khanna Ph.D

Dr. Sangeeta Khanna Ph.D 4 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Physical\Test\Grand Test\Grand test-2\+2 Grand test-2.doc

16. 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

Sol. i = 1 + 4 17. 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 18. 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

19. 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 20. 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 21. 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.

A

B

C

Reaction coordinate

Potential energy

T1

T2

T3

Pressure

m

x

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22. 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. +2.80C d. -2.80C D Sol. i = 3

23. 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

Sol. An electrolyte with more valency of cation

24. 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 coagulate the gold sol more readily than Sodium chloride

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

25. 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

26. 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

27. 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 Sol. Compound with biggest alkyl gp will have lowest CMC 28. 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

Sol. In non-polar medium it will dimerise

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29. 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

30. 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

Sol. millimole = 50 × 0.06 = 3.0 = 50 × 0.042 = 2.1 After millimole adsorbed = 0.9 (mg) mass = 0.9 × 60 = 54.0 mg Adsorbed on 3 gm charcoal

Per gm = 3

54 = 18

31. The boiling points of C6H6, CH3OH, C6H5NH2 and C6H5NO2 are 80°C, 65°C, 184°C and 212°C, respectively. Which of the following will have the highest vapour pressure at the room temperature?

(a) C6H6 (b) CH3OH (c) C6H5NH2 (d) C6H5NO2 B

Sol. B.pt

1 p.v

32. The vapour pressure of water at 300 K in a closed container is 0.4 atm. If the volume of the container is doubled, its vapour pressure at 300 K will be (a) 0.8 atm (b) 0.2 atm (c) 0.4 atm (d) 0.6 atm C

33. Benzene and toluene form an ideal solution. The vapour pressures of benzene and toluene are 75 mm and 25 mm respectively, at 20°C. If the mole fractions of benzene and toluene in vapour are 0.75 and 0.25, respectively, the vapoure pressure of the ideal solution is (a) 62.5 mm (b) 50 mm (c) 30 mm (d) 40 mm B

Sol. B

B

B

B

BT

BB

T

B

X2525

75X3 ;

)X1(25

X75

25.0

75.0

)X1(P

XP

; XB = 20.5; Pt = 50

34. The vapour pressure of a solution of two liquids, A (P° = 80 mm, X = 0.4) and B (P° = 120 mm, X = 0.6) is found to be 100 mm. It shows that the solution exhibits (a) negative deviation from ideal behaviour (b) positive deviation from ideal behavior (c) ideal behavior (d) positive deviation at lower concentration A

Sol. Pcal = BBAA XPXP ; Pcal is more than Pobserved, So –ve deviation

35. When 25 ml of CCl4 and 25 ml of toluene is mixed, the total volume of the solution will be (a) 50 ml (b) > 50 ml (c) < 50 ml (d) Indefinite B

Sol. Positive deviation V = +ve

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Dr. Sangeeta Khanna Ph.D 7 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Physical\Test\Grand Test\Grand test-2\+2 Grand test-2.doc

36. The relationship between osmotic pressure at 273 K when 10 g glucose (1), 10 g urea (2) and 10 g

sucrose (3) are dissolved in 250 ml of water, is

(a) 1 > 2 > 3 (b) 3 > 1 > 2 (c) 2 > 1 > 3 (d) 2 > 3 > 1 C

Sol. O.P. Wt.M

1

37. An aqueous solution of a non-volatile, non-electrolyte solute (molecular mass = 150) boils at 373.26 K.

The composition of solution, in terms of mass per cent of solute, is (Kb of water = 0.52)

(a) 50% (b) 7.5% (c) 6.98% (d) 75%

C

Sol. Tb = Kbm; m52.0

26.0 ; m = 0.5; Wsolute = 0.5 × 150 = 75; Wsolvent = 1000 gm

100075

10075% mass

= 6.98%

38. When the depression in freezing point is carried out, the equilibrium exist between (a) liquid solvent and solid solvent (b) liquid solute and solid solvent (c) liquid solute and solid solute (d) liquid solvent and solid solute A

39. Among the colligative properties of solution, which one is the best method for the determination of molecular masses of proteins and polymers? (a) osmotic pressure (b) lowering in vapour pressure (c) lowering in freezing point (d) elevation in boiling point A

40. At the same temperature, each of the following solution has the same osmotic pressure except (a) 0.140 M-sucrose (b) 0.07 M-KCl (c) 0.070 M-Ca(NO2)2 (d) 0.140 M-urea C

