Question Bank (Solutions): Some Basic Concepts Of Chemistry | Chemistry Class 11 - NEET PDF Download

 Page 1


2 
 
 
1. (b) : Pressure ?
 Force
 
Area 
Therefore, dimensions of pressure ? ?
?
MLT
?2 
L
2 
= ML
–1
T
–2 
Therefore, vapour density 
? 
Weight of certain volume of substance 
Weight of same volume of hydrogen 
0.24  
and dimensions of energy per unit volume 
?
 Energy 
? 
ML
2
T
? 2
 
? ML
?1
T
?2
 
? 
4.005 ?10
?3 
? 59.93
 
Volume L
3
 6. (c) : If 1 L of one gas contains N molecules, 2 L of any 
2. (d) : Zeros placed left to the number are never 
significant, therefore the no. of significant figures for the 
numbers 161 cm , 0.161 cm and 0.0161 cm are sam e, i.e., 3. 
3. (c) : According to Avogadro’s hypothesis, ratio of 
the volumes of gases will be equal to the ratio of their no. 
gas under the same conditions will contain 2N molecules. 
7. (c) : C 2H 4 + 3O 2 ? 2CO 2 + 2H 2O 
28 g 96 g 
For complete combustion, 
2.8 kg of C 2H 4 requires ?
 96 
? 2.8 ?10
3 
g 
of moles. 
28 
3
 
So, no. of moles = 
   Mass 
 
= 9.6 × 10 g = 9.6 kg of O2 
Mol. mass 
8. (d) : Average isotopic mass of X 
n
H 
? 
w 
; n
O
 
?
 w 
; n
CH 
?
 w 
? 
200 ? 90 ?199 ? 8 ? 202 ? 2 
2 
2 
2 
32 
4 
16 90 ? 8 ? 2 
So, the ratio is 
w 
:
 w
 
:
 w 
or 16 :1: 2. 
? 
18000 ?1592 ? 404 
? 199.96 amu ? 200 amu 
2 32 16 
4. (a) : C 3H 8 + 5O 2 3CO 2 + 4H 2O 
1 vol. 5 vol. 3 vol. 4 vol. 
100 
9. (a) : Average atomicmass ??
19 ?10 ? 81 ?11 
100 
? 10.81 
According to the above equation, 10. (d) : 1 mole of substance = NA atoms 
1 vol. or 1 litre of propane requires 5 vol. or 5 litres of O 2
to burn completely. 
5. (b) : Weight of gas = 0.24 g, 
Volume of gas = 45 mL = 0.045 litre and density of 
108 g of Ag = NA atoms ??
24 g of Mg = NA atoms ??
1g of Ag ? 
N
A 
atoms 
108 
1g of Mg ? 
N
A 
atoms 
24 
H 2 = 0.089 g/L 32 g of O 2 = NA molecules = 2 NA atoms 
Weight of 45 mL of H 2 = density × volume 
= 0.089 × 0.045 = 4.005 × 10
–3 
g 
? 1g of O ? 
N
A 
atoms 
16 
Hints & Explanations 
Page 2


2 
 
 
1. (b) : Pressure ?
 Force
 
Area 
Therefore, dimensions of pressure ? ?
?
MLT
?2 
L
2 
= ML
–1
T
–2 
Therefore, vapour density 
? 
Weight of certain volume of substance 
Weight of same volume of hydrogen 
0.24  
and dimensions of energy per unit volume 
?
 Energy 
? 
ML
2
T
? 2
 
? ML
?1
T
?2
 
? 
4.005 ?10
?3 
? 59.93
 
Volume L
3
 6. (c) : If 1 L of one gas contains N molecules, 2 L of any 
2. (d) : Zeros placed left to the number are never 
significant, therefore the no. of significant figures for the 
numbers 161 cm , 0.161 cm and 0.0161 cm are sam e, i.e., 3. 
3. (c) : According to Avogadro’s hypothesis, ratio of 
the volumes of gases will be equal to the ratio of their no. 
gas under the same conditions will contain 2N molecules. 
7. (c) : C 2H 4 + 3O 2 ? 2CO 2 + 2H 2O 
28 g 96 g 
For complete combustion, 
2.8 kg of C 2H 4 requires ?
 96 
? 2.8 ?10
3 
g 
of moles. 
28 
3
 
So, no. of moles = 
   Mass 
 
= 9.6 × 10 g = 9.6 kg of O2 
Mol. mass 
8. (d) : Average isotopic mass of X 
n
H 
? 
w 
; n
O
 
?
 w 
; n
CH 
?
 w 
? 
200 ? 90 ?199 ? 8 ? 202 ? 2 
2 
2 
2 
32 
4 
16 90 ? 8 ? 2 
So, the ratio is 
w 
:
 w
 
