MOLE CONCEPT (MAIN)

FOUNDATION BUILDER (OBJECTIVE)

1. (D)

2. (D)

3. (B)

4. (A)
(Most stable isotope of carbon)

5. (D)

6. (C)

7. (A)
5.6
Moles of gas =  0.25
22.4
7.5
Molecular weight of gas =  30
0.25
Hence NO.

8. (A)
Molecular weight of C60 H122  60 12  122  842.
842
Weight of a molecule = 23
 1.39 10 21 g .
6.022 10
9. (A)
1 mole contains Avogadro number of atoms.

10. (A)
1.4
Moles of N 2   0.05.
28
Number of atoms = 0.05  2  6.02  1023 .
= 6.02  1022 .

11. (D)
22.4  10 3
(A)  NA  6.022  10 23
22400
22
(B)  6.022  10 23  3.011  10 23
44
11.2
(C)  6.022  10 23  3.011 10 23
22.4
(D) 0.1 6.022  10 23  6.022  10 22

12. (C)
Number of gms of H 2SO4  0.25  98  24.5

13. (D)

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 
Moles of H 2   0.5
2
Volume of H 2 in l = 0.5  22.4  11.2l .

14. (D)
19.7  1000
Moles of Au =  100
197
Atoms of Au = 100  6.022 10 23  6.022 10 25 .

15. (A)
44
Mass of one molecule of CO 2  23
 7.3110 23
6.02 10

16. (C)
0.224
Number of moles of H 2   0.01
22.4

17. (B)

18. (B)
WH  3  3  9g WN  3  14  42g

19. (C)
In one H2O molecule: 10 proton, 8 neutrons, 10 electrons
36g
Hence in 36 ml, n H2O   2mols
18g / mol
 Protons = 2N A  10  20 N A

20. (A)
w
n atoms  . Hence it should be of same weight ‘W’
at.wt

21. (B)
10 3 N A
no. of moles =  10 3
NA
 wt  103  mol.wt  103 M 0 g  M 0 mg

22. (A)
1 16
A:12 g ; B   16  8g ; C : 10 g ; D   8g
2 2

23. (D)
A : 2.5  5N A  12.5 N A ; B :10NA ; C : 4  3N A  12N A ; D  1.8  8N A  14.4N A .
Hence [D]

24. (C)
52 amu
 13
4 amu

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25. (B)
One ion contains: 7 + 24 + 1 = 32 e
 total es  2 N A  32  64 NA

26. (D)
n C  0.5  6  3  wt = 36 g

27. (C)
28 46 36 54
A: ; B: ; C: ; D:
44 46 18 108

28. (D)
180
n H 2O   10
18
 no. of es  10 10 N A  100 N A

29. (C)
2.48
n Na 2S2O3 .5H 2O   0.01
248
 nH 2O  5  0.01  molecules  0.05 N A

30. (C)
90 10 1 1
n Ag     atom  N A  5 1022
100 108 12 12

31. (B)
18  333
n H 2O  = 9.
54  (96  3)  (18 18)

32. (C)
0.018
n H2O   103 . Hence, molecules = 103 NA
18

33. (A)
4.2
n N 3   0.3 .  total = 0.3  8N A  2.4N A
14

34. (D)
3.42
n C  12  n C12H 22O11  12 
 0.12
342
 atom = 0.12 N A  [D]

35. (B)
8.4
n MgCO3   0.1
84
Each contain (12 + 6 + 24) protons
Hence, total  0.1 42NA  2.5 1024

36. (B)

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 4.4 2.24
n total    0.2  molecules = 0.2N A
44 22.4

37. [D]
1 14
(i) 
1000 58
1 2
(ii) 
1000 28
1 1
(iii) 
1000 23
(iv) 1ml  1g water
1
3
18

38. (B)
w w
n gas  
mol.wt. 3a

39. (A)
558.5
n Fe   10 moles
55.85
In 60 g carbon, n C  5  twice = 10 moles

40. (B)
Say n Mg  PO   n ; then n O  8n
3 4 2

0.25
 8n = 0.25  n   3.125  10 2
8

41. (B)

nx : ny 
 w  w 
2 : 2  2 :1
10 20

42. (C)
X
46  96  180  180  X  55.9
100

43. (C)
25.4 8 1 1
n I : nO  :  :  2:5
127 16 5 2
Hence I 2O5 .

