Chapter 1

Some Basic Concepts of Chemistry

Solutions

SECTION – A
Objective Type Questions
1. A sample of ammonium phosphate (NH4)3PO4 contains 3.18 moles of hydrogen atoms. The number of moles
of oxygen atoms in the sample is
(1) 0.265
(2) 0.795
(3) 1.06
(4) 4.00
Sol. Answer (3)

In NH4 3 PO4 12 moles of 'H' are present with 4 moles of oxygen atom.
4
 3.18 moles of 'H' are present with =  3.18 = 1.06 moles of oxygen atom.
12

2. Two metallic oxides contain 27.6% and 30% oxygen respectively. If the formula of the first oxide is X 3O4, that
of the second will be
(1) XO (2) XO2
(3) X2O5 (4) X2O3
Sol. Answer (4)
This question can be solved by two methods.

Method I Method II
st
Oxide I Oxide II Formula of 1 X3O4

X3O4 O = 27.6% O = 30% wt of X

Eq. mass of X = 8
wt of O
X = 72.4% X = 70%
72.4
   21
8 20.9
O = 27.6% X = 70% 27.6
4 8
72.4% of X = 3 mol of X Positive charge of X = 2  =
3 3
3
70% of X  70 2.90 mol of X 8
72.4  Atomic mass of X   21  56 g
3

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 2 Some Basic Concepts of Chemistry Solution of Assignment

Similarly  Atomic mass = eq. mass × Valency]

70
27.6% of O = 4 mol of O Eq. mass of X   8  18.66
30
4
30% of O 
3 4.34 atom of O Atomic mass of X  56g
27.6

X : O Calculate from 1st oxide

2.9 : 4.34 Atomic mass 56
i.e., 2 : 3  Valency    3
Eq. mass 18.6
Formula will be : X2O3 Formula will be : X2O3

3. Calculate the molality of solution containing 3 g glucose dissolved in 30 g of water. (molar mass of glucose
= 180)
(1) 0.50 m (2) 0.56 m (3) 0.091 m (4) 0.05 m
Sol. Answer (2)

moles of solute nB
Molality =  wB = 3 g, wA = 30 g
wt of solvent (kg) w A (kg)
3
1
Molality (m) = 180  1000 = 1800  1000 = 0.56 m.
30

4. An element, X has the following isotopic composition 200X : 90% 199X : 8% 202X : 2.0%. The weighted average
atomic mass of the naturally occurring element X is closest to
(1) 201 amu (2) 202 amu (3) 199 amu (4) 200 amu
Sol. Answer (4)
 percentage x atomic mass
Average atomic mass =
100
 200  90   199  8    202  2 
=
100
= 199.96  200 amu

5. Which has the maximum number of molecules among the following?

(1) 8 g H2
(2) 64 g SO2
(3) 44 g CO2
(4) 48 g O3
Sol. Answer (1)
Maximum number of moles have maximum number of molecules
 calculate number of moles.
8 44
8 g H2 moles = = 4 moles 44 g of CO2 = = 1 mol CO2
2 44

64 48
64 g SO2 moles = = 1 moles 48 g of O3 = = 1 mol of O3
44 48
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Solution of Assignment Some Basic Concepts of Chemistry 3
6. What volume of oxygen gas (O2) measured at 0°C and 1 atm, is needed to burn completely 1 L of propane
gas (C3H8) measured under the same conditions?
(1) 10 L (2) 7 L (3) 6 L (4) 5 L
Sol. Answer (4)
C3H8 + 5O2 3CO 2 + 4H2O
For 1 mol propane 5 mol O2 gas is needed.
22.4 L propane = 5 × 22.4 L of O2 gas needed

 1 L propane = 5 L of O2 gas is required

7. Volume occupied by one molecule of water (density = 1 g cm–3) is

(1) 5.5 × 10 –23 cm3 (2) 9.0 × 10 –23 cm3 (3) 6.023 × 10 –23 cm3 (4) 3.0 × 10 –23 cm3
Sol. Answer (4)
As water is liquid its density = 1 g/mL
i.e., 1 g of H2O have volume = 1 mL
18
Mass of one molecule = g
6.023  1023
18 18
 23 g of H2O have volume = mL = 3.0 × 10–23 mL
6.023  10 6.022  1023

8. The total number of electrons in 1.6 g of CH4 to that in 1.8 g of H2O

(1) Double (2) Same (3) Triple (4) One fourth
Sol. Answer (2)
–
[Total number of e in CH4]
1.6
Number of e– in 1.6 g of CH4 =  10  N0 
N0
16
[Total number of e– in H2O]
1.8
Number of e– in 1.8 g of H2O =  10  N0 
N0
18

9. When x molecules are removed from 200 mg of N2O, 2.89 × 10–3 moles of N2O are left. x will be
(1) 1020 molecules (2) 1010 molecules (3) 21 molecules (4) 1021 molecules
Sol. Answer (4)
From Equation

– =

(200 mg of N2O) – x molecules = –3

2.89 × 10 moles of N2O
200
200 mg of N2O have molecule =  103  6.022  1023
44
 2.7 × 1021 molecule
 2.89 × 10-3 moles of N2O have molecule = 2.89 × 10-3 × 6.022 × 1023
= 1.7 × 1021 molecules
 200 mg of N2O – x molecule = 2.89 × 10–3 moles of N2O

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 4 Some Basic Concepts of Chemistry Solution of Assignment

[2.7 × 1021 – x = 1.7 × 1021] molecule

[x = (2.7 - 1.7) × 1021] molecule
= 1021 molecule

10. 4 g of hydrogen reacts with 20 g of oxygen to form water. The mass of water formed is
(1) 24 g (2) 36 g (3) 22.5 g (4) 40 g
Sol. Answer (3)
2H2  O2  2H2 O
4g 32 g 36 g
When 4 g of H2 reacts with 32 g of O2 gives 36 g of H2O.
Now present oxygen is 20 g
 O2 will be the limiting reagent and H2O will be calculated from O2
 32 g of O2 given = 36 g of H2O
36
20 g of O2 given =  20 = 22.5 g H O
32 2

11. Which has maximum molecules?

(1) 7 g N2O (2) 20 g H2 (3) 16 g NO2 (4) 16 g SO2
Sol. Answer (2)
Maximum number of moles have maximum Number of molecules.
7 20 16
Moles of N2O = ; moles of H2 = ; moles of NO2 =
44 2 46
16
Moles of SO2 =
64
H2 have maximum number of moles and have maximum number of molecules.

12. The number of molecules in 4.25 g of NH3 is approximately

(1) 4 × 1023 (2) 1.5 × 2023 (3) 1 × 1023 (4) 6 × 1023
Sol. Answer (2)
4.25
Moles of NH3 = = 0.25 moles
17
Number of molecule = 0.25 × 6.022 × 1023 molecule = 1.50 × 1023 molecule.

