Solution

KSR AKSHARA ACADEMY

JEE main - Chemistry

1. (a) 0.2

5.85×1000
Explanation: M = 58.5×500
 = 0.2
2. (a) 0.0093 mol

Explanation: Volume of O2 = 21

100
× 1000 mL at STP

∴ Mole of O2 = 21×1000

100×22400
= 9.375 × 10-3 mol

3. (c) 1 M, 2 N

Explanation: N =
24.5×1000

98
= 2 ;

×250
2
24.5×1000
M = = 1
98×250

4. (d) 232.4 g

Explanation: 232.4 g
5. (c) 3

Explanation: K2Cr2O7 + 6FeSO4 + 7H2SO4 ⟶  3Fe2(SO4)3 + Cr2(SO4)3 + 7H2O + K2SO4

∴  1 mole of K2Cr2O7 oxidizes 6 moles of FeSO4 completely and produces 3 moles of ferric sulphate.

6. (a) 1.05

Explanation:  I.S.  = 1

2
Σcz
2

0.1×100
[Al2(SO4)3] after mixing = 200
= 0.05 M

0.2×100
[Na2SO4] after mixing = 200
= 0.1 M

∴   I.S.  =
1

2
Σcz
2

= 0.05 × 2 × 32 + 0.05 × 3 × 22 + 0.1 × 2 × 12 + 0.1 × 1 × 22

=
2
= 1.05
2.1

7. (d) (molecular weight)/6

Explanation: The following reaction occur between S 2 O3

2−
 and 
C r2 O
2−

7
: 26H
+
+ 3S2 O
2−

3
+ 4C r2 O
2−

7
⟶ 6SO
2−

4
+ 8C r
3+
+ 13H2 O

Change in oxidation number of Cr 2 O7

2−
 per formula unit is 6 (it is always fixed for Cr 2−
2 O7 )

 Molecular weight 
Hence, equivalent weight of K2Cr2O7 =  6

8. (c) 98

Explanation: NaOH is the monoacidic base. So its molar mass is the same as equivalent mass.

In this reaction, 1g equivalent NaOH reacts with one mole H3PO4. So the molar mass of the acid is its equivalent mass. Hence
it will be =3× 1+31+4 × 16=98.
9. (c) 5 × 10
21

Explanation: 1 carat = 200 mg

0.5 carat = 100 mg

1 mole of C-atoms weighs = 12 g

⇒12 g of carbon = 6.023 x 1023 C-atoms

⇒ 100 mg of carbon = (6.023 x 1023 x 100)/12 x 1000

= 5.02 x 1021 
10. (b) 2CuBr →  CuBr2 + Cu

Explanation:

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 In disproportionation reactions, same element undergoes oxidation as well as reduction. e.g

Here, CuBr get oxidised to CuBr2 and also it get reduced to Cu. Other given reactions and their types are given below.

In the given reaction, MnO  get oxidised to Mn2+ and I- get reduced to I2. It is an example of redox reaction. The reaction
−

take place in acidic medium.

2KMnO4 ⟶ K2MnO4 + MnO2 + O2

The given reaction is an example of decomposition reaction. Here, one compound split into two or more simpler compounds,
atleast one of which must be in elemental form.

2NaBr + Cl2 ⟶  2NaCl + Br2

The given reaction is an example of displacement reaction. In this reaction, an atom (or ion) replaces the ion (or atom) of
another element from a compound.
11. (a) 0.5 N

Explanation: meq. of HCl = 100 × 0.3 = 30

meq. of H2SO4 = 200 × 0.6 = 120

30+120 1
∴  Nmixture  = =
300 2

12. (d) 1 : 0.9 : 0.93

Explanation: Let mass of all compounds is w gm, then mole ratio of Fe in FeO, Fe2O3 ​and Fe3O4​is,  w

72
:
2w

160
:
3w

232
 = 1 : 0.9 :
0.93
13. (c) Student D

Explanation: Student A and C have reported values that are neither precise nor accurate. Student B has reported values that are
precise but not accurate. Student D has reported values that are both precise and accurate.
14. (a) (iii)

Explanation: A molecular of a compound has atoms of different elements. which cannot be separated into its constituent
elements by physical methods of separation. The physical properties of a compound are different than its constituent elements.
For a given compound the ratio of different elements is fixed.
15. (b) 4 : 1

Explanation: Number of moles of H2 = 

1

Number of moles of O2 =  32

4

Hence, molar ratio =  1

2
:
4

32
 = 4 : 1

16. (c) 6.02 × 1023

Explanation: 16 g CH4 = 1 mole CH4 = N molecules of CH4

3
17. (d) 5

Explanation: The balanced redox reaction is:

