Q. No. 1 Which of the following contains equal to those in 12g Mg? (At. Wt. Mg = 24)Option 1 12 gm COption 2 7 gm N2

Option 3 32 gm O2

Option 4 None of theseCorrect Answer 2Explanation 12 1

Moles of Mg = =24 2

1Atoms = N

2 A

7 1Moles of N = =

28 42

1 NNo of atoms = N 2=

4 2 A

A

Q. No. 2If moles of oxygen combine with Al to form Al O ,the weight of Al used in

1he1

2 t2 3

reaction is (Al = 27)Option 1 27 gOption 2 54 gOption 3 40.5 gOption 4 81 gCorrect Answer 2Explanation 3

2Al+ O Al O2

2 2 3

32 mole mole

2

For mole of O , moles of Al required is 22

e3

mol2

at = 2 27 = 54gm

Q. No. 3 Which has the highest mass?Option 1 50 g of ironOption 2 5 moles of N2

Option 3 0.1 mol atom of AgOption 4 1023 atoms of carbonCorrect Answer 2Explanation 50 gm Fe

5 moles N = 5 28 = 140gm2

1 mole Ag = 0.1 108 = 10.8 gm

10 110 atomof C = moles = moles

6.0226.022 10

2323

23

1= 12= 2 gm

6.022

Q. No. 4 The number of atoms present in 0.5 mole of nitrogen is same as the atoms inOption 1 12 g of COption 2 64 g of SOption 3 8 g of OOption 4 48 g of MgCorrect Answer 1Explanation 0.5 mole N contain 0.5 N molec es ul s A2

= 0.5 N 2 = N atom A A

This is same for 12 gm of C

Q. No. 5 Which of the following weighs the least?Option 1 2 g atom of N (at. wt. of N = 14)Option 2 atoms of C (at. wt of3 10 C = 12) 23

Option 3 1 mole of S (at. wt. of S = 32)Option 4 7 g silver (at. wt. of Ag = 108)Correct Answer 2Explanation 2 gm atom of N 28 gm

3 10 13 10 atomof C = = moles

26 10

2323

23

1= 12=6gm

2

1 mole of S = 32 gm7 gm Ag

Q. No. 6 If NA is Avogadro’s number then number of valence electrons in 4.2 g of nitride ions(N3–) is

Option 1 2.4 NA

Option 2 4.2 NA

Option 3 1.6 NA

Option 4 3.2 NA

Correct Answer 1Explanation 4.2

Moles = = 0.314

0.3 mole contains 0.3 NA ionsOne ion has 8 valence electron : total valence electro 0.3Nn = 8 = 2.4 NA A

Q. No. 7 Hemoglobin contains 0.33 % of iron by weight. The molecular weight of hemoglobin isapproximately 67200. The number of iron atoms (at. wt. of Fe = 56) present in onemolecule of hemoglobin is

Option 1 6Option 2 1Option 3 4Option 4 2Correct Answer 3Explanation 100 gm of Haemoglobin contains 0.33 gm of iron. Let there are x atoms of iron present

in one molecule.100 1

Moles of Haemoglobin= =67200 672

0.33Moles of Fe = = 4

672 56x x

Q. No. 8 The number of molecules in 4.25 g of ammonia is aboutOption 1 1.0 10 23

Option 2 1.5 10 23

Option 3 2.0 10 23

Option 4 2.5 10 23

Correct Answer 2Explanation 4.25

Moles = = 0.2517

No. of molecules = 0.25 6.023 10 23

= 1.5 10 23

Q. No. 9 If 20% nitrogen is present in a compound, its minimum molecular weight can beOption 1 144Option 2 28Option 3 100Option 4 70Correct Answer 4Explanation For minimum molecular weight 1 mole of compound must contain 1 mole of atom.

Moles of compound = Moles of atom

100 20= M= 70

M 14

Q. No. 10 The weight of molecule of the compound C60H122 isOption 1 1.4 10 g -21

Option 2 1.09 10 g -21

Option 3 5.025 10 g 23

Option 4 16.023 10 g 23

Correct Answer 1Explanation (6

wt0

o1

f on2+12

e molecu2 1)

= gm6. 0

l022

e1

23

=1.4 10 gm -21

Q. No. 11 Choose the wrong statement :Option 1 6.02 11 mole means particles0 23

Option 2 Molar mass is mass of one moleculeOption 3 Molar mass is mass of one mole of a substanceOption 4 Molar mass is molecular mass expressed in gramsCorrect Answer 2Explanation Molar mass is the mass of one mole of molecules.

Q. No. 12 Which among the following is the heaviest?

Option 1 One mole of oxygenOption 2 One molecule of sulphur trioxideOption 3 100 amu of uraniumOption 4 44 g of carbon dioxideCorrect Answer 4Explanation 1 mole O2 = 32 gm

1 molecule of SO 80 1.66 10= 3-27

100 1100 .66amu of U k= 10 g -27

44 gm of CO2

Q. No. 13 Rearrange the following I to IV in order of increasing masses and choose the correctanswer [At. wt. of N = 14 u, O = 16 u, Cu = 63 u]I 1 molecule of oxygenII 1 atom of nitrogen

III mol molecule of1 10 oxygen -10

mol atIV om of1 1 ppe0 co r -10

Option 1 II I III IV Option 2 IV III II I Option 3 II I II IV Option 4 I II IV III Correct Answer 1Explanation I. 1 molecule of O2 = 32 a.m.u

= 32 1.66 10 gm -24

II. 1 atom of nitrogen = 14 a.m.u

=14 1.66 10 gm -24

III. 10 mole o = 10f O 32 gm-10-102

IV. 10 mole of = 10 63.5 gC mu -10-10

II I III IV

Q. No. 14 The number of moles of SO2Cl2 in 13.5 g isOption 1 0.1Option 2 0.2Option 3 0.3Option 4 0.4Correct Answer 1Explanation 13.5