44. (A)
mol. Wt = 2 VD = 100
71
w chlorine  100  71g
100
w metal  29 g

45. (C)
Mol.wt.  0.8  28  0.2  32  28.8
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 M
 VD   14.4
2

46. (A)
Dcl2 M cl2 71
Dcl2 wrt air   
Dair M air 29

47. (B)
30.4
Say NO X . Then 14  16x   14  x  2
100
M 46
 Doxide wrt O2  oxide   1.44
MO 2 32

48. (D)

CaCO3   CaO CO2
1 1 1

Quantity of limes tones = wt. of one mole mole of CaCO3

= 100 kg

49. (A)
Moles of H 2S  2
11.2
Moles of SO 2   0.5
22.4
SO 2  2H 2S  3S  2H 2O
moles 1 2 32
3 0.5
given 0.5 2 x  1.5
1
L.R.
50. (C)
100
Moles of Mg  OH 2   1.724
58
Moles 2H 3PO4  3Mg  OH 2  Mg 3  PO 4  2  6H 2 O
Moles 2 3 1 6
2  1.724
Given
3
2  1.724
Weight of H 3 PO 4   98  112.6g
3

51. (D)
n H2O  n CH3OH  2  4  wt  4 18  72g

52. (A)
WO  3.6769  2.0769  1.6g
with
2 mole X   5 mole'O '
with 1.6
' n ' moles   mole 'O '
16

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 0.2
n  0.04
5

53. (A)

Ag 2CO3   2Ag
2.7
WAg   2 108  2.11g
 216  60 
54. (D)
n CO2  2  n C2 H5OH  2
 WCO2  2  44  88g

55. (C)

KClO 3   KCl  3 O 2
2
48g
Hence % loss in wt =  100  39.18
122.5

56. (A)
2 2 2
n Fe   n H 2O   Wiron   56  37.39
3 3 3

57. (B)
1.62
n CaCO3  n CaO   n CaCl2  0.0289
56
0.0289 111
% of CaCl2  100  32.11%
10

58. (D)
3BaCl 2  2Na 3PO4  Ba 3  PO4  2  6NaCl
Moles 3 2 1 6
1 0.2
0.5 0.2  0.1
2
(L.R.)

59. (A)
Ca  OH  2  H 2SO 4 
 CaSO 4  2H 2O
0.2 0.5
LR
n CaSO4  n Ca  OH   0.2
2

60. (B)
A  2B 
 C  3D
5 8
LR
nB n
nC   4 ; n D  3  B  12
2 2

61. (D)

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 n : 4Al  3C 
 Al4C3
1350 2400
50  200
27 12
L.R
given
4Al 144 
given
  w  1800g
50  W 

62. (D)
2A 
 2B
B 
 2C
3C 
 4D
2 2 4
 n D  nA   
2 1 3
32

3

63. (B)
 
n
molality  1000 urea : NH 2 C NH 2 

w solvent  || 
 
 O 
18
 60 1000  0.192
1500 1.052  18
64. (D)
1
n
Molarity  1000  98 1000  0.01
V  mL  1000

65. (A)
 Al3   20  0.2  2  0.2M
  40

66. (D)
1 mole KMnO4 
 5 moles FeSO4
V  0.01 
 50  0.01
 V  10mL

67. (B)
 100  4
n H     0.001 2  2 10
 1000 
 no. of H  2 104 N A  1.2 1020

68. (A)
3 molal  3 mole NaOH in 1000g solvent
  120  1000 
 vol      1009mL
d  1.11 

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 n 3
Molarity  1000   2.97
V  mL  1.009

69. (B)
2.65  1000
Molarity of NO 2 CO 3   0.1 M.
106  250
0.1 10
After dilution of 10 mL solution =  0.001M
1000
70. (A)
n NaCl 1
X NaCl    0.0177
n NaCl  n H 2O 1  1000
18

