Test Venue: Lajpat Bhawan, Madhya Marg, Sector 15-B ... · PDF fileTest Venue: Lajpat Bhawan, Madhya Marg, ... A solution is prepared by adding 2 g of a substance A to 18g of water

Dr. Sangeeta Khanna Ph.D 1 CHEMISTRY COACHING CIRCLE D:\Important Data\2017\+1\Physical\Grand Test\GT-2\+1 Grand Test -2.doc

Test Venue:

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

Dr. Sangeeta Khanna Ph.D


GRAND TEST – 2 (02.07.2017) Test-11

READ INSTRUCTIONS CAREFULLY 1. The test is of 2 hour duration. 2. The maximum marks are 299. 3. This test consists of 66 questions. 4. Keep Your mobiles switched off during Test in the Halls.

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

This Section contains 40 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONLY ONE is correct. 40 × 4 = 160 Marks

1. X g of calcium carbonate was completely burnt in air. The weight of solid residue formed is 28 g. What is the value of X in grams?

2gm 56gm 100

3 CO CaOCaCO

a. 44 b. 200 c. 150 d. 50

D Sol. 56 g CaO is formed from CaCO3 = 100

28 g CaO is formed from CaCO3 = 56

28100 = 50

2. The volume of oxygen at NTP evolved when 1.70 g of sodium nitrate is heated to a constant mass is:

[NaNO3

NaNO2 + 2

1O2] [M.Wt. of Na = 23; N = 14]

a. 0.112 litre b. 0.224 litre c. 22.4 litre d. 11.2 litre

B

Sol. NaNO3 NaNO2 + 2

1O2

1 mole NaNO3 give 2

1mol O2

85 gm 11.2 lit O2

1.7 gm = 7.185

2.11

3. In the following reaction

B2O3 + 3H2O 2B(OH)3

how many moles of water are required to produce 5.0 moles of B(OH)3 starting with 3.0 moles of

B2O3?

a. 6.0 b. 3.0 c. 2.0 d. 7.5

D

Sol. For 2 mol B(OH)2 = 3 mol H2O required

5 mol = 52

3 = 7.5 mol

4. The number of neutron in 5g of D2O (D is )H21 are:

a. 0.25 NA b. 2.5 NA c. 1.1 NA d. 250 NA

B

Sol. = AN20

5 molecule of D2O & One D2O have 10 neutrons AN

20

105



5. A gaseous mixture contains CH4 and C2H6 in equimolecular proportion. The weight of 2.24 litres of this mixture at STP is:

a. 4.6 g b. 1.6 g c. 2.3 g d. 23 g C Sol. Equimolecular will have same volume i.e. 1.12 lit of each gas 22.4 lit CH4 = 16 gm

1.12 lit CH4 = 12.14.22

16 = 0.8

An equimolar mixture will have equal volume so volume of CH4 & C2H6 will be 1.12 lit each 6. Which of the following has the smallest number of molecules?

a. 22.4 × 103 mL of CO2 gas b. 22 g of CO2 gas b 11.2 litre of CO2 gas d. 0.1 mole of CO2 gas D 7. Four grams of hydrocarbon (CXHy) on complete combustion gave 12 gram of CO2. What is the

empirical formula of the hydrocarbon?

a. CH3 b. C4H9 c. CH d. C3H8 D

Sol. Weight of carbon in 12 g CO2 = 1244

12 = 3.2727 g

Weight of hydrogen in the hydrocarbon = 4 – 3.2727 = 0.7273 g

Number of moles of carbon = 12

2727.3 = 0.2727

Number of moles of hydrogen = 1

7273.0 = 0.7273

C : H = 0.2727 : 0.7273 = 3 : 8 Compound is C3H8

8. The maximum volume at S.T.P. is occupied by : [At. Wt. S = 32, C = 12]

a. 12.8 g of SO2 b. 6.02 × 1022 molecules of CH4 c. 0.5 mol of NO2 d. 1 gm – molecule of CO2