Moles = = 0.1135

Q. No. 15 The largest number of molecules is inOption 1 36 g of waterOption 2 28 g of carbon monoxideOption 3 46 g of ethyl alcoholOption 4 54 g of nitrogen pentoxideCorrect Answer 1Explanation 36

36 gm water = moles = 2moles18

No. of molecules = 2N A

2828 gmCO= =1 mole

28No. of molecules = N1 A

4646 gm of CH CH OH= =1mole

463 2

no of molecules = 1 N A

5454 gmof N O = moles

1082 5

1 1= mole = N molecules

2 2 A

Q. No. 16 Which of the following contains maximum number of atoms?Option 1 molecul6.023 1 es of CO0 21

2

Option 2 22.4 L of CO2 at STPOption 3 0.44 g of CO2

Option 4 None of theseCorrect Answer 2Explanation molecu6.023 les of O10 21

2

6.023 10= moles =10 moles

6.022 10

21-2

23

No of atom=10 N 3 atom -2A

22.422.4 lit of CO = =1 mole

22.42

Atom = 1 N 3 = 3N atom A A

0.440.44 gmCO = moles = 0.01 moles

442

atom = 0.01 N 3 atom A

Q. No. 17 If 0.5 mol of BaCl2 is mixed with 0.2 mol of Na3PO4, the maximum number of mole ofBa3(PO4)2 that can be formed is

Option 1 0.7Option 2 0.5Option 3 0.30Option 4 0.10Correct Answer 4Explanation 3 BaCl +2Na PO Ba (PO ) +6NaCl2 3 4 3 4 2

0.5 mole 0.2 mole

0.5 mole BaCl requir2

0.5 molees = Na PO3

32 4

= 0.33 mole Na3PO4

Na PO is the limiting reac antt 3 4

moles of Ba PO 0.( 1) mole 3 4 2

Q. No. 18 One mole of a mixture of CO and CO2 requires exactly 20 gram of NaOH in solution for

complete conversion of all the CO2 into Na2CO3. How many moles more of NaOHwould it require for conversion into Na2CO3 if the mixture (one mole) is completelyoxidised to CO2

Option 1 0.2Option 2 0.5Option 3 0.4Option 4 1.5Correct Answer 4Explanation Total moles of mixture = 1 mole

Let moles of CO = xmoles of CO = (1 - )  x 2

2

CO + 2NaOH Na CO +H O2 2 3 2

20 1Moles of NaOH= = mole

40 21 mole of CO2 requires 2 mole of NaOH

1 - mole of CO requires 2 1 - mo( ) ( le of NaOH)x x 2

12(1- ) =

2x

3=

4x

3 3 1Moles of CO= , moles of CO =1- =1- =

4 4 4x 2

On oxidation only CO converts its CO2

3Moles of CO formed=

4 2

Total moles of CO = 1 mole 2

Moles of NaOH required = 2 moleExtra moles of NaOH required = 2 - 0.5

= 1.5

Q. No. 19 The number of water molecules present in a drop of water (volume = 0.0018 ml) atroom temperature is (density of H2O = 1 g/mL)

Option 1 6.023 10 19

Option 2 1.084 10 18

Option 3 4.84 10 17

Option 4 6.023 10 23

Correct Answer 1Explanation Mass of water = v

= 1 0.0018= 0.0018 gm

0.0018Moles = = 0.0001

18

No of molecules = 0.0001 6.022 10 23

= 6.022 10 19

Q. No. 20 What is the weight of oxygen required for the complete combustion of 2.8 kg of

ethylene?Option 1 2.8 kgOption 2 6.4 kgOption 3 9.6 kgOption 4 96 kgCorrect Answer 3Explanation C H + 3O 2CO + 2H O2 4 2 2 2

2.8 10Moles of C H = = 100 moles

28 3

2 4

Moles of O required = 300 moles 2

wt of oxygen = 300 32 gm = 9.6 kg

Q. No. 21 A sample of pure calcium weighing 1.35 g was quantitatively converted to 1.88 g ofpure calcium oxide. Atomic mass of calcium would be:

Option 1 20Option 2 40Option 3 16Option 4 35.5Correct Answer 2Explanation 1

Ca+ O CaO2

2

1.35Moles of Ca=

M1.35

Moles of CaO =M

1.88Moles of CaO =

561.35 1.88

= M= 40M 56

Q. No. 22 30 g of magnesium and 30 g of oxygen are reacted, then the residual mixture containsOption 1 60 g of Magnesium oxide onlyOption 2 40 g of Magnesium oxide and 20 g of oxygenOption 3 45 g of Magnesium oxide and 15 g of oxygenOption 4 50 g of Magnesium oxide and 10 g of oxygenCorrect Answer 4Explanation 2Mg + O 2MgO2

30 gm 30 gm

30Moles of Mg = =1.25

2430

Moles of O = = 0.9375322

1.251.25 moles of Mg requires moles of O = 0.625

2 moles2

Mg is the limiting reactantMoles of MgO formed = 1.25

Mass of MgO = 1.25 40 gm = 50 gm

Moles of O2 left = 0.9375 - 0.625= 0.3125

0.3125wt o 3f O = 2gm = 10 gm 2

Q. No. 23 Silicon carbide, is produced by heating SiO2 and C to high temperatures according tothe equation :

SiO (s) + 3C(s) SiC(s) + 2CO(g)2

How many grams of SiC could be formed by reacting 2.00g of SiO2 and 2.0 g of C?Option 1 1.33Option 2 2.56Option 3 3.59Option 4 4.0Correct Answer 1Explanation SiO + 3C (s) SiC + 2CO2

2 gm2 gm

2Moles of SiO = = 0.033moles

602

2Moles of C = = 0.166 moles

12C is in excess and the limiting reactant is SiO2

Moles of SiC formed = 0.033wt of SiC formed = 0.033 40 gm

= 1.32 gm

Q. No. 24 Given the reaction

Pb(NO ) (aq) + 2KI PbI (s) + 2KNO (aq)3 2 2 3What is the mass of PbI2 that will precipitate if 10.2 g of Pb(NO3)2 is mixed with 5.73 gof KI in a sufficient quantity of H2O?

