1) Consider the following pairs:

I. CH4, C2 H6 II. CO, CO2 III. NO, NO2 IV. H2O, H2O2

In which cases, law of multiple proportion is followed?

a) I, II b) I, II, III c) I, III, IV d) I, II, III, IV

2) In which case purity of the substance is 100%?

a) 1 mol of CaCO3 gave 11.2 L CO2 (at STP)

b) 1 mol of MgCO3 gave 40.0 g MgO.

c) 1 mol of NaHCO3 gave 4g H2O

d) 1 mol of Ca(HCO3)2 gave 1 mol CO2

3) and are two isotopes of chlorine. If average atomic mass is 35.5 then ratio of

these two isotopes is:

a) 35:37 b) 1: 3 c) 3 : 1 d) 2:1

4) Ionic mass of X3- is 17. If it has 10 electrons, then number of neutrons are:

a) 10 b) 13 c) 7 d) 17

5) M2+ ion is isoelectronic of SO2 and has (Z+2) neutrons (Z is atomic number of M). Thus,

Ionic mass of M2+ is:

a) 70 b) 60 c) 68 d) 64

6) X+ , Y2+ and Z- are isoelectronic of CO2. Increasing order of protons in X+, Y2+, and Z- is :

a) X+ = y2+ = Z- b) X+ < Y2+ < Z- c) Z- < X+ < Y2+

d) Y2+ < X+ < Z-

7) Which has maximum number of H-atoms per gram of the substance?

a) CH4 b) CuSO4 . 5H2O c) H2O2 d) H2O

8) If each O-atom has two equivalents, volume of one equivalent of O2 gas at STP is

a) 22.4L b) 11.2L c) 5.6L d) 44.8L

9) If Avogadro’s number would have been 1 x 10-10, instead of 6.02 x 1023 then mass of one

atom of H would be

a) 1 amu b) 1x1010 amu c) 6 amu d) 6x1013 amu

10) A hydrocarbon has 3 g carbon per gram of hydrogen, hence, simplest formula is

a) CH4 b) C6H6 c) C3H8 d) CH2

11) Number of atoms in 20g Ca is equal to number of atoms in

a) 20g Mg b) 1.6 g CH4 c) 1.8 G H2O d) 1.7g NH3

12) Rest mass of an electron is 9.11 x 10-31 Kg. Molar mass of the electron is

a) 1.50 x 10-31 Kg mol-1 a) 9.11 x 10-31 Kg mol-1

c) 5.5 x 10-7 Kg mol-1 d) 6.02 x 1023 Kg mol-1

13) Which is / are correct about 4.25 g NH3?

a) It contains 0.25 mol of NH3 b) It contains 0.7 mol of H-atoms

c) It contains total of 1.0 mol of N and H atoms d) It contains 1.5x1023 molecules of NH3

14) A spherical ball of radius 7 cm contains 56% iron. If density is 1.4g / cm3 , number of

mol of Fe present approximately is

a) 10 b) 15 c) 20 d) 25

15) Among the following groupings which respondents the collection of isoelectronic

species?

a) NO, CN-, N2, O b) NO+, C , O , CO

c) N2, C , CO, NO b) CO, NO+, CN-, C

16) X- is isoelectronic of CO and has (Z+2) neutrons (Z=atomic number of X-). Thus,

a) Ionic mass of X- is 28 b) Ionic mass of X- is 30

c) atomic number of X- is 13 d) atomic number of X- is 14

17) 10L of hard water with temporary hardness [of Ca(HCO3)2] required 0.56 g of lime:

Ca(HCO3)2 + CaO 2CaCO3 + H2O

Temporary hardness in terms of ppm of CaCO3 is:

a) 56 b) 1/2 c) 100 d) 200

18) 2KCLO3 2KCl + 3O2 . 5.0g of KClO3 gave 0.03 mol of O2. Hence, percentage purity

of KClO3 is:

a) 49% b) 50% c) 95% d) 98%

19) 1.70g AgNO3 (aq) reacts with 5.85 g NaCl (aq) to form AgCl (white ppt):

a) 14.35 g b) 0.1435 g c) 1.435 g d) 5.85 g

20) When 10g CaCO3 reacts with 20 g BaCl2

BaCl2 + CaCO3 BaCO3 + CaCl2

then limiting reactant is:

a) CaCO3 b) BaCl2 c) CaCl2 d) None of these

21) In the following reaction.

