test venue: lajpat bhawan, madhya marg, sector 15-b ...jun 02, 2018 · (c) lowering in freezing...

Dr. Sangeeta Khanna Ph.D 1 CHEMISTRY COACHING CIRCLE D:\Important Data\2018\+2\Physical\Test\Grand Test-2\+2 Grand Test-2.doc

Test Venue:

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

Dr. Sangeeta Khanna Ph.D


Test Dated 25.6.2018

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

2. The maximum marks are 362.

3. This test consists of 86 questions.

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

Section – A (Single Correct Choice Type) Negative Marking [-1]

This Section contains 45 multiple choice questions. Each question has four choices A), B), C) and D) out

of which ONLY ONE is correct. 45 × 4 = 180 Marks

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. 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

3. What is the molality of the 870 g solution made by dissolving 120 g Br2 in CHC3. [At. Wt. Br = 80]

[M.wt of Br2= 160] a. 1 b. 1/2 c. 3/4 d. 1/4 A

Sol. 750

1000

4

3

]120870[

1000

160

120M

= 1

4. 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

5. 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

6. 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



7. 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

8. 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

9. 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

10. 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

11. 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 .

12. 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 13. 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

14. 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 15. 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



16. 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

17. 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

18. 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 19. 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

20. 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

21. 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

22. 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



23. 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

24. 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

25. 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%

26. 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

27. 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

28. 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

29. 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 + 30. 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]



31. 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

32. 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) 33. 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

34. 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 35. 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 36. 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



37. The cell reaction for the given cell is spontaneous if

Pt, C2 | C– (1 M) || C– (1 M) | Pt, Cl2

(P1) (P2) a. P1 > P2 b. P1 < P2 c. P1 = P2

d. P2 = 1 atm. B

Sol. 1

20red

0oxicell

P

P log

2

059.0EEE Pt ,Cl|Cl||Cl|Cl ,Pt

21 P

2

P

2

= 1

20cell P

Plog

2

059.0E

Now Ecell will be + ve if P2 > P1 38. The hydrogen electrode is dipped in a solution of pH = 3 at 25°C. The potential of the cell would be

(the value of 2.303 RT/F is 0.059 V)

a. 0.177 V b. 0.087 V c. -0.177 V d. 0.059 V C

Sol. H+ + e– 2H2

1

E = E° - ]H[

1log

n

059.0

= 0 - pH1

059.0 = -0.059 × 3 = -0.177 V

39. In the electrolysis of copper chloride solution using copper electrodes, the mass of cathode increases by 3.25 g. In the anode,

a. 0.05 mol of Cu will go into solution as Cu2+ ions b. 560 mL of O2 at S.T.P. will be liberated c. 112 mL of Cl2 will liberate d. 3.2 mol of copper metal will dissolve A

Sol. CuCl2 Cu+2 + 2Cl–

Copper deposited = 5.63

25.3 = 0.05 mol

The Cu anode will give same amt. of Cu+2 ions in solution. 40. Four moles of electrons were transferred from anode to cathode in an experiment on electrolysis of

water. The total volume of the two gases (dry and at STP) produced will be approximately (in litres)

a. 22.4 b. 44.8 c. 67.2 d. 89.4 C

Sol. H2O H2 + 2O2

1

or 2H+ + 2e– H2

i.e., 2 moles of e– produce 1 mole of H2 i.e., 22.4 L. Hence 4 moles of e– will give H2 = 44.8 L.

O2 produced = 2

1of H2 = 22.4 L

Total volume = 44.8 + 22.4 = 67.2 L 41. Red hot carbon will remove oxygen from the oxide AO and BO but not from MO, while B will remove

oxygen from AO. The activity of metals A, B and M in decreasing order is:

a. A > B > M b. B > A > M c. M > B > A d. M > A > B C Sol. C removes oxygen from AO and BO, C is better reducing agent than A and B; B remove oxygen from

AO, B is better reducing agent than A. Better reducing means more active. Hence the correct order is M > C > B > A.