D Sol. 12.8 g of SO2 = 4.48 L; 6.02 × 1022 molecules of CH4 = 2.24 L 0.5 mol of NO2 = 11.2 L; 1 g – molecule of CO2 = 22.4 L. 9. A compound contains atoms of three elements, A, B and C. If the oxidation numbers of A is +2, B is

+5 and that of C is -2, the possible formula of the compound is:

a. A3(BC4)2 b. A3(B4C)2 c. ABC2 d. A3(BC3)2 A

Sol. In (A), i.e., A3 (BC4)2 we have 3 (+2) + 2 (+ 5 + 4 × –2) = 6 + 2 × (–3) or 6 – 6 = 0 10. What is oxidation number of N & Carbon respectively in H C N

a. +3, – 2 b. +2, – 3 c. +3, – 3 d. – 3, + 2 D

11. What are the Actual oxidation states of Bromine in the given compound

OBrBrBrO

O

O

||

||

O

O

||

||

O

O

||

||

a. + 7 & + 6 b. + 6 & + 4 c. + 4 only d. + 3

16

B



12. In which of the following reaction H2SO4 act as oxidising agent.

a. H2SO4 + NaOH NaHSO4 + H2O

b. 2KMnO4 + 10KI + 8H2SO4 6K2SO4 + 2MnSO4 + 8H2O + 5I2

c. 2H2SO4 + Cu CuSO4 + SO2 + 2H2O d. H2SO4 + CaO CaSO4 + H2O C

13. Which of the following is correct increasing order of oxidation number of underlined atom.

a. HCl < HClO < HClO2 < HClO3 b. SO2 < SO3 < H2SO4 c. NaCl < Na2SO4 < Na3PO4 d. CrO5 < Cr2O3 < CrCl5 A

14. Number of moles of 4MnO required to oxidize one mole of ferrous oxalate completely in acidic

medium will be:

a. 7.5 moles b. 0.2 moles c. 0.6 moles d. 0.4 moles C

Sol. 3)]i....(OH4MnH8MnOe5[ 22

4

5)]ii......(Fe .CO2e3OFeC[ 32

242

32

2424

Fe5CO10Mn2OFeC5H16MnO3

As 3 moles of 4MnO required to oxidize 5 moles of Fe oxalate.

So number of moles of 4MnO required to oxidize 1 mole of oxalate = 6.0

5

3

15. A gas mixture of 3 lit. propane and butane on combustion give 10 lit. CO2 . What will be composition of gas.

a. C3H8 = 2 lit; C4H10 = 1 lit b. C3H8 = 3 lit, C4H10 = 1lit. c. C3H8 = 1.5 lit; C4H10 = 1.5 lit. d. C3H8 = 0.5 lit, C4H10 = 2.5 lit. A

16. A sample of nitric acid is 69% by mass and it has a density of 1.4 gm/ml. Its molarity is

a. 6.5 M b. 15.3 M c. 7.2 M d. 1.54 M B

Sol. 69 g acid in 100 g, volume of 100 g sample = d

100

.Wt.M

10d %M

4.1/100

1000

63

69M 4.1

100

1000

63

69 15.3 M

17. How much KClO3 is required to produce enough oxygen so that 10 gm carbon is completely burned.

2KClO3

2KCl + 3O2

C + O2 CO2

a. 68 gm b. 58 gm c. 78 gm d. 88 gm A

Sol. 2KClO3 2KCl + 3O2

(O2 + C CO2) 3 3 mol carbon = 2 mole KClO3

12

10mole carbon =

12

10

3

1 mole KClO3

= 36

5.122210 = 68 gm



18. The final molarity of a solution made by mixing 50 mL of 0.5 M HCl and 150 mL of 0.25 M HCl and water to make the volume 250 mL is

a. 0.5 M b. 1 M c. 0.75 M d. 0.25 M D

Sol. M1V1 + M2V2 = MV 0.5 × 50 + 0.25 × 150 = M × 250

M 25.0250

5.3725M

19. Which of the following gases will have least volume if 10 g of each gas is taken at same temperature and pressure?

a. CO2 b. N2 c. CH4 d. HCl A

Sol. mass molecular

1moles of Number

Molecular mass of CO2 = 44, N2 = 28, CH4 = 16, HCl = 36.5 CO2 will have least volume 20. If 1 mol is present in 1 L or 1000 mL then 0.2 mol is present in,

a. 150 mL b. 200 mL c. 175 mL d. 225 mL B

Sol. If 1 mol is present in 1 L or 1000 mL then 0.2 mol is present in

mol 2.0mol 1

mL 1000 = 200 mL

21. A solution is prepared by adding 2 g of a substance A to 18g of water. Calculate the mass per cent of the solute. a. 8% b. 9% c. 10% d. 11% C

Sol. A solution is prepared by adding 2g of a substance A to 18 g of water.

