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SRIGAYATRI EDUCATIONAL INSTITUTIONS

INDIA

STICHIOMETRY UT-03

1. How many significant figures are present in the following ; (A) 0.0025 , (B) 500.0

a)2,4 b)4,2 c) 5,4 d) 2,1

2. How many significant figures should be present in the product of 5 5.364

a) 5 b) 4 c) 3 d) 1

3. Law of multiple proportions is illustrated by one of the following pairs

a) 2 2&H S SO b)

3 2&NH NO c) 2 2&Na S Na O d)

2 &N O NO

4. 2.16 gm of Cu on reaction with 3HNO followed by ignition of the nitrate give 2.7 gm of copper

oxide. In another experiment 1.15 gm of copper oxide upon reduction with hydrogen gave

0.92 gm of copper. This data illustrate the law of

a) Multiple proportions b) Definite proportions

c) Reciprocal proportions d) Conservation of mass

5. Neon has two istopes 20Ne and

22Ne . If the atomic weight of neon is 20.2, the ratio of the

relative abundances of the isotopes is

a) 1 : 9 b) 9 : 1 c) 70% d) 80%

6. Which has the maximum no.of molecules among the following .

a) 8g H2 b) 64 g 2SO c) 44 gm

2CO d) 48g 3O

7. The total no. of electrons in 1.6 g f 4CH to that in 1.8 gm of

2H O

a) Double b) Same c) Triple d) One fourth

8. The no. of atoms in 4.25 gm of 3NH approximately

a) 234 10 b) 231.5 10 c) 231 10 d) 236 10

9. A solution is prepared by adding 2 gm of a substance ‘A’ to 18 gm of 2H O . Calculate the mass

percentage of the solute

a) 10% b) 20% c) 30% d) 40%

10. How many grams of NaOH is added to water to prepare 250 ml solution of 2M NaOH

a) 39.6 10 b) 32.4 10 c) 20 d) 24

11. The total no. of electrons in 4.2 gm of 3N ion is AN =Avagadro's number

a) 2.1 NA b) 4.2 NA c) 3 NA d) 3.2 NA

12. The density of 3M solution of NaCl is 1.25gm/ml. Calculate the molality of the solution

a) 1.79 m b) 279 m c) 2.79 m d) 2.09 m

13. Concentrated aqueous sulphuric acid is 98% by mass and has density of 1.80 gm/L . The

volume of acid required to make one litre of 0.1 M 2 4H SO solution is

a) 10.05 ml b) 22.20 ml c) 5.55 ml d) 11.10 ml

14. Mole fraction of solvent in aqueous solution of NaOH having molality of 3m is

a) 03 b) 0.05 c) 0.7 d) 0.95

15. When 100 ml of 10

M

2 4H SO is mixed with 500 ml of 10

M NaOH. Then the nature of resulting

solution and resulting normality is

a) Acidic , 5

N b) Basic ,

5

N c) Basic,

20

N d) Acidic,

10

N

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16. The oxidation number of ‘C’ in 4 3 2 2 3, , ,CH CH Cl CH Cl CHCl and CCl4 is respectively

a) -4, -2, 0, +2, +4 b) +2, +4, 0, -2 , -4 c) 4, 2, 0, -2, 4 d) 0 , -2, 2, 4 ,4

17. Which of the following involves the reduction of copper

a) 2

1

2s g s

Cu O CuO b) 2 2 2aq aq aq

Cu I CuI

c) 2 2 22aq

CuCl F CuF Cl d) 2 2CuO H O Cu OH

18. The oxidation number of sulphur in 2 2 8H S O is

a) +7 b) +6 c) -6 d) +4

19. When 2SO is passed in acidified potassium dichromate solution. The oxidation number of ‘S’

is changed from

a) + 4 to zero b) +4 to +2 c) +4 to +6 d) +6 to +4

20. Oxidation state of ‘Fe’ in 3 4Fe O is

a) 5

4 b)

4

5 c)

3

2 d)

8

3

21. In acidic medium, 2 2H O changes from

2

2 7Cr O to

5CrO , which has two (-O-O-) bonds.