MnO2 + 4HCl MnCl2 + 2H2O + Cl2

2 mol MnO2 reacts with 4 mol of HCl to form 11.2 L Cl2 at STP. Thus, per cent yield of Cl2

is:

a) 25% b) 50% c) 100% d) 75%

22) Mixture of 1 mol of Na2 CO3 and 2 mol of NaHCO3 forms 1 mol of CO2. Thus per cent

yield of CO2 is:

a) 25% b) 50% c) 75% d) 100%

23) 100mL solution of NaOH (containing 4 g NaOH per litre) and 50mL of HCl (containing

7.3g HCl per litre) react as:

NaOH(aq) + HCl (aq) NaCl(aq) + H2O (l)

0.5 g of NaCl of formed. Thus, unreacted NaOH is

a) 0.058 g b) 3.66 g c) 10.8g d) 0.63g

24) Al and KClO3 react together to form Al2O3 according to

2KClO3 2KCl + 3O2

4Al + 3O2 2Al2O3

4 mol of KClO3 (50% pure) on reaction with excess of Al forms Al2O3.

25) Mole of fraction of ethanol (C2H5OH) in ethanol-water system is 0.25. Thus, it has

a) 25% ethanol by weight of solution.

b) 75% water by weight of solution

a) 2 mol b) 4 mol c) 6 mol d) 8 mol

26) A certain compound has the molecular formula X4O6 having 57.2% X. thus:

a) atomic mass of X is 32 b) X may contain five valence electrons

c) X is an electropositive metal d) X can be a non-metal.

27) An aqueous solution of glucose (C6H12O6) contains 1 mol glucose in 1.8L solution. Thus.

a) it is 10% aqueous solution b) its density is 0.1g mL-1

c) Solution contains 3.33 x 1020 glucose molecules per mL.

d) solution contains 24g atoms.

28) A solution of solute X contains 40% X by weight of solution. 800g of this solution was

cooled when 100g of solute is precipitated. Thus, percentage composition of the remaining

solution is:

a) 31.4% b) 20.0% c) 23.0% d) 24.0%

29) Which reactant is limiting and which is in excess?

Limiting Excess

a) K2PtCl4 NH3

b) NH3 K2PtCl4

c) Exact molar ratio

d) Would be decided by the quantity of the product formed

30) Number of moles of K2PtCl6 consumed is

a) 0.048 b) 0.024 c) 0.012 d) 0.096

31) Number of moles of NH3 consumed is

a) 0.048 b) 0.024 c) 0.096 d) 0.192

32) Number of moles of cis-platin formed is

a) 0.012 b) 0.024 c) 0.048 d) 0.192

33) Number of moles of excess reactant which remains unreacted is

a) 0.024 b) 0.34 c) 0.54 d) 0.56

A. Assertion / Reason Type Questions

Codes:

a) Both A and R are true and R is the correct explanation of A.

b) Both A and R are true but R is not the correct explanation of A

c) A is true but R is false

d) A is false but R is true

1) Assertion (A) : In the following reaction:

SO2 + 2H2S 3S + 2H2O

2) Assertion (A) : 20 g of 50% pure CaCO3 on heating gave 2.24 L CO2 at STP and 15.6 g

non-volatile residue.

CaCO3 CaO + CO2

3) Assertion (A) : H2 + O2 2H2O.

2g H2 react with 1 g O2 to form 2 g H2 O.

N2 + 3H2 2NH3

2KClO3 2KCL + 3O2

CHAPTER – IV

1 mol SO2 and 1 mol H2S form 3 mol sulphur.

Reaction (R) : 1 mol SO2 and 2 mol H2S form 3 mol sulphur and 2 mol H2O.

Reaction (R) : Pure CaCO3 is 10 g which gave 4.4 g CO2 (=0.1 mol = 2.24 Lat STP )

and 15.6 g residue is left.

Reaction (R) : 2 mol N2 react with 1 mol O2 to form 2 mol H2O.