:
 w 
or 16 :1: 2. 
? 
18000 ?1592 ? 404 
? 199.96 amu ? 200 amu 
2 32 16 
4. (a) : C 3H 8 + 5O 2 3CO 2 + 4H 2O 
1 vol. 5 vol. 3 vol. 4 vol. 
100 
9. (a) : Average atomicmass ??
19 ?10 ? 81 ?11 
100 
? 10.81 
According to the above equation, 10. (d) : 1 mole of substance = NA atoms 
1 vol. or 1 litre of propane requires 5 vol. or 5 litres of O 2
to burn completely. 
5. (b) : Weight of gas = 0.24 g, 
Volume of gas = 45 mL = 0.045 litre and density of 
108 g of Ag = NA atoms ??
24 g of Mg = NA atoms ??
1g of Ag ? 
N
A 
atoms 
108 
1g of Mg ? 
N
A 
atoms 
24 
H 2 = 0.089 g/L 32 g of O 2 = NA molecules = 2 NA atoms 
Weight of 45 mL of H 2 = density × volume 
= 0.089 × 0.045 = 4.005 × 10
–3 
g 
? 1g of O ? 
N
A 
atoms 
16 
Hints & Explanations 
46 
Mol. wt. 
N
A 
32 
3 
Some Basic Concepts of Chemistry  
7 g of Li = NA atoms ??
1g of Li ? 
N
A 
atoms 
7 
5 L N 2 
?
?
6.023 ?10
23 
? 5 
22.4 
? 1.344 ?10
23
 
molecules 
Therefore, 1 g of Li(s), has maximum number of atoms. 
2 g H 2 = 6.023 × 10
23 
molecules 
11.   (a) : (a) Mass of water = V × d = 18 × 1 = 18 g 
Molecules of water = mole × NA ? 
18 
N
A 
? N
A
 
0.5 g H2 ? 
6.023 ?10
23 
? 0.5 
? 1.505 ?10
23 
molecules
2 
18 
(b) Molecules of water = mole × NA ??
0.18 
N
18 
32 g O 2 = 6.023 × 10
23 
molecules 
6.023 ?10
23 
?10 
23
 
= 10
–2 
NA 
10 g of O 2 ? ?
32
 
? 1.882 ?10 
molecules 
(c) Moles of water ?
 0.00224 
? 10
?4
 19. (b) : Number of molecules = moles × NA 
22.4 
–4 
Molecules of N 2 =
 7 
NA = 0.5 NA 
Molecules of water = mole × NA = 10 NA
 
14
 
(d) Molecules of water = mole × NA = 10
–3 
NA 
12. (a) : Let atomic weight of element X is x and that of 
Molecules of H 2 = NA 
element Y is y. 
For XY , n ?
??w 
 
Molecules of NO 2 = 
16
 NA = 0.35 NA 
2 
  10  
Mol. wt. 
10 
Molecules of O2 
16 
= 0.5 NA 
0.1 ? 
x ? 2 y 
? x + 2y = 
For X3 Y 2, n ?
??w 
 
? 100 
0.1  
...(i) 
? 2 g H 2 (1 mole H 2) contains maximum molecules. 
20. (a) : Specific volume (vol. of 1 g) of cylindrical virus 
particle = 6.02 × 10
–2 
cc/g 
0.05 ?
??9 
 
3x ? 2 y 
? 3x ? 2 y ?
 9 
? 180 
0.05 
...(ii) 
Radius of virus, r = 7 Å = 7 × 10
–8 
cm 
Volume of virus = ?r
2
l 
On solving equations (i) and (ii), we get x = 40 
40 + 2y = 100 ? 2y = 60 ? y = 30 
13. (a) : Mass of 1 mol (6.022 × 10
23 
atoms) of carbon 
= 12 g 
If Avogadro number is changed to 6.022 × 10
20 
atoms 
then mass of 1 mol of carbon 
? 
22 
? (7 ?10
?8 
)
2 
?10 ?10
?8 
= 154 × 10
–23 
cc 
7 
wt. of  one virus particle ?
??Volume (cc) 
 
Specific volume (cc/g) 
? 
154 ?10
?23 
g
 
6.02 ?10
?2
 
? 
12 ? 6.022 ?10
20 
? 12 ?10
?3 
g
 
? Molecular wt. of virus = wt. of N particles 
6.022 ?10
23
 
A 
?23 
14. (c) : 1.8 gram of water ? 
6.023 ?10
23 
?1.8 
18 
= 6.023 × 10
22 
molecules 
18 gram of water = 6.023 × 10
23 
molecules 
18 moles of water = 18 × 6.023 × 10
23 
molecules 
15. (d) : Number of moles of H 2 = 1/2 
? 
154 ?10 
? 6.02 ?10
23 
g / m ol 
6.02 ?10
?2
 
= 15400 g/mol = 15.4 kg/mol 
21. (d) : 17 g of NH 3 = 4NA atoms 
4.25 g of NH ? 
4N
A 
? 4.25 atoms
17 
Number of moles of O2 ?
 4  
32 
= NA atoms = 6 × 10
23 
atoms 
22. (a) :  Quantity of iron in one molecule 
Hence, molar ratio ? 
1 
:
 4 
? 4 : 1 
= 
67200 
× 0.334 = 224.45 amu 
16. (c) : 
2 32 
100 
each. 
8 g H 2 has 4 moles while the others has 1 mole 
No. of iron atoms in one molecule of haemoglobin 
17. (b) : No. of atoms = N A × No. of moles × 3 
23 23 
? 
224.45 
? 4
 
56 
= 6.023 × 10 × 0.1 × 3 = 1.806 × 10 
23. (a) : Volume of oxygen in one litre of air 
18. (a) : At STP, 22.4 L = 6.023 × 10
23 
molecules 
?
 21 
?1000 ? 210 mL 
15 L H 
? 
6.023 ?10
23 
?15 
? 4.033 ?10
23 
molecules 
100 
  210  
2
 22.4 
Therefore, no. of moles = 
22400 
= 0.0093 mol 
= 
A 
Page 3


