test venue: lajpat bhawan, madhya marg, sector …...20 ml of co was mixed with 50 ml of oxygen and...

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

Test Venue:

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

Dr. Sangeeta Khanna Ph.D


READ THE INSTRUCTIONS CAREFULLY

1. The test is of 2 hours 15 min. duration.

2. The maximum marks are 268

3. This test consists of 65 questions.

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

Single Answer 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. (Mark only One choice) 45 × 4 = 180 Marks

1. One of the following combinations illustrate law of reciprocal proportions?

a. N2O3, N2O4, N2O5 b. NaCl, NaBr, NaI c. CS2, CO2, SO2 d. PH3, P2O3, P2O5 C

2. When 1 mol of a gaseous unsaturated hydrocarbon is burnt in excess of O2, a contraction of volume equal to triple the volume of hydrocarbon burnt. The hydrocarbon is

a. CH4 b. C2H6 c. C3H4 d. C4H8 D

Sol. CH4 + 2O2 CO2 + 2H2O() contraction in volume = 2

C2H6 + 2

7O2 2CO2 + 3H2O contraction in volume = 2.5

C3H4 + 4O2 3CO2 + 2H2O contraction in volume = 2

C4H8(g) + 6 O2 4CO2(g) + 4H2O contraction in volume = 3 3. If 20 mole SO2; 20 mole O2; 20 mole H2O are used and 18 mole of H2SO4 is produced. What is the

percentage yield of reaction 2SO2 + O2 + 2H2O 2H2SO4

a. 98% b. 80% c. 90% d. 100% C

Sol. Theoretical yield = 20 mole, as SO2 is limiting reagent

Percentage yield = 10020

18 = 90%

4. For the reaction Ba(OH)2 + 2HClO3 Ba(ClO3)2 + 2H2O, Calculate the no. of mole of H2O formed when 0.1 mole of Ba(OH)2 is treated with 0.250 mole HClO3.

a. 0.1 b. 0.125 c. 0.25 d. 0.2

D Sol. Limiting reagent is Ba(OH)2 5. The following diagrams represent the reaction of A2

(shaded spheres) with B2 (unshaded spheres). Identify the limiting reactant and write a balanced equation for the reaction.

a. A2 is the limiting reactant; A + 3 B AB3

b. A2 is the limiting reactant; A2 + 3 B2 2 AB3

c. B2 is the limiting reactant; A + 3 B AB3.

d. B2 is the limiting reactant; A2 + 3 B2 2 AB3 D 6. What is the % of free SO3 in a sample of oleum labelled as 109%

a. 20% b. 9% c. 40% d. 80%

saturated hydrocarbon

unsaturated hydrocarbon



C 7. What is equivalent mass of carbon in CO2

a. 3 b. 4 c. 6 d. 12 A

Sol. 32gm oxygen combines with 12 gm C

8 gm oxygen = 3832

12

8. Number of atoms in 1 ml of ammonia gas at STP is

a. 2.7 × 1019 b. 1.08 × 1020 c. 10.8 × 1020 d. 5.4 × 1019 B

Sol. 22400 m have No. molecule = 22400

N0 ;

No. of atom = 4 × 22400

N0

9. 20 mL of CO was mixed with 50 mL of oxygen and the mixture was exploded. On cooling, the resulting mixture was shaken with KOH. Which gas and what volume of it will be present finally?

a. CO, 5 mL b. O2, 5 mL c. CO, 10 mL d. O2, 40 mL D

Sol. CO + 2

1O2 CO2

20ml 50 ml Unused O2 = 40 10. Ammonia is produced by the following sequence of reactions

Step 1 CH4 + H2O 3H2 + CO

Step 2 N2 + 3H2 2NH3

How many gms of CH4 must be used to produce 25 gm NH3 assuming 100% yield in each step. a. 10.8 gm b. 11.7 gm c. 13.8 gm d. 14.1 gm B

Sol. Moles NH3 = 47.117

25 Moles of CH4 =

2

47.1

Mass = 162

47.1 = 11.7 gm

11. Three volumes of a gaseous hydrocarbon containing, hydrogen and sulphur is burnt in an excess of oxygen to yield three volumes of CO2, three volumes of SO2 and six volumes of water vapour. The formula of the compound is

a. C6H6S b. C4H4S c. CH4S d. C2H6S

C 12. Given the numbers 786, 0.786 and 0.0786. The number of significant figures for the three numbers is

