1

CHAPTER 01- SOME BASIC CONCEPTS OF

CHEMISTRY

Single Correct Type

1. Among the following the balanced redox reaction of 33AsO +

4MnO 3

4AsO + MnO2 (in alkaline medium) is

(A) 3 33AsO +

4MnO2 + H2O 3 3

4AsO + 2MnO2 + 2OH–

(B) 3 33AsO +

4MnO + H2O 3 3

4AsO + MnO2 + 2OH–

(C) 2 33AsO +

43MnO + H2O 2 3

4AsO + 3MnO2 + 2OH–

(D) 33AsO +

4MnO + H2O 3

4AsO + MnO2 + 2OH–

2. Among the following the balanced redox reaction of Zn + N3O Zn2+ +N2O + H2O (in Acidic Medium) is

(A) Zn + N 3O + 10H+ Zn2+ + N2O + 5H2O

(B) 4Zn + 2N 3O + 10H+ 4Zn2+ + N2O + 5H2O

(C) 4Zn + 2N 3O + H+ 4Zn2+ + N2O + H2O

(D) 4Zn + N 3O + 10H+ 4Zn2+ + N2O + H2O

3. 1 mole of equimolar mixture of ferric oxalate and ferrous

oxalate will require x mole of 4KMnO in acidic medium for

complete oxidation, x is

(A) 0.6 mole (B) 0.9 mole

(C) 0.3 mole (D) 1.8 mole

4. An evacuated glass vessel weighs 50 g when empty, 148 g

when filled with a liquid of density 0.98 g/ml and 50.5 g

when filled with an ideal gas at 760 mm Hg at 300 K. The

molecular weight of the gas is

2

(A) 132.5 (B) 123.15

(C) 120 (D) 64

5. 7.35 g of a dibasic acid was dissolved is water and diluted to

250 ml. 25 ml of this solution was neutralised by 15 ml of N

NaOH solution. The molecular weight of the acid will be.

(A) 49 (B) 98

(C) 35 (D) 50.

6. Density of some aq. solution (60% by wt.) is lowered upto

1.2 gm/cc by addition of water. If initial density was 1.6

gm/cc then new approximate percentage by wt. will be (Take

density of water = 1 gm/cc)

(A) 45% (B) 39%

(C) 27% (D) 19%

7. 0.5 g sample of containing 2MnO is treated with HCl. The

liberated 2Cl is passed into KI solution, where 2I is liberated.

Now liberated 2I requires 30 mL M

10 2 2 3Na S O for complete

reaction. Then what is the percentage purity of 2MnO in the

sample?

(A) 90% (B) 80%

(C) 26.1% (D) 60%

8. A 35.0cm solution of 2 2

H O liberates 0.508 g of 2

I from an

acidified KI solution. The strength of 2 2

H O solution in terms

of volume strength at STP is

(A) 2.24 vol (B) 8.96 vol

(C) 1.12 vol (D) 4.48 vol

3

9. A sample of pure 2 4 2 2 4 2

. .2KHC O H C O H O (three replaceable

hydrogen) requires 4.62 mol of NaOH for titration. How

many moles of 4

KMnO will the same sample react with ?

(A) 1.232 (B) 2.232

(C) 3.232 (D) 4.232

10. A 150ml of solution of 2

I is divided into two unequal parts.

Part – I reacts with hypo solution in acidic medium where

15ml of 0.4M hypo was consumed. Part – 2 was added to

100ml of 0.3M hot and conc. NaOH solution. Residual base

required 10ml of 2 4

0.3M H SO solution for complete

neutralization. What was the initial concentration of 2

I .

(A) 0.08 M (B) 0.1 M

(C) 0.2M (D) 0.04M

Multiple Correct Type

11. In which of the following statements is/are correct?

(A) Weight of 12.044 × 1023 atoms of carbon is 24g.

(B) Weight of 236.022 10 molecules of 3CaCO is 100 g.

(C) Number of moles of 0.635 g of Cu is 2. (Atwtof Cu =

63.5)

(D) The number of molecules in 11.2 litre of 2SO gas at NTP

is 0.5.

12. 4g of hydrogen is ignited with 4 g oxygen. Following

reaction takes place

2 2 22H O 2H O

Select the correct statement(s).

4

(A) Oxygen is limiting reactant

(B) Hydrogen is limiting

reactant

(C) 4g of hydrogen reacts with 2g oxygen

(D) 4.5 g of water will be formed

13. 0.2 mole of 3 4Na PO and 0.5 mole of 3 2Ba NO are mixed in

one litre of solution. Which of the following is/are correct.