100solution of Mass

Aof Mass Aof cent per Mass

100 w aterof g 18 A of g2

g2

100g 20

g2 = 10%

22. In a compound C, H and N are present in the ratio of 9 : 1 : 3.5 by weight. If molecular weight of the compound is 108, then the molecular formula of the compound is

a. C2H6N2 b. C3H4N2 c. C6H8N2 d. C9H12N3 C

Sol. C H N 9 1 3.5

Molar ratio 75.012

9 1

1

1 25.0

14

5.3

Simpler molar ratio 325.0

75.0 4

25.0

1 1

25.0

25.0

So, empirical formula = C3H4N Empirical formula weight = (3 × 12 + 4 × 1 + 14) = 54

254

108n

Molecular formula = (C3H4N)2 = C6H8N2



23. 30gm Mg, react with 40gm O2 .what is limiting reagent and how much MgO is formed. [At wt. of Mg = 24] a. Mg ; 100 gm b. O2, 50 gm c. O2; 100 d. Mg; 50 gm D

Sol. Mg + 2

1O2 MgO

24g of Mg reacts with O2 = 16 g

30 g of Mg reacts with O2 = g 203024

16

16 g O2 react with Mg = 24

40 g O2 react with Mg = 4010

24 = 60

Mg is limting reagent

4024

30 from Mg

= 50 gm 24. 40 mL of a mixture of CO(g) and H2(g) is mixed with an excess of O2 and sparked. After the reaction,

15 mL of CO2 is obtained. The volume of CO in the mixture would be

a. 10 mL b. 25 mL c. 30 mL d. 15 mL D

25. 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%

26. What is the % of free SO3 in a sample of oleum labelled as 109%

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

Sol. SO3 + H2O H2SO4 18g H2O = 80 g SO3

9g = 18

980 = 40

27. 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 28. 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 29. 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 30. 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)

31. 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

32. The density of a liquid is 1.2 g/mL. There are 35 drops in 2 mL. The number of molecules in 1 drop is:

(molecular weigh of liquid = 70)

a. AN35

13 b. A

2

N35

1

c. A2

N)35(

2.1 d. 12NA

C

Sol. Volume of one drop = 35

2mL

Number of moles in one drop = 2)35(

2.1

7035

2.12

Number of molecules in one drop = A2N

)35(

2.1

33. Phosphine (PH3) decomposes to produce vapours of phosphorus and H2 gas. What will be the

change in volume when 100 mL of phosphine is decomposed? 4PH3(g) P4(g) + 6H2(g)

a. + 50 mL b. + 500 mL c. + 75 mL d. - 500 mL



C

Sol. 4PH3(g) P4(g) + 6H2(g)

4 mL 1 mL 6 mL For 4 volume PH3 change in volume is 3

100 mL 4

100 100

4

6 For 100 volume PH3 change will be 75

100 mL 25 mL 150 mL Volume increased by 75 mL. 34. 10 g of hydrofluoric acid gas occupies 5.6 litre of volume STP. If the empirical formula of the gas is

HF, then its molecular formula in the gaseous state will be: [At. Mass of F = 19; H = 1]

a. HF b. H2F2 c. H3F3 d. H4F4 B

35. In reaction, 332

232 HCO6BrOBr5OH3CO6Br3

a. Br2 is reduced and 2

3CO is oxidized b. Br2 is oxidized and 2

3CO is reduced

c. Br2 undergoes both oxidation and reduction d. Br2 is neither oxidized nor reduced C

Sol.