13. The maximum number of molecules is present in

(1) 15 L of H2 gas at STP (2) 5 L of N2 gas at STP
(3) 0.5 g of H2 gas (4) 10 g of O2 gas
Sol. Answer (1)
Gas have maximum number of moles will have maximum number of molecules 22.4 L at STP gas = 1 mole
15 5
Moles of H2(g) = = 0.67 mol; moles of N2 = = 0.223 mol
22.4 22.4
0.5 10
Moles of 0.5 g H2 gas = = 0.25 mol; moles of O2 10 g = = 0.1325
2 32
 maximum number of molecules = 15 L of H2 (g) at STP

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Solution of Assignment Some Basic Concepts of Chemistry 5
14. The number of atoms in 0.1 mol of a tetraatomic gas is (NA = 6.02 × 1023 mol–1)
(1) 2.4 × 1022 (2) 6.026 × 1022 (3) 2.4 × 1023 (4) 3.600 × 1023
Sol. Answer (3)
23 23
Total number of atom = 0.1 × 4 × 6.022 × 10 = 2.4 × 10 atom

[4 atom are present as gas is tetra-atomic]

15. How many grams of NaOH should be added to water to prepare 250 ml solution of 2 M NaOH?
(1) 9.6 × 103 (2) 2.4 × 103 (3) 20 (4) 24
Sol. Answer (3)
M  V(mL) 2  250
Moles of NaOH =  = 0.5 moles of NaOH
1000 1000
given mass x
Moles =  0.5 mole = given mass = 40 × 0.5 = 20 g
mol.mass 40

16. Haemoglobin contains 0.334% of iron by weight. The molecular weight of haemoglobin is approximately 67200.
The number of iron atoms (Atomic weight of Fe is 56) present in one molecule of haemoglobin is
(1) 4 (2) 6 (3) 3 (4) 2
Sol. Answer (1)
0.334
Weight of Fe in heamoglobin =  67200 = 224.48 u
100
Mass of one Fe atom = 56 u
224.48
 Total number of Fe atom = 4
56

17. In the reaction, 2SO2 + O2  2SO3

when 1 mole of SO2 and 1 mole of O2 are made to react to completion
(1) All the oxygen will be consumed (2) 1.0 mole of SO3 will be produced
(3) 0.5 mole of SO2 is remained (4) All of these
Sol. Answer (2)
2SO 2 + O 2 2SO3
2 moles of SO2 reacts with 1 mole of O2 as 1 mol of SO2 is present
 SO2 will be limiting reagent will formed  1 mol of SO3

18. If the weight of metal chloride is x gram containing y gram of metal, the equivalent weight of metal will be

x 8( y  x ) y 8( x  y )
(1) E  35.5 (2) E  
(3) E  35.5 (4) E 
y x (x  y) y
Sol. Answer (3)
Mx   x Cl 
 MClx
y (g) (x  y)(g) x (g)

weight of metal

Equivalent mass of metal in case of chlorides  35.5
weight of Cl

y
 equivalent mass E =  35.5
x  y
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 6 Some Basic Concepts of Chemistry Solution of Assignment

19. 10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in
this reaction will be
(1) 1 mol (2) 2 mol (3) 3 mol (4) 4 mol
Sol. Answer (4)
2H2  O2 
 2H2 O
4g 32 g 2 moles
As 10 g of H2 is to react with 64 g of O2
 O2 will complete consumed and will act as LR.
Calculation will be made on the basis of weight of O2
⎧⎪32 g of O2 gives H2O = 2 moles ⎫⎪
⎨ ⎬
⎪⎩ 64 g of O2 gives H2O = 4 moles ⎪⎭
20. Consider the following reaction sequence:
S8(s) + 8O2(g)  8SO2(g)
2SO2(g) + O2(g)  2SO3(g)
How many grams of SO3 are produced from 1 mole S8?
(1) 1280 g (2) 960 g (3) 640 g (4) 320 g
Sol. Answer (3)
1 mole S8  8SO3  8 × 80 g
= 640 g

SECTION – B
Objective Type Questions
1. The total number of electrons in 4.2 g of N3– ion is (NA is the Avogadro’s number)
(1) 2.1 NA (2) 4.2 NA (3) 3 NA (4) 3.2 NA
Sol. Answer (3)
4.2
Total number of moles = = 0.3 mol
14
1 mol of N3– have electrons = 10 × N0.

 Number of e in 0.3 mol = 0.3 × 10 × N0 3  N0

2. The number of mole of nitrogen in one litre of air containing 10% nitrogen by volume, under standard conditions,
is
(1) 0.03 mole (2) 2.10 moles (3) 0.186 mole (4) 4.46 × 10–3 mole
Sol. Answer (4)
10
Volume of N2 in 1 L i.e., 1000 ml of N2 = × 1000 = 100 ml
1000
22400 ml at STP = 1 mol.
1 1
 100 ml at STP = × 100 = = 4.46  10 3 mol
22400 224
3. Liquid benzene (C6H6) burns in oxygen according to 2C6H6 (l) + 15O2(g)  12CO2(g) + 6H2O(g)
How many litres of O2 at STP are needed to complete the combustion of 39 g of liquid benzene?
(1) 74 L (2) 11.2 L (3) 22.4 L (4) 84 L

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Solution of Assignment Some Basic Concepts of Chemistry 7
Sol. Answer (4)

2C6H6 (l) + 15 O2 (g) 12 CO2 (g) + 6H2O (g)

2 × 78 g 15 × 22.4 L
From equation 15 × 22.4 L of O2 is required for = 156 g of benzene
i.e., 156 g benzene for complete combustion required O2(STP) = 15 × 22.4 L
15  22.4
39 g benzene for complete combustion required O2(STP) = × 39 = 84 L of O2
156

4. 1 mol of KClO3 is thermally decomposed and excess of aluminium is burnt in the gaseous product. How many
moles of Al2O3 are formed?
(1) 1 (2) 2 (3) 1.5 (4) 3
Sol. Answer (1)
2KClO3 2KCl + 3O2
2 moles 3 moles
2 mol of KClO3 gives = 3 mol O2
3
1 mol of KClO3 gives = mol O2
2
For Al burning
3
2 Al  O2 
 Al2 O3
2  
1mole

3
As mole of O2 gives 1 mole Al2O3
2
 1 mole Al2O3 formed.

5. The amount of zinc required to produce 1.12 ml of H2 at STP on treatment with dilute HCl will be
(1) 65 g (2) 0.065 g (3) 32.5 × 10–4 g (4) 6.5 g
Sol. Answer (3)
Zn  2HCl  ZnCl2  H2 (g)
1moL 22.4 L  22400 ml
22400 ml of H2 gas is produced from Zn = 65 g
65
1.12 ml of H2 gas is produced from Zn =  1.12 g = 3.25 × 10–3 g
22400
i.e., 32.5 × 10–4 g

6. Volume of CO2 obtained at STP by the complete decomposition of 9.85 g Na2CO3 is

(1) 2.24 litre (2) Zero (3) 0.85 litre (4) 0.56 litre
Sol. Answer (2)
Na2CO3 is soda ash which does not decompose upon heating even to redness.
 CO2 will not be evolved.

