test venue: lajpat bhawan, madhya marg, sector 15-b, chandigarh · 2017-06-12 · test venue:...

Dr. Sangeeta Khanna Ph.D 1 CHEMISTRY COACHING CIRCLE D:\Important Data\2017\+1\Physical\Grand Test\GT-1\+1 Grand Test -1.doc

Test Venue:

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

Dr. Sangeeta Khanna Ph.D


GRAND TEST – 1 (11.6.2017) Test-9

READ INSTRUCTIONS CAREFULLY 1. The test is of 2 hour duration. 2. The maximum marks are 246. 3. This test consists of 55 questions. 4. Keep Your mobiles switched off during Test in the Halls.

SECTION – A (Single Answer) Negative marking [-1]

This Section contains 29 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONLY ONE is correct. 29 × 4 = 116 Marks

1. The percentage of Se in peroxidase enzyme is 0.5% by mass (atomic mass of Se = 78.4 amu). Then,

the minimum molecular mass of enzyme which contains not more than one Se atom is:

a. 1.568 × 104 amu b. 1.568 × 107 amu c. 1.568 × 103 amu d. 1.568 × 106 amu A

Sol. 0.5 g Se is present in 100 g enzyme.

78.4 g Se will be present in = 4.785.0

100 g enzyme

= 15680 amu = 1.568 × 104 amu 2. 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 = 1.657 gm

1 mole = 4.680242.0

657.1 ;

M.Wt. = 68.4 i.e. 1 + x + 32 = 68.4 x = 35.4 3. A compound has haemoglobin like structure. It has one Fe in a molecule. It contain 4.6% of Fe. The

approximate molecular mass is:

a. 1400 g mol –1 b. 1000 g mol – 1 c. 1100 g mol–1 d. 1200 g mol–1 D Sol. 1 g atom of Fe (56 g Fe) is present in 1 mole of the compound As 4.6 g Fe are present in 100 g of the compound So 56 g Fe will be present in

= g 566.4

100 = 1217 g of the compound

So approximate molecular mass = 1200. 4. Element A (atomic weight 12.01) and element B (atomic weight 16) combine to form a new substance

X. If two moles of B combine with one mole of A in this compound, the weight of one mole of X is:

a. 28.01 g b. 44.01 g c. 40.02 g d. 56.02 g B Sol. 2 B + A = AB2

mol. wt. = 2 × 16 + 12.01 5. A compound was found to contain nitrogen and oxygen in the ratio, nitrogen 28 g and 80 g of oxygen.

The formula of the compound is:

a. NO b. N2O3 c. N2O5 d. N2O4 C



Sol. g atom of N = 2 14

28

G atom of oxygen = 516

80

6. In the reaction, 4A + 2B + 3C A4B2C3, what will be the number of moles of product formed, starting from 1 mol of A, 0.6 mol of B and 0.72 mol of C?

a. 0.25 b. 0.3 c. 0.24 d. 2.32 C

Sol. 324mol 0.72mol 0.6mol 1

CB A 3C 2B A4

In the present case, reactant ‘C’ will be the limiting reactant because it will give least amount of product on being completely consumed. 3 mol C gives 1 mol product,

0.72 mol ‘C’ will give 0.24 mol of product 7. The following diagram represents the reaction of

A2 (unshaded spheres) with B2 (shaded spheres). How many moles of product can be produced from the reaction of 1.0 mole of A2 and 1.0 mol of B2.

a. 0.5 mol of product b. 1.0 mol of product c. 2.0 mol of product d. 3.0 mol of product A

Sol. 2A2 + 2B2

1 BA4

A2 is limiting Reagent 8. The weight of sulphuric acid needed to react 3 g magnesium carbonate is: [M.Wt. of MgCO3 = 84,