GET EQUIPPED FOR JEE MAIN

1. (A)

KClO3   KCl  3 O 2 , n O2  3
2 2
2Al  3 O 2   Al 2O 3
2
nO
n Al2O3  2  1
3
2

2. (A)
Consider 1 L solution
29
  d 1000   H 2SO4  3.6  98
100
 d = 1.22 g/mL

3. (A)
N2  3H2 
 2NH3
n: 10 15 
LR
10  5  10 moles
n N2  5 n H2  0

4. (C)
Fe2  SO 4 3  3BaCl2 
 3BaSO 4  2FeCl3
n: ? 1
2
n BaCl2 n FeCl3 1
  n BaCl2  2  3  0.75moles
3 2 2

5. (C)
 Al2  SO4 3  3Cu
3CuSO4  2Al 
displaces
54g Al  192 g Cu
displaces
27g Al  
  96g

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6. (A)
CH4  2O2 
 CO2  2H2O
5 8 – –
5–4=1 – 4 8
n CO2  4 ; nCH4 (remaining) = 1

7. (C)
X
C10H X  nO 2 10CO 2    H 2O
2
X
Hence, n  10 
4
with  X
1 mole C10 H X    10   moles
 4
with
2.5 moles   32.5 moles
x 32.5 1
i.e. 10    13
4 2.5
 x  13  10   4  12

8. (D)
with
100g CaCO3   2mole HCl
with  25L 
 g     0.75 M HCl
 1000 
   0.9375g

9. (B)
2.125
n AgCl  n Cl  n HCl   V  L   Molarity
143.5
2.125  1000
 Molarity   0.59
143.5  25

10. (D)
with
2X   3 16g Oxygen
with
1g   0.16g Oxygen
3 16
 X  150
0.16  2

11. (D)
3
2Al  O2   Al2O3
2
n: n 1
2
with 3
2  27g Al   mole O2
2
with 1
   mole
2
2  27
  18g
3

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12. (D)
n BaSO4  nSO2  nS (POAC on S)
8 1
 
32 4

13. (B)
n NaBr  n1 , n KBr  n 2  say 
0.97
n AgBr  n Br  n1  n 2   0.00516
108  80 
Also, n1  103  n 2  119   0.560
0.56  103  0.00516
 n2   0.00178
16
 WKBr  119n 2  0.212 g

14. (A)
16g 31.2
A : nH  4  4 ; B : nH  4  1.64
16g 76
34.2 36
C : n H  22   2.2 ; D : n H  12   2.4
342 180

15. (C)
Total atoms = 200  0.05  NA  1020  NA
 0.05 NA  31022

16. (C)
Mol. Wt of A 2 B3  150  96  246
 For 5 mol,  246  5 g = 1.23 kg

17. (A)
200 144
A :10N A ; B :11  6.43 N A ; C  N A  3  9N A
342 48
D : 2.5  3N A  7.5N A .
Hence [A]

18. (D)
60
(i) 5g (ii)  35.5 (iii) 0.1 35.5 (iv) 0.5  71
106.5

19. (A)
1 1 1 1
A:  3N A ; B :  26N A ; C :  8N A ; D :  2N A
44 114 30 26

20. (C)
9.2
 2  n 1  n  0.4  wt  0.4  30  12g
46

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21. (D)
8 1
n CO2  n , say. Then n O  2n  n
16 4

22. (A)
3 1023
A : 0.2  14 g  2.8g ; B : 12 g  6g ; C : 32 g ; D : 7 g.
6 1023

23. (D)
1 gram molecule: 44 g
1 molecule of CO 2 = 44 amu

24. (A)
n H  n  2  2n  4  10n
n C  2n 1  2n
n C : n H  1: 5

25. (D)
Total charge = 1 N A  3e  3N A e coulomb

26. (D)
69.98
 Mol.wt  2112  mol.wt  360
100

27. (D)
12g H 2O 8% H 2O

45g silica 
 45g silica
43g others 43g others
100goriginal ' w 'grams
8 % of w = water
i.e. 92 % of w = silica others
92
Hence,  w  88g  w  95.65
100
45
% of silica  100  47%
95.65