(a) 3, 4 and 5 respectively (b) 3, 3 and 3 respectively (c) 3, 3 and 4 respectively (d) 3, 4 and 4 respectively B

Sol. Zeros to the left of the first non-zero digit in a number are not significant 13. If law of conservation of mass was to hold true, then 20.8 g of BaCl2 on reaction with 9.8 g of H2SO4

will produce 7.3 g of HCl and BaSO4 equal to

(a) 11.65 g (b) 23.3 g (c) 25.5 g (d) 30.6 g

B



Sol. g 73

4g 98

42g 2082 HCl2BaSOSOH BaCl

Mass of BaSO4 produced = 20.8 + 9.8 – 7.3 = 23.3 g

14. Which one of the following pair of substances illustrates law of multiple proportions?

(a) CO, CO2 (b) NaCl, NaBr (c) H2O, D2O (d) MgO, Mg(OH)2

A

15. A sample of Na2CO3 contains 6.02 × 1023 Na+ ions. The mass of the sample is (at. Mass Na = 23, C = 12, O = 16):

(a) 53 g (b) 106 g (c) 165 g (d) 212 g A

Sol. No. of Na+ ions = 6.02 × 1023

No. of formula units of 2

1002.6CONa

23

32

= 3.01 × 1023 Molar mass of Na2CO3 = 2 × 23 + 12 + 16 × 3 = 46 + 12 + 48 = 106 g mol –1 6.02 × 1023 formula units of Na2CO3 has mass = 106 g 3.01 × 1023 formula units of Na2CO3 has mass

= g 531002.6

1001.3106

23

23

16. Two oxides of a metal contain 50% and 40% metal (M) respectively. If the formula of first oxide is MO2, the formula of second oxide will be

(a) MO2 (b) MO3 (c) M2O (d) M2O5 B

Sol. Oxide I Oxide II Metal, M 50% 40% Oxygen, O 50% 60% As first oxide is MO2 Let atomic mass of M = x

% O = 10032x

32

or

32x

32

100

50

or

32x

325.0

or 0.5x + 16 = 32

0.5 x = 16 x = 32 At. mass of metal M, x = 32 2nd oxide M O 40 60

Mole 32

40

16

60

1.25 3.75 1 3 Formula = MO3 17. 60 g of a compound on analysis gave 24 g C, 4 g H and 32 g O. The empirical formula of the

compound is:

(a) C2H4O2 (b) C2H2O2 (c) CH2O2 (d) CH2O D Sol.

Mass in g % age M. Ratio S.R

C 24 40 33.3

12

40

1.0



H 4 6.66 66.6

1

66.6

2.0

O 32 53.33 33.3

16

33.53

1.0

Empirical formula = CH2O 18. A metal nitride M3N2 contains 28% of nitrogen. The atomic mass of metal M is (a) 24 (b) 54 (c) 9 (d) 87.62 A

Sol. % N = 10028x3

142

100x328

2828

x = 24 19. The mass of CaCO3 is required to react with 25 mL of 0.75 M HCl is (a) 0.94 g (b) 9.4 g (c) 0.094 g (d) 0.49 g A

Sol. CaCO3 + 2HCl CaCl2 + CO2 + H2O 25 mL of 0.75 M HCl

)L mol 75.0(L1000

25 1-

= 0.01875 mol

Moles of CaCO3 required = 2

HCl of Moles

mol 10 375.92

01875.0 3-

Mass of CaCO3 required = 9.375 × 10-3 mol × 100 g mol–1 = 0.9375 g = 0.94 g

20. 50 mL solution of BaCl2 (20.8% w/v) and 100 mL solution of H2SO4 (9.8% w/v) are mixed (Ba = 137, Cl = 35.5, S = 32)

BaCl2 + H2SO4 BaSO4 + 2HCl BaSO4 formed is: a. 23.3 g b. 46.6 g c. 29.8 g d. 11.65 g