M.Wt. of H2SO4 = 98] [H2SO4 + MgCO3 MgSO4 + CO\2 + H2O]

a. 3.5 g b. 7.0 g c. 1.7 g d. 17.0 g

A Sol. Meq. of MgCO3 = Meq. of H2SO4

;100049

w1000

2/84

3 or 84 gm MgSO3 react with 98 gm H2SO4

w = 3.5 g 3 gm will react = gm 5.3384

98

9. The vapour density of gas A is three times that of gas B. If the molecular weight of A is M, the molecular weight of B is:

a. 3M b. M 3 c. M/3 d. 3/M C Sol. Vapour density of A = 3 × Vapour density of B. mol. wt. of A = 3 × mol. wt. of B. 10. If the specific heat of a metallic element is 0.214 cal/ g, the atomic weight will be closest to: a. 66 b. 12 c. 30 d. 65 C

Sol. At. wt. × specific heat 30214.0

6.4

11. 4.48 litre of methane at S.T.P. corresponds to:

a. 1.2 × 1022 molecules of methane b. 0.5 mole of methane c. 3.2 g of methane d. 0.1 mole of methane



C

Sol. 4.48 litre mol 2.05

1

4.22

48.4CH4

= 16 × 0.2 = 3.2 g CH4 = 0.2 × 6.02 × 1023 = 1.2 × 1023 molecules.

12. How many Al2(SO4)3 formula units can be produced by using 3 × 1020 SO 24 ions

a. 9 × 1020 b. 1020 c. 6 × 1020 d. 3 × 10–20

B

Sol. 24SO3 will produce = 1 Al2(SO4)3

3 × 1020 will product = 1020

13. If 30 mL of H2 and 20 mL of O2 reacts to form water, what is left at the end of the reaction?

a. 10 mL H2 b. 5 mL H2 c. 10 mL O2 d. 5 mL O2 D

Sol. )g(OH)g(O2

1)g(H 2

mL 15

v ol21

2

mL 30v ol 1

2

Remaining O2 at the end = 20 – 15 = 5 mL 14. Which of the following will have maximum number of atoms

a. 2 mole CO2 b. 10 mole H2 c. 5 mole NH3 d. 4 mole H2O B,C

Sol. (A) 2 × N0 × 3 (B) 10 × N0 × 2 (C) 5 × N0 × 4 (D) 4 × N0 × 3 15. Which of the following will have maximum mass

a. 44.8 lit SO2 at STP b. 100 lit. He (at STP)

c. 100 lit H2O() (at STP) d. 22.4 lit N2O4(g) at STP

C Sol. (A) 44.8 lit is 2 mole

2 mole × 64 = 228 gm

(B) mole 46.44.22

100 = 4.46 × 4 gm

= 17.85 gm (C) 100 lit H2O

= 105 m = 105 gm

(d = 1 gm/m) (D) 22.4 lit. N2O4 = 1 mole = 92 gm 16. Mole fraction of ethanol in ethanol-water system is 0.25. What is the percentage concentration of

ethanol by weight of the solution?

a. 46% b. 56% c. 66% d. 86% A

Sol. Mole fraction of ethanol (xB) = 0.25 Mole fraction of water (xA) = 0.75

By definition, xB = 3

1

25.0

75.0

x

x

n

n ;

nn

n x;

nn

n

A

B

A

B

AB

AA

AB

B



3

1

W

M

M

W

A

A

B

B or 3

1

)mol g46(

)mol g 18(

W

W

1-

-1

A

B or 54

46

3

1

18

46

W

W

A

B

This means that if weight of ethanol is 46 g, that of water will be 54 g.

Percent concentration of ethanol = 100100

46 = 46%

17. What is molality, molarity and mole fraction of H2SO4 solution which is 10% by mass with density 1.8 gm/ml.

m M mole fraction m M mole fraction a. 1.13 1.8 0.02 b. 1.43 1.40 0.08 c. 2.13 2.08 0.02 d. 0.0113 0.018 0.02 A

Sol. Molality 10 gm H2SO4 in 90 gm H2O

m = molal 13.190

1000

98

10

Molarity

Vsolution = ml8.1

100

M = molar 8.198

108.110

Mole fraction

18

90n OH2

= 5 mole 102.098

10n

42SOH

02.0102.05

102.042SOH

18. Which of the following contains the maximum number of oxygen atoms?

a. 1g O b. 1g O2 c. 1 g O3 d. all contains same

D

Sol. (a) 16

1 (b) atom 2

32

1 but molecule

32

1 (c)