28. (C)
M3N2 . 28 % nitrogen
28
   3M  28   28  M  24
100

29. (D)
0.014%  mol.wt  2  at.wt of N
0.014
i.e.  M  2 14  28
100
2800
M  3
 2 105
14 10

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30. (A)
90  20  21x  22  10  x 
Average atomic mass =  20.11
100
x = 9%

31. (B)
A. A. M = Mole fraction of O18 18  Mole fraction of O16 16

32. (C)
6.023  10 23
Moles of Ca  OH  2  1
6.023  10 23
3.01 10 22
Moles of HCl =  0.05
6.02  1023
3.01  10 22
HCl   0.05
6.02  10 23
Ca  OH  2  2HCl  CaCl 2  2H 2 O
1 2 1
0.05  1
1 0.05  0.025
2
 L.R.
33. (A)
1.595
Moles of CuSO 4   0.01
1595
Weight of solvent = 100 – 1.595 = 98.505
98.505
Volumes of solvent   82  10 3 L
1.2  1000
0.01
Molarity   0.12M
82  103
34. (B)
28
(A) atoms of O 2   6.022  10 23 ~ 3  10 23
32
3
(B) atoms of Be   6.022  10 23 ~ 2  10 23
9
8
(C) atoms of C   6.022  10 23  4  10 23
12
19
(D) atoms of F2   6.022  10 23  1 10 23
19

35. (C)
X Y X Y
20 80
: 1 : 2  XY2
10 200

36. (C)
Avogadro hypothesis

37. (A)
3 2.68
Moles of magnesium =
  0.00335
24 100
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 Number of magnesium atoms = 0.00335  6.022  10 23
 2.011021 atoms.
38. (D)
25  10 3
Moles of comphon   0.164  10 3
10  12  16  16
Number of atoms  0.164  103  6.022  10 23  27 (1 Molecule has 27 atoms).
 2.67 1021
39. (D)
Moles of e   52  2  54 .

40. (B)
1
Moles of Ag = .
107
1
Moles of Ag 2S required =
107  2
Mass of Ag 2S 
107  2  32   1.1495
107  2
1.1495
Mass of ore required =  100  85.78g
1.34

41. (D)
Moles of Al = 27 1
27
2Al  2NaOH  2H 2 O  2NaAlO 2  3H 2
Moles 2 2 2 2 3
31
Given 1 excess  1.5
2
(L.R.)
Vol. of H 2 evolved = 1.5  22.4  33.6 L .

WINDOW TO JEE MAIN

1. (A)
n solub le
Molarity 
Vso lub le Lt 
Vsolution is affected by Temperatuere.

2. (C)
560
n Fe   10
56
No. of atoms = 10 NA
70
In 70 g of N no. of atoms =  N A  5 NA
14
20
In 20 g of H no. of atoms =  N A  20 N A
1

3. (A)
Mole ratio of C : H : N
9 1 3.5
: :
12 1 14

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 3 1 1
: :
4 1 4
3 : 4 : 1
Empirical formula  C3 H 4 N
Empirical formula mass  36  4  14
 54
108
n 2
54
Molecular formula  C3 H 4 N  2
 C 6 H8 N 2

4. (D)
2BCl3  3H 2  2B  6HCl
no.of moles of H 2 3

no. of moles of B 2
3 21.6
No. of moles of H 2   3
2 10.8
Volume of H 2  3  22.4 L  67.2L

5. (B)
6.02 10 20
NA
Molarity   0.01
0.1

6. (C)
Refer theory

7. (C)
V=1L
Wtotal  11.021000  1020g
nsoluble = 2.05
352.8
Wtotal  100  1216.55g
29
= 1020 – 123 = 897 g
2.05
molality   2.28
0.897

8. (B)
no. of moles of oxygen atom 8

no. of moles of Mg 3  PO 4  2 1
0.25
No. of moles of Mg 3  PO 4 2 
8
 0.03125

9. (A)
V = 1L
nsoluble = 3.6
wsoluble = 3.6  98  352.8
352.8
w total  100  1216.55g
29
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 1216.55
density 
1000
= 1.22 g/ml