4) Assertion (A) : 2 mol N2 react with 3 mol H2 to form 2 mol NH3.

1 mol SO2 and 1 mol H2S form 3 mol sulphur.

Reaction (R) : N2 is excess and H2 is the limiting reactant.

5) Assertion (A) : 4 mol of KClO3 (50% pure) gave 3 mol of O2 on heating strongly and thus

yield is 100%

Type-I.

4.Which has maximum number of millimoles of Cl- ion?

(a) 0.208 g Bacl2 (b) 100 ml. M Bacl2

(c) 0.745 g (d) Equal

5. 100 ml. of solution of CaCl2 is evaporated to dryness; residue obtained is 0.111 g. Molarity

of CaCl2 solution is:

(a) 0.1 M (b) 1.0 M

(c) 0.01 M (d) 0.001 M

6. Hardness in water is 200 ppm CaCO3. Molarity of CaCO3 is:

(a) 2 x 10-3 M (b) 1 x 10-3 M

(c) 2 x 10-2 M (d) 2 x 10-4 M

7. Mole fraction of CaCO3 in hard water having hardness 200 ppm CaCO3 is:

(a) 0.1 (b) 3.6 x 10-5

(c) 27.78 x 10-3 (d) 0.035

15. 2.86 g of Na2CO3. xH2O in 100 ml. solution is 0.2 N. Hence, x is:

(a) 5 (b) 10

(d) 20 (d) 2

16. Urea solution is one molal. Urea present in one Kg solution is:

(a) 60 g (b) 56.6 g

(c) 10.60 g (d) 10.0 g

17. H2SO4 is 98% by weight of solution. Hence it is:

(a) 1 molal (b) 10 molal

(c) 50 molal (d) 500 molal

Chapter – 3

1) Formal charge on the middle oxygen in O3 is

a) -1 b) +1 c) 0 d) -3

7) Oxidation numbers of Cl atoms in CaOCl2 (bleaching powder) are

a) zero on each

b) -1 on Cl* and +1 on Cl**

c) +1 on Cl* and -1 on Cl**

d) 1 on each

8) In which case oxidation number of Cr has been affected?

a) 2CrO +2H+ Cr2O + H2O

b) Cr2O + 2OH- 2CrO + H2O

c) (NH4)2Cr2O7 N2 + Cr2O3 + 4H2O

d) CrO2Cl2 + 2OH- CrO + 2HCl

11) Which is intramolecular oxidation reduction reaction?

a) (NH4)2Cr2O7 N2 + Cr2O3 + H2O

b) NH4NO3 N2O + 2H2O

c) 2KClO3 2KCl + 3O2

d) All of the above

12) Which is not the disproportionation reaction?

a) 3H3PO2 2H3PO3 + PH3

b) HCHO + OH- HCOO- + CH3OH

c) NH4NO3 H2O + 2H2O

d) 3Cl2 + 6OH- 5Cl- 3H2O

13) Upto what products can water be oxidised?

a) to O2 and H2 b) to OH-

c) to OH- d) to O2

15) In the reaction Cu + H2SO4 CuSO4 + H2O + SO2:

a) H+ is the oxidizing agen

b) SO is the oxidizing agent

CaCl*

ClO**

O O O.. .. .... ..:

c) both are correct

d) none is correct

16) In the following redox reaction,

Cr2O + Fe2+ Fe3+ + Cr3+

1 mol of Cr2O oxidizes.

a) 1 mol of Fe2+ b) 3 mol of Fe2+

c) 4 mol of Fe2+ b) 6 mol of Fe2+

17) One mol of ferrous oxalare is oxidized by x mol of MnO in acidic medium. x is.

a) 0.6 b) 0.1 c) 0.3 d) 1.0

18) Values of p, q, r, s and t are in the following redox reaction.

pBr2 + qOH- rBr- sBrO + tH2O

p q r s t p q r s t

a) 3 6 1 5 3 b) 3 6 5 3 1

c) 3 6 5 1 3 d) 3 5 1 6 3

19) Cr2O + 2I - + 14H+ I2 + 2Cr3+ 7H2O which are not in balanced position?