3M nO
−

4
+ 5F eC2 O4 + 24H
+
→ 3M n
2+ 3+
+ 5F e + 10C O2 + 12

∵  5 moles of FeC2O4 require 3 moles of KMnO4

∴  1 mole of FeC2O4 will require 3

5
 mole of KMnO4

18. (c) Base = Weak, Acid = Strong, Endpoint = Yellow to pinkish-red

Explanation:
Methyl orange show Pinkish colour towards more acidic medium and yellow orange colour towards basic or less acidic media.
Its working pH range is

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 Weak base have the pH range greater than 7. When methyl orange is added to this weak base solution it shows yellow orange
colour.

Now when this solution is titrated against strong acid the pH move towards more acidic range and reaches to end point near 3.9
where yellow orange colour of methyl orange changes to Pinkish red resulting to similar change in colour of solution as well.

19. (a) Option (iii)

Explanation: Except (iii) all postulates was given by the Dalton.

20. (d) 0, +1 and -2

Explanation: In S8, oxidation number of S is 0, elemental state.

In S2F2, F is in -1 oxidation state, hence S is in + 1 oxidation state.

In H2S, H is in +1 oxidation state, hence S is in -2 oxidation state.

21. (d) 8.5 g of NH3

Explanation: Total number of atoms in a given amount of substance = n × NA × Atomicity

Atomicity of NH3, O2, CO2, and He molecules are 4, 2, 2, and 1 respectively.

8.5 g of NH3 = ( 8.5

17
)× 4 ×  NA = 2 NA atoms

NΛ
4 g of O2 = 4

32
× 2 ×  NA = 4
 atoms

22 g of CO2 = ( 22

44
)× 3 ×  NA  = 3

2
NA atoms

4 g of He = NA atoms

Thus, the system that contains 8.5 g of NH3 has the maximum number of atoms.

22. (a) different molecular weight

Explanation: Since the molecular formula is n times the empirical formula, therefore, different compounds having the same
empirical formula must have different molecular weights.
23. (c) 3 : 1

Explanation:

  35Cl 37Cl

Molar ratio x (1 - x)
Mavg = 35 + x + 37 (1 - x) = 35.5

= 35x + 37 (1 - x) = 35.5

⇒  2x = 1.5

x = 

4
3

So, ratio of 35Cl : 37Cl =   = 3 : 1

4

24. (a) 70 g

Explanation:

Thus, 70% H2SO4 ​required = 49 × 100

70
​= 70 g

25. (b) 4 : 1

nH wH mO wH
1×32 4 1
Explanation:
2 2 2 2
= × = = [ = ]
n mH w 2×4 1 w 4
O2 2 O2 O2

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26. (a) 27 g

Explanation: 27 g
27. (d) 1

9.108×6.023
× 10
8

Explanation: 9.108 × 10-31 kg = 1 electron

1
∴ 1 kg =
−31
electron
9.108×10

=
1

−31
×
1

23
mole electron
9.108×10 6.023×10

28. (c) 84.5

−3 −3

Explanation: Concentration of Cl−  

300mg 300× 10  g 300× 10 Eq

−1
=
−1
 =  −1
 

10  L 10  L 35.5× 10  L

As Cl− have eq. wt of 35.5 so, concentration in 

Eq.

L
 = 0.0845 × 1000meq./L = 84.5 meq./L

29. (d) 2.05 M

Explanation: Mass of solute = 120 g

Mass of water = 1000 g

Mass of solution = 1120 g

1120
∴ Volume of solution ( m

d
) =
1.15
 mL

milli mole = M × Vin mL

120

60
× 1000 = M ×
1120

1.15

∴ M = 2.05
30. (c) 9.108×6.023
1
× 10
8

Explanation: Mass of an electron = 9.108 ×  10-31 kg

∵  9.108 ×  10-31 kg = 1.0 electron

31
1 10 1
∴ 1kg =
−31
 electrons  =
9.108
×
23

9.108×10 6.023×10

=
1

9.108×6.023
× 10
8
 mole of electrons

31. (d) 135.5 ×  1022

Explanation: 4.4 g of CO2 = 4.4

44
 = 0.1 mol

= 0.1 ×  6.022 ×  1023 molecules

= 6.022 ×  1022 molecules

= 6.022 ×  1022 × 22

= 132.5 ×  1022 electrons (∵ 1 molecule of CO2 contain 22 electrons)

32. (d) 1.66 M

Explanation: Mole and millimole do not change on dilution.