7. One litre of CO2 is passed through red hot coke. The volume becomes 1.4 litres at same temperature and
pressure. The composition of products is
(1) 0.8 litre of CO2 and 0.6 litre of CO (2) 0.7 litre of CO2 and 0.7 litre of CO
(3) 0.6 litre of CO2 and 0.8 litre of CO (4) 0.4 litre of CO2 and 1.0 litre of CO

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 8 Some Basic Concepts of Chemistry Solution of Assignment

Sol. Answer (3)

CO2 (g)  C(s)   2CO(g)
Initial 1L 0 0
Final volume (1  x) 2x
Final volume = 1 – x + x + 2x = 1.4 L
 x  0.4 L
 Volume of CO = 2x = 2 × 0.4 = 0.8 L
Volume of CO2 = (1 – x) = 1 – 0.4 = 0.6 L

8. Suppose that A and B form the compounds B 2 A 3 and B 2 A. If 0.05 mole of B 2 A 3 weighs 9 g and
0.1 mole of B2A weighs 10 g, the atomic weight of A and B respectively are
(1) 30 and 40 (2) 40 and 30 (3) 20 and 5 (4) 15 and 20
Sol. Answer (2)
Let the atomic mass of B = y g ; A = x g
In B2A3
given weight
2y + 3x = mol. mass of B2 A3 =
mole
9
2y + 3x = g
0.05
In B2A
10
 2y + x = g
0.1
Solving x and y
⎧ x  40 ⎫
⎨ ⎬
⎩ y  30 ⎭

M M
9. When 100 ml of H SO is mixed with 500 ml of NaOH then nature of resulting solution and normality of
10 2 4 10
excess of reactant left is
N N N N
(1) Acidic, (2) Basic, (3) Basic, (4) Acidic,
5 5 20 10
Sol. Answer (3)

M M
100 ml of H2 SO4 500 ml of NaOH
10 10

N N
H2SO4 [because N factor is 1]
5 10

⎛ 1⎞ ⎛ 1⎞
 meq of H2SO4 
⎜ 100  ⎟  20 meq  meq of NaOH  ⎜ 500  ⎟  50 meq
⎝ 5 ⎠ ⎝ 10 ⎠

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Solution of Assignment Some Basic Concepts of Chemistry 9
larger meq – smaller meq
For Neutralisation Reaction =
Total volume
50  20 1
=  N NaOH [because larger meq of NaOH will remain]
600 20
 Solution will be basic

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

(1) 0.3 (2) 0.05 (3) 0.7 (4) 0.95
Sol. Answer (4)
⎧m = molality ⎫
1000 . x B ⎪ ⎪
m = x .m ⎨ xB = molality fraction of solute ⎬
A A ⎪ x = molality fraction fo solvent ⎪
⎩ A ⎭
x A  xB 
1
 xA = (1 – xB)
1000 . xB
m = 1  x  M
B A
Putting m = 3
MA = 18 because aqueous solution is present
1000 . xB
3 = 1  x  18  54 (1 – xB) = 1000 xB
B
= 54 – 54 xB = 1000 xB
54
xB =  xB = 0.05. `
1054

 xA 1  xB  1  0.05  0.95

11. Concentrated aqueous sulphuric acid is 98% H2SO4 by mass and has a density of 1.80 gmL–1. Volume of acid
required to make one litre of 0.1 M H2SO4 solution is
(1) 16.65 mL (2) 22.20 mL (3) 5.55 mL (4) 11.10 mL
Sol. Answer (3)
Molarity of 98% H2SO4 by mass having density 1.80 g/ml will be
% w/w  d  10 98  1.80  10
M=  = 18 M.
M. mass 98

M1 = 18 M M2 = 0.1 m
V1 = ? V2 = 1000 ml
Applying M1V1 = M2V2
18 × V1 = 1000 × 0.1
100
V1 = = 5.55 ml
18
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10 Some Basic Concepts of Chemistry Solution of Assignment

12. Number of significant figures in 6.62 × 10–34.

(1) Two (2) Three (3) Four (4) One
Sol. Answer (2)
Number of significant figures = Three
i.e., 6.62
13. Ammonia gas is passed into water, yielding a solution of density 0.93 g/cm3 and containing 18.6% NH3 by
weight. The mass of NH3 per cc of the solution is
(1) 0.17 g/cm3 (2) 0.34 g/cm3 (3) 0.51 g/cm3 (4) 0.68 g/cm3
Sol. Answer (1)

18.6  0.93  10
Molarity of NH3 solution = =10.17 M
17

Strength (g/L) = molarity × mol. mass

= 10.17 × 17 = 172.9 g/L
i.e., 1000 mL of solution contain NH3 = 172.9 g

172.9
1 mL or 1 cm3 of solution contain NH3 = = 0.172 g  0.17 g
1000

14. A certain amount of a metal whose equivalent mass is 28 displaces 0.7 L of H2 at S.T.P. from an acid hence
mass of the element is
(1) 1.75 g (2) 0.875 g (3) 3.50 g (4) 7.00 g
Sol. Answer (1)
Weight of metal whcih can displace 11.2 L of H2 gas is equivalent mass.
 11.2 L of H2 (g) have mass = 28 g

28
0.7 L of H2 have mass =  0.7 = 1.75 g
11.2

15. Number of Fe atoms in 100 g Haemoglobin if it contains 0.33% Fe. (Atomic mass of Fe = 56)
(1) 0.035 × 1023 (2) 35 (3) 3.5 × 1023 (4) 7 × 108
Sol. Answer (1)

0.33
Mass of Fe = 100 × = 0.33 g
100

0.33
 Moles of Fe = = 5.89 × 10–3 mole
56

 Number of atom of Fe = 5.89 × 10-3 × 6.022 × 1023 = 0.035 × 1023 atom

16. An organic compound containing C and H gave the following analysis C = 40%, H = 6.7%. Its empirical formula
would be
(1) CH4 (2) CH2O
(3) C2H4O2 (4) C2H4
Sol. Answer (2)
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Solution of Assignment Some Basic Concepts of Chemistry 11

% age Atomic mass Moles Simple ratio

40
C 40% 12  3.33 1
12
6.7
H 6.7% 1  6.7 2
1
53.3
O 53.3% 16  3.33 1
16

 EF  CH2O

17. The number of electrons in 1.6 g of CH4 is approximately

(1) 25 × 1024 (2) 1.5 × 1024 (3) 6 × 1023 (4) 3.0 × 1024
Sol. Answer (3)