10. (C)
2Al  s   6HCl  aq   2Al 3  aq   6Cl   aq   3H 2  g 
Per mole of HCl, no. of moles of
1
H 2 formed 
2
1
Volume of H 2 at STP   22.4
2
 11.2L

11. (B)
Molality  5.2 m .
i.e. if wt. of H 2 O  1000gm
then no. of moles of CH 3OH  5.2
5.2
X CH 3OH   0.0856
1000
5.2 
18

12. (C)
Volume of solution 
1000  120  m
1.15
1120
 m
1.15
120 1.15 1000
Molarity   2.05M
60 1120

13. (C)
Molarity 
 750  0.5   250  2   0.875M
750  250
14. (A)
weight
Number of atoms   N A  species
atomic weight
 In 4 g of hydrogen
4
Number of atoms   N A  2  4N A
2
[Here species = 2 because hydrogen is present as H2]
In 71 g of chlorine = 2NA
71
Number of atoms =  N A  2  2N A
71
In 127 g of iodine,
127
Number of atoms =  N A  2  2N A
127
In 48 g of magnesium,
48
Number of atoms =  N A 1  2N A
24
[Here Mg is present as Mg so species = 1]
Thus, the number of atoms are largest in 4 g of hydrogen.

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15. (B)
Heavy water is D2O
In it,
Number of p   1 2  8  10
Number of e  1 2  8  10
Number of n 0  1 2  8  10
(D have1 n0 because it is actually, 1H2)

16. (D)
18 g H2O contains 2 g H
 0.72 g H2O contains 0.08 g H.
44 g CO2 contains 12 g C
 3.08 g CO2 contains 0.84 g C
0.84 0.08
 C:H  :  0.07 : 0.08  7 : 8
12 1
 Empirical formula = C7H8

17. (C)
3 M solution means 3 moles of solute (NaCl) are present in 1000 L of solution.
Mass of solution = volume of solution  density
 1000 1.252
= 1252 g
Mass of solute = No. of mole  molar mass of NaCl
 358.5g
= 175.5 g
Mass of solvent = (1252 – 175.5)g
= 1076.5 g
= 1.076 kg
moles of solute
Molality 
mass of solvent in kg
3
  2.79m
1.076

18. (A)
M1V1  M 2V2
Final concentration, M 
V1  V2
10  2  200  0.5

200  10
20  100

210
120
  0.57 M
210

19. (B)
N O2 n O2 WO 2 M O 2 WO 2 M N2 1 28
     
N N2 n N2 WN 2 M N 2 WN 2 M O2 4 32
7

32

20. (B)
BaCl2  H 2SO 4  BaSO4  2HCl

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 20.8gm 4.9gm 0.05 mole
20.8 4.9
 mole  mole  0.05  233gm
208 98
 0.1mole  0.05mole  11.65gm

21. (B)
1000  120
Volume of solution  ml
1.12
 1000ml
120 1000
Molarity   2M
60 1000

22. (D)
Molecular mass of compound  16  2  32gm
4
% of H in N 2 H 4  100
32
12.5%

23. (A)
No. of moles of acetic acid absorbed by 3gm charcoal
  0.6  0.042   50  10 3
 9 104 mole
Wt. absorbed by  9 104  64gm
3gm  0.054gm
0.054
Wt. absorbed per gram   0.018gm
3
 18mg
24. (C)
A  2B  3C  AB2C3
6
mole 1mole 0.036
60
 0.1mole mole
C is limiting reagent
No. of moles of AB2 C3 formed  0.012
Wt. of AB2 C3 formed  4.8gm
4.8
Molecular wt. of AB2C3   400
0.012
60  2x   3  80   400
x  50

25. (B)
BaCl2 .xH 2 O
18x 9

208  18x 61
208  18x  122x
x2

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26. (B)
1 32
8 100
x
x  400

27. (B)
 y y
C x H y   x   O 2  xCO 2  H 2O
 4 2
5 25
y
x
4  25  5
1 5
y
x 5
4
C3 H 8

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