41. Aqueous solutions of 0.004 M – Na2SO4 and 0.01 M – Glucose are isotonic. The percentage

dissociation of Na2SO4 is

(a) 25% (b) 60% (c) 75% (d) 40%

C

Sol. 01.0004.0i ; 5.2004.0

01.0i ; 21i ; = 0.75

42. If 0.1 molal aqueous solution of sodium bromide freezes at – 0.3348°C at one atmospheric pressure,

the per cent dissociation of salt in the solution is (Kf = 1.86)

(a) 90 (b) 80 (c) 60 (d) 20 B

Sol. Tf = iKfm; 0.3348 = i × 1.86 × 0.1; i = 1 + 43. In KI solution, mercuric iodide is added. The osmotic pressure of resultant solution will

(a) increase

(b) decrease

(c) remains unchanged (d) increase or decrease, depending on amount B

Sol. Due to association to form K2[HgI4]

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Dr. Sangeeta Khanna Ph.D 8 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Physical\Test\Grand Test\Grand test-2\+2 Grand test-2.doc

44. Which of the following solution will show minimum osmotic effect? (a) KCl solution (b) Colloidal gold sol (c) CaCl2 solution (d) Na3PO4 solution B

45. The vapour pressure of a saturated solution of sparingly soluble salt (XCl3) was 17.20 mm Hg at 27°C. If the vapour pressure of pure water is 17.25 mm Hg at 27°C, what is the solubility of the sparingly soluble salt XCl3, in mole per litre? (a) 4.04 × 10–2 (b) 8.08 × 10–2 (c) 2.02 × 10–2 (d) 4.04 × 10–3 A

Sol. 45.55m

m

25.17

2.1725.17

i

5.55m

m

25.17

05.0

M = 4.04 × 10–2 same is M

46. PtCl46H2O can exist as a hydrated complex. 1.0 molal aqueous solution has depression in freezing point of 3.72°C. Assume 100% ionization and Kf of water = 1.86°C mol–1 kg. The complex is

(a) [Pt(H2O)6]Cl4 (b) [Pt(H2O)4]Cl22H2O

(c) [Pt(H2O)3Cl3]Cl3H2O (d) [Pt(H2O)2Cl4]4H2O C

Sol. i = 2 Tf = ikfm ; (give two ions in solution) 47. pH of a 0.1 M solution of a monobasic acid is 2.0. Its osmotic pressure at a given temperature, T K is

(a) 0.1 RT (b) 0.11 RT (c) 0.09 RT (d) 0.01 RT B

Sol. HA H+ + A–

10–2 = C i = 1 +

10–2 = 0.1 ×

= 0.1 i = 1.1 o.p. = 1.1 × 0.1 × RT

48. A solution of ‘x’ mole of sucrose in 100 g of water freezes at – 0.2°C. As ice separates out, the freezing point goes down to – 0.25°C. How many grams of ice would have separated? (a) 18 g (b) 20 g (c) 80 g (d) 25 g B

Sol. 100

1000

342

W2.0

gm 8.610

3422.0W

solv ent

fW

1000

Wt.M

8.6T

W

1000

342

8.625.0

W = 79.5 80 Ice = 100 – 80 = 20 49. Equal volumes of M/20 glucose solution at 300 K and M/20 sucrose solution at 300 K are mixed

without change in temperature. If the osmotic pressure of glucose solution, sucrose solution and the

mixture of two solutions are 1, 2 and 3, respectively, then

(a) 1 = 2 = 3 (b) 1 > 2 > 3 (c) 1 < 2 < 3 (d) 1 = 2 < 3

A

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Dr. Sangeeta Khanna Ph.D 9 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Physical\Test\Grand Test\Grand test-2\+2 Grand test-2.doc

50. A cylinder fitted with a movable piston contains liquid water in equilibrium with water vapour at 25°C. Which of the following operation(s) results in a decrease in the equilibrium vapour pressure at 25°C? (a) Moving the piston downward for a short distance (b) Removing small amount of vapour. (c) Removing a small amount of liquid water. (d) Dissolving some salt in the water. D

SECTION – B (Assertion and Reason) Negative Marking

This Section contains 10 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONLY ONE is correct. Marks: 10 × 4 = 40

(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 dependent 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