42. Point out the correct statement: 1. Electrolysis of aqueous solution of LiCl shows pH > 7 2. Oxidation number of C in carbon suboxide is +2/3 3. Oxidation number of P in sodium hypophosphite is +1 4. The oxidation number and covalency of oxygen in O2 molecule are 0 and 2

a. 1, 3 b. 1, 2, 4 c. 1, 2, 3, 4 d. 1, 3, 4 D

Sol. Carbon suboxide is 2

2

3

x

OC

3x – 4 = 0 3x = +4 or x = +4/3 43. In the diagram given below the value of X is:

a. 0.325 V b. 0.65 V c. -0.35 V d. -0.65 V A

Sol. 02

01

03

GGG

= -0.15 F – 0.5 F = -0.65 F

03

03

nFEG

-0.65 F = 03

FE2

03

E = 0.325 V

44. For the cell, Pt; H2 (1 atm) | HCl (0.1M)|| CH3COOH (0.1M)|H2 (1 atm); Pt, the emf will not be zero because

a. the temperature is constant b. acids used in two half cells are different in basicity c. emf depends upon molarities of acids used d. pH of 0.1 M HCl & 0.1 M CH3COOH is not the same D

Sol. CH3COOH (0.1 M) will not give 0.1 M H+ -ion concentration because of being weak acid while HCl (0.1 M) will give 0.1 M H+ -ion concentration.

45. In H2 – O2 fuel cell, which of the following is correct when acid is used as electrolyte?

a. anode reaction : H2(g) 2H+ + 2e– b. Cathode reaction : 2H+ + 2e– + )g(O2

12 H2O()

c. Net reaction: H2(g) + 2

1O2(g) H2O () d. All of the above

D

Cu2+

Cu+ Cu

+ 0.15 V + 0.50 V

E0 = X volt



Level – II SECTION – B (Assertion and Reason) Negative Marking [-1]

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

(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: 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

3. 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

4. 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

5. 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

6. Assertion: 0.1 m aqueous solution of glucose has higher depression in freezing point than 0.1m aqueous solution of urea Reason: kf has same value in both a. (a) b. (b) c. (c) d. (d) D

7. Assertion: The boiling point of 0.1 m-urea solution is less than that of 0.1 m-KCl solution. Reason: 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

8. Assertion: Galvanic cell Zn| Cu|Cu||Zn 2M

2M 21

; emf of cell is more than E°(1.1V) if M1 > M2

Reason: According to Nernst’s equation as concentration of cathodic half cell increases, emf will increase a. (a) b. (b) c. (c) d. (d) D



9. Assertion: Except osmotic pressure, all other colligative properties depend on the nature of solvent Reason: Osmotic pressure does not depend on temperature a. (a) b. (b) c. (c) d. (d)

C 10. Assertion: In electrolysis, the quantity of electricity needed for depositing 1 mole of silver is different

from that required for 1 mole of copper. Reason: The molecular weights of silver and copper are different. a. (a) b. (b) c. (c) (d). d B

SECTION – C (Paragraph Type) Negative marking [-1] This Section contains 3 Comprehension. Each of these questions has four choices A), B), C) and D) out of which ONLY ONE is correct. 9 × 4 = 36 Marks

Comprehension – 1 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.

1. 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 2. 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 3. 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 – 2

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.

4. 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 + Toluene A,B,C



5. 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

6. 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

Comprehension – 3

Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately 6.023 × 1023) are present in a few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept. A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes. (Atomic masses: Na = 23, Hg = 200; 1 Faraday = 95600 C)

7. The total number of moles of chlorine gas evolved is

a. 0.5 b. 1.0 c. 2.0 d. 3.0 B

Sol. No. of mole of NaCl = mol 21000

5004

& mol of Cl2 = 1 mol

8. If the cathode is a Hg electrode, the maximum mass (gram) of amalgam formed from this solution is

a. 200 b. 225 c. 400 d, 446 D

Sol. Na deposited is 2 mol; Mass of Na(Hg) = 2 × 23 + 400 = 446 9. The total charge (Coulombs) required for complete electrolysis is

a. 24125 b. 48250 c. 96500 d. 193000 D

Sol. 2 F required for 2 mol NaCl

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. 10 × 5 = 50 Marks 1. 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.