1.6
Moles of CH4 = = 0.1 mol
16
Number of e– of CH4 = 0.1 × 10 × N0

= 6  1023

18. 6.025 × 1020 molecules of acetic acid are present in 500 ml of its solution. The concentration of solution is
(1) 0.002 M (2) 10.2 M (3) 0.012 M (4) 0.001 M
Sol. Answer (1)

6.022  1020
Moles of oxalic acid = = 10–3 moles
6.022  1023

10 3
Molarity =  1000
500

= 2  103 M

= 0.002 M

19. How many litre of oxygen at STP is required to burn 60 g C2H6?

(1) 22.4 L (2) 11.2 L (3) 22.4 × 7 L (4) 8.5 L
Sol. Answer (3)

2C2H6  7O2  4 CO2  6 H2O

 
2 moles 7 moles

For 2 moles of C2H6 = 7 moles of O2 required.

i.e., for 60 g of C2H6 7  22.4 L of O2 at STP required

20. For the formation of 3.65 g of HCl gas, what volume of hydrogen gas and chlorine gas are required at NTP
conditions?
(1) 1 L, 1 L (2) 1.12 L, 2.24 L (3) 3.65 L, 1.83 L (4) 1.12 L, 1.12 L

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12 Some Basic Concepts of Chemistry Solution of Assignment

Sol. Answer (4)

H2 (g)  Cl2 (g)  2HCl (g)

1mol 1mol 36.5 g  2
22.4 L 22.4 L 36.5 g  2
1.12 L 1.12 L 3.65 g
For (36.5 × 2) g of HCl volume of H2 and Cl2 required will be 22.4 L H2 and 22.4 L of CL2

 For 3.65 g and 1.12 L of H2 ⎫

⎬
1.12 L of Cl2 ⎭

21. Specific volume of cylindrical virus particle is 6.02 × 10–2 cc/gm whose radius and length are 7 Å and 10 Å
respectively. If NA = 6.02 × 1023, find molecular weight of virus.
(1) 15.4 kg/mol (2) 1.54 × 104 kg/mol
(3) 3.08 × 104 kg/mol (4) 3.08 × 103 kg/mol
Sol. Answer (1)
One molecule (gm)
M.wt.
d
V

1 M.wt.
2

6.02  10 r 2  h

  (7  10 8 )2  10  10 8
M.Wt. (One molecule in gm) =
6.02  10 2

22 7  7  6.02  10 3
M.Wt. (One mole in kg) =  = 2.2 × 7 = 15.4 kg mol–1
7 6.02  10 2
22. The crystalline salt Na2SO4.xH2O on heating loses 55.9% of its mass ad becomes anhydrous. The formula
of crystalline salt is
(1) Na2SO4.5H2O (2) Na2SO4.7H2O
(3) Na2SO4.2H2O (4) Na2SO4.10H2O
Sol. Answer (4)

SECTION – C
Previous Years Questions

1. Suppose the elements X and Y combine to form two compounds XY2 and X3Y2. When 0.1 mole of XY2 weighs
10 g and 0.05 mole of X3Y2 weighs 9 g, the atomic weights of X and Y are [NEET-Phase-2-2016]
(1) 40, 30 (2) 60, 40 (3) 20, 30 (4) 30, 20
Sol. Answer (1)
For XY2,
∵ 0.1 mole XY2  10 g
 1 mole XY2  100 g
and X + 2Y = 100 …(i)

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Solution of Assignment Some Basic Concepts of Chemistry 13
For X3Y2,
∵ 0.05 mole X3Y2  9 g
 1 mole X3Y2  180 g
and 3X + 2Y = 180 …(ii)
On solving,
X = 40
and Y = 30

2. What is the mass of the precipitate formed when 50 mL of 16.9% solution of AgNO3 is mixed with
50 mL of 5.8% NaCl solution? (Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5) [Re-AIPMT-2015]
(1) 7 g (2) 14 g
(3) 28 g (4) 3.5 g
Sol. Answer (1)

50  16.9

nAgNO3  0.05 Mole
100  169.8

50  5.8
nNaCl  0.05 Mole
100  58.5

 AgNO3 + NaCl 
 AgCl + NaNO3
t = 0; 0.05 mole 0.05 mole 0
t = t; 0 0 0.05 mole
 Mass of AgCl = 0.05  143.3
= 7.16
7g

3. If Avogadro number NA, is changed from 6.022 × 1023 mol–1 to 6.022 × 1020 mol–1, this would change
(1) The ratio of chemical species to each other in a balanced equation [Re-AIPMT-2015]
(2) The ratio of elements to each other in a compound
(3) The definition of mass in units of grams
(4) The mass of one mole of carbon
Sol. Answer (4)
Fact.

4. 20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium
oxide. What will be the percentage purity of magnesium carbonate in the sample? (At. wt. : Mg = 24)
[Re-AIPMT-2015]
(1) 60 (2) 84
(3) 75 (4) 96
Sol. Answer (2)

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14 Some Basic Concepts of Chemistry Solution of Assignment


MgCO3  MgO  CO2
84 g 40 g
xg 8g

84  8

 x  16.8 g
40
16.8
 % purity of MgCO3 =  100
20
= 84%

5. A mixture of gases contains H2 and O2 gases in the ratio of 1 : 4 (w/w). What is the molar ratio of the two
gases in the mixture? [AIPMT-2015]
(1) 2 : 1 (2) 1 : 4
(3) 4 : 1 (4) 16 : 1
Sol. Answer (3)

nH2 12 32 4
=  
nO2 4 32 2  4 1

6. 1.0 g of magnesium is burnt with 0.56 g O2 in a closed vessel. Which reactant is left in excess and how much?
(At. wt. Mg = 24; O = 16) [AIPMT-2014]
(1) Mg, 0.16 g (2) O2, 0.16 g
(3) Mg, 0.44 g (4) O2, 0.28 g
Sol. Answer (1)

2Mg  O2 
 2MgO
2 × 24 g 32 g

∵ O2 is limiting reagent

2  24
 32 g of O2 will react with, g
32
2  24  0.56
 0.56 g of O2 will react with, = 0.84 g
32
 Excess of Mg = 1 – 0.84 = 0.16 g

7. When 22.4 litres of H2(g) is mixed with 11.2 litres of Cl2(g), each at STP, the moles of HCl(g) formed is equal to
[AIPMT-2014]
(1) 1 mol of HCl(g) (2) 2 mol of HCl(g)
(3) 0.5 mol of HCl(g) (4) 1.5 mol of HCl(g)
Sol. Answer (1)

H2  Cl2 
 2HCl

nCl2 nHCl 11.2 nHCl

 
1 2 22.4 2
nHCl  1

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Solution of Assignment Some Basic Concepts of Chemistry 15
8. 6.02 × 1020 molecules of urea are present in 100 mL of its solution. The concentration of solution is
[NEET-2013]
(1) 0.01 M (2) 0.001 M (3) 0.1 M (4) 0.02 M
Sol. Answer (1)