3

50

2 OBrBrBr

36. For the redox reaction:

OHCOMnHOCMnO 2222

424

the correct coefficients of the reactants for the balanced equation are:

4MnO

242OC

H

a. 2 5 16 b. 16 5 2 c. 5 16 2 d. 2 16 5

A

Sol. OHCOMnHOCMnO 2222

424

2x – 8 = - 2 2x = 6 x = 2

5 ] e2CO 2OC48

2

36

242

2 ]MnOMne5 24

OH8CO10Mn2H16O5Cr MnO2 2222

424

37. 0.16 g of a dibasic acid required 25 mL of N/10 NaOH for complete neutralization. The molecular mass of acid is:

a. 32 b. 64 c. 128 d. 256

C Sol. ngm Acid = ngm Base

xWt.M

W =

1000

NV

1000

251.02

.Wt.M

16.0



38. A mixture containing 20 mL H2, 10 mL CO and 20 mL O2 was exploded in a eudiometer tube. After

cooling the final volume of gas mixture will be:

a. 15 mL b. 5 mL c. 25 mL d. 50 mL

A

Sol. H2 + 2O2

1 H2O

20 ml 10 ml

ml 10

2

ml 5

2

ml 10

COO2

1 CO

Oxygen unused = 20 – 15 = 5 ml

CO2 formed = 10 ml

Volume of gases = 15 ml

39. 10 mL of a gaseous hydrocarbon was burnt completely in 80 mL O2 at NTP. The volume of gases

after explosion and cooling was 70 mL. The volume reduced to 50 mL on treatment with KOH. The

formula of the hydrocarbon is:

a. C3H6 b. C2H2 c. CH4 d. C2H4

D Sol. Volume of CO2 = 70 – 50 = 20 10 x = 20 x = 2

Volume of O2 used = 80 – 50 = 30

304

yx10

y = 4 40. 0.58 g of a hydrocarbon on combustion gave 0.9 g water. The % C in the compound is:

a. 17.24 b. 82.7 c. 27.85 d. 58.6 B

Sol. % H = 10058.0

9.0

18

2 = 17.24

% C = 100 – 17.24 = 82.75



Level – 2

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

This Section contains 5 questions. Each question has four choices A), B), C) and D) out of which ONLY ONE is correct. 5 × 4 = 20 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): In redox reactions, H2O2 may behave as oxidizing as well as reducing agent, depending upon the other reactants. Reason (R): In H2O2, oxygen is in –1 oxidation state, which may be increased to zero or decreased to –2.

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

2. 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. 1 N H2SO4 have 49 gm H2SO4 whereas 1 M will have 98 gm. 3. Assertion (A): Vapour density of CH4 is half of O2

Reason (R): 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 4. Assertion (A): In compounds, equivalent mass of an element always combines with equivalent mass

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

5. Assertion (A): In the process : MnO2 → Mn2O3, the equivalent mass of MnO2 is equal to its molecular mass. Reason (R): The oxidation state of Mn Changes from +4 to +3. a. (A) b. (B) c. (C) d. (D) A



SECTION – C (Paragraph Type) Negative Marking

This Section contains 5 questions. Each question has four choices A), B), C) and D) out of which ONLY ONE is correct. 5 × 5 = 25 Marks

Comprehension – 1 Human hair consists largely of protein such as keratin. Keratin is unusual in that it contains a very large amount of sulfur – containing amino acid cystein. The sulfur content of hair may be found by complete alkaline hydrolysis followed by oxidation to convert all sulfur to sulfate.