(A) 0.2 mole of 3 4 2Ba PO is obtained

(B) 0.1 mole of 3 4 2Ba PO is obtained

(C) Molarity of 3 4 2Ba PO in solution is 0.1 M

(D) Molarity of 3 4 2Ba PO in solution is 0.2 M

14. 1 mole of 2 4H SO will exactly neutralise

(A) 2 moles of 4NH OH (B) 1 mole of 2

Ca OH

(C) 2 moles of NaOH (D) 0.5 mole of 2

Ba OH

15. 200 ml of 0.3 M Ca(OH)2 will be completely neutralized by

(A) 1200 ml of 0.1 M HCl (B) 600 ml of 0.1 M H2SO4

(C) 400 ml of 0.1 M H3PO4 (D) 600 ml of 0.2 M HNO3

16. 100 ml of 0.1 M H2SO4 is mixed with 200 ml of 0.2 M HCl

then resulting mixture will be neutralized by

(A) 600 ml of 0.1 M NaOH (B) 300 ml of 0.1 M Ca(OH)2

(C) 200 ml 0.1 M of Al(OH)3 (D) 400 ml of 0.1 M KOH

17. 1 molar of 1 litre of 2 4H SO will exactly neutralize

(A) 2 molar of 1 litre 3NH (B) 1 molar of 1 litre Ca 2

OH

(C) 0.5 molar of 1 litre Ba 2

OH (D) 2 molar of 1 litre NaOH

5

18. 10 moles of 2SO and 4 moles of

2O are mixed in a closed

vessel. Following reaction takes place 2 2 32SO g O g 2SO g .

Select the correct statement(s).

(A) 2SO is limiting reactant (B)

2O is limiting reactant

(C) 8 moles of 3SO are formed (D) 10 moles of

3SO are

formed

19. Which of following contains the same number of moles?

(A) 1g of 2O and 2 g of 2SO (B) 1 g of 2O and 1 g of 3O

(C) 1 g of 2CO and 1g of 2N O (D) 11.2 L of 2CO at STP and

1 g 2H

20. A sample of H2O2 solution labelled as “28 volume” has

density of 26.5 g/L. Mark the correct option(s) representing

concentration of same solution in other units:

(A) 2 2H OM 2.5 (B) w

% 17v

(C)Mole fraction of H2O2 =

0.2 (D) 2 2H Om 13.88

Integer Type

21. Calculate the amount of calcium oxide approximately in kgs

required to react completely with 852g of P4O10.

22. SO2Cl2 + 2H2O H2SO4 + 2HCl . If 2 moles of SO2Cl2 is

mixed with excess of water then moles of NaOH needed for

complete neutralization will be

23. 5 ml of 8 N nitric acid, 4.8 ml of 5N HCl and a certain

volume of 17 M H2SO4 are mixed together and made up to

6

2l. 30 ml of this acid solution exactly neutralises 42.9 ml of

Na2CO3 solution containing 1 g of Na2CO3. 10H2O in 100

ml of water, then the volume of H2SO4 solution in ml was

24. 0.50g of a mixture of K2CO3 and Li2CO3 required 30 mL of

0.25 N HCl solution for neutralization. What is %

composition of Li2CO3?

25. 1.575 g of oxalic acid (COOH)2 . xH2O are dissolved in

water and the volume made upto 250ml. On titration 16.68

gm of this solution requires 25 ml of N/15 NaOH solution

for complete neutralization. Calculate x.

26. 0.7g 2 3 2Na CO .xH O were dissolved in water and the volume was

made to 100 ml, 20 ml of this solution required 19.8 ml of

N/10 HCl for complete neutralisation. The value of x is

27. A solid mixture (4.69g) consisting of lead nitrate and sodium

nitrate was heated below 600°C until the mass of the residue

was constant. If the loss in mass is 28% then the mass of lead

nitrate in the mixture was (in grams)

28. A 25 ml portion of a phosphate mixture (NaH2PO4 & H3

PO4) requires 10.87 ml of 0.1 N NaOH for methyl orange

end point, another 25 ml portion requires 46.43 ml of 0.1 N

NaOH for titration to the phenolpthalein end point. Then ten

times of molar concentration of NaH2 PO4 will

approximately be

29. A 1.0g sample of a pure organic compound containing

chlorine is fused with Na2O2 to convert chlorine to NaCl.