D

Sol. 2HCl BaSO SOH BaCl

mol 14

mol 142

mol 12

Mass of BaCl2 = g100

8.2050

Moles of BaCl2 = 05.0100208

8.2050

mol

Mass of H2SO4 = g 8.9100

8.9100

Moles of H2SO4 = mol 1.098

8.9

Here BaCl2 is the limiting reactant

1 mol of BaCl2 gives 1 mol of BaSO4

Mass of BaSO4

= 0.05 mol × [ (137 + 32 + 64) g mol – 1] = 0.05 mol × 233 g mol –1 = 11.65 g



21. 2 mol of H2S and 11.2 L of SO2 at N.T.P. react to form x moles of sulphur, x is

SO2 + 2H2S 3S + 2H2O (a) 1.5 (b) 3 (c) 11.2 (d) 6 A Sol. 22.4 L of SO2 at N.T.P. = 1 mol

11.2 L of SO2 at N.T.P. = 0.5 mol

OH2S3S2H SO 2mol 22

mol 1 2

Here SO2 is the limiting reactants

Moles of sulphur formed = 3 × moles of SO2 = 3 × 0.5 mol = 1.5 mol 22. A certain amount of a metal whose equivalent mass is 28 displaces 20 lit of H2 at S.T.P. from an acid.

Hence, mass of the element is: (a) 50.0 gm (b) 5.08 g (c) 3.50 g (d) 7.00 g A Sol. 11.2 lit H2 displaced by 28 gm

20 lit H2 by = 202.11

28 = 50 gm

23. 1 mol of CH4 contains

(a) 6.02 × 1023 atoms of H (b) 4 g-atom of hydrogen (c) 1.81 × 1023 molecules of CH4 (d) 3.0 g of carbon

B Sol.1 mol of CH4 = 1 mole of C + 4g atoms of H. 24. Which of the following has the smallest number of molecules?

a. 0.1 mol of CO2 gas b. 11.2 L of CO2 gas at N.T.P. c. 22 g of CO2 gas d. 22.4 × 103 ml of CO2 gas A

Sol. 22 CO mol 5.022.4

11.2 ;CO mol 1.0 ;

22 CO mol 122400

22400 ;CO mol 5.0

44

22

25. The volume of water to be added to 100 cm3 of 0.5 NH2SO4 to get decinormal

N

10

1 concentration

is:

a. 400 cm3 b. 500 cm3 c. 450 cm3 d. 100 cm3 A

Sol. N1V1 = N2V2; 100 × 0.5 = V10

1 ; V = 500;

Volume of water added = 400 mL 26. How many significant figures does the result of the following calculation have?

3950.2

6500.0235.000.635

a. 2 b. 3 c. 4 d. 5



B

Sol. 40.999 (Minimum significate No = 3; As = 40.5

has been significant figures ‘3’ corresponding to the term 0.235.

27. How many atoms do 70 amu and 70 g of Li73

respectively have?

a. 10, 10 b. 10, 10 NA c. 10 NA, 10 NA d. 10 NA, 10

B

Sol. 70 amu Li = 107

70 atoms

70 g Li = 7

70mol = 10 mol = 10 NA atoms

28. H3AO4 has 31.6% of ‘A’ by mass. What is the atomic mass of the atom ‘A’?

a. 17 b. 31 c. 35.5 d. 40

B

Sol. Let the mass of ‘A’ = x

Molar mass of H3AO4 = 3 + x + 64 = 67 + x

According to the question,

6.31x67

100x

x = 30.95 ~ 31

29. What volume of hydrogen gas at 273 K and 1 atm pressure will be consumed in obtaining 21.6 g

elemental boron (atomic mass = 10.8) from the reaction of boron trichloride by hydrogen? [2BCl3 +

3H2 6HCl + 2B]

a. 67.2 L b. 44.8 L c. 22.4 L d. 89.6 L

A

Sol. HCl62B 3H BCl2

g 21.610.82

L 67.2L 22.43

mol 3 23

At NTP volume of 3 mole H2 is 67.2 lit

30. The total charge in coulombs on 19 g of 34

PO ions is : [At. Wt. of P = 31]

a. 6.02 × 1023 b. 1.8 × 103 c. 5.79 × 104 d. 57.9

C

Sol. 95 g i.e., 1 mol 34

PO 3F charge

19g 34

PO 95

193 × 96500 C = 57900 C = 5.79 × 104 C

31. O2 gas produced by the reaction of 2.45 g of KClO3 is completely reacted with H2 gas obtained by

the reaction of zinc and dil. H2SO4. What weight of zinc is required for the production of this H2 gas?