Option 1 2.06 gOption 2 4.13 gOption 3 7.96 gOption 4 15.9 gCorrect Answer 3Explanation Pb(NO ) + 2KI PbI +2KNO3 2 2 3

5.73 gm10.2 gm

10.2Moles of Pb(NO ) = = 0.0307

3323 2

5.73Moles of KI = = 0.0345

166L.R is K.IMoles of PbI formed = 0.0172 2

wt of PbI = 0.0172 462 2

= 7.946 gm

Q. No. 25 If 9 moles of O2 and 14 moles of N2 are placed in a container and allowed to react

according to the equation :

3O + 2N 2N O2 2 2 3The reaction proceeds until 3 moles of O2 remain, how many moles of N2O3 arepresent at that instant?

Option 1 6Option 2 3Option 3 4Option 4 12Correct Answer 3Explanation 3O + 2N 2N O2 2 2 3

9 moles 14 moles

O2 is the limiting reactant ideally. When 3 moles of O2 remains at that instant moles ofO2 reacted = 6 moles

2moles of N O formed= 6 = 4mole

3 2 3

Q. No. 26 Iron (III) oxide can be reduced with CO to form metallic iron as described byunbalanced chemical reaction

Fe O + CO Fe + CO2 3 2The number of moles of CO required to form one mole of Fe from its oxide is

Option 1 1Option 2 1.5Option 3 2Option 4 3Correct Answer 2Explanation Fe O + 3CO 2 Fe + 3CO2 3 2

2 mole of Fe is formed from 3 moles of CO

1 mole of Fe is formed from moles o32

f CO

Q. No. 27 The mass of CaO that shall be obtained by heating 20 kg of 90 % pure lime-stone(CaCO3) is

Option 1 11.2 kgOption 2 8.4 kgOption 3 10.08 kgOption 4 16.8 kgCorrect Answer 3Explanation CaCO CaO + CO (g)3 2

20,000 90wt of pure CaCO =

100

3

= 18000 gm18000

Moles of CaCO = =180 moles1003

moles of CaO = 180wt of CaO = 180 56gm

= 10.08 kg

Q. No. 28 If potassium chlorate is 80 % pure, then 48 g of oxygen would be produced from

(atomic mass of K = 39)Option 1 153.12 g of KClO3

Option 2 122.5 g of KClO3

Option 3 245 g of KClO3

Option 4 98.0 g of KClO3

Correct Answer 2Explanation 2KClO 2KCl +3O (g)3 (s) 2

48Moles of O = =1.5

322

3 moles of O2 is formed from 2 mole KClO3

1.5 moles of O is form2

1.5ed mo fro les KClO3

m 2 3

= 1 mole KClO3

Q. No. 29 Antimony reacts with sulphur according to the equation

2Sb(s) + 3S(s) Sb S (s) 2 3The molar mass of Sb2S3 is 340 g mol–1

What is the percentage yield for a reaction in which 1.40 g of Sb2S3 is obtained from1.73 g of antimony and a slight excess of sulphur?

Option 1 80.9%Option 2 58.0%Option 3 40.5%Option 4 29.0%Correct Answer 2Explanation 2Sb(s) + 3S Sb S 2 3

1.73 gm

1.73Moles of Sb = = 0.0141

122Moles of Sb2S3 ideally formed

0.0141= = 0.00705

2Ideal wt of Sb S = 0.00705 [340]gm2 3

= 2.397But actual wt of Sb2S3 = 1.40 gm

1.40yield= 100= 58

2.397 % %

Q. No. 30 NH3 is produced according to the following reaction :

N (g)+3H (g) 2NH (g)2 2 3In an experiment 0.25 mol of NH3 is formed when 0.5 mol of N2 is reacted with 0.5 molof H2. What is % yield?

Option 1 75%Option 2 50%Option 3 33%Option 4 25%Correct Answer 1Explanation N (g) +3H (g) 2NH (g)2 2 3

0.5 mole 0.5 mole

H2 is the limiting reactant ideally the moles of NH3 formed

2= 0.5= 0.34

3

Actual moles of NH3 formed = 0.250.25

yield= 100 = 750.34

Q. No. 31 What is the weight % sulphuric acid in an aqueous solution which is 0.502 M insulphuric acid? The specify gravity of the solution is 1.07.

Option 1 4.77%Option 2 5.67%Option 3 9.53%Option 4 22.0%Correct Answer 1Explanation Molarity = 0.502 M

502 moles of H SO is present in 1 lit of solution i.e. 100 ml of solution 2 4

Mass of solution V= = 1.07 1000= 1070 gm

= 0.5Mass 02 of sol 98 = 49.ut me  196 g

49.096wt = 100= 4.77

1070 % %

Q. No. 32 Mole fraction of ethanol in ethanol-water mixture is 0.25. Hence, percentageconcentration of ethanol (C2H6O) by weight of mixture is

Option 1 25Option 2 75Option 3 46Option 4 54Correct Answer 3Explanation 1

X = 0.25=4C H OH2 5

1 mole of C H OH is present in 4 mole of solution 2 5

Moles of solvent = 4 - 1 = 3 molewt of C2H5OH = 46 gmwt of solven = 3 18 mt = 54 gTotal wt of solution = 46 + 54 = 100 gm

46by wt = 100 = 46

100% %

Q. No. 33 A molal solution is one that contains one mole of a solute inOption 1 1000 g of the solventOption 2 One litre of the solventOption 3 One litre of the solutionOption 4 22.4 litres of the solutionCorrect Answer 1Explanation In a molal solution one mole of solute is present in 1000 gm of solvent.

