INDIA STICHIOMETRY UT-03
Transcript of INDIA STICHIOMETRY UT-03
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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
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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