2 
 
 
1. (b) : Pressure ?
 Force
 
Area 
Therefore, dimensions of pressure ? ?
?
MLT
?2 
L
2 
= ML
–1
T
–2 
Therefore, vapour density 
? 
Weight of certain volume of substance 
Weight of same volume of hydrogen 
0.24  
and dimensions of energy per unit volume 
?
 Energy 
? 
ML
2
T
? 2
 
? ML
?1
T
?2
 
? 
4.005 ?10
?3 
? 59.93
 
Volume L
3
 6. (c) : If 1 L of one gas contains N molecules, 2 L of any 
2. (d) : Zeros placed left to the number are never 
significant, therefore the no. of significant figures for the 
numbers 161 cm , 0.161 cm and 0.0161 cm are sam e, i.e., 3. 
3. (c) : According to Avogadro’s hypothesis, ratio of 
the volumes of gases will be equal to the ratio of their no. 
gas under the same conditions will contain 2N molecules. 
7. (c) : C 2H 4 + 3O 2 ? 2CO 2 + 2H 2O 
28 g 96 g 
For complete combustion, 
2.8 kg of C 2H 4 requires ?
 96 
? 2.8 ?10
3 
g 
of moles. 
28 
3
 
So, no. of moles = 
   Mass 
 
= 9.6 × 10 g = 9.6 kg of O2 
Mol. mass 
8. (d) : Average isotopic mass of X 
n
H 
? 
w 
; n
O
 
?
 w 
; n
CH 
?
 w 
? 
200 ? 90 ?199 ? 8 ? 202 ? 2 
2 
2 
2 
32 
4 
16 90 ? 8 ? 2 
So, the ratio is 
w 
:
 w
 
:
 w 
or 16 :1: 2. 
? 
18000 ?1592 ? 404 
? 199.96 amu ? 200 amu 
2 32 16 
4. (a) : C 3H 8 + 5O 2 3CO 2 + 4H 2O 
1 vol. 5 vol. 3 vol. 4 vol. 
100 
9. (a) : Average atomicmass ??
19 ?10 ? 81 ?11 
100 
? 10.81 
According to the above equation, 10. (d) : 1 mole of substance = NA atoms 
1 vol. or 1 litre of propane requires 5 vol. or 5 litres of O 2
to burn completely. 
5. (b) : Weight of gas = 0.24 g, 
Volume of gas = 45 mL = 0.045 litre and density of 
108 g of Ag = NA atoms ??
24 g of Mg = NA atoms ??
1g of Ag ? 
N
A 
atoms 
108 
1g of Mg ? 
N
A 
atoms 
24 
H 2 = 0.089 g/L 32 g of O 2 = NA molecules = 2 NA atoms 
Weight of 45 mL of H 2 = density × volume 
= 0.089 × 0.045 = 4.005 × 10
–3 
g 
? 1g of O ? 
N
A 
atoms 
16 
Hints & Explanations 
46 
Mol. wt. 
N
A 
32 
3 
Some Basic Concepts of Chemistry  
7 g of Li = NA atoms ??
1g of Li ? 
N
A 
atoms 
7 
5 L N 2 
?
?
6.023 ?10
23 
? 5 
22.4 
? 1.344 ?10
23
 
molecules 
Therefore, 1 g of Li(s), has maximum number of atoms. 
2 g H 2 = 6.023 × 10
23 
molecules 
11.   (a) : (a) Mass of water = V × d = 18 × 1 = 18 g 
Molecules of water = mole × NA ? 
18 
N
A 
? N
A
 
0.5 g H2 ? 
6.023 ?10
23 
? 0.5 
? 1.505 ?10
23 
molecules
2 
18 
(b) Molecules of water = mole × NA ??
0.18 
N
18 
32 g O 2 = 6.023 × 10
23 
molecules 
6.023 ?10
23 
?10 
23
 
= 10
–2 
NA 
10 g of O 2 ? ?
32
 
? 1.882 ?10 
molecules 
(c) Moles of water ?
 0.00224 
? 10
?4
 19. (b) : Number of molecules = moles × NA 
22.4 
–4 
Molecules of N 2 =
 7 
NA = 0.5 NA 
Molecules of water = mole × NA = 10 NA
 
14
 
(d) Molecules of water = mole × NA = 10
–3 
NA 
12. (a) : Let atomic weight of element X is x and that of 
Molecules of H 2 = NA 
element Y is y. 
For XY , n ?
??w 
 
Molecules of NO 2 = 
16
 NA = 0.35 NA 
2 
  10  
Mol. wt. 
10 
Molecules of O2 
16 
= 0.5 NA 
0.1 ? 
x ? 2 y 
? x + 2y = 
For X3 Y 2, n ?
??w 
 
? 100 
0.1  
...(i) 
? 2 g H 2 (1 mole H 2) contains maximum molecules. 
20. (a) : Specific volume (vol. of 1 g) of cylindrical virus 
particle = 6.02 × 10
–2 
cc/g 
0.05 ?
??9 
 