Thus 500 × 5 = 1500 × M

∴  M = = 1.66 M 5

33. (d) 1 : 8

Explanation: Both O2 and N2 are diatomic.

∴  ratio of their number of molecules will be equal to ratio of their number of moles.

 mass of O2
number of moles of O2 =  32

 mass of N2
number of moles of N2 =  28

∴  number of moles of O2 : number of moles of N2

 mass of O2  mass of N2

=  32
: 28

=  1

32
:
7

28
=
1

32
:
1

∴ ratio is 4 : 32 = 1 : 8
34. (d) 1000 g of the solvent

Explanation: Molality is the number of moles of solute dissolved in 1000 grams of solvent.

So, 1 molal solution contains one mole of solute in 1000 grams of solvent.

1 mole
1M =
1000 grams

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35. (b) 36 g of water

Explanation: Number of molecules present in 36 g of water = 36

18
× NA = 2NA

Number of molecules present in 28 g of CO = 28

28
× NA = NA

Number of molecules present in 46 g of C2H5OH = 46

46
× NA = NA

Number of molecules present in 54 g of N2O5 = 54

108
× NA = 0.5NA

Here, NA is Avogadro’s number. Hence, 36 g of water contain the largest (2NA) number of molecules.

36. (a) 2

Explanation: m eq. of Na2CO3 ⋅ xH2O in 20 mL = 19.8 × 1

10

∴ m eq. of Na2CO3 ⋅ xH2O in 100 mL = 19.8 × 1

10
× 5

∴  
w

E
× 1000 = 19.8 ×
1

10
× 5

or 0.7
× 1000 =
19.8

M/2

∴ M = 141.41

∴ 23 × 2 + 12 + 3 × 16 + 18x = 141.41
∴  x = 2
37. (d) the mass of one mole of carbon

Explanation: Mass of 1 mole carbon = Mass of N molecules of carbon

= mass of carbon in amu × N

= 12 × 6.023 × 1020 amu

= 12 × 6.023 × 1020 × 1.66 × 10-24g

= 1.2 × 10-3g

Thus mass of 1 mole carbon will be 1.2 × 10-3 g in place of 12 g.

38. (b) 84.3

Explanation: M2CO3 + 2HCl → 2MCl + H2O + CO2

Mole of M2CO3 = mole of CO2

Mole of M 2
CO3 (
w

M
) = 0.01186

Molar mass of M 2 CO3 (M ) =

1

0.01186
= 84.3 g mol-1

39. (b) 2.0 M

22
6.02×10
Explanation: Mole of glucose = 23
 = 0.1 = 2

6.02×10

∴  Mglucose  =
0.1×1000

50
 = 2
40. (d) 3, 3 and 4 respectively

Explanation: 3, 3 and 4 respectively

41. (b) 3.7 ×  104

Explanation: 1 metric ton of Al = 103 kg = 106 g

1 mole of Al = 27 g of Al

 106 g of Al =   moles = 3.7 ×  104

6
10
∴
27

42. (c) 40g

Explanation: 0.5 mole of H 3O
+
= 20g

Also H O  is monovalent,

3
+

Thus, mol.wt.=Eq.wt.

∴1mole of H3O+=40g

43. (c) 0.125 mol

Conc.HCl

Explanation: 2H 3 AsO4 + 5H2 S −−−−−−→ As2 S5 + 8H2 O

2 mol 1 mol

1 mol 1/2 mol

∵  The molar mass of H3AsO4 is 3 ×  1 + 79 + 4 ×  16 = 142g/mol

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 ∴  Number of moles of H3AsO4 =  35.5

142
 = 0.25 mol

∴  Number of moles of As2S5 =  0.25

2
= 0.125 mol

44. (a) one gram of iron

Explanation:
i. 1 mole of calcium

= Atomic mass in gram = 40.0 g

ii. 1 mole of Ag atoms = 108 g

= 6.022 ×  1023 atoms

i.e., mass of 6.022 ×  1023 atoms of Ag = 108 g

∴  Mass of 1 atom of Ag

=  108

23
 = 1.793 ×  10-22 g
6.022×10

iii. 1 mole of carbon atoms = 12.0 g

∴  6.022 ×  1023 atoms of carbon = 12.0 g

Mass of 1023 atoms of C

23

= 12×10

23
 = 1.993 g
6.022×10

iv. Mass of iron = 1 g (given)

So, correct order of increasing mass: one atom of silver < 1 gram of iron < 1023 atoms of carbon < one mole of calcium.
45. (b) 84.3

Explanation: M 2 C O3 + 2HC l → 2M C l + H2 O + C O2

1 g 0.01186 mole

Number of moles of M2CO3 reacted = Number of moles of CO2 evolved

M
1
 = 0.01186 [M = molar mass of M2CO3]

M =  1

0.01186
 = 84.3 g mol-1

46. (d) 26

Explanation: The number of equivalents of H2SO4 ​is equal to the number of equivalents of SO3. It is also equal to the number
of equivalents of NaOH.