(molar mass of KClO3 = 122.5 g, Zn = 65 g) Reaction involved are :

2KClO3

KCl + 3O2

3O2 + 6H2 6H2O

6Zn + 6H2SO4 6H2 + 6ZnSO4

a. 1.3 g b. 2.6 g c. 3.9 g d. 5.2 g

C

Sol. 2KClO3 2KCl + 3O2

3O2 + 6H2O 6H2O



6 Zn + 6H2SO4 6H2 + 6 ZnSO4 2 × 122.5 g KClO3 = 6 × 65 g Zn

g5.1222

Zn g 65 6 KClO 45.2 3

Zn = 3.9 g Zn

32. Density of a 2.05 M solution of acetic acid (molecular mass 60) in water is 1.02 g/mL. The molality of the solution is:

a. 3.28 mol kg – 1 b. 2.28 mol kg -1 c. 0.44 mol kg–1 d. 1.14 mol kg–1 B

Sol.

BMM-1000d

M 1000Molality

6005.202.11000

05.21000

1- kg mol 285.2897

2050

1231020

2050

2.05 moles in 1000 ml solution

mass of solution = 1000 × 1.02

= 1020 gm

mass of solvent = 1020 – 2.05 × 60

= 1020 – 123 = 897

m = 1000897

05.2

= 2.28 m

33. The density of a 3.60 M soluphuric acid solution that is 29% H2SO4 (molar mass = 98 g mol –1) by

mass, will be:

a. 1.45 gm/m b. 1.65 gm/m c. 1.88 gm/m d. 1.22 gm/m D

Sol. BM

d % x10M

22.1216.12910

986.3

%x10

MMd B

34. 20 mL of 0.1 M, 30 mL of 0.2 M and 30 mL of 0.3 M solutions of oxalic acid are mixed and the volume

is made 100 mL. The molarity of the resulting solution is:

a. 0.21 M b. 8.51 M c. 5.67 M d. 0.17 M

D

Sol. VolumeTotal

VMVMVMMR 332211

M 17.0100

17

100

962

100

)3.030()2.030()1.020(

35. A hypothetical metal oxide has 32% metal. The V.D. of Chloride salt of this metal is 79. The atomic

mass of the metal is:

a. 16.16 b. 46.67 c. 15.04 d. 50.78

C

Sol. Eq. Wt. Of metal = 76.368

832

oxygen of Weight

8Weight

Valency = 435.53.76

792

35.5E

V.D.2

At. wt. = E × V = 3.76 × 4 = 15.04 36. Which of the following contains the greatest number of atoms.

a. 1.0 g of butane (C4H10) b. 1.0 g of nitrogen (N2) c. 1.0 g of silver (Ag) d. 1.0 g of water (H2O) A



Sol. Number of atoms = Number of molecules × number of atoms in one molecule Number of atoms in 1.0 g butane (C4H10)

23A 1045.114N

58

1

(1 molecule contains 14 atoms) Number of atoms in 1.0g N2

2223 103.4210023.628

1

Number of atoms in 1.0 g silver (Ag)

2123 105.510023.6108

1

Number of atoms in 1.0 g water (H2O)

2323 10310023.618

1

This butane has maximum number of atoms 37. Specific heat of metal is 0.6 and its chloride contains 87% of chlorine. What is the molecular weight of

the metal oxide?

a. 24 b. 54 c. 62 d. 27 D

Sol. Weight of metal chloride = 100 g Weight of chlorine = 87 g Weight of metal = 100 – 87 = 13g

Eq. wt. of metal = 5.35chlorine of tWeigh

metal of Weight

30.55.3587

13

Approx. at. wt. = 6.106.0

4.6

heat .sp

4.6

Approx. valency = 23.5

6.10

At. wt. of metal = 5.30 × 2 = 10.60

Mol. wt. of metal oxide, MO = 10.60 + 16 = 26.6 27 38. The number of atoms in 0.1 mole of triatomic gas is (NA = 6.022 × 1023 mol-1)

a. 6.026 × 1022 b. 1.806 × 1023 c. 3.600 × 1023 d. 1.800 × 1022 B

Sol. Number of atom = No. of moles × NA × atomicity = 0.1 × 6.022 × 1023 × 3 = 1.806 × 1023 39. What will be equivalent mass of H3PO4 in the given reaction

2NaOH + H3PO4 Na2HPO4 + 2H2O

a. Molar mass (m) b. 2

M c.