(a) H+ and H2O (b) Cr2O72- and Cr3+

(c) I- and I2 (d) All are balanced

22) If the following is balanced reaction.

4O - + 2H2O 4 OH - + 3O2

then x is and O2x- is:

(a) -1 and species is superoxide

(b) -2 and species is peroxide

(c) -4 and species is oxide

(d) -1 and species is peroxide

23) O3 is estimated by I- in iodometric method. Balanced reaction is:

(a) 2H+ + O3 + 2I- I2 + O2 + H2O

(b) 2I- + O3 + H2O I2 + 2OH-

(c) both (a) and (b)

(d) none of the above

24) I- reduces IO to I2 and itself oxidized to I2 in acidic medium. Final reaction is:

(a) I- + IO + 6H+ I2 + 3H2O

(b) I- + IO I2 + O3

(c) SI- + IO + 6H+ 3I2 + 3H2O

26) 0.05 mol of Ca(OH)2 can neutralise H2SO4. This H2SO4 can be neutralised by:

(a) 0.05 mol of NaOH (b) 0.10 mol of NaOH

(c) 0.05 mol of POH(OH)2 (d) none is correct

27) N2 + 3H2 2NH3. In this reaction equivalent weight of N2 is:

(a) 4.67 (b) 28

(c) 14 (d) 2.33

28) 0.05 equivalent of H3PO4 is neutralised by:

(a) 0.05 equivalent of NaOH

(b) 0.05 mol of Al(OH)3

(c) both are correct

(d) none is correct

29) H3PO4 can be neutralised by NaOH, Ca(OH)2 and Al(OH)3, 1 equivalent of H3PO4 will

require mol of each in the ratio:

(a) 1 : 1 : 1 (b) 1 : 0.5 : 0.33

(c) 1: 2 : 3 (d) 2 : 3 : 6

30) 1.575 g of a dibasic and required 0.025 equivalent of NaOH. Hence, molar mass of

dibasic acid (mol-1) is:

(a) 15.75 g (b) 25 g

(c) 63 g (d) 126 g

31) NaHC2O4 is neutralized by NaOH and can also be oxidized by KMnO4 (in acidic

medium). Equivalent weight is related to molecular weight (M) of NaHC2O4 in these two

reactions as:

(a) M,M (b) 2M,2M

(c) M/2,M (d) M, M/2

32) Cl2 changes to Cl- and ClO- in cold NaOH. Equivalent weight of Cl2 will be:

(a) M (b) M/2

(c) m/3 (d) 2M/3

33) Equivalent weight of KMnO4 in acidic medium, neutral medium and concentrated alkaline

medium respectively are M/5, M/1, M/3. Reduced products can be:

(a) MnO2, MnO , Mn2 + (b) MnO2, Mn2+, MnO

(c) Mn2+, MnO , MnO2 (d) Mn2+, MnO2, MNO

36)Which of the following changes requires a reducing agent?

(a) CrO Cr2O

(b) BrO BrO-

(c) H3AsO3 HAsO

(d) Al(OH)3 Al(OH)

37) For the reaction between MnO4- + C2O in basic solution, the unbalanced equation is:

MnO4- + C2O MnO2(s) + CO

In balanced equation, the number of OH- ions is:

(a) 0 (b) 4 on the right

(c) 4 on the left (d) 2 on the left

3. 0.1 mol of MnO (in acidic medium) can oxidise:

(a) 0.5 mol of Fe2+ (b) III>II>I

© 0.6 mol of Cr2O (d) III>I>II

4. Which of the following represent redox reactions?

(a) Cr2O + 2OH- 2CrO + H2O

(b) 2CrO + 2H+ Cr2O + H2O

© 2MnO + 3Mn2 + 4OH- 5MnO2 + 2H2O

(d) 2Cu+ Cu + Cu2+

7. Consider following reaction H3PO + Ca(OH)2 CaHPO4 + 2H2O and selectg true

statements:

(a) Eqivalent weight of H3PO4 is 49

(b) Resulting mixture is neutralised by 1 mol of KOH

© CaHPO4 is an acid salt

(d) 1 mol of H3PO4 Can be completely neutralised by 1.5 mol of Ca(OH)2

10. A compound contains atoms of three elements A,B and C with following oxidation

number

A = +2

B = +5

C = -2

Possible formula of the compound is:

(a) A2(BC3)2 (b) A3(BC4)2

(c) A3(B4C)2 (d) ABC2

Aqueous solution of sodium hypochlorite (NaOCI) is a household belach and is a strong

oxidising agent that reacts with chromite ion [Cr(OH) ] in basic solution to yield chromate

(CrO ) and chloride ion.