3x ? 2 y 
? 3x ? 2 y ?
 9 
? 180 
0.05 
...(ii) 
Radius of virus, r = 7 Å = 7 × 10
–8 
cm 
Volume of virus = ?r
2
l 
On solving equations (i) and (ii), we get x = 40 
40 + 2y = 100 ? 2y = 60 ? y = 30 
13. (a) : Mass of 1 mol (6.022 × 10
23 
atoms) of carbon 
= 12 g 
If Avogadro number is changed to 6.022 × 10
20 
atoms 
then mass of 1 mol of carbon 
? 
22 
? (7 ?10
?8 
)
2 
?10 ?10
?8 
= 154 × 10
–23 
cc 
7 
wt. of  one virus particle ?
??Volume (cc) 
 
Specific volume (cc/g) 
? 
154 ?10
?23 
g
 
6.02 ?10
?2
 
? 
12 ? 6.022 ?10
20 
? 12 ?10
?3 
g
 
? Molecular wt. of virus = wt. of N particles 
6.022 ?10
23
 
A 
?23 
14. (c) : 1.8 gram of water ? 
6.023 ?10
23 
?1.8 
18 
= 6.023 × 10
22 
molecules 
18 gram of water = 6.023 × 10
23 
molecules 
18 moles of water = 18 × 6.023 × 10
23 
molecules 
15. (d) : Number of moles of H 2 = 1/2 
? 
154 ?10 
? 6.02 ?10
23 
g / m ol 
6.02 ?10
?2
 
= 15400 g/mol = 15.4 kg/mol 
21. (d) : 17 g of NH 3 = 4NA atoms 
4.25 g of NH ? 
4N
A 
? 4.25 atoms
17 
Number of moles of O2 ?
 4  
32 
= NA atoms = 6 × 10
23 
atoms 
22. (a) :  Quantity of iron in one molecule 
Hence, molar ratio ? 
1 
:
 4 
? 4 : 1 
= 
67200 
× 0.334 = 224.45 amu 
16. (c) : 
2 32 
100 
each. 
8 g H 2 has 4 moles while the others has 1 mole 
No. of iron atoms in one molecule of haemoglobin 
17. (b) : No. of atoms = N A × No. of moles × 3 
23 23 
? 
224.45 
? 4
 
56 
= 6.023 × 10 × 0.1 × 3 = 1.806 × 10 
23. (a) : Volume of oxygen in one litre of air 
18. (a) : At STP, 22.4 L = 6.023 × 10
23 
molecules 
?
 21 
?1000 ? 210 mL 
15 L H 
? 
6.023 ?10
23 
?15 
? 4.033 ?10
23 
molecules 
100 
  210  
2
 22.4 
Therefore, no. of moles = 
22400 
= 0.0093 mol 
= 
A 
3 
A A 
3 
23 22 
24. (c) : Each nitrogen atom has 5 valence electrons, 
therefore total number of valence electrons in N
–
 ion 
78.4 g Se will be present in
 100
 ? 78.4 ? 1.568 ? 10
4
 
is 16. Since the molecular mass of N
–
 
3 
is 42, therefore 
0.5  
? Minimum molecular weight of enzyme is 
total number of valence electrons in 4.2 g of N
–
 ion 1.568 × 10
4
. 
= 
4.2 
?16 ? N ? 1.6 N 
42 
31. (c) : Urea (NH 2 CONH 2), % of N ? 
28 
? 100 ? 46.66% 
60 
25. (b) : A vogadro’s no., N A = 6.02 × 10
23 
molecules = 1 m ole 
? 6.02 × 10
24 
CO molecules = 10 moles CO 
= 10 g atoms of O = 5 g molecules of O 2 
26. (a) : Here, Cp/Cv = 1.4, which shows that the gas is 
diatomic. 
22.4 L at NTP = 6.02 × 10
23 
molecules 
? 11.2 L at NTP = 3.01 × 10
23 
molecules 
Since gas is diatomic, 
? 11.2 L at NTP = 2 × 3.01 × 10
23 
atoms 
= 6.02 × 10
23 
atom 
27. (a) : 1 mol of CO 2 = 44 g of CO2 
Similarly, % of N in other compounds are : 
(NH 4 ) 2 SO 4 = 21.2%; CaCN 2 = 35.0% and NH 4 NO 3 = 35.0% 
32. (d) : Haber’s process, N 2 + 3H 2 ? 2NH 3 
2 moles of NH 3 are formed by 3 moles of H 2. 
? 20 moles of NH 3 will be formed by 30 moles of H 2. 
33. (c) : Density = 1.28 g/cc, 
Conc. of solution = 2 M 
Molar mass of NaOH = 40 g mol
–1 
Volume of solution = 1 L = 1000 mL 
Mass of solution = d × V = 1.28 × 1000 = 1280 g 
Mass of solute = n × Molar mass = 2 × 40 = 80 g 
? 4.4 g CO2 = 0.1 mol CO 2 = 6 × 10
22 
molecules 
23 
Mass of solvent = (1280 – 80) g = 1200 g 
80 
[Since, 1 mole CO 2 = 6 × 10 molecules] Number of moles of solute = 
40 
= 2 
= 2 × 6 × 10
22 
atoms of O = 1.2 × 10
23 
atoms of O 
28. (d) : As we know, 
? Molality = 
2 ?1000 
1200 
? 1.67 m 
22400 cc of N 2O contain 6.02 × 10
23 
molecules 34.   (c) : HCOOH 
D
?
eh
?
yd
?
ra
?
tin
?
g a
?
ge
?
nt
? 
CO + H O 
? 1 cc of N 2O contain 
6.02 ?10
23
 