Oxidation state of ‘Cr’ in CrO5 is

a) +5 b) +3 c) +6 d) +10

22. Oxidation number of sodium in sodium amalgam is

a) +1 b) 0 c)-1 d) +2

23. In alkaline solution KMnO4 reacts as follows

4 2 4 2 22 2 2 1/ 2KMnO KOH K MnO H O O

Therefore, its equivalent weight will be

a) 31.6 b) 52.7 c) 79.0 d) 158.0

24. Given oxidation number of Sulphur is -2. The equivalent weight of Sulphur is

a) 16 b) 32 c) 9 d) 4

25. In the reaction, 2 2 2 3 2 4 62I Na S O NaI Na S O equivalent weight of iodine is

a) M.Wt. b) . .

2

M Wt c)

. .

4

M Wt d)

. .

3

M Wt

26. In balancing the reaction , 2

3 4 2 , , &Zn NO H Zn NH zH O x y x x y z are

a) 4,10,3 b) 3,8,3 c) 3,10,3 d) 4,3,10

27. When an oxide 2 3M O is oxidized to 2 5M O . Its equivalent weight is

a) . .

1

WtM b)

. .

2

M Wt c)

. .

4

M Wt d)

. .

8

M Wt

28. The relative number of atoms of different elements in a compound are as follows : A=1.33,

B=1 and C=1.5.The empirical formula of the compound is

a) 2 2 3A B C b) ABC c) 8 6 9A B C d) 3 3 4A B C

29. Determine the empirical formula of an oxide of Iron, which has 69.9%. Iron and 30.1%

dioxygen by mass

a) FeO b) 2 3Fe O c) 4 3Fe O d) 3 4Fe O

30. A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass

is 98.96 g. What is its molecular formula.

a) 2 4 2C H Cl b) 2 3 3C H Cl c)

2CH Cl d) 2 2CH Cl

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31. The empirical formula of an organic compound is 2CH O . Its vapour density is 42 gm. Its

molecular formula is

a) 2CH b)

2 2C H c) 3 6C H d)

3 8C H

32. An oxide of nitrogen contains 36.8% by weight of nitrogen . The formula of the compound is

a) 2N O b)

2 3N O c) NO d) 2NO

33. 60 gms of limestone on heating produces 22 gm of 2CO . The percentage of

3CaCO in limestone

is

a) 80% b) 60% c) 83.3% d) 87.6%

34. How many moles of methane are required to produce 22gm of 2 gCO after combustion

a) 0.5 b) 1 c) 0.05 d) 2

35. 50.0 kg of 2 gN and 10.0 kg of 2 g

H are mixed to produce 3 gNH . Calculated amount of

3NH

produced

a) 3.33 kg b) 1.78 kg c) 56.66 kg d) 56.66 gm

36. Calculate the amount of water (g) produced by the combustion of 16 gm of methane

a) 18 gm b) 16 gm c) 32 gm d) 36 gm

37. 20 ml of a hydrocarbon requires 100 ml of oxygen under the same conditions for complete

combustion, 60 ml of 2CO is formed. The formula of hydrocarbon is

a) 2 4C H b) 3 6C H c) 3 8C H d) 2 6C H

38. 0.01 ml of Iodoform reacts with excess of Ag power to produce a gas whose volume at STP is

a) 224 ml b) 112 ml c) 336 ml d) 448 ml

39. One litre of 2CO is passed through hot coke. The volume becomes 1.4 litre at same

temperature & pressure. The composition of product is

a) 20.8litreof CO and0.6litreof CO b)

20.8litreof CO and0.7litreof CO

c) 20.6litreof CO and0.8litreof CO d)

