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Std. XI Sci. Success Chemistry - I
Formulae
Mass of elements
Atomic mass
1.
One mole of atoms =
2.
Mass of one atom =
3.
Number of moles (n)
Mass of substance
=
Molar mass of substance
Atomic mass
6.022 × 1023
4.
Number of molecules =
n × Avogadro number
5.
Volume of gas at STP = n × 22.414 L
6.
Percentage (By weight) =
Mass of the element in 1 mole of the compound
× 100%
Gram molecular weight of the compound
7.
Molecular formula = r × empirical formula
where r is ratio of molecular mass to
empirical mass.
Molecular mass
Mass of one molecule =
6.022 × 1023
8.
Solved Problems
*1.
Calculate the volume of hydrogen required
for complete hydrogenation of 0.25 dm3 of
ethyne at STP.
Solution :
The hydrogenation of ethyne (C2H2) is
represented as,
C2H2(g) + 2H2(g) → C2H6(g)
ethyne
ethane
1 vol
2 vol
1 vol
3
1 volume C2H6 ≡ 1 dm C2H2 requires
2 dm3H2 gas
∴ 0.25 dm3C2H2 will require
= 2 × 0.25
= 0.5 dm3 H2 gas
Volume of H2(g) required = 0.5 dm3.
*2.
3.49 g of ammonia at STP occupies volume
of 4.48 dm3 .Calculate molar mass of
ammonia.
Some Basic Concepts of Chemistry
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Solution :
22.4 dm3 of a gas at STP ≡ 1 mol of gas
≡ molar mass of the gas
Now,
 4.48 dm3 of NH3 at STP ≡ 3.49 g
∴ 22.414 dm3 of NH3 at STP
3.49 × 22.414
≡
4.48
= 17.46
∴ Molar mass of NH3 = 17.46 g mol–1.
*3.
Calculate the volume of oxygen required
for complete combustion of 0.25 mole of
methane at STP.
Solution :
1 mol of methane (CH4) at STP ≡ 22.414 dm3
∴ 0.25 mol methane at STP ≡ 22.414 × 0.25
= 5.6 dm3
The combustion of CH4 is represent as,
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
1 vol
2 vol
1 vol CH4 ≡ 1 dm3 CH4 ≡ 2 dm3O2
∴ 5.6 dm3CH4 ≡ 2 × 5.6 = 11.2 dm3
Volume of O2 required = 11.2 dm3 at STP
*4.
Calculate the number of moles and
molecules of ammonia present in 5.6 dm3 of
its volume at STP.
Solution :
22.4 dm3 of NH3 ≡ 1 mol NH3
5.6
∴ 5.6 dm3 of NH3 ≡
= 0.25 mol NH3
22.4
1 mol NH3 contains
= 6.022 × 1023 molecules of NH3
∴ 0.25 mol NH3 contains
= 6.022 × 1023 × 0.25
= 1.505 × 1023 molecules of NH3
Number of moles of NH3 = 0.25
Number of molecules of NH3 = 1.505 × 1023
*5.
Calculate the mass of potassium chlorate
required to liberate 6.72 dm3 of oxygen at
STP, molar mass of potassium chlorate is
122.5 g/mol.
Solution :
Molar mass of potassium chlorate (KClO3)
= 122.5 g mol–1
22.4 dm3 of O2 at STP = 1 mol O2
∴ 6.72 dm3 of O2 at STP
1.16
HORIZON Publications
Std. XI Sci. Success Chemistry - I
=
KClO3 → KCl +
1 mol
∴
∴
6.72
= 0.3 mol O2
22.4
3
O2
2
3
mol
2
3
mol O2 requires ≡ 1 mol KClO3
2
≡ 1 mol KClO3 ≡ 122.5 g KClO3
122.5 × 0.3
0.3 mol O2 requires =
3/ 2
= 24.5 g KClO3
Mass of potassium chlorate required = 24.5 g
*6.
Calculate number of atoms of hydrogen
present in 5.6 g of urea (molar mass of urea
= 60 gmol1). Also calculate number of
atoms of N, C and O.