3

M d. Can’t be predicted

B 40. What will be molarity of H2O2 solution having volume strength ‘20 volume’.

a. 2 M b. 1.78 M c. 1.34 M d. 0.178 M B

Sol. 78.12034

3.03strength Volume

34

03.3M



41. 1 lit oxygen at NTP is allowed to react with three times of CO at NTP. Volume of each gas after the reaction

a. CO2 = 2 lit ; CO = 1 lit b. CO2 = 3 lit ; CO = 2 lit c. CO2 = 3 lit ; CO = zero d. CO2 = 4 lit; CO = zero A

Sol. lit. 2

2lit.

2

unused) lit. (1lit. 2

CO2 O CO2

42. 1 lit. of a solution of 0.7 M HCl was heated in a beaker & it was found that when the volume of the solution was reduced to 500 ml 3.4 gm of HCl was lost. What will be molarity of resulting solution?

a. 1.82 M b. 1.21 M c. 2.42 M d. 0.7 M B

Sol. M = V

1000

.Wt.M

mass

5.36

Mass7.0

Mass = 25.55 gm mass lost = 3.4 g Mass left = 25.55 – 3.4 = 22.15 gm V = 500 ml

M = M 2.1500

1000

5.36

15.22

43. Two students performed the same experiment separately and each one of them recorded two readings of mass which are given below. Correct reading of mass is 3.0 g.

Student Readings

I II

A B

3.00 3.05

2.99 2.95

On the basis of given data, mark the correct option out of the following statement. a. Results of both the students are neither accurate not precise b. Results of student A are both precise and accurate

c. Results of student B are neither precise nor accurate d. Results of student B are both precise and accurate B

Sol. Results of student A are both precise and accurate. The values are precise (3.01 and 2.99) as they are close to each other. The values are accurate as they are close to the true value (3.00 g)

44. Match the following Column I and Column II and choose the correct codes from the option gives below.

Column – I Column - II

A. 46 g of Na 1. 0.01 mol

B. 6.022 × 1023 molecules of H2O 2. 2 mol

C. 0.224 L of O2 STP 3. 1 mol

D. 84 g of N2 4. 22.7 lit at STP

E. 1 mol of any gas 5. 3 mol

Codes A B C D E A B C D E a. 2 3 1 5 4 b. 1 2 3 4 5 c. 4 2 1 3 4 d. 5 4 3 1 2



A 45. The number of moles of BaCO3, which contain 1.5 moles of oxygen atom are

a. 0.5 b. 1 c. 3 d. 6.02 × 1023 A

Sol. Formula of the compound BaCO3 suggested that 3 moles of oxygen atoms are contained in one mole of BaCO3.

1.5 mole will be contained in 0.5 mole of BaCO3

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

This Section contains 5 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONLY ONE is correct. Marks: 5 × 4 = 20

(A) Statement -1 is True ; Statement-2 is True ; Statement-2 is a correct explanation for Statement-1.

(B) Statement-1 is True; Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement -1 is True ; Statement-2 is False. (D) Statement-1 is False ; Statement-2 is True 1. Statement–1: When 7.0 g nitrogen and 3.0 hydrogen are allowed to react to form ammonia as a

single product, 10.0 g ammonia is formed. Statement-2: Chemical reactions follows the Law of conservation of mass. a. (A) b. (B) c. (C) d. (D) D