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Dr. Sangeeta Khanna Ph.D 10 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Physical\Test\Grand Test\Grand test-2\+2 Grand test-2.doc

7. Statement I: 0.1 m aqueous solution of glucose has higher depression in freezing point than 0.1m aqueous solution of urea Statement II: kf has same value in both a. (a) b. (b) c. (c) d. (d) D

8. Statement – I: The boiling point of 0.1 m-urea solution is less than that of 0.1 m-KCl solution. Statement – II: Elevation of boiling point is directly proportional to the number of species present in the solution and KCl solution will have more number of particles. a. (a) b. (b) c. (c) d. (d) A

9. Statement – I: Addition of a non-volatile solute to a volatile solvent increases the boiling point. Statement – II: Addition of non-volatile solute to a volatile solvent results decreases in the vapour pressure a. (a) b. (b) c. (c) d. (d) A/B

10. Statement – I: Except osmotic pressure, all other colligative properties depend on the nature of solvent Statement – II: Osmotic pressure does not depend on temperature a. (a) b. (b) c. (c) d. (d)

C

SECTION – C (Paragraph Type) This Section contains 3 Comprehension. Each of these questions has four choices A), B), C) and D) out of which ONLY ONE is correct. 10 × 4 = 40

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).

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1. The phenomenon involve in the separation of mixture of pigments is based on a. surface phenomenon b. Chemical phenomenon c. solubility phenomenon d. surface & solubility phenomenon 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 7. Correct order of Flocculation power of effective ion for colloidal gold sol is

a. K4[Fe(CN)6] > Al2(SO4)3 > BaCl2 b. Al2(SO4)3 > BaCl2 > K4[Fe(CN)6] c. BaCl2 >Al2(SO4)3 > K4[Fe(CN)6] d. BaCl2 = Al2(SO4)3 = K4[Fe(CN)6] B

Comprehension – 3 A solution is said to be ideal if each of its components obey Raoult’s law for the entire composition

range. The law states that the vapor pressure of any component in the solution depends on the mole fraction of that component in the solution and vapor pressure of that component in the pure state. Solutions are non-ideal if they do not obey Raoult’s law over the entire composition range. The vapor pressure of the solution is either higher or lower than that predicted by Raoult’s law. Depending on the type of deviation from ideal behavior, non-ideal solutions may be classified as showing negative deviation (lower vapor pressure than predicted) or positive deviation (higher vapor pressure than predicted). However, in either case, corresponding to a particular composition, they form constant boiling mixtures called azeotropes.

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8. Which of the following mixtures do you expect will show positive deviation from Raoult’s law? (a) Benzene + Acetone (b) Benzene + Chloroform (c) Benzene + Carbon tetrachloride (d) Benzene + Ethanol A,D

9. An azeotropic solution of two liquids has boiling point higher than either of the two liquids when it (a) is saturated (b) shows a negative deviation from Raoult’s law (c) shows a positive deviation from Raoult’s law (d) shows no deviation from Raoult’s law. B

10. A solution has a 1 : 4 mole ratio of heptanes to hexane. The vapor pressure of the pure hydrocarbons at 20°C are 440 mm Hg for heptanes and 120 mm Hg for hexane. The mole fraction of heptane in the vapor phase would be (a) 0.200 (b) 0.478 (c) 0.549 (d) 0.786 B

SECTION – D (More than One Answer Type) No Negative Marking

This Section contains 10 multiple choice questions. Each question has four choices A), B), C) and D) out of which One or More than one answer may be correct. Marks = 10 × 5 = 50 1. Identify the incorrect statements regarding enzymes:

a. Enzymes are specific biological catalysts that can normally function at very high temperature (T~1000K)

b. Enzymes are highly stereospecific protein c. Enzymes are specific biological catalysts that cannot be poisoned d. Enzymes are specific biological catalysts that possess well defined active sites

A,C 2. At critical micelle concentration:

a. surfactant particles ion undergo association b. Osmotic Pressure decreases. c. oil, grease, dirt etc., get dissolved d. osmotic pressure increases A,B,C Sol. At critical micelle concentration surfactant particles undergo association so osmotic pressure of

solution decreases because number of particles in the solution decreases. (a), (b) and (c) are correct. (d) is incorrect because osmotic pressure decreases when association takes place.