Solution :
Molecular formula of mass = CO(NH2)2
= CON2H4
Molar mass of urea = 60 g mol–1
W
Number of moles of urea = n =
M
5.6
=
= 0.09333 mol
60
Now,
 1 mol urea contains 4 gram atoms of
hydrogen (H)
∴ 0.09333 mol urea contains = 4 × 0.09333
= 0.3733 gram atom of H
Now,
1 gm atom of H contains
= 6.022 × 1023 atoms of H
0.3733 gm atom of H contains
= 6.022 ×1023 × 0.3733
= 2.248 × 1023 atoms of H
1 mol of urea contains = 2 gram atoms of
nitrogen (N)
∴ 0.09333 mol of urea contains
= 2 × 0.09333
= 0.1866 gram atom of N
∴ Number of N atoms
= 0.1866 × 6.022 × 1023
= 1.123 × 1023 N atoms
1 mol of urea contains
= 1 gram atom of carbon (C)
∴ 0.09333 mol urea contains
= 0.09333 gram atom of C
Some Basic Concepts of Chemistry
∴
∴
Number of C atoms
= 0.09333 × 6.022 × 1023
= 0.562 × 1023
1 mol of urea contains
= 1 gram atom of oxygen (O)
0.09333 mol urea contains
= 0.09333 gram atom of O
Number of oxygen atoms (O)
= 0.09333 × 6.022 × 1023
= 0.562 × 1023
Number of H atoms = 2.248 × 1023
Number of N atoms = 1.123 × 1023
Number of C atoms = 0.562 × 1023
Number of O atoms = 0.562 × 1023
*7.
Calculate the number of atoms of C, H and
O in 72.5 g isopropanol, C3H7OH (molar
mass 60).
Solution :
Molar mass of isopropanol,
C3H7OH = 60 g mol–1
Mass of isopropanol = 72.5 g
Number of moles of isopropanol
mass of isopropanol
=
molar mass
W
∴ n=
M
72.5
=
60
= 1.208 mol
1 mol of C3H7OH ≡ 3 gram atom of C
≡ 8 gram atom of H
≡ 1 gram atom of O.
∴ Gram atom of C = 3 × 1.208 = 3.624
∴ Number of atoms of C
= 3624 × 6.022 × 1023
= 2.182 × 1024
Gram atom of H = 8 × 1.208 = 9.664
∴ Number of atoms of H
= 9.664 × 6.022 × 1023
= 5.82 × 1024
Gram atom of O = 1.208
∴ Number of atoms of O
= 1.208 × 6.022 × 1023 = 7.274 × 1023
Number of atoms of C = 2.182 × 1024
Number of atoms of H = 5.82 × 1024
Number of atoms of O = 7.274 × 1023
1.17
Std. XI Sci. Success Chemistry - I
Phosphoric acid is widely used in
carbonated
beverages,
detergents,
toothpastes and fertilizers. Calculate the
mass percentage of H, P and O in
phosphoric acid if atomic masses are H = 1,
P = 31 and O = 16.
Solution :
Phosphoric acid has molecular formula
H3PO4.
∴ Molecular (molar) mass of H3PO4
= 3 × 1 + 31 + 4 × 16 = 98 g mol–1
Percentage composition of H in H3PO4
Atomic mass of H
=
× 100
Molecular mass of H 3PO 4
3
=
× 100 = 3.06 %
98
Percentage composition of P in H3PO4
Atomic mass of P
=
× 100
Molecular mass of H 3PO 4
31
=
× 100 = 31.63 %
98
∴ Percentage of O = 100 – (3.06 + 31.63) =
65.31 %
Percentage of H in H3PO4 = 3.06 %
Percentage of P in H3PO4 = 31.63 %
Percentage of O in H3PO4 = 65.31 %
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Percentage composition of N in HNO3
= 22.22 %
Percentage composition of O in HNO3
= 76.193 %
*8.
*9.
Calculate the mass percentage composition
of the elements in nitric acid.
(H = 1, N = 14, O = 16).
Solution :
The molecular formula of nitric acid is NHO3.