48

1 molecule but atom 3

48

1

19. 1.520 of hydroxide of a metal on ignition gave 0.995 g of oxide. The equivalent weight of the metal is:

a. 0.995 b. 190 c. 1.90 d. 9

D

Sol. oxide metal of w eightEquivalent

hydroxide metal of w eightEquivalent

oxide metal of Weight

hydroxide metal of Weight

8E

17E

995.0

520.1

m

m

0.528 Em = 4.78

On solving, we get Em = 9

20. Percentage of free SO3 present in a oleum sample labeled as 118%:

a. 80% b. 40 % c. 30% d. 95%

A

Sol. H2O + SO3 H2SO4

18 gm 80 gm

118% means, 18 gm water is added to convert SO3 to H2SO4.



21. A compound contains 69.5% oxygen and 30.5% nitrogen and its molecular weight is 92. The formula

of compound is

a. N2O b. NO2 c. N2O4 d. N2O5

C

Sol.

Element % %/At. wt. Ratio

N

O

30.5

69.5

18.214

5.30

34.416

5.69

1

2

Empirical formula = NO2

Empirical formula weight = 46

246

92n

Molecular formula = (NO2)2 = N2O4

22. An organic compound containing C and H has 49.3% carbon, 6.84% hydrogen and its vapour density

is 73. Molecular formula of the compound is

a. C3H5O2 b. C4H10O2 c. C6H10O4 d. C3H10O2

C

Sol.

Element % Relative number of atom Simplest ratio

C H O

49.3 6.84 43.86

1.412

3.49

84.61

84.6

74.216

86.43

325.174.2

1.4

525.274.2

84.6

22174.2

74.2

The empirical formula is C3H5O2 Empirical formula weight = 3 × 12 + 5 × 1 + 2 × 16 = 36 + 5 + 32 = 73 Molecular weight of the compound = 2 × VD = 2 × 73 = 146

w t.formula empirical

w t..moln

273

146

Molecular formula = Empirical formula × 2 = (C3H5O2) × 2 = C6H10O4

23. CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g) How many moles of methane are required to produce 22 g CO2 (g) after combustion?

a. 1 mol b. 0.5 mol c. 0.25 mol d. 1.25 mol B Sol. 0.5 mol According to the chemical equation



CH4(g) + 2O2(g) CO2(g) + 2H2O(g) 44 g CO2(g) is obtained from 16 g CH4(g) [ 1 mol CO2(g) is obtained from 1 mol of CH4(g)]

Mole of CO2(g) = )g(CO g 44

)g(CO g 22

2

2 = 0.5 mol CO2 (g)

Hence, 0.5 mol CO2(g) would be obtained from 0.5 mol CH4(g) or 0.5 mol of CH4(g) would be required to produce 22g CO2(g).

24. Match the Column I with Column II and choose the correct option from the codes given below

Column – I Column – II A.

10 g CaCO3 iondecomposit

CaO + CO2

1. 0.224 L CO2

B. 1.06 g Na2CO3 HCl Excess 2NaCl + CO2 + H2O 2. 4.48 L CO2

C. 2.4 g C

combustion

O Excess 2 CO2 3. 0.448 L CO2

D. 0.56g CO

combustion

O Excess 2 CO2 4. 2.24 L CO2

5. 22.4 L CO2

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

Sol. (A) L 22.42

iondecomposit

g 1003 CO CaOCaCO

100 g CaCO3 on decomposition gives = 22.4 L

10 g CaCO3 on decomposition will give = 2CO L100

104.22

= 2.24 L CO2

(B) L 4.2222

HCl Excess

g 1063 COOHNaCl2NaNO

106 g Na2CO3 gives = 22.4 L CO2

232 CO L106

1.0622.4give w illCONa g 06.1

= 0.224 L CO2

(C) L 4.222

combustion

O Excess

g 12COC 2

12 g carbon on combustion gives = 22.4 L CO2

2.4 g carbon on combustion gives = 2CO L12

4.24.22

= 2 × 2.24 L CO2 = 4.48 L CO2

(D) L 22.42

2Combustion

O Excess

g 56)1612(2

2CO CO2 2

56 g carbon monoxide on combustion gives = 2 × 22.4 L CO2

0.56 g carbon monoxide on combustion will give = 2COL56

56.04.222

= 0.448 L CO2 25. What will be the molarity of chloroform in the water sample which contains 15 ppm chloroform by