7

The sample is then dissolved in water, and the chloride

precipitated with AgNO3, giving 1.96g of AgCl. If the

molecular weight of organic compound is 147, how many

chlorine atoms does each molecule contain?

30. A solution of H2O2 is titrated against a solution of KMnO4.

The reaction is

2

4 2 2 2 22MnO 5H O 6H 2Mn 5O 8H O

If it requires 46.9 mL of 1.45 M KMnO4 to oxidize 20 g of

H2O2, find the approximate mass percentage of H2O2 in this

solution.

8

SOLUTIONS

Single Correct Type

1. (A)

(i) Identify the oxidation & reduction halves

Reduction half reaction: 24 MnOMnO

Oxidation half reaction: 34

33 AsOAsO

(ii) Atoms of the element undergoing oxidation and

reduction are already balanced.

(iii) Balancing O atoms,

Reduction half reaction: 2H2O + 4MnO MnO2 + 4OH–

Oxidation half reaction: 2OH– + 33AsO 3

4AsO + H2O

(iv) Balancing H atoms,

H atoms are already balanced in both the half-reactions.

(v) Balancing charge,

Reduction half reaction: 3e– + 2H2O + 4MnO MnO2 +

4OH– … (B)

Oxidation half reaction: 2OH– + 33AsO 3

4AsO + H2O +

2e– … (A)

(vi) Multiply equation (A) by 3 and equation (B) by 2 and

then add (A) and (B).

3e– + 2H2O + 4MnO MnO2 + 4OH–] 2

2OH– + 33AsO 3

4AsO + H2O + 2e–] 3

3 33AsO +

4MnO2 + H2O 3 3

4AsO + 2MnO2 + 2OH–

2. (B)

Find the oxidation states of atoms undergoing redox

changes.

9

ONZnONnZ 2

12

3

50

Increase in O.S = 2 4

decrease in O.S = 8

4Zn + 2N 3O 4Zn2+ + N2O

Charge on LHS = –2

Charge on RHS = + 8

Difference in charge = +8 – (–2) = 10

As the medium of reaction is acidic, adding 10H+ on LHS to

equalize the charges

4Zn + 2N 3O + 10H+ 4Zn2+ + N2O

To equalize the O and H atoms, add 5H2O on RHS to get the

balanced equation.

4Zn + 2N 3O + 10H+ 4Zn2+ + N2O + 5H2O

3. (B)

2

3

43

3 4 2

2

4

32Fe C O Vs MnO

CO Mn

n factor 56mole ratio 5 = 6

0.5_______? = 0.6

(1) (5)Fe

2 4

3 4 2

2

4Fe C O Vs MnO

CO Mn

n factor 53mole ratio 5 = 3

0.5_______? = 0.3

Total moles require

Fe

d is 0.9

4. (B)

Weight of liquid 148 50 98g

10

Volume of liquid 98100ml

0.98

= Volume of vessel

Weight of gas = 50.5 -50 = 0.5 g

WPV RT

M

WRT 0.5 0.0821 300M 123.15

PV 1 0.1

760 100P 1 atm, V 0.1L

760 1000

5. (B)

Let the Eq. weight of the acid be E.

Eq. weight of acid = 7.35/E.

m.e. of the acid 7.35 1000

E

= 7350/E

Now, 250 ml of the acid contains 7350m.e

E

25 ml of the acid contains 735

Em.e

m.e of 25 ml of the acid = m.e. of NaOH

7351 15, Eq.weight 49

E

molecular weight = 2 49 98 .

6. (C)

mass of 1 litre of solution = 1600 gms.

mass of solute = 960 gm

Let V litre of water is added.

new vol = (1+v) litre = (1+v) 1000 cc

new mass = 1600 + V(1000) gms

1600 1000V 1.6 V1.2

1 V 1000 1 V

11

1.6 V 1.2 1.2V

0.2V 0.4

V 2 litre

new wt. % = 960 10027% approx

1600 2000

7. (C)

2 2 3 2 2 4 62Na S O I 2NaI Na S O

(n 1)

2 2 3 2 2 3

M N30ml Na S O 30ml Na S O

10 10

meq. of 2 2 3Na S O is 3

So meq. of 2MnO 3 meq.

Wt. of 2MnO = 3 873 10 0.1305

2

gm

0.1305% 100 26.1%

0.5

8. (D)

meq 2 2 2

H O meq I

/17 1000 0.508 / 254 / 2 1000w

0.068w g

34 g of 2 2

H O gives 11.2 lit of 2

O

2

11.2 0.0680.068 22.4

34g ml of O

5cc of 2 2

H O solution gives = 22.4 ml of 2

O

1cc gives = 22.44.48 .