Q. No. 34 An aqueous solution of ethanol has density 1.025 g/mL and it is 2 M. What is themolality of this solution?

Option 1 1.79Option 2 2.143Option 3 1.951Option 4 None of theseCorrect Answer 2Explanation 2M C2H5OH

2 moles of C2H5OH is present in 1 lit or 1000 ml of solution.Mass of solution = V

= 1.025 1000= 1025 gm

Mass of solute = 2 46 gm = 92 gmMass of solvent = 1025 - 92 = 933 gm

2 1000Molality = =2.14

933

Q. No. 35 What volume of 0.4 M FeCl3.6H2O will contain 600 mg of Fe3+?Option 1 49.85 mLOption 2 26.78 mLOption 3 147.55 mLOption 4 87.65 mLCorrect Answer 2Explanation Let volume V lit

moles of solute = 0.4 V

moles of Fe = 0.4V 3+

600moles of Fe = 0.4V 56 =

1000 3+

V = 26.78 ml

Q. No. 36 A sample of H2SO4 (density 1.8 g/ml) is 90% by weight. What is the volume of the acidthat has to be used to make 1 litre of 0.2 M H2SO4?

Option 1 16 mLOption 2 10 mLOption 3 12 mLOption 4 18 mLCorrect Answer 3Explanation 100 gm of solution contain 90 gm of H2SO4

100 1Vol. of solution= lit

1.8 1000

1= lit

1890

Moles of H SO =982 4

90 18Molarity = =16.53

98 1

Let volume of solution = V lit16.53 V = 1 0.2 [Moles of solute will be same] V 12 ml

Q. No. 37 The density (in g mL–1) of a 3.60 M sulphuric acid solution that is 29% H2SO4 (molarmass = 98 g mol–1) by mass will be

Option 1 1.45Option 2 1.64Option 3 1.88Option 4 1.22Correct Answer 4Explanation 3.60 moles of H2SO4 is present in 100 ml of solution. Let density = d gm/ml

M ass of solu = d 0tion 100Mass of solut = 3.60e  98 gm

= 352.8 gm352.8

by wt = 100 =291000d

d = 1.22 gm/ml

Q. No. 38 An antifreeze mixture contains 40% ethylene glycol (C2H6O2) by weight in the aqueoussolution. If the density of this solution is 1.05 g mL, what is the molar concentration?

Option 1 6.77 MOption 2 6.45 MOption 3 0.0017 MOption 4 16.9 MCorrect Answer 1Explanation 100 gm of solution contain 40 gm ethylene glycol (C2H6O2).

100Vol. of solution= ml

1.051

= lit10.540

Moles of solute =62

40 10.5Molarity = =6.77M

62 1

Q. No. 39 What is the molarity of ion in aqueous solution that contain 34.2 ppSO m of2-4

Al2(SO4)3? (Assume complete dissociation and density of solution 1 g/mL)Option 1 3 10 M-4×Option 2 2 10 M-4×Option 3 10-4 MOption 4 None of theseCorrect Answer 1Explanation 34.2 gm of Al2(SO4)3 is present in 106 gm of solution

10Vol. of solution = ml

1

6

= 103 lit34.2

moles of solute = = 0.1 mole342

Al (SO ) 2Al + 3SO 3+ 2-2 4 3 4

0.3 mole0.1 mole

0.3Molarity of SO =

10 2-

4 3

= 3 10 M-4×

Q. No. 40 The mole fraction of a given sample of I2 in C6H6 is 0.2. The molality of I2 in C6H6 isOption 1 0.32Option 2 3.2Option 3 0.032Option 4 0.48Correct Answer 2Explanation 2 1

X = 0.2 = =10 5I2

i.e. 1 mole of I2 is present in 5 mole of solution.mole of C H = 5 - 1 = 4 6 6

wt of C H = 4 78 gm6 6

= 312 gm1 1000

molality = = 3.20312

Q. No. 41 In which mode of expression, the concentration of a solution remains independent oftemperature?

Option 1 MolarityOption 2 NormalityOption 3 FormalityOption 4 MolalityCorrect Answer 4Explanation Molality in values the wt. of solvent which does not changes with temperature.

Q. No. 42 With increase of temperature, which of these changes?Option 1 MolalityOption 2 Weight fraction of soluteOption 3 Fraction of solute present in unit volume of waterOption 4 Mole fractionCorrect Answer 3Explanation With increase of temperature volume changes.

Q. No. 43 Molarity and Normality changes with temperature because they involve:Option 1 MolesOption 2 EquivalentsOption 3 WeightsOption 4 VolumesCorrect Answer 4Explanation But molarity and normality changes with temperature because they involve volume

and volume changes with temperature.

Q. No. 44 When 500.0 mL of 1.0 M LaCl3 and 3.0 M NaCl are mixed. What is molarity of Cl– ion?Option 1 4.0 M

Option 2 3.0 MOption 3 2.0 MOption 4 1.5 MCorrect Answer 2Explanation M illimoles of = 5La 1Cl 00 3

= 500

m. moles of Cl = 1500 -

M illim oles of N = 500 3C 0a l = 150-m. moles of Cl = 1500

Total m. moles of Cl- = 3000Total volume = 100 ml

Molarity of C3000

= = 310

l00

-

Q. No. 45 When 50 mL of 2.00 M HCl, 100 mL of 1.00 M HCl and 100 mL of 0.500 M HCl aremixed together, the resulting HCl concentration of the solution is

Option 1 0.25 MOption 2 1.00 MOption 3 3.50 MOption 4 6.25 MCorrect Answer 2Explanation 50 2+100 1+100 0.5

M =50+100+100

net

= 1 M

Q. No. 46 A sample of H2SO4 (density 1.8 g mL–1) is 90% by weight. What is the volume of theacid that has to be used to make 1 L of 0.2 M H2SO4?