mass? (M.Wt. of CHCl3 = 119.5)

a. 1.25 × 10–4 M b. 2.5 × 10–4 M c. 1.5 × 10–3 M d. 1.25 × 10–5 M A



Sol. 106 gm solution have = 15 gm CHCl3

103 gm will have = 3

610

10

15

= 15 × 10-3 gm/lit

lit/mol105.119

15 3

= 1.25 × 10-4 mol/lit 26. 500 mL of CO2 is passed over red-hot coke. The volume becomes 700 mL. The composition of the

product is

a. 250 mL CO2 and 450 mL CO b. 700 mL CO c. 300 mL CO and 400 mL CO2 d. 300 mL CO2 and 400 mL CO D

Sol. CO2 + C 2CO After Reaction 500 –x 2x Total volume of gases 500 –x + 2x = 700 x = 200 CO = 2 × 200 = 400 CO2 = 300 27. 15 mL of a gaseous hydrocarbon on complete combustion yielded 45 mL CO2 and 60 mL steam. The

formula of the hydrocarbon is

a. C3H6 b. C4H10 c. C3H4 d. C3H8 D

Sol. C3H8 + 5O2 3CO2 + 4H2O(g) 1ml 3 ml 4 ml 15 ml 45 ml 60 ml 28. Liquid benzene burns in oxygen according to the following equation.

2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O(g) How many litres of oxygen (At NTP) is required for the complete combustion of 39 g of liquid benzene? a. 11.2 b. 22.4 c. 42.0 d. 84.0

D Sol. 2C6H6 + 15O2 12CO2 + 6H2O 78 × 2gm 15 × 22.4 lit.

39gm 278

394.2215

= 84 lit. 29. How many millilitres of a 9 N H2SO4 solution will be required to neutralize completely 20 mL of a 3.6 N

NaOH solution?

a. 18.0 mL b. 8.0 mL c. 16.0 mL d. 80.0 mL B

Sol. ngm (H2SO4) = ngm NaOH 9 × V = 20 × 3.6

V = mL 89

6.320



SECTION – B (ASSERTION & REASON) Negative Marking [-1]

This Section contains 6 multiple choice questions. Each question has four choices A), B), C) and

D) out of which ONLY ONE is correct. (6 × 4 = 24 Marks)

(a) Both A and R are correct; R is the correct explanation of A (b) Both A and R are correct; R is not the correct explanation of A (c) A is correct; R is incorrect (d) A is incorrect and R is correct

1. Assertion (A): The molality of the solution is independent of temperature. Reason (R): The molality of the solution is expressed in unit of moles per g of solvent. a. (a) b. (b) c. (c) d. (d) C

Sol. Molality is not expressed as volume of the solution as molarity or normality. So, it does not depend upon temperature.

2. Assertion (A): A solution which contains one gram equivalent of solute per litre of solution is known as molar solution.