3. Which of the following statements are correct?

a. The small the gold number of a lyophilic colloid, larger will be its protective power b. Lyophilic sols are easily coagulated by small amount of electrolyte

c. FeCl3 coagulates blood d. The flocculation value for As2S3 sol is independent of the anion of coagulating electrolyte

A,C,D Sol. (a) is correct because it will be most effective if it has lowest gold number.

(c) is correct because FeCl3 coagulates blood by neutralizing the charge. (d) is correct because As2S3 is –vely charged colloid and coagulated by +ve ions (cations).

its floacculation is independent of charge on anion. 4. Select the correct statements among following:

a. At 83 K, N2 is physisorped on the surface of iron b. At 773 K and above N2 is chemisorped on the iron surface c. Activation energy is +ve in case of physisorption and zero in case of chemisorption d. Activation energy is zero in case of physisorption and +ve in case of chemisorption A,B,D

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5. -

)excess(I |AglKIAgI

With respect to the above reaction (a) the fixed layer is formed by I– ions (b) the diffused layer is formed by K+ ions (c) the negative charge on the above colloid is due to preferential adsorption of I– ions (d) fixed and diffused layers lead to the development of zero potential A,B,C

6. What is not true for Brownian movement of colloidal particles?

a. This motion depends on nature of colloid b. It is one of the factors responsible for stability of sols c. It is independent of size of particles d. This motion is faster when dispersion medium is highly viscous A,C,D

7. Blood cells in the human body have semipermeable membrane and depending upon concentration of solution inside blood cells and outside (in the blood), ‘Lysis’ (expansion of blood cells) and ‘Crenation’ (contraction of blood cells) may occur. Kidneys are responsible for keeping solution inside blood cell and blood at same concentration. Identify the correct information(s). (a) Lysis will occur when blood cells are kept in a solution which is isotonic with blood (b) Crenation will occur when blood cells are kept in a solution which is hypertonic with blood. (c) Blood cells will have normal shape when placed in a solution isotonic with blood (d) Lysis will occur when blood cells are kept in a solution which is hypotonic with blood B,C,D

8. Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s) is(are)

(a) G = +ve (b) Ssystem = +ve (c) Ssurrounding = 0 (d) H = 0 B,C,D

9. 1 mol benzene Hg) mm 42p( benzene and 2 mol toluene toluenep( = 36 mm Hg) will have:

(a) total vapor pressure 38 mm Hg. (b) mole fraction of vapors of benzene above liquid mixture is 7/19. (c) positive deviation from Raoult’s law. (d) negative deviation from Raoult’s law A,B

Sol. Pt = 363

242

3

1

= 14 + 24 = 38 mm/g

19

7

38

14VBenzene

10. Identify the correct statements (a) The solution formed by mixing equal volumes of 0.1 M urea and 0.1 M glucose will have the same

osmotic pressure. (b) 0.1 M K4[Fe(CN)6] and 0.1 M Al2(SO4)3 are isotonic solutions (c) For association of a solute in a solution, i > 1. (d) The ratio of van’t Hoff factors for 0.2 M glucose and 0.1 M sucrose is 2 : 1. A,B

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SECTION – E (Matrix Type) No Negative Marking

This Section contains 4 questions. Each question has four choices (A, B, C and D) given in Column I and five statements (p, q, r, and s) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. 8 × 4 = 32 Marks

1. Match the entries of column I with appropriate entries of column II. (One or More than one match)

Column – I Column – II (A) Butter (B) Rain cloud (C) Soap solution (D) fog

(p) dispersed phase is liquid (q) Medium is gas (r) Gel (s) micelles

Sol. A p, r; B p, q; C s; D p, q 2. Match the column – I with column – II. (One or More than one match)

Column I Column II

(A) Acetone + CHCl3 (p) Smix. > 0

(B) Ethanol + Water (q) Vmix. > 0

(C) C2H5Br + C2H5I (r) Hmix. < 0

(D) Acetone + Benzene (s) Maximum boiling azeotropes

(t) Minimum boiling azeotropes

Sol. A p, s, r, B p, q, t; C p; D p, q, t

A -ve deivation; B = +ve deviation; C = Ideal solution; D = Positive deviation

3. Match the solutions with their characteristics. (One or More than one match)

Column – I Column – II

(A) CH3COOH in H2O (P) Neither association nor dissociation

(B) CH3COOH in benzene (Q) When a non-volatile solute is added

(C) Polymer in water (R) Molecular mass observed greater than molecular actual

(D) Vapor pressure of a liquid decreases (S) Tf(obs.) > Tf(calc.)