Option 1 16 mLOption 2 18 mLOption 3 12 mLOption 4 10 mLCorrect Answer 3Explanation 100 gm sample contain 90 gm of H2SO4

1V

0ol

0 1= ml= lit

1. of sampl

18e

.890 18

Molarity of solution= =16.5398 1

Let the vol. of solution taken = V litMoles of acid = V 16.53 = 1 0.2 or 0.12 ml V=0.012

Q. No. 47 What is the concentration of nitrate ions if equal volumes of 0.1 M AgNO3 and 0.1 MNaCl are mixed together?

Option 1 0.1 MOption 2 0.2 MOption 3 0.05 MOption 4 0.25 MCorrect Answer 3Explanation AgNO +NaCl AgCl+NaNO

3 3

0.1 V0.1 V0 0.1V0 0.1V

0.1V[NO ]= = 0.05M

2V -

3

Q. No. 48 How many grams of NaBr could be formed if 14.2 g of NaI are reacted with 40.0 mL ofa 0.800 M Br2?

2NaI+Br 2NaBr+ I2 2

Option 1 3.30Option 2 4.80Option 3 6.59Option 4 9.75Correct Answer 3Explanation 2NaI + Br 2NaBr+ I2 2

14.2 gm 40 ml of0.8M

14.2Moles of NaI= = 0.0946

1500.8 40

Moles of Br = = 0.0321000

2

0.0946 moles of NaI will be reacting with mole of Br i.e. 0.00

4.09

73 moles46

2 2

Br is limiting reactant 2

Moles of NaBr =formed 0.32 2 = 0.064

= 0.064wt of 103 =NaBr 6.59

Q. No. 49 If AgBr is assumed to be completely insoluble, What mass of AgBr precipitates when30.0 mL of a 0.500 mol/L solution of AgNO3 is added to 50.0 mL of an 0.400 mol/Lsolution of NaBr?

Option 1 3.76 gOption 2 1.28 gOption 3 2.82 gOption 4 3.76 kgCorrect Answer 3Explanation AgNO +NaBr Ag Br+NaNO3 3

50ml,30ml,0.4M0.500M

M illimoles of = 30AgN 5O 0.3

= 15Millimoles of =5Na 0Br   0.4

= 20Limiting reactant is AgNO 3

Millimoles of AgBr = 15

15M = [188] gm

1ass of AgBr  

000   

= 2.82

Q. No. 50 In a titration, 15.0 cm3 of 0.100 M HCl neutralizes 30.0 cm3 of Ca(OH)2 . What is themolarity of Ca(OH)2 solution?

Option 1 0. 0125Option 2 0.0250Option 3 0.0500Option 4 0.200Correct Answer 2Explanation Milli eq. of HCl = mlli. Eq of Co(OH)2

(M V 'n'factor) =(M V 'r'factor) HCl Ca(OH)2

0.1 15 1=30 M 2 M = 0.025

Q. No. 51 10 mL of 1 M BaCl2 solution and 5 mL 0.5 M K2SO4 are mixed together to precipitateout BaSO4 . The amount of BaSO4 precipitate will be

Option 1 0.005 molOption 2 0.00025 molOption 3 0.025 molOption 4 0.0025 molCorrect Answer 4Explanation BaCl +K SO BaSO + 2KCl2 2 4 4

m. moles of BaCl2 = 10m. moles of K2SO4 = 2.5

K SO is the limiting reac antt 2 4

m. moles of BaSO = 2.5 4

= 0.0025 mole

Q. No. 52 Mg of a substance when vaporised occupy a volume of 5.6 litre at NTP. The molecularmass of the substance will be:

Option 1 MOption 2 2 MOption 3 3 MOption 4 4 MCorrect Answer 4Explanation Mass of 5.6 lit at N.T.P is M gm

Mass of 22.4 lit at N.T.P isM

22.45.6

= 4 M

Q. No. 53 Number of molecules in 1 litre of oxygen at NTP is :Option 1 6.02 10

32 23

Option 2 6.02 1022.4

23

Option 3 32 22.4

Option 4 3222.4

Correct Answer 2Explanation 1

Moles of O =22.42

1No. of molecular = 6.022 10

22.4 23

Q. No. 54 The number of molecules in 89.6 litre of a gas at NTP are :Option 1 6.02 10 23

Option 2 2 6.02 10 23

Option 3 3 6.02 10 23

Option 4 4 6.02 10 23

Correct Answer 4Explanation 89.6

Moles = = 422.4

No. of molecular = 4 6.022 10 23

Q. No. 55 The mass of 112 cm3 of CH4 gas at STP isOption 1 0.16 gOption 2 0.8 gOption 3 0.08 gOption 4 1.6 gCorrect Answer 3Explanation 112 1

Moles = =22400 200

1Mass = 16 = 0.08 gm

200

Q. No. 56 An oxide of metal (M) has 40% by mass of oxygen. Metal M has atomic mass of 24. Theempirical formula of the oxide is

Option 1 M2OOption 2 M2O3

Option 3 MOOption 4 M3O4

Correct Answer 3Explanation Let the valency of metal = x

100 gm of oxide contain 40 gm of oxygenwt of metal = 100 - 40 = 60 gm40 gm of oxygen combines with 60 gm of metal