22400 
molecules 
n
i 
? 
2.3 
?
 1  
conc. H
2
SO
4 
2
 
0 0 
Since in N 2O molecule there are 3 atoms 
? 1 cc N O ? 
3 ? 6.02 ?10 
atoms ? 
1.8 ?10 
atoms 
46 20 
nf = 0 
H C O 
?
c
?
on
?
c. 
?
H
2
?
SO
?
4
?
?
1  1 
20 20 
2
 22400 224 
2 2 4 
No. of electrons in a molecule of N 2O = 7 + 7 + 8 = 22 
n
i 
? 
4.5 
?
 1  
Hence, no. of electrons in 1 cc of N 2O 
23 
90 20 
nf =0 
? 
6.02 ?10 
? 22 electrons ? 
1.32 
?10
23 
electrons 
H 2O gets absorbed by conc. H2 SO 4. Gaseous mixture 
22400 
29. (c) : 
224 
(containing CO and CO 2) when passed through KOH 
pellets, CO 2 gets absorbed. 
Moles of CO left (unabsorbed) ?
 1 
?
 1 
?
 1 
 
20 20 10 
Mass of CO = moles × molar mass ?
 1 
? 28 ? 2.8 g 
35. (b) : 
10 
16.9% solution of AgNO 3 means 16.9 g of AgNO3
in 100 mL of solution. 
= 8.45 g of AgNO 3 in 50 mL solution. 
Similarly, 5.8 g of NaCl in 100 mL solution 
? 2.9 g of NaCl in 50 mL solution. 
The reaction can be represented as : 
Hence, empirical formula of the compound would be 
CH 3O. 
Initial 
mole 
AgNO 3 + NaCl AgCl? + NaNO 3 
8.45/170   2.9/58.5 0 0 
= 0.049 = 0.049 
30.  (a) : In peroxidase anhydrous enzyme, 0.5% Se is 
present means, 0.5 g Se is present in 100 g of enzyme. In a 
molecule of enzyme one Se atom must be present. Hence, 
Final moles 0 0 0.049 0.049 
? Mass of AgCl precipitated = 0.049 × 143.3 
= 7.02 ? 7 g 
Element % Atomic 
mass 
Mole Simple 
ratio 
C 38.71 12 
38.71 
? 3.22
12 
3.22 
? 1 
3.22 
H 9.67 1 
9.67 
? 9.67
1 
9.67 
? 3 
3.22 
O 51.62 16 
51.62 
? 3.22
16 
3.22 
? 1 
3.22 
CO + CO 2 + H 2O 
0 0 0 
1  1  1  
20 20 20 
Page 4


2 
 
 
1. (b) : Pressure ?
 Force
 
Area 
Therefore, dimensions of pressure ? ?
?
MLT
?2 
L
2 
= ML
–1
T
–2 
Therefore, vapour density 
? 
Weight of certain volume of substance 
Weight of same volume of hydrogen 
0.24  
and dimensions of energy per unit volume 
?
 Energy 
? 
ML
2
T
? 2
 
? ML
?1
T
?2
 
? 
4.005 ?10
?3 
? 59.93
 
Volume L
3
 6. (c) : If 1 L of one gas contains N molecules, 2 L of any 
2. (d) : Zeros placed left to the number are never 
significant, therefore the no. of significant figures for the 
numbers 161 cm , 0.161 cm and 0.0161 cm are sam e, i.e., 3. 
3. (c) : According to Avogadro’s hypothesis, ratio of 
the volumes of gases will be equal to the ratio of their no. 
gas under the same conditions will contain 2N molecules. 
7. (c) : C 2H 4 + 3O 2 ? 2CO 2 + 2H 2O 
28 g 96 g 
For complete combustion, 
2.8 kg of C 2H 4 requires ?
 96 
? 2.8 ?10
3 
g 
of moles. 
28 
3
 
So, no. of moles = 
   Mass 
 
= 9.6 × 10 g = 9.6 kg of O2 
Mol. mass 
8. (d) : Average isotopic mass of X 
n
H 
? 
w 
; n
O
 
?
 w 
; n
CH 
?
 w 
? 
200 ? 90 ?199 ? 8 ? 202 ? 2 
2 
2 
2 
32 
4 
16 90 ? 8 ? 2 
So, the ratio is 
w 
:
 w
 
:
 w 
or 16 :1: 2. 
? 
18000 ?1592 ? 404 
? 199.96 amu ? 200 amu 
2 32 16 
4. (a) : C 3H 8 + 5O 2 3CO 2 + 4H 2O 
1 vol. 5 vol. 3 vol. 4 vol. 
100 
9. (a) : Average atomicmass ??
19 ?10 ? 81 ?11 
100 
? 10.81 
According to the above equation, 10. (d) : 1 mole of substance = NA atoms 
1 vol. or 1 litre of propane requires 5 vol. or 5 litres of O 2
to burn completely. 
5. (b) : Weight of gas = 0.24 g, 
Volume of gas = 45 mL = 0.045 litre and density of 
108 g of Ag = NA atoms ??
24 g of Mg = NA atoms ??
1g of Ag ? 
N
A 
atoms 
108 
1g of Mg ? 
N
A 
atoms 
24 
H 2 = 0.089 g/L 32 g of O 2 = NA molecules = 2 NA atoms 
Weight of 45 mL of H 2 = density × volume 
= 0.089 × 0.045 = 4.005 × 10
–3 
g 
? 1g of O ? 
N
A 
atoms 
16 
Hints & Explanations 
46 
Mol. wt. 
N
A 
32 
3 
Some Basic Concepts of Chemistry  
7 g of Li = NA atoms ??
1g of Li ? 
N
A 
atoms 
7 
5 L N 2 
?
?
6.023 ?10
23 
? 5 
22.4 
? 1.344 ?10
23
 