20.4litreof CO and1.0litreof CO

40. No.of Fe atoms in 100 gm of Hemoglobin if it contain 0.33% Fe (atomic mass of Fe = 56)

a) 230.0035 10 b) 35 c) 233.5 10 d) 87 10

41. KCl is used as an electrolyte in salt bridge because

a) K and Cl are iso electronic b) Movements ions are required

c) Both the ions almost same velocity d) They have similar site

42. Electrode potential depends upon

a) Size of electrode b) surface area of electrode

c) temperature d) Shape of electrode

43. Three metals A, B and C are arranged in the increasing order of standard reduction potential,

hence their chemical reactivity will be

a) A<B<C b) A>B>C c) B>C>A d) A=B=C

44. Standard reduction electrode potential of three metals x, y and z are -1.2 V, +0.5V and -3.0V

respectively. The reducing power will be

a) X>Y>Z b) Y>Z>X c) Y>X>Z d) Z>X>Y

45. Oxidation number of carbon in carbon suboxide.

a) 2

3 b)

4

3 c) +4 d)

4

7

INTEGER TYPE QUESTION

46. The density the solution prepared by dissolving 120 gm of urea (Molar mass = 60u) in 1000

gm of water is 1.15 g/ml The molarity of the solution is ______

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47. The Mass of potassium dichromate crystals required to oxidise 7.50 3cm of 0.6M salt solution

is _______

48. Haemoglobin contains 0.334 % of Iron by weight. The molecular weight of Haemoglobin is

approximately 67200. The number of atoms present in one molecule Haemoglobin (atomic

mass Fe = 56)

49. 10 gm of hydrogen and 64 gm of oxygen were filled in a steel vessel and exploded Amount of

water produced in this reaction will be ….…

50. At S.T.P the density of CCl4 vapuor in g/L will be nearest to…………

51. 10 ml of 2 2H O solution ( volume strength is = x ) required 10 ml of

1

0.56 N

4MnO solution in

Acid medium Hence ‘X’ is ________

52. Oxidation number of central atom in 2

4Ni CN

is

53. In 3N H Oxidation number of nitrogen is __________

54. ________ weight of oxygen ( in gms ) required to completely react with 27 gm of ‘Al’

55. In the reaction , 3 4 4 222H PO Ca OH CaHPO H O the equivalent weight of phosphoric

acid is ________

KEY SHEET

1) 1 2) 2 3) 4 4) 2 5) 2 6) 1 7) 2 8) 4 9) 1 10) 3

11) 3 12) 3 13) 3 14) 4 15) 3 16) 1 17) 2 18) 2 19) 3 20) 4

21) 3 22) 2 23) 4 24) 1 25) 2 26) 1 27) 3 28) 3 29) 2 30) 1

31) 3 32) 2 33) 3 34) 1 35) 3 36) 4 37) 3 38) 2 39) 3 40) 1

41) 3 42) 3 43) 2 44) 4 45) 2

46) 2.05 47) 22.05 48) 4 49) 4 50) 6.875

51) 10 52) +2 53) 1

3 54) 24 55) 49

SOLUTIONS

1. 0.0025 has Two significant figures 500.0 has 4 significant figures.

2. The answer to the calculation 5 5.364 will contain 4 significant figures as 5.364 contains. Four

significant figures. Here exact figure 5 is not considered .

3. Two elements chemically combine to give two (or ) more compounds

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4. Weight of copper = 2.16 g

Weight of copper oxide obtained on ignition = 2.70 g

Percentage of copper in copper oxide 2.16

100 80%2.70

% of oxygen in copper oxide = 100 – 80 = 20%

Weight of copper left after reeducation = 0.92 gm

Weight of copper oxide = 1.15 gm

% in Copper oxide = 0.92

100 80%1.25

% of oxygen = 100 – 80 = 20 %

Thus the data illustrates the law of define proportions

5. Let W % of 20Ne x

W % 22 100Ne x

20.2 = 20 100 22

100

x x

2020 = 20x + 2200 -22x

2x = 180

X = 90 %

20% 90%w of Ne

22% 100 90:10w of Ne

20 22: 9:1Ne Ne

6. Number of molecules = mole × NA

8 gm H2 = 8

42

moles

2

6464 1

64gmSO mole

2

4444 1

44gmCO mole

3

4848 1

48g O mole

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7. Number of e in 1.6 gm CH4 = 0 0

1.610

16e N N

Number e in 1.8 gm of 2 0 0

1.810

18H O e N N

8. Number of atoms = mole No atomicity

23 234.256 10 4 6 10

17

9. Mass percent of A : 100Mass of A

Mass of solution

2

10020

= 10 %

10. Molarity (M) = 1000

. . .