Reason (R): Normality = molarity × solute of w t..eq

solute of w t.mol

a. (a) b. (b) c. (c) d. (d) D

Sol. A solution which contains one gram mole of solute per litre of solution is known as molar solution (M).

solute of w t..eq

solute of w t..molmolarityNormality N = M × n

3. Assertion (A): If 100 ml of 0.1 N HCl is mixed with 100 ml of 0.2 N HCl, the normality of the final solution will be 0.15 N Reason (R): Gram equivalents of similar solution like HCl can be obtained by N1V1 + N2V2 = NV a. (a) b. (b) c. (c) d. (d) A

Sol. 100 × 0.1 + 100 × 0.2 = 200

200

30 = N

4. Assertion (A) : 1 Normal H2SO4 solution is more concentrated than 1 M Solution Reason (R): 1 N has less mass of solute than 1 M H2SO4. a. (a) b. (b) c. (c) d. (d) D

Sol. 1N have 49 gm H2SO4 whereas 1 M have 98 gm. 5. Assertion (A): Both 12 g of carbon and 27 g of aluminium will have 6.02 × 1023 atoms

Reason (R) : Gram atomic mass of an element contains Avogadro number of atoms.

a. (a) b. (b) c. (c) d. (d) A 6. Assertion (A): Molarity and molality for very dilute aqueous solution is approximately equal.

Reason (R): For all aqueous solution, total mass of solvent is equal to total volume of solution. a. (a) b. (b) c. (c) d. (d)

C Sol. All solutions are not dilute solutions



SECTION – D (More than One Answer) No Negative Marking This Section contains 8 multiple choice questions. Each question has four choices A), B), C) and

D) out of which more than one answer is correct. (8 × 5 = 40 Marks)

1. In which of the following pairs do 1 g of each have an equal number of molecules? a. N2O and CO b. N2 and C3O2 c. N2 and CO d. N2O and CO2 C,D Sol. Both have same molecular mass 2. Which of the following is a redox Reaction

a. H2 + Cl2 2HCl b. CH4 + O2 CO2 + 2H2O

c. NaCl + AgNO3 AgCl + NaNO3 d. Na + 2

1H2 NaH

A,B,D Sol. In C no change in oxidation number of all elements 3. 10 g of NaCl is dissolved in 250g water. The correct way to express concentration of NaCl in solution are:

a. mass fraction = 0.04 b. mole fraction = 0.0122 c. molality = 0.684 d. molarity = 0.884 A,B,C

Sol. Mass fraction = 04.0250

10

Mole fraction = 0121.005.14

1709.0

88.1317.0

1709.0

68.01000250

17.0M

Molarity cannot be calculated as density is not given 4. A solution is prepared by dissolving 5.3 g of Na2CO3 in 250 cm3 of solution. The solution can be

described as

a. Decinormal solution b. Decimolar solution c. 0.4 N solution d. 0.2 M solution C,D

Sol. N 4.0250

100

3.5

3.5N ;

x = 2 so M = 2.02

N

5. 24.0 g carbon and 96.0 g O2 reacts according to the equation 2C(s) + O2(g) 2CO(g)

Which of the following statements are correct? a. carbon gives least amount of the product b. 1.204 × 1023 molecules of O2 will be left unreacted c. 5.6 moles of CO is formed d. 72 g of carbon should be taken to consume the oxygen completely A,D Sol. Carbon is limiting reagent;

2C + O2 2CO

12

24

36

96

= 2 mole 3 mol 2 mole



2 mole O2 will be unused. 3 mole O2 require = 6 mole carbon i.e. 6 × 12 = 72 gm Carbon 6. Which of the following compounds have same empirical formula?

a. Formaldehyde HCHO b. Glucose(C6H12O6) c. Sucrose (C12H22O11) d. Acetic acid (CH3COOH) A,B,D

7. Which of the following Reaction is intramolecular Redox reaction

a. 23 O3KCl2KClO2

b. OH2ONNONH 2234

c. 23 COMgOMgCO

d. 223 O2

1NOAgAgNO

A,B,D 8. Which of the following disproportionation?

a. 2Cu+ Cu + Cu2+ b. 3Cl2 + 6OH– ClO3– + 5Cl– + 3H2O

c. 2H2S + O2 2H2O + 2S d. Na + 2Cl2

1 NaCl

A,B

Sol. A = 2

2

01Cu CuCu2

B = OH3Cl5ClOOH6Cl3 215

30

2

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 Marks

1. Match the Column-I with Column-II. (Single Match)

Column I (Molarity) Column II (Normality)