Molecular mass of HNO3 = 1 + 14 + 3 × 16
= 63 g mol–1
Percentage composition of H in HNO3
Atomic mass of H
=
× 100
Molecular mass of HNO3
1
=
× 100 = 1.587 %
63
Percentage composition of N in HNO3
Atomic mass of N
=
× 100
Molecular mass of HNO3
14
× 100 = 22.22%
63
Percentage composition of O in HNO3 = 100
– (1.587 + 22.22) = 76.193 %
Percentage composition of H in NHO3
= 1.587 %
=
Some Basic Concepts of Chemistry
*10. Analysis of vitamin C shows that it
contains 40.92 % carbon by mass 4.58 %
hydrogen and 54.50 % oxygen. Determine
the empirical formula of vitamin C.
Solution :
Element
% by
mass
Gram
atom
Atomic
ratio
C
40.92
40.92
12
3.41
3.41
= 3.41
=1
4.58
1
4.58
3.41
= 4.58
= 1.34
54.50
16
3.41
3.41
= 3.41
=1
H
O
4.58
54.50
Whole
No. ratio
(Atoms)
1×3
=3
1.34 × 3
=4
1×3
=3
Hence the empirical formula of vitamin C is
C3H4O3.
Empirical formula of vitamin C is C3H4O3.
11.
An inorganic compound contained 24.75%
(w/w) potassium and 34.75% (w/w)
manganese and some other common
elements. Give the empirical formula of the
compound. (K = 39u, Mn = 59u, O = 16u).
[Intext Question text book page no. 17]
Solution :
Given : Atomic mass of K = 39 u, Mn = 59 u
and O = 16 u
Percent mass of potassium
= 24.75 %
Percent mass of manganese
= 34.75 %
Total percent mass
= 59.50 %
∴ Remaining mass must be that of oxygen
∴ Percent mass of oxygen = 100 – 59.50
= 40.50%
To find : The empirical formula of the given
inorganic compound = ?
% of potassium
Moles of potassium =
Atomic mass of potassium
=
Moles of manganese =
24.75
= 0.635
39
% of manganese
Atomic mass of manganese
1.18
HORIZON Publications
Std. XI Sci. Success Chemistry - I
=
∴
34.75
= 0.589
59
Moles of oxygen =
% of oxygen
Atomic mass of oxygen
40.50
= 2.53
16
Ratio of K : Mn : O =
0.635 0.589 2.53
: =
:
1.08 : 1 : 4.29 ≈ 1 : 1 : 4
0.589 0.589 0.589
The empirical formula of given inorganic
compound is KMnO4.
=
∴
Phosphoric acid used in carbonated
beverages contain 3.086% (w/w) hydrogen,
31.61% (w/w) phosphorous and remaining
oxygen. If atomic masses of hydrogen,
phosphorous and oxygen are 1.01 u, 31.o u
and 16 u respectively and if molar mass of
phosphoric acid is 98.03 g mol-1, what is the
molecular formula of phosphoric acid ?
[Intext Question text book page no. 17]
Solution :
Given : Atomic mass of H = 1.01 u,
P = 31.0 u and O = 16 u
Percent mass of hydrogen
= 3.086 %
Percent mass of phosphorous = 31.61 %
∴ Total percent mass
= 34.696 %
∴ Remaining amount is oxygen
∴ Percent mass of oxygen = 100 – 34.696
= 65.304 %
To find : The
molecular
formula
of
phosphoric acid = ?
Moles of hydrogen = % mass of hydrogen
∴
12.
Multiple Choice Questions
1.
2.
3.
4.
Atomic mass of hydrogen
=
3.086
= 3.05
1.01
Moles of phosphorous =
=
% mass of phosphorous
Atomic mass of phosphorous
31.61
= 1.019
31
Moles of Oxygen =
% mass of oxygen
Atomic mass of oxygen
65.304
= 4.08
16
Ratio of H : P : O
3.05 1.019 4.08
=
:
:
= 3 :1: 4
1.019 1.019 1.019
Hence, the empirical formula of phosphoric
acid = H3PO4
=
∴
Some Basic Concepts of Chemistry
Empirical formula mass
= 3 (1.01) + 1 (31) + 4 (16)
= 3.03 + 31 + 64
= 98.03 g mol–1
Molar mass
98.03
r=
= = 1
Empirical formula mass 98.03
Molecular formula = r × Empirical formula
= l × H3PO4
= H3PO4
The molecular formula of phosphoric acid is
H3PO4.
5.
6.