Wt

G M Wt Vol in ml

= 1000

40 250

wt

Wt = 20 gm

11. Number of e in 4.2 g 3N = 4.2

1014

Ae N

= 3 NA

12. Molarity (m) = 1

100

100 .

molarity

d M M

M = Molarity

1M = molecule wt of solute

1000 3

1.25 100 3 58.5m

M = 2.79

13.

% 1098 1.80 10

18. 98

wd

wM M

Mol wt

Then 1 1 2 2M V M V

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1

1

18 1000 0.1

5.5

V

V ml

14. Mole fraction of solute XNaOH = 2

NaoH

NaOH

X

X nH O

But for one molal solution,

NaoHX molarity

2

NaOH

mX

m nH O

3

3 55.55NaOHX

3

58.55NaOHX

0.05NaOHX

But 2 1 0.05XH O

= 0.95

15. Number of M.eq. of 2 4H SO N V

= 2

10010

= 20 M.eq.

Number of M.eq. of NaOH = N V

= 1

50010

= 50 M.eq.

∴ Solution shows basic nature due to excess of M.eq. of base

Resulting normality (N) = b b a a

a b

N V N V

V V

50 20

600N

1

20N

16. 4 4 0 4CH x x

3 3 1 0 2CH Cl x x

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2 2 2 2 0 0CH Cl x x

4 4 0 4CCl x x

3 1 3 0 2CHCl x x

17. 2 1 1

Re2

ductionCu I Cu I

18.

+2 + 2x – 12 -2 = 0

2x = 12

x = +6

19.

20. 3 4Fe O

3x – 8 = 0

3x = +8

8

3x

21.

X – 4 - 2 = 0

X = +6

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22. An amalgam is a homogeneous mixture

The oxidation state of sodium and mercury in sodium amalgam is zero

23. 7 2 6

4 2 42e

KMnO K MnO

For Two moles of 4 , 2KMno e are required one mole of 4 ,1KMno e is required

Eq .wt of 4KMnO = 4.

1

Mol wt of KMnO

24. Eq , wt of ‘s’ = . 32

162

At wt

Volency

25. In the above reaction

.

.2

mol wtEq wt

26. 2

3 4 24 10 4 3Zn NO H Zn NH H O

27.

28.

: :

(1.33 : 1 : 1.5) 6

A B C

8 6 9A B C

29. The relative moles of Iron in Iron oxide

% of Iron by mass

Atomic mass of Fe

69.91.25

55.85

The relative mass of oxygen in Iron oxide

% 30.1

1.8816.00

of oxygen by mass

Atomic mass of oxygen

Simplest molar existing of Iron to oxygen

2 3

1.25 :1.88

1:1.5

2 : 3

emf Fe O

30. 2

4.074.04

1.008Hn

24.27

2.02112.01

Cn

71.65

2.02135.453

Cln

Divide each value by smallest value

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Since 2.021 is smallest value, division by it gives a ratio of 2:1:1 for H:C:Cl

E.m.f : CH2Cl

Molecular formula (MF) = (CH2Cl) × n

2 2 4 22

98.962

49.48n CH Cl C H Cl

31. Mol.wt = 2×vapour density = 2×42

Mol.wt = 44

∴ Mol.wt C3H6 is 44 gram 32. Relative no.of atoms of nitrogen and oxygen

36.8 63.2

: :14 16

N O

33. 3 2CaCO CaO CO

100 gm 3CaCO - 44 gm CO2

? - 22 gm

100 22

5044

gm

3

50% 100

60CaCO

83.3 %

34. 24 2 22

g g g gCH O CO H O

1 mole of 4CH …………. 1 mole of

2CO (44 gm)

½ mole of C ……………… ½ mole of 2CO (22 gm)

0.5 mole of CH4 is required to produce 22 gm CO2

35. 2 2 33 2

g g gN H NH

No. of moles of 3

2

50 101.78 10

28N

No. of moles of H2 =

3310 10

5 102

According to equation , 1 moles of N2 requires 3 mole of H2.