8 gm of oxygen combine60

8 gm of metal40

s with

Equivalent wt of metal = 12 gm24

Eq wt = 12 = Valency = 2Valency

Formula = MO

Q. No. 57 What is the empirical formula of a compound composed of O and Mn in equal weightratio?

Option 1 MnOOption 2 MnO2

Option 3 Mn2O3

Option 4 Mn2O7

Correct Answer 4Explanation The wt. of Mn and O are equal

Moles of Mn=55x

Moles of O=16x

Mole ratio of Mn : O= ;55 16x x

It empirical formula = Mn O 2 7

Q. No. 58 Determine the empirical formula of Kelvar, used in making bullet proof vests, is70.6% C, 4.2% H, 11.8% N and 13.4% O :

Option 1 C7H5NO2

Option 2 C7H5N2OOption 3 C7H9NOOption 4 C7H5NOCorrect Answer 4Explanation Given :

C = 70.6%, H = 4.2%,N = 11.8%, O =13.4%

Moles Simplest of ratioC 70.6

= 5.8812

5.88= 7.02

0.8375H 4.2

= 4.21

4.2=5.01

0.8375N 11.8

= 0.84214

0.842=1.005

0.8375O 13.4

= 0.837516

0.8375=1

0.8375Empirical formula C H NO 7 5

Q. No. 59 A compound contains atoms of three elements A, B and C. If the oxidation number ofA is +2, B is +5 and C is -2, the possible formula of the compound is:

Option 1 A(BC3)2 ☒

Option 2 A3(BC4)2 ☒

Option 3 A3(B4C)2 ☐

Option 4 ABC2 ☐

Explanation The sum of oxidation number of all the element, in a compound is equal to zero.In (a) i.e. A(BC ) ,sum of oxidation number + 2 + [5 - 6] =is 2 03 2

In (b) i.e. A3(BC4)2 sum of oxidation = 0

Q. No. 60 The carbonate of a metal is isomorphous (similar formula) with magnesium carbonate

and contains 6.091% of carbon. The atomic weight of metal isOption 1 24Option 2 56Option 3 137Option 4 260Correct Answer 3Explanation The formula of carbonate of metal = MCO3

12C = 100 =6.091

M+60%

M = 137

Q. No. 61 The Ew of an element is 13. It forms an acidic oxide which with KOH forms a saltisomorphous with K2SO4. The atomic weight of element is

Option 1 13Option 2 26Option 3 52Option 4 78Correct Answer 1Explanation Formula of salt will be M2SO4

Valency of element is 1

Equivalent w1

tM

=

M13 =

1M = 13

Q. No. 62 A hydrate of Na2SO3 losses 22.2% of H2O by mass on strong heating. The hydrate isOption 1 Na2SO3.4H2OOption 2 Na2SO3.6H2OOption 3 Na2SO3.H2OOption 4 Na2SO3.2H2OCorrect Answer 4Explanation Let the hydrate is Na2SO3. xH2O

100 gm of hydrate looses 22.2 gm of H2O on heating

Na SO .xH O Na SO + xH O2 3 2 2 3 222.2100 gm

100Moles of hydrate =

126+18x100

It will gives moles of H O126+18

xx 2

22.2mole of H O=

182

100 22.2= = 2

126+18 18x xx

Na SO .2H O 2 3 2

Q. No. 63 One of the following combinations illustrate law of reciprocal proportionsOption 1 N2O3, N2O4, N2O5

Option 2 NaCl, NaBr, NalOption 3 CS2, CO2, SO2

Option 4 PH3, P2O3, P2O5

Correct Answer 3Explanation C S (in CS )2

12 gm 64 gm

C O (in CO )212 gm 32 gm

wt ratio of S : O = 64 : 32 = 2 : 1 when S and O combines directly then they form SO in 2

which weight ratio of S : O is 32 : 32 = 1 : 1. This illustrate the law of reciprocalproportion.

Q. No. 64 If water samples are taken from sea, river, clouds, lake or snow, they will be found tocontain H2 and O2 in the approximate ratio 1 : 8. This indicates the law of

Option 1 Multiple proportionOption 2 Definite proportionOption 3 Reciprocal proportionsOption 4 None of theseCorrect Answer 2Explanation Any chemical compound always contain fixed ratio by wt of element in it no matter

they are prepared from any source or by any chemical method. This is law of definiteproportion.

Q. No. 65 The law of multiple proportion is illustrated byOption 1 Carbon monoxide and carbon dioxideOption 2 Potassium bromide and potassium chlorideOption 3 Water and heavy waterOption 4 Calcium hydroxide and barium hydroxideCorrect Answer 1Explanation When two elements combine to for more than one compound then with a fixed wt. of

one element the wt. ratio of other elements combining bears a simple whole nomultiple. This is the law of multiple proportion.

Q. No. 66 The percentage of copper and oxygen in samples of CuO obtained by differentmethods were found to be the same. This illustrates the law of

Option 1 Constant proportionsOption 2 Conservation of massOption 3 Multiple proportionsOption 4 Reciprocal proportionsCorrect Answer 1Explanation

Q. No. 67 Two samples of lead oxide were separately reduced to metallic lead by heating in acurrent of hydrogen. The weight of lead from one oxide was half the weight of leadobtained from the other oxide. The data illustrates.

Option 1 Law of reciprocal proportionsOption 2 Law of constant proportionsOption 3 Law of multiple proportionsOption 4 Law of equivalent proportionsCorrect Answer 3Explanation With a fixed mass of oxygen the wt of lead combining bears a simple whole no

multiple. This is the law of multiple proportion.

Q. No. 68 One part of an element A combines with two parts of another element B. Six part ofthe element C combine with four parts of the elements B. If A and C combine togetherthe ratio of their weights will be governed by

Option 1 Law of definite proportionsOption 2 Law of multiple proportionsOption 3 Law of reciprocal proportionsOption 4 Law of conservation of massCorrect Answer 3Explanation If A and B combine directly to form a compound and if C and B combine to form and

compound if A and C combine directly to form a 3rd compound then wt ratio of A and Cwill be governed by law of reciprocal proportion.