(A) 1 M Al2(SO4)3 (p) 3 N

(B) 1 M H2SO4 (q) 1 N

(C) 1 M HNO3 (r) 2 N

(D) 1 M Al(OH)3 (s) 6 N

Sol. A s ; B r; C q ; D p 2. Match the Column-I with Column-II. (More than One Match)

Column I Column II

(A) Change of HNO2 HNO3 (p) Oxidation Reaction

(B) P PH3 (q) Oxidation number per atom increases by two

(C) KMnO4 MnO2 (r)

3

Wt.M.Wt.E

(D) H2O2 O2 (s) Reduction Reaction

Sol. A P, Q 5

33

2 HNOHNO

D P 02

122 OOH

B R, S 33

0PHP

C R, S 4

2

7

4 MnOKMnO



SECTION – F (Integer Type) No Negative Marking

This Section contains 10 questions. The answer to each question is an integer ranging from 0 to 10. 10 × 5 = 50 Marks

1. The hydrated salt Na2SO4. nH2O, undergoes 55.9% loss in weight on heating and becomes anhydrous. The value of n will be:

Sol. 10

Na2SO4nH2O 55.9

21.44

42 OnH SONa

44.1 combine with 55.9 gm

142 will combine with = gm1421.44

9.55

= 180 gm Mole of H2O = 10 2. CaCO3 is decomposed by HCl (density 1.825 g/cc)

CaCO3 + 2HCl CaCl2 + H2O + CO2 Volume of HCl in (ml) required to decompose 10 g of 50% pure CaCO3 is: Sol. 2

Weight of CaCO3 = 5 gm

Mole = 05.0100

5

Mole of HCl = 0.05 × 2 = 0.1

Weight = 0.1 × 36.5 = 3.65 V

Md

Volume of HCl 825.1

65.3 = 2 ml

3. 10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Moles of water

produced in this reaction will be:

OHO

2

1H 222

Sol. 4 mol Mole ratio of H2 : O2 : H2O :: 2 :1 : 2 [oxygen is limting Reagent] 4. 2N; 100 ml H2SO4 is mixed with 200 ml solution of NaOH of unknown Normality. Normality of NaOH

will be Sol. 1 2 × 100 =, N × 200 5. How many times the molecular mass of glucose (C6H12O6) is multiple of empirical formula mass? Sol. 6 6. How many of the following compounds have at least one atom in +6 oxidation state

H2SO4; SO2; SO3; H2SO5; XeF6; XeO8; HNO3; H3PO4; K2Cr2O7 Sol. 5 H2SO4; SO3; H2SO5; XeF6; K2Cr2O7 7. What is oxidation number of Iron in the given compound K3[Fe(CN)6] Sol. +3 8. How many significant figures will be present in 0.0025 × 105 Sol. 2 9. What is the equivalent mass of CH4 in its combustion reaction Sol. 2

O2H CO2O CH 242

8e

244



28

16.Wt.E

10. How many of the following statements are correct

(a) H2SO4 + NaOH NaHSO4 + H2O; E. Wt. of H2SO4 = 49

(b) Al(OH)3 + HCl Al(OH)2Cl + H2O; Acidity of Al(OH)3 = 1

(c) B(OH)3 + NaOH Na[B(OH)4]; E. Wt. of B(OH)3 = M. Wt

(d) 2Na2S2O3 + I2 Na2S4O6 + 2NaI; E. Wt. of Na2S2O3 = M. Wt.

(e) 2FeS2 + 2

11O2 Fe2O3 + 4SO2 E.Wt. of FeS2 =

11

Wt.M

(f) In Redox reaction KMnO4 will always act as oxidising agent and H2S will act as Reducing agent.

(g) OH2NNONH 2224

This is an example of conproportionation

(h) Na + 2

1H2 NaH

H2 act as Reducing agent Sol. 6 [b, c, d, e, f, g]

a = wrong as n-factor of H2SO4 = 1 h = H2 act as oxidising agent as it oxidizes Na to Na+

test venue: lajpat bhawan, madhya marg, sector 15-b, chandigarh · 2017-06-12 · test venue:...

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