5vol

9. (A)

2 4 2 2 4 2

. .2KHC O H C O H O has three replaceable H

1 1NaOH H

4.62 4.62NaOH H

12

4.62

3mol of salt

1.54 mol

1 mol of salt has = 2 mol of 2

2 4C O

1.54 mol of salt has = 3.08 mol of 2

2 4C O

2 2

2 4 4 25 2 2 10HC O MnO Mn CO

5 mol of 2

2 4 42C O mol MnO

3.08 mol of 2

2 4 4

23.08 1.232

5C O mol MnO

10. (B)

2 2 2 3 2 4 6I 2Na S O 2NaI Na S O

Millimoles of 2 2 3

Na S O consumed = 6

Millimoles of 2

I consumed = 3

2 3 2

3I 6NaOH 5NaI NaIO 3H O

Millimole of 2

I reacts with 30 6NaOH 12

2

Total millimole of 2

I consumed in the reaction = 15

Initial concentration of 2

15I 0.1M

150

Multiple Correct Type

11. (A, B, D)

For (A) The number of moles of 23

23

12.044 10C 2

6.02 10

Wt of C-atm 2 12 24g

For (B) weight of 3CaCO = number of moles × mol.wt 23

23

6.02 10100 100g

6.02 10

13

For (C) number of moles of Cu wt of Cu 0.635

at.wt of Cu 63.5 = 0.01

Hence it is wrong.

For (D) number of 2

volat NTP 11.2SO 0.5

22.4 22.4

12. (A, D) 2 2 2

4g 32g 36g

2H O 2H O

4 g of 2H will completely react with 32 g of

2O , but only 4

g of 2O is present. Hence,

2O is limiting reactant.

Since, 32 g of 2O produces 36g of 2H O

Hence 4 g of 2O will produce 36

4 4.5g32

of 2H O

13. (B, C)

3 4 3 3 4 32 22Na PO 3Ba NO Ba PO 6NaNO

Moles taken 0.2 0.5

Moles reacted 0.2 0.3

Moles left 0 0.2

3 4Na PO is limiting reactant.

Moles of 3 4 2Ba PO formed = 0.1

Molarity of 3 4 2

0.1Ba PO 0.1M

1

14. (A,B,C)

Species No. of equivalent (no. of

moles x n – factor)

1 mole of 2 4H SO 1 2 2

2 moles of 4NH OH 2 1 2

1 mole of 2

Ca OH 1 2 2

14

2 mole of NaOH 2 1 2

0.5 mole of 2

Ba OH 0.5 2 1

Equal number of equivalents of acid and base neutralises.

15. (A, B, C, D)

m eq of Ca(OH)2 200 0.3 2 120 (nf = n-factor)

(A) 1200 0.1 = 120 M eq of HCl HCl nf = 1

(B) 600 0.1 2 = 120 M eq of H2SO4 H2SO4 nf = 2

(C) 400 0.1 3 = 120 M eq of H3PO4 H3PO4 nf = 3

(D) 600 0.2 1 = 120 M eq of HNO3 HNO3 nf = 1

16. (A, B, C)

m eq of H2SO4 + m eq of HCl

100 0.1 2 + 200 0.2 1

20 + 40 = 60 m eq of Acid

(A) 600 0.1 = 60 m eq of NaOH

(B) 300 0.1 2 = 60 m eq of Ca(OH)2

(C) 200 0.1 3 = 60 m eq of Al(OH)3

(D) 400 0.1 1 = 40 m eq of KOH

17. (A, B, D)

n-factor for 2 4H SO 2

Equivalent of 2 4H SO 2

Equivalence of 3NH 2 n factor

2 1 2

Equivalence of 2Ca(OH) 1 2 2

Equivalence of 2Ba(OH) 0.5 2 1

15

Equivalence of NaOH =2×1 =2

18. (B, C)

2 2 32SO O 2SO

Moles taken initially 10 moles 4 moles 0

Moles involved in reaction 8 4 8

Final moles 2 0 8

2O is limiting reactant.

2SO is excess reactant as 2 moles

of 2SO remains unreacted. 8 moles of 3SO are produced.