Q. No. 69 n g of substance X reacts with m g of substance Y to form p g of substance R and q g ofsubstance S. This reaction can be represented as follows.X + Y = R + sThe relation which can be established in the amounts of the reactants and theproducts will be

Option 1 n - m = p - qOption 2 n + m = p + qOption 3 n = mOption 4 p = qCorrect Answer 2Explanation X + Y = R + S

n gm m gm p gm g gm

Total wt of reactants = Total wt of productsn + m = p + q

Q. No. 70 Which one is the best example of law of conservation of mass?Option 1 6 g of carbon is heated in vacuum, there is no change in massOption 2 6 g of carbon combines with 16 g of oxygen to form 22 g of CO2

Option 3 6 g water is completely converted into steamOption 4 A sample of air is heated at constant pressure when its volume increases but there is

no change in massCorrect Answer 2Explanation During any chemical reaction total mass is conserved.

Q. No. 71 SO2 gas was prepared by (i) burning sulphur in oxygen, (ii) reacting sodium sulphitewith dilute H2SO4 and (iii) heating copper with conc. H2SO4. It was found that in eachcase sulphur and oxygen combined in the ratio of 1 : 1. The data illustrates the law of :

Option 1 Conservation of massOption 2 Multiple proportionsOption 3 Constant proportionsOption 4 Reciprocal proportionsCorrect Answer 3Explanation

Q. No. 72 A sample of CaCO3 has Ca = 40% C = 12% and O = 48% If the law of constantproportions is true, then the mass of Ca in 5 g of CaCO3 from another source will be:

Option 1 2.0 gOption 2 0.2 g

Option 3 0.02 gOption 4 20.0 gCorrect Answer 2Explanation Any chemical compound always has same % of elements no matter it is obtained from

any source.40

Mass of Ca = 5100

= 2 gm

Q. No. 73 H2S contains 5.88% hydrogen, H2O contains 11.11% hydrogen while SO2 contains 50%sulphur. These figures illustrate the law of:

Option 1 Conservation of massOption 2 Constant proportionsOption 3 Multiple proportionsOption 4 Reciprocal proportionsCorrect Answer 4Explanation In H S, H = 5.88 S = 94.122 %

In H O, H = 11.11 ,O = 88.892 %wt ration S : O = 1 : 1

In SO , S = 50 ,O = 50 2

wt ratio of S : O = 1 : 1This illustrate the law of reciprocal proportion.

Q. No. 74 Hydrogen combines with chlorine to form HCl. It also combines with sodium to formNaH. If sodium and chlorine also combine with each other, they will do so in the ratioof their masses as:

Option 1 23 : 35.5Option 2 35.5 : 23Option 3 1 : 1Option 4 23 : 1Correct Answer 1Explanation In HCl, wt ratio of H : Cl = 1 : 35.5

In NaOH, wt ratio of H : Na = 1 : 23If Na and Cl combine directly then wt ratio = 23 : 35.5

Q. No. 75 x g of Ag was dissolved in HNO3 and the solution was treated with excess of NaCl when2.87 g of AgCl was precipitated. The value of x is

Option 1 1.08 gOption 2 2.16 gOption 3 2.70 gOption 4 1.62 gCorrect Answer 2Explanation Ag + HNO AgNO AgCl NaCl

3 3Mole ratio is 1 : 1

Moles of Ag = Moles of AgCl

2.87=

108 143.5x

= 2.16 gmx

Q. No. 76 A 1.50 g sample of an ore containing silver was dissolved, and all the Ag+ wasconverted to 0.125 g Ag2S. What was the percentage of silver in the ore?

Option 1 14.23%Option 2 10.8%Option 3 8.27%Option 4 720%Correct Answer 4Explanation 0.125

Moles of Ag S =2482

2 0.125moles of Ag in Ag S =

248

2

= 0.001wt of Ag = 0.001 108

= 0.10880.1088

of Ag = 100 = 7.21.50

% %

Q. No. 77 NaOH is formed according to the reaction1

2Na + O Na O2

2 2

Na O + H O 2NaOH2 2

To make 4 g of NaOH, Na required isOption 1 4.6 gOption 2 4.0 gOption 3 2.3 gOption 4 0.23 gCorrect Answer 3Explanation 4

Moles of NaOH= = 0.1 mole40

Mole ratio of Na and NaOH is 1 : 1Moles of Na = 0.1 molewt of Na = 2.3 gm

Q. No. 78 2H PO + 3Ca(OH) Ca PO + 6 H O Equivalent weight of H PO in( t) his reaction3 4 2 3 4 2 2 3 4

isOption 1 98Option 2 49Option 3 32.66Option 4 24.5Correct Answer 3Explanation 98

Equal= = 32.66 or H PO has lost all root it3 3 4

Q. No. 79 The Ew of H3PO4 in the reaction is

Ca(OH) + H PO CaHPO + 2H O2 3 4 4 2(Ca = 40, P = 31, O = 16)

Option 1 49

Option 2 98Option 3 32.66Option 4 147Correct Answer 1Explanation Ca(OH) +H PO CaHPO +2H O2 3 4 24

In the reaction H3PO4 has lost 2Hits 'n' factor = 2

M 98Eq.wt = = = 49

2 2

Q. No. 80 What weight of a metal of equivalent weight 12 will give 0.475 g of its chloride?Option 1 0.12 gOption 2 0.24 gOption 3 0.36 gOption 4 0.48 gCorrect Answer 1Explanation x

M + Cl MCl2

2 X

n factor of metal = valency of metalLet the wt of metal = y gmGm - eq of metal = gm - eq of MClX

y 0.475=

M M + x(35.5)x x

M=12x

On solving y = 0.12 gm

Q. No. 81 How many grams of phosphoric acid would be needed to neutralise 100 g ofmagnesium hydroxide? (The molecular weights are : H3PO4 = 98 and Mg (OH)2 = 58.3)