6.02  1020
Moles of urea =  10 3 moles
6.02  1023

moles 10 3
 Molarity =  1000   1000 = 10–2 M = 0.01 M
volume (mL) 100

9. How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2 M HNO3? The
concentrated acid is 70% HNO3? [NEET-2013]
(1) 90.0 g conc. HNO3 (2) 70.0 g conc. HNO3
(3) 54.0 g conc. HNO3 (4) 45.0 g conc. HNO3
Sol. Answer (4)

nHNO3
∵ M=  1000 ,
VmL

2  250
 nHNO3 =  0.5 mole
1000

0.5  63  100
 Mass of concentrated acid required = = 45 g
70

10. Mole fraction of the solute in a 1.00 molal aqueous solution is [AIPMT (Prelims)-2011]
(1) 1.7700 (2) 0.1770
(3) 0.0177 (4) 0.0344
Sol. Answer (3)

1
solute = = 0.0177
1  55.55

11. Which has the maximum number of molecules among the following? [AIPMT (Mains)-2011]
(1) 8 g H2 (2) 64 g SO2
(3) 44 g CO2 (4) 48 g O3
Sol. Answer (1)
Number of molecules in

8 64
H2 = N = NA, SO2 = N = NA
2 A 64 A

44 48
CO2 = N = NA, O3 = N = NA
44 A 48 A

 Maximum number of molecules is present in 8 g H2

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16 Some Basic Concepts of Chemistry Solution of Assignment

12. The number of atoms in 0.1 mol of a triatomic gas is (NA = 6.02 × 1023 mol–1) [AIPMT (Prelims)-2010]
(1) 6.026 × 1022 (2) 1.806 × 1023
(3) 3.600 × 1023 (4) 1.800 × 1022
Sol. Answer (2)
Number of atoms = 3 × 0.1 × 6.022 × 1023 = 1.806 × 1023

13. 25.3 g of sodium carbonate, Na2CO3 is dissolved in enough water to make 250 mL of solution. If sodium
carbonate dissociates completely, molar concentration of sodium ion, Na+ and carbonate ions, CO32 are
respectively (Molar mass of Na2CO3 = 106 g mol–1) [AIPMT (Prelims)-2010]
(1) 0.955 M and 1.910 M (2) 1.910 M and 0.955 M
(3) 1.90 M and 1.910 M (4) 0.477 M and 0.477 M
Sol. Answer (2)

moles of Na2 CO3

Molarity of Na2CO3 =  1000
volume (mL)

25.3
Moles of Na2CO3 =
106

25.3
25.3
 Molarity of Na2CO3 = 106  1000 =  4 = 0.955 M
250 106

 2Na  CO32

Na2 CO3 

As one mole Na2CO3 gives 2 mol Na and 1 mol CO32



 Molarity of Na = 2 × molarity of Na2CO3 = 2  0.955 

1.910 M

Molarity of CO32 = 1 × molarity of Na2CO3 = 0.955 M.

14. 10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in
this reaction will be [AIPMT (Prelims)-2009]
(1) 3 mol (2) 4 mol
(3) 1 mol (4) 2 mol
Sol. Answer (2)

 2H2 (g)  O2 (g) 

 2H2 O(l)
4g 32 g 2 mole
8g 64 g 4 mole

Amount of water produced = 4 mol

15. How many moles of lead (II) chloride will be formed from a reaction between 6.5g of PbO and 3.2 g of HCl?
[AIPMT (Prelims)-2008]
(1) 0.029 (2) 0.044 (3) 0.333 (4) 0.011

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Solution of Assignment Some Basic Concepts of Chemistry 17
Sol. Answer (1)

PbO  2HCl  PbCl2  H2 O

224 g 73 g 1 mol

As 73 g HCl reacts with 224 g of PbO

224
 On reaction for 3.2 g of HCl with PbO the required amount of PbO = × 3.2 = 9.81 g
73
But only 6.5 g PbO is present
 PbO will be the LR and calculation will be according to PbO.
224 g of PbO gives = 1 mol PbCl2

1
6.5 g of PbO gives = × 6.5 = 0.029 mole
224

16. Volume occupied by one molecule of water (density = 1 g cm–3) is [AIPMT (Prelims)-2008]
–23 3 –23 3
(1) 5.5 × 10 cm (2) 9.0 × 10 cm
(3) 6.023 × 10–23 cm3 (4) 3.0 × 10–23 cm3
Sol. Answer (4)
∵ 18 g H2O  18 mL

18
 1 molecule   3.0 × 10–23 cm3
6.022  1023

17. What volume of oxygen gas (O2) measured at 0C and 1 atm, is needed to burn completely 1 L of propane
gas (C3H8) measured under the same conditions? [AIPMT ((Prelims)-2008]
(1) 10 L (2) 7 L (3) 6 L (4) 5 L
Sol. Answer (4)

C3 H8 (g)  5 O2 (g) 
 3 CO2 (g)  4H2 O(l)
1L 5L

18. An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H,
9.67%. The empirical formula of the compound would be [AIPMT (Prelims)-2008]
(1) CH4O (2) CH3O (3) CH2O (4) CHO
Sol. Answer (2)

Element %/At. wt. Simplest Ratio No. of Atoms

C (38.71) (3.22)
12 3.22 = 1 1

H (9.67) (9.67)
1 3.22 = 3 3

O (51.62) (3.22)
16 3.22 = 1 1

Empirical formula = CH3O

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18 Some Basic Concepts of Chemistry Solution of Assignment

200 199 202

19. An element, X has the following isotopic composition ; X : 90% ; X : 8.0% ; X : 2.0%
The weighted average atomic mass of the naturally occurring element X is closest to :
[AIPMT (Prelims)-2007]
(1) 199 amu (2) 200 amu
(3) 201 amu (4) 202 amu
Sol. Answer (2)
200  90  199  8  202  2
Average atomic mass = = 199.96  200 amu
100

20. Concentrated aqueous sulphuric acid is 98% H2SO4 by mass and has a density of 1.80 g mL–1. Volume of acid
required to make one litre of 0.1 M H2SO4 is [AIPMT (Prelims)-2007]
(1) 5.55 mL (2) 11.10 mL
(3) 16.65 mL (4) 22.20 mL
Sol. Answer (1)
10 d 10  98  1.8

M=  18M
MB 98
 M1V1 = M2V2  18 × V1 = 0.1 × 1000
100
 V1 = = 5.55 mL
18

21. How many grams of CH3OH should be added to water to prepare 150 ml solution of 2 M CH3OH?
(1) 9.6 × 103 (2) 2.4 × 103
(3) 9.6 (4) 2.4
Sol. Answer (3)
M  V mL 2  150
Moles of CH3OH =  = 0.3 mole
1000 1000
 weight of CH3OH = moles × mol. mass
= 0.3  32 
9.6 g