 = 54 × 0.4 × 10−3

x (1−x)×2
Hence,  98
× 2 +
80

or x = 0.74

1−0.74
The percentage of free SO3 =  1
× 100  = 26%

47. (a) 0.051

Explanation: Molality of the glucose solution = 3 m

⇒  3 mol of glucose is present in 1 kg of water.

Number of moles of water present in 1 kg of water

1000g
=  18g/mol
 = 55.55 mol.

3
xglucose =  3+55.55
 = 0.051

48. (c) 59.73

Explanation: 45 mL = 0.24 g

22400 mL =
0.24×22400

45
= 119.47 g mol
−1

∴  Vapour density  =
 Molar mass 

2
=
119.47

2
 = 59.73

49. (d) All of these

Explanation: The terms which include volume will be temperature-dependent otherwise all are temperature independent as,

Molality: It is the number of moles present in a kilogram of solution. Mathematically, the expression for molality is as follows:
      Molality =   no. of moles

mass in kg

Mole fraction: The ratio of the number of moles of one component to the total number of all the components present in the
solution, is called the mole fraction of that component.

Percentage by mass: The number of grams of solute dissolved in one gram of solution is called the mass fraction of the solute.

So, all are temperature independent.

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50. (c) M

Explanation: It is an acid salt; Only 1 H is replaceable.

51. (a) 12.044 × 1020 molecules

Explanation: Milli mole = M × VmL

= 0.02 × 100 = 2

∴ molecules = 2N × 10-3

= 2 × 6.02 × 1023 × 10-3 = 12.044 × 1020

52. (b) CH4

Explanation: ∵ 2.24 L of gas has mass = 1.6 g at STP

∴  22.4 L of gas has mass = × 22.4

1.6

2.24

= 16 gat STP

So, given gas is CH4 because CH4 has molecular mass = 16 g

53. (a) 40

Explanation: Let the metal be M with atomic wt. be x.

So, 3M+N2​→M3​N2​

3x                  3x+28
So, = 12

3x
14.8
  

(3x+28)

Which in solving for x gives 40.

The metal is calcium and its nitride is Ca3​N2​.

54. (b) (i)

Explanation: 11.2 litre H2 at STP is produced for every mole of HCl(aq) used

55. (a) S2

Explanation: As molar mass is lowest for S2, so number of moles(n) = m

w
 is highest for S2

56. (c) 6

Explanation:
n-factor of dichromate is 6.

Also, n-factor of Mohr’s salt is 1 as : 

∵  1 mole of dichromate = 6 equivalent of dichromate

∴  6 equivalent of Mohr’s salt would be required.

Since, n-factor of Mohr’s salt is 1, 6 equivalent of it would also be equal to 6 moles.

Hence, 1 mole of dichromate will oxidise 6 moles of Mohr’s salt.

57. (b) 1.61

Explanation: Wt. of H2SO4 = 15g

Wt. of solution = 100 × 1.1 = 110g

Wt. of solvent (water) = 110 - 15 = 95g

15
∴ m =
95
 = 1.61
98×
1000

58. (c) 47%

Explanation:
Silica Water Clay Mineral

45 12 43 Initial %

a 8 (92-a) % after heating

The % ratio of silica and clay remains constant on heating

45
i.e., =
43

92−a
a

a = 47%

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59. (a) 3.68 mol Na

Explanation: For a lever: W1d1 = W2d2

2 mol Cu = 127 g

∴  127 ×  200 = W2 × 300

∴  W2 = 84.67 g

Now, 3.68 mol Na ≡  84.64 g

60. (a) 2240 ml

Explanation: Mg(HCO3)2 → MgO + 2CO2 + H2O

This means that 146 g of Mg(HCO3)2 gives 2×44, i.e. 88 g of CO2.

88×7.3
So 7.3 g will give  146
= 4.40 g.

No of moles of CO2 in 4.4 g = 4.4

44
= 0.1 moles

Now, 

1 mole of a gas at STP occupies 22.4 l, so 0.1 moles will occupy 0.1×22.4 = 2.24L or 2240 ml.

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