7.
8.
The gas law was formulated by
(a) Avogadro
(b) Gay-Lussac
(c) Aristotle
(d) Joseph Priestley
The branch of chemistry which deals with
carbon compound is called …… chemistry.
(a) organic
(b) inorganic
(c) carbon
(d) bio
…… can not be carried out in a lab.
(a) Photosynthesis
(b) Reduction
(c) Oxidation
(d) Hydration
How may moles of electron weigh one
kilogram?
(a) 6.022 × 1022
1
(b)
×1031
9.108
6.022
(c)
×1054
9.108
1
(d)
×108
9.108 × 6.022
What is the mass of 0.5 mole of ozone
molecules ?
(a) 8 g
(b) 16 g
(c) 24 g
(d) 48 g
Percentage of nitrogen in urea is about
(a) 46 %
(b) 85 %
(c) 18 %
(d) 28 %
Which of the following has same moleculear
formula and empirical formula ?
(b) C6H12O6
(a) CO2
(c) C2H4
(d) all of these
The empirical formula of C2H2 is…….
(a) C2H4
(b) CH
(c) CH4
(d) all of these
1.19
Std. XI Sci. Success Chemistry - I
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
…… is the quantitative relationship between
the reactants and products in a balanced
chemical equation.
(a) Stiochiometry
(b) Complexometry
(c) Chemistry
(d) Reactions
If 0.5 mol of BaCl2 is mixed with 0.2 mole of
Na3PO4, the maximum number of moles of
Ba3(PO4)2 that can be formed is
(a) 0.7
(b) 0.5
(c) 0.3
(d) 0.1
When oxygen forms ozone, the atomicity
changes from
(a) 1 to 2
(b) 1 to 3
(c) 2 to 3
(d) 2 to 4
3
If 0.0224 dm of a gas at STP weigths 0.06g ,
the moleuclar mass of the gas will be
(a) 6
(b) 600
(c) 30
(d) 60
6.4 mg of sulphur on comustion will give
number of molecules of SO2 equal to
(a) 1.2044 × 1020 (b) 6.4 × 1023
(c) 0.2
(d) 3.2 × 1021
0.01 mol of NH4NO3 will weigh (Atomic
mass, H = 1m N = 14, O = 16)
(a) 6.4 × 10−3 kg
(b) 3.4 × 10−4 kg
(c) 7.2 × 10−3 kg
(d) 8 × 10−4 kg
The number of atoms of He present in 104 u
of He is
(a) 13
(b) 26
–3
(c) 3.5 × 10 kg
(d) 8 × 10–4 kg
Mole is the SI unit of ………
(a) Volume
(b) Pressure
(c) Amount of substance
(d) Density
Mixture of solids constitute …… system.
(a) Homogeneous
(b) Heterogenous
(c) Either (a) or (b)
(d) Neither (a) not (b)
Two elements, A and B, combine to form a
compound in which ‘a’ g of A combines with
‘b1’ and ‘b2’ g of B respectively. According to
law of multiple proportion ……..
(a) b1 = b2
(b) b1 and b2 bear a simple whole number
ratio
(c) a1 and b1 bear a whole number ratio
(d) no relation exists between b1 and b2
Some Basic Concepts of Chemistry
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19.
20.
21.
22.
23.
Molecule is the smallest particle of …..
(a) compound
(b) substance
(c) mixture
(d) element
One mole of oxygen weighs ………
(a) 8 g
(b) 32 g
(c) 16 g
(d) 6.023 × 1023 g
The number of atoms present in a molecule of
a substance is called ……….
(a) Atomicity
(b) Volume
(c) Density
(d) Mass
The number of molecules present in 8 g of
oxygen gas are ………
(b) 3.011 × 1023
(a) 6.022 × 1023
23
(c) 12.044 × 10
(d) 1.505 × 1023
Atomicity of mercury vapour is ……..
(a) 1
(b) 2
(c) 3
(d) 4
1.
6.
11.
16.
21.
b
a
c
c
a
2.
7.
12.
17.
22.
a
a
d
b
d
Answers
3. a
8. b
13. a
18. b
23. a
4.
9.
14.
19.
d
a
d
a
5.
10.
15.
20.
c
d
b
b
1.20