22. The total number of valence electrons in 4.2 g of N3– ion is (NA is the Avogadro’s number)
(1) 2.1 NA (2) 4.2 NA
(3) 1.6 NA (4) 3.2 NA
Sol. Answer (3)
4.2
Moles of N3 = = 0.1 mol
42
–
 Total number of valence electrons in N3 = 0.1 NA × 16 = 1.6 NA

23. The number of mole of oxygen in one litre of air containing 21% oxygen by volume, under standard conditions,
is
(1) 0.0093 mole (2) 2.10 moles
(3) 0.186 mole (4) 0.21 mole

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Solution of Assignment Some Basic Concepts of Chemistry 19
Sol. Answer (1)

21
Volume of O2 = × 1000 = 210 mL
100

10
Moles of O2 at STP = = 0.0093 mole
22400

24. The amount of zinc required to produce 224 ml of H2 at STP on treatment with dilute H2SO4 will be (Zn = 65)
(1) 65 g (2) 0.065 g
(3) 0.65 g (4) 6.5 g
Sol. Answer (3)

Zn  H2 SO 4  ZnSO 4  H2
1moL 22400 mL at STP

22400 mL of H2 is produced by 1 mol Zn i.e., = 65 g

65
224 mL of H2 is produced by 1 mol Zn i.e., = × 224 = 0.65 g
22400

25. Given the numbers : 161 cm, 0.161 cm, 0.0161 cm. The number of significant figures for the three numbers is
(1) 3, 3 and 4 respectively (2) 3, 4 and 4 respectively
(3) 3, 4 and 5 respectively (4) 3, 3 and 3 respectively
Sol. Answer (4)
All have same significant figures.

26. Change in volume when 100 mL PH3 decomposed to solid phosphorus and H2 gas.
(1) Increase in 50 mL (2) Decrease in 50 mL
(3) Increase in 150 mL (4) Decrease in 200 mL
Sol. Answer (1)

4 PH3 (g) 
P4 (s)  6H2 (g)
4 moles 6 moles

As 4 moles of PH3 (g) converts into 6 moles H2 (g)

 4 × 22.4 L of PH3 (g) will produce = 6 × 22.4 L of H2

89.6 L of PH3 (g) will produce = 134.4 L

134.4
1 L of PH3 (g) will produce = = 1.5 L
89.6

134.4
100 mL of PH3 (g) will produce = × 100 = 150 mL
89.6

 Increase in volume wil be 50 mL

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20 Some Basic Concepts of Chemistry Solution of Assignment

27. In the reaction,

4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l )
When 1 mole of ammonia and 1 mole of O2 are made to react to completion
(1) All the oxygen will be consumed (2) 1.0 mole of NO will be produced
(3) 1.0 mole of H2O is produced (4) All the ammonia will be consumed
Sol. Answer (1)

4NH3 (g)  5O 2 (g)  4NO (g)  6H2 O (l)

4 mol NH3 reacts with 5 mol O2

5
1 mol NH3 reacts with = 1.25 mol of O2
4

as 1 mol of O2 is taken therefore all the O2 will be consumed.

28. An organic compound containing C, H and N gave the following analysis C = 40%, H = 13.33%,
N = 46.67%. Its empirical formula would be
(1) CH4N (2) CH5N
(3) C2H7N2 (4) C2H7N
Sol. Answer (1)

Percentage At mass Moles Simple ratio

40 3.33
C 40% 12  3.33 1
12 3.33
6.7 13.33
H 6.7% 1  6.7 4
1 3.33
53.3 3.33
O 53.3% 16  3.33 1
16 3.33

 Empirical formula = CH4N

29. How many g of dibasic acid (mol. weight 200) should be present in 100 ml. of the aqueous solution to give
strength of 0.1 N?
(1) 10 g (2) 2 g
(3) 1 g (4) 20 g
Sol. Answer (3)

N  E  VmL
Amount of acid (in gram) =
1000

200
Here, E = = 100
2

0.1  100  100

 Amount of acid = =1g
1000

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Solution of Assignment Some Basic Concepts of Chemistry 21
30. The number of atoms in 4.25 g of NH3 is approximately
(1) 4 × 1023 (2) 2 × 2023
(3) 1 × 1023 (4) 6 × 1023
Sol. Answer (4)

4.25
moles of NH3 = = 0.25 mol
17
As 1 NH3 have 4 atoms

 Total number of atoms = 0.25  4  6.022  1023 atoms

 6.022  1023 atoms

31. Volume of CO 2 obtained at STP by the complete decomposition of 9.85 gm BaCO3 is (Mol. wt. of
BaCO3 = 197)
(1) 2.24 litre (2) 1.12 litre
(3) 0.85 litre (4) 0.56 litre
Sol. Answer (3)

BaCO3   BaO  CO2
1mol 22.4 L

mol mass of BaCO3 197 + 12 + 48 = 257 g

9.85
moles of BaCO3 = g = 0.038 mol
257
1 mol of BaCO3 gives CO2 = 22.4 L
0.038 moL of BaCO3 gives CO2 = 22.4 × 0.038 = 0.85 L
32. Percentage of Se in peroxidase anhydrous enzyme is 0.5% by weight (at. wt. = 78.4) then minimum molecular
weight of peroxidase anhydrous enzyme is
(1) 1.568 × 104 (2) 1.568 × 103
(3) 15.68 (4) 2.136 × 104
Sol. Answer (1)
0.5% of Se by weight is present
 0.5% of enzyme have weight = 78.4 g

78.4
100% of enzyme have wt = × 100
0.5

= 15680 g = 1.568 × 104 g

33. 2.5 litre of 1 M NaOH solution mixed with another 3 litre of 0.5 M NaOH solution. Then find out molarity of
resultant solution.
(1) 0.80 M (2) 1.0 M
(3) 0.73 M (4) 0.50 M

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22 Some Basic Concepts of Chemistry Solution of Assignment

Sol. Answer (3)

for same solution


M1V1  M2 V2  M3 (V1  V2 )

(2.5 × 1) + (3 × 0.5) = M3 (2.5 + 3)

+ =

M1 V1 M2 V2 M3 (V1 + V2)

2.5 + 1.5 = M3 × 5.5

4
M3 = = 0.727  0.73 M
5.5

34. Which has maximum molecules?

(1) 7 gm N2 (2) 2 gm H2 (3) 16 gm NO2 (4) 16 gm O2
Sol. Answer (2)
Maximum number of molecules of gas are present which are having maximum number of moles

7 16
for N2 = = 0.28 mol, for NO2 = = 0.34 mol
28 46

2 16
for H2 = = 1 mol, for O2 = = 0.5 mol
2 32

As H2 have maximum number of moles

 H2 will have maximum number of molecules.