molecules 
Therefore, 1 g of Li(s), has maximum number of atoms. 
2 g H 2 = 6.023 × 10
23 
molecules 
11.   (a) : (a) Mass of water = V × d = 18 × 1 = 18 g 
Molecules of water = mole × NA ? 
18 
N
A 
? N
A
 
0.5 g H2 ? 
6.023 ?10
23 
? 0.5 
? 1.505 ?10
23 
molecules
2 
18 
(b) Molecules of water = mole × NA ??
0.18 
N
18 
32 g O 2 = 6.023 × 10
23 
molecules 
6.023 ?10
23 
?10 
23
 
= 10
–2 
NA 
10 g of O 2 ? ?
32
 
? 1.882 ?10 
molecules 
(c) Moles of water ?
 0.00224 
? 10
?4
 19. (b) : Number of molecules = moles × NA 
22.4 
–4 
Molecules of N 2 =
 7 
NA = 0.5 NA 
Molecules of water = mole × NA = 10 NA
 
14
 
(d) Molecules of water = mole × NA = 10
–3 
NA 
12. (a) : Let atomic weight of element X is x and that of 
Molecules of H 2 = NA 
element Y is y. 
For XY , n ?
??w 
 
Molecules of NO 2 = 
16
 NA = 0.35 NA 
2 
  10  
Mol. wt. 
10 
Molecules of O2 
16 
= 0.5 NA 
0.1 ? 
x ? 2 y 
? x + 2y = 
For X3 Y 2, n ?
??w 
 
? 100 
0.1  
...(i) 
? 2 g H 2 (1 mole H 2) contains maximum molecules. 
20. (a) : Specific volume (vol. of 1 g) of cylindrical virus 
particle = 6.02 × 10
–2 
cc/g 
0.05 ?
??9 
 
3x ? 2 y 
? 3x ? 2 y ?
 9 
? 180 
0.05 
...(ii) 
Radius of virus, r = 7 Å = 7 × 10
–8 
cm 
Volume of virus = ?r
2
l 
On solving equations (i) and (ii), we get x = 40 
40 + 2y = 100 ? 2y = 60 ? y = 30 
13. (a) : Mass of 1 mol (6.022 × 10
23 
atoms) of carbon 
= 12 g 
If Avogadro number is changed to 6.022 × 10
20 
atoms 
then mass of 1 mol of carbon 
? 
22 
? (7 ?10
?8 
)
2 
?10 ?10
?8 
= 154 × 10
–23 
cc 
7 
wt. of  one virus particle ?
??Volume (cc) 
 
Specific volume (cc/g) 
? 
154 ?10
?23 
g
 
6.02 ?10
?2
 
? 
12 ? 6.022 ?10
20 
? 12 ?10
?3 
g
 
? Molecular wt. of virus = wt. of N particles 
6.022 ?10
23
 
A 
?23 
14. (c) : 1.8 gram of water ? 
6.023 ?10
23 
?1.8 
18 
= 6.023 × 10
22 
molecules 
18 gram of water = 6.023 × 10
23 
molecules 
18 moles of water = 18 × 6.023 × 10
23 
molecules 
15. (d) : Number of moles of H 2 = 1/2 
? 
154 ?10 
? 6.02 ?10
23 
g / m ol 
6.02 ?10
?2
 
= 15400 g/mol = 15.4 kg/mol 
21. (d) : 17 g of NH 3 = 4NA atoms 
4.25 g of NH ? 
4N
A 
? 4.25 atoms
17 
Number of moles of O2 ?
 4  
32 
= NA atoms = 6 × 10
23 
atoms 
22. (a) :  Quantity of iron in one molecule 
Hence, molar ratio ? 
1 
:
 4 
? 4 : 1 
= 
67200 
× 0.334 = 224.45 amu 
16. (c) : 
2 32 
100 
each. 
8 g H 2 has 4 moles while the others has 1 mole 
No. of iron atoms in one molecule of haemoglobin 
17. (b) : No. of atoms = N A × No. of moles × 3 
23 23 
? 
224.45 
? 4
 