2. Statement – 1: Vapour density of CH4 is half of O2 Statement – 2: 1.6 g CH4 contains same number of electrons as in 3.2 g of O2

a. (A) b. (B) c. (C) d. (D) C

Sol. Vapour density of methane = 8 Vapour density of oxygen = 16 1.6 g CH4 has 6.023 × 1023 electrons while 3.2 g O2 has 0.1 × 16 × 6.023 × 1023 electrons. 3. Statement-1: When a hydrocarbon is burnt and the products of combustion are cooled to the original

temperature and pressure, a contraction in volume occurs. Statement-2: The contraction in volume is solely due to the liquefaction of water vapours. a. (A) b. (B) c. (C) d. (D)

C 4. Statement-1: In compounds, equivalent mass of an element always combines with equivalent mass

of other element Statement-2: It is according to Dalton’s atomic theory a. (A) b. (B) c. (C) d. (D) B

5. Statement–I: Concentration of any solution is independent from the amount of solution, but it depends on the relative amount of solute and solvent. Statement–II: Concentration of a solution has same magnitude in any unit to express concentration.

a. (A) b. (B) c. (C) d. (D) C



SECTION – C (Comprehension Type) Negative Marking [-1]

This Section contains 3 paragraphs. Each of these questions has four choices A), B), C) and D) out of which ONLY ONE is correct. 7 × 4 = 28 Marks

Comprehension – 1 Isotopes are the atoms of same element; they have same atomic number but different mass numbers. Isotopes have different number of neutrons in their nucleus. If an element exists in two isotopes having atomic masses ‘a’ and ‘b’ in the ratio m : n, then average atomic mass will be

nm

bnam

.

Different isotopes of same element have same position in the periodic table. The elements which have single isotope are called monoisotropic elements. Greater is the percentage composition of an isotope, more will be its abundance in nature.

1. Naturally occurring Bromine is a mixture of 79Br & 81Br. What is isotopic composition of naturally

occurring Bromine (81) [Average Atomic mass of Bromine 80]

a. 50% b. 25% c. 75% d. 20% A

So. Average mass = 100

)x100(81x79 ; x = 50

2. The ratio of the mass of C – 12 atom to that of an atom of elements X (whose atomicity is four) is 1 : 9. The molecular mass of elements X is:

a. 480 g mol–1 b. 432 g mol–1 c. 36 g mol–1 d. 84 g mol–1 B

Sol. 108

12

9

1

X

C12 i.e. atomic mass of X = 108

Molar mass of X4 = 108 × 4 = 432 3. Average atomic mass of magnesium is 24.31 a.m.u. This magnesium is composed of 79 mole % of

24Mg and remaining 21 mole % of 25Mg and 26Mg. Calculate mole % of 26Mg.

a. 10 b. 11 c. 15 d. 16 A Sol. Let mole % of 26Mg be x

31.24100

79(24)x(26)x)25-(21

x = 10%

Comprehension – 2 Chemists work with standardised solution, a solution whose concentration is known. The requirement

to standarise the solution are:

1. the volume of the solution.

2. the number of moles of solute in that volume.

A primary standard solution is used in determining the molarity of a solution. To find the molarity of

HCl, 0.317 g of Na2CO3 is used in titrating the solution of HCl. 23 mL of acid are required to neutralise

the sodium carbonate. The stoichiometric equation used is

2HCl(aq) + Na2CO3(aq) 2NaCl + H2O + CO2



[M.Wt. of Na2CO2 = 106 & M.Wt of HCl = 36.5]

4. What is the molarity of HCl in the above case?

a. 0.261 M b. 0.522 M c. 0.1 M d. 1 M

A

Sol. ngm HCl = ngm Na2CO3 E.Wt. of Na2CO3 = 2

106

53

317.0

1000

N23

N = 0.261 N; same will be its molarity

5. Equivalent mass of Na2CO3 if neutralisation equation is:

HCl + Na2CO3 NaHCO3 + NaCl

a. 106 b. 5 3 c. 26.5 d. 13.25

A

Comprehension-3

18 mL of a gaseous organic compound, A, and just sufficient amount of oxygen required for complete combustion yielding on burning 36 mL of CO2, 54 mL of water vapour and 18 mL of N2. All volumes are measured at the same temperature and pressure. The compound A contains only carbon, hydrogen and nitrogen.