A 1 gm sample of hair is hydrolysed and the sulfur content converted to sulfate. Aqueous BaCl2 solution is added until no more precipitation takes place, taking the total volume to 2.5 lit. On filtration of the mixture 260 mg BaSO4 is recovered after drying. [M.Wt. of BaSO4 = 233]

1. Calculate the % by mass of sulfur in the sample of human hair to 3 significant figures. Assume that all

the sulfur in the sample is converted to BaSO4 ppt.

a. 35.7 % b. 3.57% c. 6.85% d. 3.570% B

Sol. Moles of S = moles of BaSO4 = mole 1014.1233

1

1000

260 3

Mass of S = 1.14 × 10–3 × 32 Percentage of S = 1.14 × 10–3 × 32 × 100 = 3.57%

% S = 100Sample of Mass

BaSO of Mass

BaSO of Mass Molecular

Sulphur of Mass .At 4

4

1001

260.0

233

32

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 using methyl orange indicator. 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 Na2CO3 = 106 & M.Wt of HCl = 36.5] 2. 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



3. Equivalent mass of Na2CO3 if neutralisation equation is:

HCl + Na2CO3 NaHCO3 + NaCl (In presence of phenolphthalein indicator)

a. 106 b. 53 c. 26.5 d. 13.25

A

Comprehensions – 3 An element forms a hydride containing 5.928% hydrogen. 0.18 g of a gaseous hydride occupies 120 mL at 1 atm NTP.

4. What is the equivalent weight of the element?

a. 1.587 b. 15.87 c. 7.93 d. 31.74 B

Sol. Metal hydride

94.19.5M H

1 gm 87.159.5

1.94

5. What is the formula mass of the hydride? a. 15.87 b. 40.00 c. 16.43 d. 33.6 D

Sol.120 ml have mass 0.18 gm

22400 ml should have 6.3322400120

18.0

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

This Section contains 5 multiple choice questions. Each question has four choices A), B), C) and

D) out of which MORE THAN ONE ANSWER is correct. 5 × 6 = 30 Marks

1. In the following redox reaction OH8Cl5 Mn2H16Cl10MnO2 222

4

Pick up the correct statements.

a. 4MnO is reduced b. Cl– is oxidising agent

c. 4MnO is an oxidizing agent d. Cl– is reduced

A, C 2. Which of the following act both oxidising as well as reducing agents?

a. HNO2 b. H2O2 c. H2S d. SO2 A,B,D

Sol. H2S acts only as a reducing agent. 3. Which of the following is/are a redox reaction?

a. NaCl + KNO3 NaNO3 + KCl b. CaC2O4 + 2HCl CaCl2 + H2C2O4

c. H2 + Cl2 2HCl d. Zn + 2AgCN 2Ag + Zn(CN)2 C,D

Sol. 2

20

2

10

)CN(ZnAg2])CN(Ag2[ Zn

and 110

202 ClH2Cl H

Only in this reaction, we see a change in oxidation number



4. 272OCr is reduced to Cr3+ by Fe2+ in acidic medium. Identify the incorrect statement from the

following:

[Cr2O7–2 + Fe+2 + H+ Fe+3 + Cr+3 + H2O]

a. 6 moles of Fe2+ are oxidized to Fe3+ ions by 1 mole of K2Cr2O7

b. Equivalent weight of Fe+2 is M/2

c. Equivalent weight of Cr2O7–2 is M/6

d. 3 moles of Fe2+ get oxidized to Fe3+ by 1 mole of K2Cr2O7

B,D

5. Which of the following is correct statement

a. M0.98O MO2 (Change in oxidation state is 1.92)

b. If 0.1 mol compound have 1 gm Hydrogen with empirical formula CH2O. Its molecular formula

can be C5H5O5.

c. 400 ml water is to be added to 100 ml 0.5 M NaOH to make it 0.1 M.

d. 1 molal aqueous solution is more concentrated than 1 molar solution.

A,C

Sol. (A) In product for one ‘M’ oxidation state = +4 (C) 100 × 0.5 = 0.1 × V for 0.98 it will be = 4 × 0.98 = 3.92 V = 500 ml

Change in oxidation state = 3.92 – 2 Water added = 400 ml

= 1.92

(B) 0.1 mole have 1 gm H-atom of 1 mol

1 mole will have = 10 m0l

SECTION – E (Matrix Type) No Negative Marking

This Section contains 3 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 × 3 = 24 Marks

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)