56 
= 6.023 × 10 × 0.1 × 3 = 1.806 × 10 
23. (a) : Volume of oxygen in one litre of air 
18. (a) : At STP, 22.4 L = 6.023 × 10
23 
molecules 
?
 21 
?1000 ? 210 mL 
15 L H 
? 
6.023 ?10
23 
?15 
? 4.033 ?10
23 
molecules 
100 
  210  
2
 22.4 
Therefore, no. of moles = 
22400 
= 0.0093 mol 
= 
A 
3 
A A 
3 
23 22 
24. (c) : Each nitrogen atom has 5 valence electrons, 
therefore total number of valence electrons in N
–
 ion 
78.4 g Se will be present in
 100
 ? 78.4 ? 1.568 ? 10
4
 
is 16. Since the molecular mass of N
–
 
3 
is 42, therefore 
0.5  
? Minimum molecular weight of enzyme is 
total number of valence electrons in 4.2 g of N
–
 ion 1.568 × 10
4
. 
= 
4.2 
?16 ? N ? 1.6 N 
42 
31. (c) : Urea (NH 2 CONH 2), % of N ? 
28 
? 100 ? 46.66% 
60 
25. (b) : A vogadro’s no., N A = 6.02 × 10
23 
molecules = 1 m ole 
? 6.02 × 10
24 
CO molecules = 10 moles CO 
= 10 g atoms of O = 5 g molecules of O 2 
26. (a) : Here, Cp/Cv = 1.4, which shows that the gas is 
diatomic. 
22.4 L at NTP = 6.02 × 10
23 
molecules 
? 11.2 L at NTP = 3.01 × 10
23 
molecules 
Since gas is diatomic, 
? 11.2 L at NTP = 2 × 3.01 × 10
23 
atoms 
= 6.02 × 10
23 
atom 
27. (a) : 1 mol of CO 2 = 44 g of CO2 
Similarly, % of N in other compounds are : 
(NH 4 ) 2 SO 4 = 21.2%; CaCN 2 = 35.0% and NH 4 NO 3 = 35.0% 
32. (d) : Haber’s process, N 2 + 3H 2 ? 2NH 3 
2 moles of NH 3 are formed by 3 moles of H 2. 
? 20 moles of NH 3 will be formed by 30 moles of H 2. 
33. (c) : Density = 1.28 g/cc, 
Conc. of solution = 2 M 
Molar mass of NaOH = 40 g mol
–1 
Volume of solution = 1 L = 1000 mL 
Mass of solution = d × V = 1.28 × 1000 = 1280 g 
Mass of solute = n × Molar mass = 2 × 40 = 80 g 
? 4.4 g CO2 = 0.1 mol CO 2 = 6 × 10
22 
molecules 
23 
Mass of solvent = (1280 – 80) g = 1200 g 
80 
[Since, 1 mole CO 2 = 6 × 10 molecules] Number of moles of solute = 
40 
= 2 
= 2 × 6 × 10
22 
atoms of O = 1.2 × 10
23 
atoms of O 
28. (d) : As we know, 
? Molality = 
2 ?1000 
1200 
? 1.67 m 
22400 cc of N 2O contain 6.02 × 10
23 
molecules 34.   (c) : HCOOH 
D
?
eh
?
yd
?
ra
?
tin
?
g a
?
ge
?
nt
? 
CO + H O 
? 1 cc of N 2O contain 
6.02 ?10
23
 
22400 
molecules 
n
i 
? 
2.3 
?
 1  
conc. H
2
SO
4 
2
 
0 0 
Since in N 2O molecule there are 3 atoms 
? 1 cc N O ? 
3 ? 6.02 ?10 
atoms ? 
1.8 ?10 
atoms 
46 20 
nf = 0 
H C O 
?
c
?
on
?
c. 
?
H
2
?
SO
?
4
?
?
1  1 
20 20 
2
 22400 224 
2 2 4 
No. of electrons in a molecule of N 2O = 7 + 7 + 8 = 22 
n
i 
? 
4.5 
?
 1  
Hence, no. of electrons in 1 cc of N 2O 
23 
90 20 
nf =0 
? 
6.02 ?10 
? 22 electrons ? 
1.32 
?10
23 
electrons 
H 2O gets absorbed by conc. H2 SO 4. Gaseous mixture 
22400 
29. (c) : 
224 
(containing CO and CO 2) when passed through KOH 
pellets, CO 2 gets absorbed. 
Moles of CO left (unabsorbed) ?
 1 
?
 1 
?
 1 
 
20 20 10 
Mass of CO = moles × molar mass ?
 1 
? 28 ? 2.8 g 
35. (b) : 
10 
16.9% solution of AgNO 3 means 16.9 g of AgNO3
in 100 mL of solution. 
= 8.45 g of AgNO 3 in 50 mL solution. 
Similarly, 5.8 g of NaCl in 100 mL solution 
? 2.9 g of NaCl in 50 mL solution. 
The reaction can be represented as : 
Hence, empirical formula of the compound would be 
CH 3O. 
Initial 
mole 
AgNO 3 + NaCl AgCl? + NaNO 3 
8.45/170   2.9/58.5 0 0 
= 0.049 = 0.049 
30.  (a) : In peroxidase anhydrous enzyme, 0.5% Se is 
present means, 0.5 g Se is present in 100 g of enzyme. In a 
molecule of enzyme one Se atom must be present. Hence, 
Final moles 0 0 0.049 0.049 
? Mass of AgCl precipitated = 0.049 × 143.3 
= 7.02 ? 7 g 
Element % Atomic 
mass 
Mole Simple 
ratio 
C 38.71 12 
38.71 
? 3.22
12 
3.22 
? 1 
3.22 
H 9.67 1 
9.67 
? 9.67
1 
9.67 
? 3 
3.22 
O 51.62 16 
51.62 
? 3.22
16 
3.22 
? 1 
3.22 
CO + CO 2 + H 2O 
0 0 0 
1  1  1  
20 20 20 
3 
Some Basic Concepts of Chemistry  
36. (c) : MgCO 3( s ) ?
?
??   
MgO
(s) 
+ CO
2(g) ? In this reaction, oxygen is the limiting reagent so 
84 g 40 g 
84 g of MgCO 3 ? 40 g of MgO 
amount of H 2O produced depends on the amount of O 2. 
Since 0.5 mol of O 2 gives 1 mol of H 2O 
? 20 g of MgCO 3 
? 
40 
? 20 = 9.52 g of MgO 
84 
? 2 mol of O 2 will give 4 mol of H 2O 
43. (b) : PbO + 2HCl ? PbCl 2 + H 2O 
Actual yield = 8 g of MgO 
6.5 
mol 
 3.2 
mol
 