19. (A, C, D)

Moles in 1 g 2O 1

32

Moles in 2 g 2

2 1SO

64 32

Moles in 1 g 3

1O

48

Moles in 1 g 2

1CO

44

Moles in 1g 2

1N O

44

Moles in 11.2 L 2CO (STP) = 11.2 1

22.4 2

Moles in 1g 2

1H

2

20. (A, C, D)

Vstrength = 56

28M 2.5

11.2

1 L contain 2.5 moles of H2O2

Or 2.5 34 = 85g H2O2

Wt. of 1 litre solution = 265 g ( d = 265g/L)

2H Ow =180g or moles of H2O = 10

16

2 2H O

2.5x 0.2

2.5 10

% w 2.5 34100 8.5

v 1000

2.5m 1000 13.88

180

Integer Type

21. (1)

4 10 3 26CaO P O 2Ca (P04)

4 10 2 3 4P O 4H O 4H PO

1mole 4mole

4 10

W 852(1)

56 / 2 M / W

M 284E P O

12 12

CaO

W 852by Equation (1);

56 / 2 284 /12

W 1008g 1kg

22. (8)

SO2Cl2 + 2H2O H2SO4 + 2HCl

So 2 moles of SO2Cl2 will give 2 moles of H2SO4 and 4

moles of HCl.

2 moles of H2SO4 will require 4 moles of NaOH

4 moles of HCl will require 4 moles of NaOH

so 8 moles of NaOH will be required.

23. (4)

2 3

2 3

30 42.9 ml Na CO1000 10

N of Na CO143 100 14342.9 10

N of acid sol 0.1N143 30

Now N1V1 + N2 V2 + N3 V3 = NV

17

40 + 24 + 34 V = 0.1 × 2000

200 64V 4ml

34

24. (4)

Weight of K2CO3 = ag

Weight of Li2CO3 = b g

a + b = 0.5 … (1)

Also for reaction:

Meq. of K2CO3 + Meq. of Li2CO3 = Meq. of HCl

a 1000 b 100030 0.25

138 / 2 74 / 2

74a + 138b = 38.295 … (2)

By Eqs. (1) and (2),

a = 0.48 g

b = 0.02 g

% of K2CO3 0.48

100 96%0.5

% of Li2CO3 0.02

100 4%0.5

25. (2)

Meq. Of oxalic acid = Meq. Of NaOH

Meq. Of oxaliac acid in 16.68 ml = Meq. Of NaOH = 25 1

15

Meq. Of oxalic acid in 250 ml = 1 25025 24.98

15 16.68

Molarity Volume in cm-3 = 24.98

1.575 1000250 24.98

(90 18x) 250

2

18

x = 2

26. (2)

Meq of HCl used 2 3 2meq of20 ml Na CO .xH O

119.8 1.98

10

Meq of 2 3 2Na CO .xH O in 100 ml = 1.98 5

Normality of 2 3 2

1.98 5Na CO XH O

100

W 1000N

E V

W 1000E

N V 0.7 1000 100

1.98 5 100

E 70.707 , mw 141.414

23 2 12 3 16 18x 141.41

x 2

27. (3)

3 2 2 2

x

3 2 24.6 x

2Pb(NO ) 2PbO 4NO O

2NaNO 2NaNO O

x223

331 = mass of PbO

4.6 x69

85

= mass of NaNO2

mass of residue = 223x 317.4 69x4.6 0.72

331 85

85 × 223x + 331 × (317.4 – 69x) = 331 × 85 × 4.6 × 0.28

18955x + 105059.4 – 22839 x = 93183.12

– 3884 x = – 11876.28

x = 3g

28. (1)

19

Me of base for methyl orange end point = mm of H3PO4 =

10.87 × 0.1 = 1.087

me of base for phenolphthalein end point = 2 × mm of H3PO4

+ mm of NaH2PO4

= 46.43 × 0.1 = 4.643

mm of NaH2PO4 = 4.643 – 2 × 1.087 = 2.469

2 4

2.469NaH PO 0.0988 0.1

25

ten times = 0.1× 10 = 1 moles / l

29. (2)

Moles of AgCl = moles of chloride = 1.960.0136

143.5

Moles of organic compound = 316.8 10

147

Chlorine atoms in each molecule of organic compound =

3

0.01362

6.8 10

30. (3)

Number of moles of KMnO4 = 3MV 0.145 46.96.8 10

1000 1000

Number of moles of H2O2 = 6.8 10-3 2.5 = 0.017

Mass of H2O2 = 0.017 34 = 0.578

Mass % of H2O2 = 0.578100 2.9

20