Al2(Cr2O7)3 Cr+3 (p) eq. =

18

.Wt.M

(B) H2O2 O2 (q) Reduction Reaction

(C) Ba(MnO4)2 BaMnO4 (r) eq. Wt. = 2

Wt.M

(D) As2O5 As2O3 (s) Oxidation Reaction

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

(A) 6e– + 272OCr 2Cr+3

For three 272OCr X = 18

(B) H2O2 O2 + 2e–

(C) 2e– + 2MnO 4 2MnO 2

4

(D) 6

3210

52 OAsOAs e4



2. Match column – I with column – II (More than One Match) [At. Wt. S = 32; N = 14; P = 31]

Column – I Column – II

(A) 30 gm of 23

CO (P) 1 Faraday charge

(B) 96 gm of 24

SO (Q) 3/2 moles of oxygen atoms

(C) 62 gm of 3NO (R) 4 moles of oxygen atoms

(D) 95 gm of 34

PO (S) 3 Faraday charge

(T) 1 gm ion (C = 12, O = 16, S = 32, N = 14, P = 31) [1-faraday is 1 mol charge] Sol. (A) P, Q; (B) R, T; (C) P, T; (D) R, S, T 1 F is charge of 1 mol of electrons. 1 mol of an ion (unipositive/uninegative) also carry 1 F charge.

(A) 30 gm of 3CO

22

1CO mol

2

1

60

30 --3 mol charge = 1 F charge (P); mole of oxygen =

2

33

2

1

(B) 96 gm of 4SO

96

96 = 1 mol

4SO = 1 gm ion (T)

= 2 mol charge = 2 F charge

1 mol 4SO = 4 moles of oxygen atoms (R)

(C) 62 gm 3NO =

62

62 = 1 mol

3NO

= 1 gm ion (T) = 1 × 1 = 1 mol charge = 1 F (P)

1 mol 3NO = 3 mol oxygen atoms

(D) 95 gm 34

PO 195

95 mol 3

4PO = 1 gm ion (T)

= 1 × 3 mol of charge = 3 F charge (S)

1 mol 34

PO = 4 moles of oxygen atoms (R)

3. Match the Column – I with Column – II. (More than One Match)

Column I Column II

(A) N2 (p) 40% Carbon by mass

(B) CO (q) Empirical formula CH2O

(C) C6H12O6 (r) Vapour density = 14

(D) CH3COOH (s) 14 NA(NA = 6.023 × 1023)electrons in a mole

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

(A) & (B) N2 = V.D. = 142

28

No. of electron in 1 mole = N0 × 14

(C) & (D) % C = 100180

72 = 40%



SECTION-F (Integer Type) No Negative Marking

This Section contains 8 questions. The answer to each question is a single digit integer ranging from 0 to 9. 8 × 5 = 40 Marks

1. 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

2. 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 3. How many of the following compounds have at least one atom in +5 oxidation state NH4NO3;

K2MnO4; N2O5; IF5; XeO3; H2SO5; NH3; PH 4 ; PCl3

Sol. 3 NH4NO3; N2O5; IF5 4. An oxybromate compound, KBrOx, where x is unknown, is analysed and found to contain 52.92 % Br.

What is the value of x? [At. Wt. of Br = 80; K = 39; O = 16] Sol. 2

Percentage Br = 100x168039

80

x = 2 5. How many significant figures does the result of the following calculation have?

3950.2

6500.0235.000.635

Sol. 3 40.999 (Minimum significate No = 3; As = 40.5 has been significant figures ‘3’ corresponding to the term 0.235. 6. A hypothetical metal oxide has 32% metal. The V.D. of Chloride salt of this metal is 79. Valency of the

metal is: Sol. 4

Eq. Wt. of metal = 76.368

832

oxygen of Weight

8Weight

Valency = 435.53.76

792

35.5E

V.D.2

7. 50 mL anhydrous oxalic acid requires 50 mL 0.2 N KMnO4 for complete oxidation. Normality of oxalic

acid is x

1; X is [M.Wt. of oxalic acid is 90]

Sol. 5 1/5 N N1V1 = 50 × 0.2

i.e. x = 5 N1V1 = 50

10

8. If the correct relation of Normality and Molarity of H3PO4 in the given reaction

H3PO4 + 2NaOH Na2HPO4 + 2H2O is x

1

N

M ; What is x

Sol. 2

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