2. 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

3. -

)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

4. 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

5. 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

6. 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

7. 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



8. 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

9. Using the standard potential values given below, decide which of the statements I, II, III, IV are correct. Choose the right answer from (a) , (b), (c) and (d)

Fe2+ + 2e– = Fe, E0 = -0.44 V Cu2+ + 2e– = Cu, E0 = + 0.34 V Ag+ + e– = Ag, E0 = + 0.80 V I. Copper can displace iron from FeSO4 solution II. Iron can displace copper from CuSO4 solution III. Cu can displace Ag from AgNO3 solution IV. Iron can displace silver from AgNO3 solution a. I and II b. II and III c. II and IV d. I and IV B, C 10. The values of E0 of some of the reactions are given below:

I2 +2e–2I– ; E0 = +0.54 volt Cl2 + 2e–

2Cl– ; E0 = +1.36 volt

Fe3+ + e– Fe2+; E0 = +0.76 volt Ce4+ + e–Ce3+; E0 = +1.60 volt

Sn4+ + 2e–Sn2+; E0 = +0.15 volt

Which of the following statement is correct? a. Fe+3 will oxidise Ce+3 b. Fe+2 will be oxidised by Ce+4 c. I2 does not displace Cl2 from KCl d. FeCl3 can oxidise SnCl2 B,C,D

SECTION – E (Matrix Type) No Negative Marking

This Section contains 2 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 × 2 = 16 Marks

1. 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 2. Match the Column – I with Column – II. [Single Match]

Column – I Column – II

(A) No. of Faradays of electricity produced by combustion of 1 mol benzene (p) 10

(B) No. of electrons gained by 2 mol of KMnO4 in acidic medium (q) 20

(C) Eq. Wt. of NH4NO3 in given reaction

OHONNONH 2234

(r) 30

(D) No. of electron exchanged in the given reaction:

3Cl2 + 6NaOH 5NaCl + NaClO3 + 3H2O

(s) 5

Sol. A R; B P; C Q; D S



A. C6H6 + 2

15O2 6CO2 + 3H2O x = 30

B. 2KMnO4 + 16H+ + 10e– 2Mn+2 + 8H2O + 2K+ x = 10

C. eq. NH 4

= 5.44

18 ; eq. mass of NO

3 = 5.154

62

eq. mass NH4NO3 = 4.5 + 15.5 = 20

D. 2e– + Cl2 2Cl–) × 5

Cl2 2ClO3– + 10e–

6Cl2 10Cl– + 2ClO3–

3Cl2 5Cl– + ClO3– (5e–)

SECTION – F (Integer Type) No Negative Marking

This Section contains 10 questions. The answer to each question can be between from 0 to 100. 10 × 4 = 40 Marks

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. 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 3. 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. 4. 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 Sol. 1 Mass in milligm = 0.02 × 103 = 20 = 2 × 101 5. 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

6. 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) 7. 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 8. Liquids A and B form ideal solution over the entire range of composition. At temperature T, equimolar

binary solution of liquids A and B has vapour pressure 45 Torr. At the same temperature, a new solution of A and B having mole fractions xA and xB, respectively, has vapour pressure of 22.5 Torr. The value of xA/xB in the new solution is ____. (given that the vapour pressure of pure liquid A is 20 Torr at temperature T)

Sol. 19

PT = BAXp

45 = 20(0.5) + )5.0(pB

70pB

22.5 = 20XA + 70(1 – XA) 50XA = 47.5

XA = 95.05

75.4

XB = 0.05

19X

X

B

A

9. How many of the following Half cell have Reduction electrode potential more than standard Reduction

potential (E°).

(a) )M 1(atm 2

2 H|H|Pt (b) )M 2(atm 1

2 H|H|Pt (c) Cu|Cu 2)M01.0(

(d) Pt|Fe ,Fe||Zn|Zn2M

2

M1

3

M2

2 (e) 2pHatm 1

2 H|H|Pt

(f) )M1(atm 2

2 Cl|Cl|Pt

Sol. (2)

b, f

(a) 2H+ + 2e– H2

E E

22

Hatm 1PH ;

]H[

plog

2

059.0E 2

(b) H+ > 1 M E > E°

(c) Cu+2 + 2e– Cu ; )Culog(2

059.0EE 2

Cu+2 < 1 M ; E < E°

(d) Zn + 2Fe+3 Zn+2 + 2Fe+2

23

222

]Fe[

]Fe][Zn[log

2

059.0EE



[Zn+2] & [Fe+2] > 1 M

E < E°

(e) atm 1

210

He2H22

E E ;1

]10[log

2

059.0E

2

(f) Cl2 + 2e– Cl–

2

1log

2

059.0EE

E > E°

10. How many Faraday’s will be required to convert nitrobenzene to nitrosobenzene Sol. (2)

NO2 +3 N = O

+1

2e– +

test venue: lajpat bhawan, madhya marg, sector 15-b ...jun 02, 2018 · (c) lowering in freezing...

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