35. In Haber process 30 litres of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only
50% of the expected product. What will be the composition of gaseous mixture under the aforesaid condition
in the end?
(1) 20 litres ammonia, 20 litres nitrogen, 20 litres hydrogen
(2) 10 litres ammonia, 25 litres nitrogen, 15 litres hydrogen
(3) 20 litres ammonia, 10 litres nitrogen, 30 litres hydrogen
(4) 20 litres ammonia, 25 litres nitrogen, 15 litres hydrogen
Sol. Answer (2)

N2  3H2 
 2NH3
at STP 10 L 30 L 20 L
for 30 L 30 L

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Solution of Assignment Some Basic Concepts of Chemistry 23
At STP 10 L of N2 reacts with 30 L of H2. gives 20 L NH3
H2 will be the LR because 30 L N2 and 30 L H2 are taken
Now the expected product will be = 50% i.e. 10 L NH3

N2  3H2 
 2NH3 Final conc n
of NH3 
10 L because 50% of product

⎪⎧Final conc of N2  (30  5)  25 L because 10 L

n
 Initial conc n 30 L 30 L 0 ⎨
⎪⎩ of NH3 is formed from 5 L of N2

⎪⎧Final conc of H2  (30  15)  15 L because 10 L

n

Final conc n (30  5 L) (30  15 L) 10 L ⎨

⎪⎩ of NH3 is formed from 15 L of H2

36. The maximum number of molecules is present in

(1) 15 L of water at STP (2) 15 L of H2O gas at STP
(3) 15 g of ice (4) Same in all
Sol. Answer (1)
As of H2O is 1 g/mL or 1 kg/L
 15 L of H2O  15 kg of H2O

15000
 nH2O = moles.
18

15
Moles of 15 L of H2O gas at STP = mole
22.4

15
Moles of 15 g of ice = mole
18
 Maximum number of moles are present in 15 L of H2O

37. Concentrated aqueous sulphuric acid is 98% H2SO4 (w/v) and has a density of 1.80 gmL–1. Molarity of solution
(1) 1 M (2) 1.8 M (3) 10 M (4) 1.5 M
Sol. Answer (3)

w
% 10
v 98  10
M=  = 10 M
mol. mass 98

38. An element, X has the following isotopic composition 56X : 90% 57X : 8% 59X : 2.0%. The weighted average
atomic mass of the naturally occurring element X is closest to
(1) 56.14 amu (2) 56.8 amu (3) 60 amu (4) 55 amu
Sol. Answer (1)

∑ percentage  atomic mass  ∑ percentage abundance of each  isotopic  atomic mass

100 100

 56  90    57  8   59  2
= = 56.14 amu
100

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24 Some Basic Concepts of Chemistry Solution of Assignment

39. 10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Volume of gaseous product
after reaction
(1) 1 × 22.4 L (2) 2 × 22.4 L
(3) 3 × 22.4 L (4) 4 × 22.4 L
Sol. Answer (1)

1
H2 (g)  O2 (g)  H2O (l)
2
10 g 64 g
From the reaction 2 g of H2 (g) combine with 16 g of O2
2
 64 g of O2 will combine with H2 = × 64 = 8 g of H2(g)
16
After the reaction 2 g of H2 (g) i.e., 1 mol of H2 (g) will remain unreacted
 volume will be 1 × 22.4 L gaseous product.

Note: Volume of H2O will not be considered as only volume of gas is asked in the question.

40. What is the [OH–] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of 0.10 M
Ba(OH)2?
(1) 0.12 M (2) 0.10 M
(3) 0.40 M (4) 0.0050 M
Sol. Answer (2)
It is neutralisation reaction
HCl Ba(OH)2
20 mL × 0.05 M 30 mL × 0.1 M
or 20 mL × 0.05 N or 30 mL × 0.2 N
meq of H = 1 meq of OH(–) = 6

6 1 5
 [OH(–)] =   0.1M
50 50

41. The number of atoms in 0.1 mol of a triatomic gas is (NA = 6.02 × 1023 mol–1)
(1) 1.800 × 1022 (2) 6.026 × 1022
(3) 1.806 × 1023 (4) 3.600 × 1023
Sol. Answer (3)
As triatomic gas means 3 atoms are present in a molecule
Number of atoms = 0.1 × 3 × 6.022 × 1023
= 1.806 × 1023 atoms

42. The total number of electrons in 2.0 g of D2O to that in 1.8 g of H2O
(1) Double (2) Same
(3) Triple (4) One fourth

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Solution of Assignment Some Basic Concepts of Chemistry 25
Sol. Answer (2)
Both have same number of e– [Both 2.0 g D2O and 1.8 g H2O have same number of atom]

2.0
Moles of D2O = = 0.1
20
Number of e– = 0.1 × 10 × N0 = 1 × N0

1.8
Moles of H2O = = 0.1
18

Number of e– = 0.1 × 10 × N0 = 1 × N0

43. From 200 mg of CO2 when x molecules are removed, 2.89 × 10–3 moles of CO2 are left. x will be

(1) 1020 molecules (2) 1010 molecules

(3) 21 molecules (4) 1021 molecules

Sol. Answer (4)

– =

(200 mg of CO2) – x molecules = –3

2.89 × 10 moles of CO2

From Equation

200
200 mg of CO2 have molecule = × 10-3 × 6.022 × 1023
44
= 2.7 × 1021
 2.89 × 10-3 moles of CO2 have molecule = 2.89 × 10-3 × 6.022 × 1023
= 1.7 × 1021 molecule
 200 mg of CO2 – x molecule = 2.89 × 10–3 moles of CO2
2.7 × 1021 – x molecule = 1.7 × 1021
x = (2.7 – 1.7) × 1021 molecue
= 1021 molecule
 The value of x will be 1021.

44. If the weight of metal oxide is x g containing y g of oxygen, the equivalent weight of metal will be

8x 8(y  x)
(1) E  (2) E 
y x

y 8(x  y)
(3) E  (4) E 
8 y

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26 Some Basic Concepts of Chemistry Solution of Assignment

Sol. Answer (4)

weight of metal
Equivalent mass of metal in oxide = weight of oxygen  8

M  xO  MO A
(x  y) yg xg
xy
Equivalent weight of metal = 8
y

8x  y
E
y

45. The number of significant figures in 2.653 × 104 is

(1) 8 (2) 4 (3) 7 (4) 1
Sol. Answer (2)

2.653 × 104 have only 4 significant figure

46. Mole fraction of solute in aqueous solution of 30% NaOH.

(1) 0.16 (2) 0.05 (3) 0.25 (4) 0.95
Sol. Answer (1)
30% NaOH means 30% by mass

30 g of NaOH
i.e., 100 g of solution

Weight of NaOH = 30 g
Mol. mass NaOH = 40
Weight of H2O = 100 – 30 = 70
Mol. mass = 18
30
moles of NaOH 40
Mol. fraction of NaOH =   0.16
moles of H2 O  moles of NaOH 70 30

18 40

SECTION – D
Assertion–Reason Type Questions

1. A : 1 a.m.u. = 1.66 × 10–24 gram.

R : Actual mass of one atom of C–12 is equal to 1.99 × 10–23 g.
Sol. Answer (2)
Both are correct but R is not explanation of A because 1 amu = 1.66 × 10–24 g

12
and mass of 1 atom of C = 1.99 × 10–23 = = 1.99 × 10–23 = g
6.022  1023
Both A and R are correct.
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Solution of Assignment Some Basic Concepts of Chemistry 27
2. A : Unit of specific gravity is gram–cc–1.
R : Specific gravity is same as density of a liquid in normal conditions.
Sol. Answer (4)

density of substance
Specific gravity have number units becuase it specific gravity = density of H O at 4C
2

Both A and R are incorrect.