Q.I. Select correct statement(s):

(a) OCI- has been oxidised and Cr(OH) has been reduced

(b) OCI- has been reduced and Cr(OH) has been oxidised.]

(c) It is simply a neutralization reaction.

(d) It is simply a displacement reaction.

Q.II. Complete balanced equation (only redox species)

(a) CIO- + Cr(OH) CrO + Cl-

(b) 3 CIO- + Cr(OH) CrO + 3Cl-

(c) 3CIO- +2 Cr(OH) 2CrO + 3Cl-

(d) CIO- + 2Cr(OH) 2CrO + Cl-

Q.III. Equivalents of CIO- used in this reaction are:

(a) 2 (b) 3

(c) 4 (d) 6

Q.IV. Equivalents of Cr(OH) used in this reaction are:

(a) 1 (b) 4

(c) 6 (d) 8

Q.V. Number of moles of NaOCl required by 1 mol of Cr(OH) is

(a) 1 (b) 1.5

(c) 2.0 (d) 2.5

Chapter – IV

4) Which has maximum number of millimoles of Cl- ion?

a) 0.208 g BaCl2 b) 100 mL of 0.1M BaCl2

c) 0.745 g KCl d) equal

5) 100 mL of solution of CaCl2 is evaporated to dryness; residue obtained is 0.111 g.

Molarity of CaCl2 solution is:

a) 0.1M b) 1.0 M c) 0.01 M d) 0.001 M

6) Hardness in water is 200 ppm CaCO3. Molarity of CaCO3 is

a) 2 x 10-3 M b) 1 x 10-3 M c) 2x10-2 M d) 2x10-4 M

7) Mole fraction of CaCO3 in hard water having hardness 200 ppm CaCO3 is

a) 0.1 b) 3.6 x 10-5 c) 27.78 x 10-3 d) 0.035

8) Ethyl alcohol is 46% by weight of solution. Hence, mole fraction of ethyl alcohol is

a) 0.46 b) 0.54 c) 0.75 d) 0.25

9) Which is temperature independent?

a) mass per cent b) Volume per cent

c) Mass/volume per cent d) Molarity

10) HNO3 is 0.001 M. Hence, concentration in ppm is

a) 1x10-3 b) 100 c) 212 d) 63

12) Volume of oxygen gas occupied by 1 equivalent at STP is

a) 22.4L b) 11.2L c) 5.6 L d) 44.8L

15) 2.86 g of Na2CO3. xH2O in 100 mL solution is 0.2N. Hence, x is:

a) 5 b) 10 c) 20 d) 2

16) Urea solution is one molal. Urea present in one Kg solution

a) 60g b) 56.6g c) 10.60g d) 10.0g

17) H2SO4 is 98% by weight of solution. Hence, it is:

a) 1 molal b) 10 molal

c) 50 molal d) 500 molal

19) H3PO4 (d=1.8g/mL) is 18M. Hence, mass percentage and molality are

a) 18, 32.4 b) 98, 32.4 c) 98, 500 d) 98, 18

22) Resultant molarity of H+ ion in a mixture of 100 mL of 0.1 m H2SO4 and 200mL of 0.1 M