? % purity =
 8 
?100 = 84% 224 36.5 
9.52 
37. (a) : 1 mole ? 22.4 litres at STP. 
= 0.029 mol = 0.087 mol 
Formation of moles of lead(II) chloride depends upon 
n
H
2
 
? 
22.4 
? 1 mol; n 
22.4 
? 
11.2 
? 0.5 mol 
2 
22.4 
the no. of moles of PbO which acts as a limiting reagent 
here. So, no. of moles of PbCl 2 formed will be equal to the 
Reaction is as, 
H
2(g)    
+ Cl
2(g) 
2HCl
(g) 
Initial 1 mol 0.5 mol 0 
Final (1 – 0.5) (0.5 – 0.5) 2 × 0.5 
= 0.5 mol = 0 mol 1 mol 
Here, Cl 2 is limiting reagent. So, 1 mole of HCl ( g ) is formed. 
no. of moles of PbO i.e. 0.029. 
44.   (c) : 3C + 2Al 2O 3 4Al + 3CO 2 
(From bauxite) 
4 moles of Al is produced by 3 moles of C. 
1 mole of Al is produced by
 3 
mole of C. 
270 ? 1000 
4 
4 
3 
4
 
38. (a) : n
Mg 
?
 1 
= 0.0416 moles 
27 
= 10 moles of Al is produced by × 10 
4 
n
O
2 
??
0.56 
32 
24 
= 0.0175 mole 
moles of C. 
Amount of carbon used = 
3 
× 10
4 
× 12 g 
4 
The balanced equation is    
2Mg + O 2 2MgO 
Initial 0.0416 mole 0.0175 mole 0 
= 
3 
4 
45. (c) : Density = 1.17 g/cc. 
× 10 × 12 kg = 90 kg 
Final    (0.0416 – 2 × 0.0175) 0 2 × 0.0175 
= 0.0066 mole 
? 1 cc. solution contains 1.17 g of HCl 
1.17 ?1000 
Here, O 2 is limiting reagent. ? Molarity = 
36.5 ?1 
= 32.05 
? Mass of Mg left in excess = 0.0066 × 24 = 0.16 g 
6.02 ?10
20
 46. (b) : BaCO 3 
197.3 g 
? BaO + CO 2 
39. (d) : Moles of urea ??
6.02 ?10
23
 
? 0.001 
22.4 L at N.T.P. 
9.85 g 
 22.4 
× 9.85 
Concentration of solution ? 
0.001 
? 1000 ? 0.01 M 
100 
40. (b) : Millimoles of solution of chloride 
= 0.05 × 10 = 0.5 
Millimoles of AgNO 3 solution = 10 × 0.1 = 1 
So, the millimoles of AgNO 3 are double than the chloride 
solution. 
197.3 
= 1.118 L 
? 9.85 g of BaCO 3 will produce 1.118 L of CO 2 at 
N.T.P. on the complete decomposition. 
47. (a) : 4NH 3( g ) + 5O 2( g ) ? 4NO ( g ) + 6H 2O ( l ) 
4 moles 5 moles 4 moles 6 moles 
? 1 mole of NH 3 requires = 5/4 = 1.25 mole of oxygen 
? XCl 2 + 2AgNO 3 ? 2AgCl + X(NO 3) 2 
41. (b) : Given that molar mass of Na 2CO 3 = 106 g 
25.3 ? 1000 
while 1 mole of O 2 requires =4/5 = 0.8 mole of NH 3. 
Therefore, all oxygen will be consumed. 
48. (c) : Zn + H 2SO 4 ? ZnSO 4 + H 2 
? Molarity of solution ? ?
Na 2CO 3 ? 2Na
+ 
+ CO 
2–
 
106 ? 250 
= 0.955 M 
65 g 22400 mL 
Since 65 g of zinc reacts to liberate 22400 mL of H 2 at 
STP, therefore amount of zinc needed to produce 224 mL 
[Na
+
] = 2[Na 2CO 3] = 2 × 0.955 = 1.910 M 
[CO 
2–
] = [Na CO ] = 0.955 M 
of H
2 
at STP ?
 65 
? 224 ? 0.65 g 
3 2 3 22400 
42. (b) : H 2 + 1/2O 2 ? H 2O 
2 g 16 g 18 g 
49. (a) : Weight of 1 mol of CCl 4 vapour 
= 12 + 4 × 35.5 = 154 g 
1 mol 0.5 mol 1 mol 
10 g of H 2 = 5 mol and 64 g of O 2 = 2 mol ? Density of CCl 4 vapour 
??? ?
?
 154 
22.4 
g L
? 1 
= 6.875 g L
–1
 
Cl