3. A : Number of atoms in 2 mole of NH3 is equal to number of atoms in 4 mole of CH4.
R : Both are chemically similar species.
Sol. Answer (4)
Number of atoms in NH3 = 2 × 4 × N0 = 8 N0
Number of atoms in CH4 = 4 × 5 × N0 = 20 N0
Both are chemically different
Both A and R are incorrect.
4. A : In the reaction

2NaOH  H3PO4  Na2HPO4  2H2O,

M
equivalent weight of H3PO4 is , where M is its molecular weight.
2

Molecular weight
R : Equivalent weight = .
n  factor

Sol. Answer (1)

2 NaOH  H3PO 4  Na2HPO 4  2H2 O

M
For above reaction n-factor = 2  equivalent mass 
2

As tow ‘H’ are replaced.

mol. mass
Equivalent mass =
n-factor

A and R are correct R is correct explanation of A.

5. A : Mass of 1 gram molecule of H2SO4 is 98 gram.
R : One gram atom contains NA atoms.
Sol. Answer (2)
mass of 1 g molecule means 1 mol of H2SO4 = 98 g
1 g atom = 1 mol atom = N0
Both are correct but R is not explanation of A.
6. A : One mole of sucrose reacts completely with oxygen produces 268.8 litre of carbon dioxide at STP.
R : Amount of oxygen required for reaction is 268.8 litre.

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28 Some Basic Concepts of Chemistry Solution of Assignment

Sol. Answer (2)

C12H22 O11  12O2 (g) 
 12 CO 2 (g)  11H2O
(Sucrose)

⎡1mol sucrose produce CO2 = 12  22.4 268.8 L ⎤

⎢ ⎥
⎣1mol sucrose requre O2 = 12  22.4 
268.8 L ⎦

Both are correct but reason is not correct explanation of Assertion.

7. A : In the reaction

2NaOH  H2SO4  Na2SO4  2H2O ,

equivalents of NaOH, Na2SO4 and H2SO4 are equal.
R : Number of equivalents = number of moles × n–factor.
Sol. Answer (1)
All are having same equivalents


2NaOH
 
 H 2 SO 4 Na 2 SO 4
2
 
1 2 1 2
n-factor = 1 2 2

Both A and R are correct R is correct explanation of A.

8. A : When 4 moles of H2 reacts with 2 moles of O2, then 4 moles of water is formed.
R : O2 will act as limiting reagent.
Sol. Answer (3)

2H2  O2 
 2H2 O
2 mol 1mol 2 mol [4 mol of H when reacts with 2 moles O produces moles of H O]
2 2 2
4 mol 2 mol 4 mol

When moles are taken in above reaction H2 will act on LR.

A is correct but R is wrong.

9. A : 50 ml, decinormal HCl when mixed with 50 ml, decinormal H2SO4, then normality of H+ ion in resultant
solution is 0.1 N.
R : Here, MV = M1V1 + M2V2.
Sol. Answer (3)
meq of HCl = 50 × 0.1 = 5
meq. of H2SO4 = 50 × 0.1 = 5

Total meq 10
Normality =   0.1N
Total volume 100

when difference solutions of different n-factors are takne than N1V1  N2 V2  N3 (V1  V2 )

A is correct but R is wrong.

10. A : 50 ml, decimolar H2SO4 when mixed with 50 ml, decimolar NaOH, then normality of resultant solution
is 0.05 N.
R : Here, NV = | N1V1 – N2V2 |.

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Solution of Assignment Some Basic Concepts of Chemistry 29
Sol. Answer (1)
Meq of H2SO4 = N × V = 50 × 0.2 = 10
Meq of NaOH = N × V = 50 × 0.1 = 5
N = m × n = 0.1 × 2 = 0.2 N H2SO4
N = m × n = 0.1 × 1 = 0.1 N NaOH

10  5 5
Nsolution =  = 0.05 N
100 100

larger NV - smaller NV N1V1  N2 V2

for neutralisation = 
Total volume Total volume
R is correct exp. of A.
11. A : Ratio of empirical formula mass and molecular formula mass may be a whole number.
R : Molecular formula mass = n × empirical formula mass, where n is the simplest whole number.
Sol. Answer (2)
= mol. formula = n × Emprical formula.
= mol. mass = n × formula mass (for ionic solids).
Both A and R are correct but R is not correct explanation of A
12. A : For a given solution (density = 1 gm/ml), molality is greater than molarity.
R : Molarity involves volume of solution while molality involves mass of solvent.
Sol. Answer (1)
If density u is 1 g/mL
 mass of solution > mass of solvent

moles
Molality = volume of solution

moles
Molality = mass of solvent

As weight of solvent is less therefore molality will be more.

Both A and R are correct R is the correct explanation of A.
13. A : 1 gram of salt in 1 m3 of solution has concentration of 1 ppm.
R : ppm is defined as number of parts by mass of solute per million parts of solution.
Sol. Answer (1)

weight of solvent 1
ppm = weight of solution  10 
6
106 = ppm  1 ppm
1

Both are A and R correct R is correct explanation of A.

14. A : Total charge on NA ions of CO32  is 1.93 × 105 coulomb.

R : Charge on one electron in 96500 coulomb.

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30 Some Basic Concepts of Chemistry Solution of Assignment

Sol. Answer (3)

CO32  as one CO32  ion have two unit charge

 1 mole i.e., = NA CO32  have charge = 2 × 96500 C = 1.93 × 105 C

Charge on one mole electron = 1.602 × 10-19 × 6.022 × 1023  96478 = 96500 C
A is true but R is false.

15. A : Number of ions in 9 gram of NH4 is equal to Avogadro's number (NA).

R : Number of ions is equal to number of atoms.

Sol. Answer (4)

9
Formula units = = 0.5 × NA
18

No formula units = moles = Number of molecules.

Both A and R are incorrect.

  

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