Hence 31.78 10 moles of N2 requires 35.358 10 moles of Hydrogen.

But we have only 35 10 moles H2.

Hence dihydrogen is limiting reagent

3 moles of Hydrogen - 2 moles of NH3

35 10 moles of H2 ……….. ? 3

35 10 23.33 10

3

moles of NH3

Wt of NH3 33.33 10

17

wt

Wt of NH3 356.66 10 gm

56.66 Kg

36. 4 2 2 22 2CH O CO H O

4 21 2 18mole of CH gm H O produced

16 gm of CH4 – 36gm of H2O

37. 2 2 24 2

x y

y yC H x O xCO H O

20 60x C

3x

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No of carbons = 3

1 mole 24

x y

yC H x O

20 mole = ?

20 1004

yx

240 20 400y

y = 8

Hydrogen = 8

Formulae = 3 8C H

38. 3 2 22 6 6CHI Ag C H AgI

2 moles of CHI3 --------- 1 mole of C2H2 gas

2 moles of CHI3 --------- 2 21 22400ml of C H gas

30.01 ?mole of CHI

0.01 22400

1122

ml

39. 22

g s gCO C CO

1 0 0

1 2x x

1 2 1.42x x

0.42x

Volume of CO = 2x = 2 × 0.4 = 0.8 L

Volume of CO2 =1 – x = 1- 0.4 = 0.6L

40. 100 gm of Heamoglobin = 0.33 gm Fe

∴ Moles of 30.335.89 10

56Fe mole

No. of ‘Fe’ atoms 3 235.89 10 6 10

230.035 10 atoms

41. &K Cl are have almost similar size

42. 0G nFE

Where G , depends on temperature

43. As the standard reduction potential (SRP) values increase, reactivity decreases(oxidizing capacity)

44. Least SRP value gets oxidized and acts as good reducing agent

45. Carbon suboxide = 3 2C O

3 4 0

3 4

4

3

x

x

x

46. Molality (m) = 1

1000

1000

M

d MM

M = Molarity, M1 = Mol. Wt. of solute

Molality (m) = 120 1000

60 1000

2m molal

1000

( )1000 1.15 60

Mmolality m

M

1000

1150 60

M

M

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2300 120 1000M M

2.05m

47. G.E Wt. of mohr’s salt = 0.6 750

0.451000

N

2 2 7. . . . 'G E wt K Cr O G E wt mohr s salt

0.45294

6

X

Mass of 2 2 7 22.05K Cr O x gm

48. 100 gm Haem. = 0.334gm of Fe

? = 56 gm

56 100

16766.460.334

g

no. of Fe atoms = 67200

416766.46

49. 2 2 2

1

2H O H O

No. of moles 2H is 5 moles and no. of moles of

2O is 2 moles.

As per equation, half mole of 2O requires one mole of dihydrogen. Hence two moles of oxygen

requires four moles of dihydrogen. But 5 moles of dihydrogen is given.

oxygen is a limiting reagent

2 2

11

2O mole H O

2 22 4M O H O

50. STP 154gm CCl4 = 22.4 Litres

= 1 Litre

154

6.87522.4

51. 1 1 2 2N V N V

1

110 10

0.56N

1

1

0.56N Normal

But vol. strength (x) = 5.6 Normality

1

5.6 100.56

x

52. 2

4Ni CN

4 2x

2x

53. 1

3 0x

N H

3 1 0x

3 1x

1

3x

54. 2 2 34 3 2Al O A lO

24 3moles of Al moles of O

24 27 3 32gm Al gm of O

27gm Al

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27 3 32

4 27

24gm

55. 3 4 4 222H PO Ca OH CaHPO H O

3 4

98.Eq wt of H PO

n factor of base

98

492