(T) Van’t Hoff factor, i > 1

Sol. A s, t; B r; C p; D q

A dissociation; B Association (c) i = 1

4. For a 5% by mass solution of H2SO4 ( = 1.01 g mL–1), match the quantities with their values (One match only)

Column – I Column – II

(A) Molarity of the solution (P) 0.537

(B) Molality of the solution (Q) 0.0093

(C) Mole fraction of H2SO4 (R) 0.05

(D) Mass fraction of H2SO4 (S) 0.515

Sol. A s; B p; C q; D r

537.095

1000

98

5m 05.0

98

5SOH mole ;X 42SOH 42

M 515.001.1100

1000

98

5M 27.5

18

95 w atermole

0092.032.5

05.0

05.027.5

05.0X

42SOH

05.010

5fraction mass

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SECTION – F (Integer Type) Negative Marking [-1]

This Section contains 10 questions. The answer to each question is a single digit integer ranging from 0 to 9. 10 × 4 = 40

1. Equimolal solution of NaCl and MgCl2 are prepared in water. Freezing point of NaCl is found to be – 2°C. What is freezing point depression of MgCl2 solution.

Sol. 3 2. 1 litre hard water contains 2 × 10–3 g of CaCO3. Its degree of hardness in ppm will be equal to ______. Sol. 2 3. When a platinum rod is dipped in 11,200 ml of H2 gas at STP. Volume of H2 reduces to 5600 ml. The

no. of mole of H2 adsorbed in pt is 25 × 10–x. Calculate x? Sol. 2

ninitial = 2

1mol = 0.5

nfinal = 0.25 mols adsorbed = 0.25 i.e. 25 × 10–2

4. Molarity of an aqueous solution of urea having mole fraction of urea 6

1. Density of solution is 1.2

gm/ml. Sol. (8)

Mole fraction of urea ‘xu’ = 51

1

6

1

nu = 1 nw = 5

Wt. of urea = 1 × 60 Wt. of water

= 60 gm = 5 × 18 = 90 gm

Wt. of solution = 60 + 90 = 150 gm

density

solution of w t.solution of Volume

ml2.1

150

2.1/150

10001

ml V

1000nM u

mol/lit 8150

1200

5. A solute ‘S’ undergoes a reversible trimerization when dissolved in a certain solvent. The boiling point elevation of its 0.1 molal solution was found to be identical to the boiling point elevation in case of a 0.08 molal solution of a solute which neither undergoes association nor dissociation in same solvent. If percent of the solute ‘S’ that undergone trimerization is x × 10; What is x.

Sol. 3 3S S3

1 0

1- 3

i = 1 -

3

2

Now 08.03

211.0

= 0.3. Hence 30% trimerization. 6. The coagulation of 10 ml colloidal solution of gold is completely prevented by addition of 0.02 gm of a

substance X to it before addition of 1 ml of 10% NaCl solution. The gold number of X is 2 × 10n n is

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Sol. 1 Mass in milligm = 0.02 × 103 = 20 = 2 × 101

7. Graph between log

m

x and log P is straight line at angle of 45° with the intercept of 0.6020.

The extent of adsorption

m

x at a pressure of 1 atm is:

Sol. Plogn

1Klog

m

xlog

= 0.602 + 1 log 1 = 0.602

m

x = 4

8. At 20°C, the osmotic pressure of urea solution is 400 mm Hg. The solution is diluted and the temperature is raised to 35°C, when the osmotic pressure is round to be 105.3 mm. Hg. Determine extent of dilution.

Sol. 4

2

1

1

2

2

1

T

T

V

V

P.O

P.O

1

2

V

V

293

308

3.105

400

4 99.3V

V

1

2

9. The depression in freezing point for 1 M urea, 0.5 M glucose, 1 M NaCl and 1 M K2SO4 are in the ratio x : 1 : y : z. The value of x + z is __________.

Sol. 8 Glucose = 1; urea = 2 (x); NaCl = 4(y); K2SO4 = 6 (z) 10. The elevation in boiling point for 0.3 molal Al2(SO4)3 solution as compared to elevation in boiling point

of 0.1 molal solution of Na2SO4 is __________ times. Sol. 5

2

1

2

1

2

1

m

m

i

i

T

Tf

1.03

3.05

= 5

45°

0.6020

m

xlog

log p