Option 1 66.7 gOption 2 252 gOption 3 112 gOption 4 168 gCorrect Answer 3Explanation Gm - eq of H3PO4 = gm - eq of Mg(OH)2

1003 = 2

98 58x

= 112 gmx

Q. No. 82 0.116 g of C4H4O4 (A) is neutralised by 0.074 g of Ca(OH)2. Hence, protonic hydrogen

( ) in (A) wilH l be

Option 1 1Option 2 2

Option 3 3Option 4 4Correct Answer 2Explanation Gm - equivalents of C4H4O4

= gm - eq of Ca(OH)2

G = moles 'n'factorm - eq of C H O 4 4 4

0.116= ( )

116x

[where x is the no of H+ released]

Gm - eq of Ca(O0.074 2

=H  74

)

2

0.116 0.074= 2 = 2

116 74x x

Q. No. 83 4.2 g of metallic carbonate MCO3 was heated in a hard glass tube and CO2 evolved wasfound to have 1120 mL of volume at STP. The Ew of the metal is

Option 1 12Option 2 24Option 3 18Option 4 15Correct Answer 1Explanation MCO MO + CO3 2

1120 ml4.2 gmof S.T.P.

1120Moles of CO = = 0.05

224002

4.2Moles of MCO =

M + 603

4.2= 0.05 M=24

M + 60

24Eq. wt of metal= =12

2

Q. No. 84 1.0 g of a monobasic acid when completely aceted upon Mg gave 1.301 g of anhydrousMg salt. Equivalent weight of acid is

Option 1 35.54Option 2 36.54Option 3 17.77Option 4 18.27Correct Answer 2Explanation Let the monobasic acid = HA

Mg + 2HA MgA + H 2 21 gm 1.301 gm

1 1.301Moles of HA = ,Moles of MgA =

A+1 24+2A

2

As 2 moles of HA gives 1 mole of MgA2

1 1 1.301moles of HA gives =

A + 1 2(A + 1) 24 + 2A

A = 35.54Molecular wt = 35.54 + 1 = 36.54Eq. wt = 36.54 /1

Q. No. 85 0.1 g of metal combines with 46.6 mL of oxygen at STP. The equivalent weight of metalis

Option 1 12Option 2 24Option 3 6Option 4 36Correct Answer 1Explanation 46.6

Moles ofoxygen= = 0.002022400

wt of oxygen = 0.0020 32 = 0.064 gm0.064 gm of O combines with 0.1 gm of metal

0.18 gm of O combines with= 8 =12 gm

0.064

Q. No. 86 When 100 ml of 1 M NaOH solution and 10 ml of 10 N H2SO4 solution are mixedtogether, the resulting solution will be :

Option 1 AlkalineOption 2 AcidicOption 3 Strongly acidicOption 4 NeutralCorrect Answer 4Explanation milli - eq of = 1aOH 0N 1 1 0

= 100m. eq of H S = 1O    0 102 4

= 100Solution is neutral

Q. No. 87 Normality of 0.74 g Ca(OH)2 in 5 mL solution isOption 1 8 NOption 2 4 NOption 3 0.4 NOption 4 2 NCorrect Answer 2Explanation Gm - equivalents of Ca(OH)2

0.74 1= =

74 502

1 1000N= = 4N

50 5

Q. No. 88 Normality of a 2 M sulphuric acid isOption 1 2Option 2 4 NOption 3 N/2Option 4 N/4Correct Answer 2

Explanation N = M 'n' factor

= 2 2 = 4N

Q. No. 89 1 L of a normal solution is diluted to 2000 ml. The resulting normality is :Option 1 N/2Option 2 N/4Option 3 NOption 4 2NCorrect Answer 1Explanation Gm - eq. of so =l 1ution 1 = 1

1 1000New normality =

2000

1 N= i.e

2 2

Q. No. 90 What volume of 0.232 N solution contains 3.17 milliequivalent of solute?Option 1 137 mLOption 2 13.7 mLOption 3 27.3 mLOption 4 12.7 mLCorrect Answer 2Explanation Milliequivalents of solute

= N V= 0.232 V = 3.17 (mL)

V = 13.7 mL

Q. No. 91 1 L solution of NaOH contains 4.0 g of it. What shall be the difference betweenmolarity and the normality?

Option 1 0.10Option 2 ZeroOption 3 0.05Option 4 0.20Correct Answer 2Explanation For NaOH ‘n’ factor = 1

N and M are sameDifference is zero

Q. No. 92 100 mL of 0.3 N HCl is mixed with 200 ml of 0.6 N H2SO4. The final normality of theresulting solution will be

Option 1 0.1 NOption 2 0.2 NOption 3 0.3 NOption 4 0.5 N

Correct Answer 4Explanation m. eq of HCl = 30

m. eq of H SO = 200 0.6 = 1202 4

Total gm - eq = 150Total volume = 300 ml

150 1N = = = 0.5N

300 2 Final

Q. No. 93 Normality of a mixture of 30 mL of 1 N H2SO4 and 20 mL of 4 N H2SO4 isOption 1 1.0 NOption 2 1.1 NOption 3 2.0 NOption 4 2.2 NCorrect Answer 4Explanation Total m. eq= 30 1 + 20 4

= 30 + 80= 110

Total volume = 50 ml110

N= =2.2N50

Q. No. 94 Normality of solution obtained by mixing 10 mL of 1 N HCl, 20 mL of 2 N H2SO4 and30 mL of 3 N HNO3 is

Option 1 1.11 NOption 2 2.22 NOption 3 2.33 NOption 4 3.33 NCorrect Answer 3Explanation 10 1+20 2+30 3

N =60

net

= 2.33 N