6. What is the molecular formula of the compound?

a. CH5N b. C2H5N c. C2H6N2 d. C4H10N2 C

Sol. 222zyx N2

z OH

2

y xCO

4

yx NHC

18 ml

4

yx18 18x 18

2

y

2

z18

18x = 36 9y = 54 9z = 18 x = 2 y = 6 z = 2 Formula is C2H6N2 7. How many volumes of oxygen are required for complete combustion of 18 ml organic compound

a. 35 mL b. 63 mL c. 7 mL d. 75 mL B

Sol. C2H6N2 + 2

7O2 2CO2 + 3H2O + N2

1 ml 2

7ml

18 ml 63 ml



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. 2 × 8 = 16

1. Match Column - I with Column - II and select the correct answer using codes given below the lists.

(One or More than One Match) Column – I Column – II

(A) 5.4 g of Al (p) 0.5 NA electron (B) 1.2 g of Mg2+ (q) 0.5 NA proton (C) 1 g of oxygen gas (r) 0.2 mole atom (D) 0.9 ml of H2O (d = 1 g/ml) (s) 0.05 moles atom/ions

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

(A) 2.027

5.4 Mole

(B) 05.024

2.1Mole

Atom = 0.05 × N0 e– = 0.05 × N0 × 10 = 0.5 N0

(C) 0315.032

1Mole

e– = 0.0315 × N0 × 16 e– = 0.504 same proton (D) 0.9 gm water

18

9.0Mole

= 0.05 H2O e– = 0.05 × 10 × N0

2. Match the Column-I with Column-II. (One or More than One Match)

Column I (Reaction)

Column II (Correct for underline Reactant)

(A) CaO + 2HCl OHCaCl 22 (p) eq. mass =

1

.Wt.M

(Reactant)

(B) COOHCH3 + NaOH COONaCH3 + H2O (q) eq. mass =

2

.Wt.M

(Reactant)

(C) 43POH + 2NaOH 42HPONa + H2O (r) Product is an acid salt

(D) 3NH + H2SO4 44HSONH (s) All hydrogen are acidic (in Reactant)

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



SECTION-F (Integer Type) No Negative Marking

This Section contains 6 questions. The answer to each question is a single digit integer ranging from 0 to 9. The correct digit below the question number in the OMR is to be bubbled.

6 × 4 = 24 Marks

1. 500 ml of 1.5 M Ca(NO3)2 solution, 500 ml of 1.5 M Al(NO3)3 solution and 500 ml of 1.5 M NaNO3

solution were mixed together. Molarity of 3NO in final solution is:

Sol. (3)

1500

)5.1500()35.1500()25.1500(NO3

= 3 M

2. One litre solution containing 490 g of sulphuric acid is diluted to 10 litres with water. What is the normality of the resulting solution?

Sol.1 N As N1V1= N2V2

10N149

4902

1049

490N2

= 1N

3. A solution of 0.1 mole of a metal chloride MClx requires 500 mL of 0.6 M AgNO3 solution for complete precipitation. The value of x is:

Sol.

xAgCl)M(NOxAgNOMCl

x:1

x3

::x:

3

1

x ratio Molar

Moles of AgNO3 given = 3.05001000

6.0

Moles of MClx = 0.1 Since 0.3 moles of AgNO3 are precipitated by 0.1 mol of salt

Value of x = 3

ngm chloride = ngm eq of AgNO3

1000

5000.6x0.1

0.1 x = 0.3; x = 3

4. One mole of potassium chlorate is thermally decomposed and excess of aluminium is burnt in the

gaseous product. How many moles of aluminium oxide are formed reaction involved is

2KClO3 2KCl + 3O2

4Al + 3O2 2Al2O3 Sol. 1

2KClO3 2KCl + 3O2

4Al + 3O2 2Al2O3 2 moles of KClO3 give 2 moles of Al2O3

1 mol of KClO3 gives 1 mol of Al2O3 5. How many of following are extensive properties; area; length; volume; Boiling point; melting point;

density; temperature; specific heat. Sol. 3 (area, length, volume) 6. How many significant figures will be present in 0.0025 × 105 Sol. 2

test venue: lajpat bhawan, madhya marg, sector …...20 ml of co was mixed with 50 ml of oxygen and...

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