H3PO3 is

a) 0.1 M b) 0.2M c) 0.267 M d) 0.133 M

23) 1.575 g of a dibasic acid is neutralized by 25 ML of 1 M NaOH solution. Hence, molar

mass of dibasic acid is:

a) 126 g mol-1 b) 63 g Mol-1 c) 12.6 g mol-1 d) none of these

24) 1.06 g Na2CO3 is dissolved in 100 mL solution. 10mL of this solution can be neutralized

by:

a) 10 mL of 0.1 N HCl

b) 10 mL of 0.1 M H3PO4

c) 20 mL of 0.1 M H2SO4

d) 20 mL of 0.1 M HCl

25) 100 mL of 1 M BaF2 solution is mixed with 100 mL of 2 M H2SO4. Resulting mixture

contains:

a) 0.1 mol of BaSO4 b) 2 M H+ c) both are correct d) none of

correct

28) 0.31 g of N-containing compound on reaction with NaOH gave NH3 which required 100

mL of 0.1 N HCl. Hence, of N is:

a) 45.16% b) 90.32% c) 22.58% d) 11.29%

29) 0.6g NH2 CONH2 is treated with NaOH and NH3 formed is passed into 300 mL of 0.1 N

H2SO4. Unused acid is:

a) 50 mL b) 25 mL c) 100 mL d) 150 mL

31) 1g of X % H2O2 required x mL of KMnO4 solution in acidic medium. Thus, molarity of

KMnO4 solution is

a) 0.12 M b) 0.60 M c) 0.024 M d) none of these

33) 50 mL of FeSO4 . 7H2O solution required 20 mL of 0.1 M Cr2O solution in acidic

medium. Hence, FeSO4. 7H2O is

a) 0.24 M b) 0.24 N c) 66.72 g L-1 d) all correct

34) In a titration H2O2 is oxidized to O2 by MnO . 24 mL of 0.1 M H2O2 require(s) 16 mL of

0.1 M MnO solution. Hence MnO changes to:

a) MnO b) MnO2 c) MnO d) Mn2 O7

37) In iodometric titration, hypo is oxidised to:

a) S4O b) SO c) SO d) SO2

38) Iodometric method can be used to estimate:

a) MnO2, Cr2I , H2O2, CuSO4, Cl2

b) CaOCl2 , MnO2, CuSO4, Cl2

c) Cr2O , CaOCl2, H2O2

d) all are correct sets

43) In hot alkaline solution Br2 disproportionates to Br- and BrO

2 Br2 + 6OH- 5Br- + 3H2O

hence, equivalent weight of Br2 is (molecular weight = M)

a) b) c) d)

47) 2H2O2 (l) 2H2O(l) + O2 (g)

100 mL of X molar H2O2 gives 3 L of O2 under the condition when 1 mol occupies 1

mol occupies 24 L, value of X is:

a) 2.5 b) 1.0 c) 0.5 d) 0.25

55) In the following reaction

O3 + 6I- + 6H+ 3I2 + 3H2O

Equivalent weight of O3 (with molecular weight M) is

a) b) c) d)

Type – II

1) 1g equivalent of a substance is the weight of that amount of a substance which is

equivalent to

a) 0.25 mol O2 b) 8g O2 c) 16g O2 d) 0.50 mol O2

Type – IV

Example 1: Following experiment is designed to determine calcium content in the blood.

Read the experiment and answer the questions at the end of it.

Calcium oxalate (CaC2O4) is insoluble in water. For this reason it can be used to

determine the amount of Ca2+ ions in fluids such as blood. The calcium oxalate isolated from

blood is dissolved in acid and titrated against a standardized KMnO4 solution.

Q1) In one test it is found that the calcium oxalate isolated from a 10.0 mL sample of blood

requires 25 mL of 1 x 10-3 M KMnO4 for titration. Calculate the number of milligrams of

calcium per mL of blood:

a) 0.50 b) 10.0 c) 0.25 d) 5.0

Q.II. Milliequivalents of KMnO2 used were

a) 0.125 b) 0.025 c) 0.250 d) 0.050

Q.III. Calcium content expressed in ppm is

a) 250 b) 200 c) 150 d) 100

Q.IV. Above titration takes place in acidic medium. Acid chosen is:

a) H2SO4 b) HCl c) CH3COOH d) HCOOH

Q.V. Indicator in the above titration is

a) phenolphthalein b) methylene blue c) methyl orange

d) KMnO4 (Self-indicator)

2) Match the half-reaction (in List I) with equivalent mass (molar mass =M) (in List II).

List I List II

A.

B.

C.

D.

Cr2O Cr3+

C2O CO2

MnO MnO2

HC2O C2O

1

2

3

4

M

M/2

M/6

M/3