Class-XI-CBSE-Chemistry
Some Basic Concepts Of Chemistry
CBSE NCERT Solutions for Class 11 Chemistry Chapter 1
Back from chapter Questions
1.
Calculate the molar mass of the following:
(i) H2 O
(ii) CO2
(iii) CH4 s
Solution:
(i) The molecular mass of water( H2 O)
We know : Atomic mass of hydrogen =1.008 u
Atomic mass of oxygen = 16.00 u
molecular mass of H2 O = (2 × Atomic mass of hydrogen)+(1 × Atomic mass of
oxygen)
= [2(1.008 u) + 1(16.00 u)]
= 2.016 u + 16.00 u
= 18.016 u
(ii) The molecular mass of carbon dioxide (CO2 )
Atomic mass of carbon =12.0 u
Atomic mass of oxygen = 16.00 u
=(1 × Atomic mass of carbon)+(2 × Atomic mass of oxygen)
= [1(12.0u) + 2(16.00u)]
= 12.0u + 32.00u
= 44.0u
(iii) The molecular mass of methane ( CH4 )
Atomic mass of carbon =12.0 u
Atomic mass of hydrogen = 1.008 u
=(1 × Atomic mass of carbon)+(4 × Atomic mass of hydrogen)
= [1(12.0u) + 4(1.008 u)]
= 12.0u + 4.032 u
Page 1 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
Some Basic Concepts Of Chemistry
= 16.032 u
2.
Calculate the mass percent of different elements present in sodium sulphate (Na2 SO4 ).
Solution:
We know : Atomic mass of sodium =23.0 u
Atomic mass of Sulphur = 32.0 u
Atomic mass of carbon =16.00 u
Calculate : The molecular formula of sodium sulphate is Na2 SO4
Molar mass of Na2 SO4 = [(2 × 23.0) + (32.0) + 4(16.00)]
= 142 u or 142 g
Formula : Mass percent of an element =
∴ Mass percent of sodium==
=
Mass of that element in the compound
×
Molar mass of the compound
Mass of sodium element in the compound
×
Molar mass of the Na2SO4
100
100
46.0 g
× 100
142.0 g
= 32.39
= 32.4%
Mass percent of Sulphur =
=
Mass of sulphur element in the compound
×
Molar mass of the Na2 SO4
100
32.0 g
× 100
142.0 g
= 22.54 %
Mass percent of oxygen =
=
Mass of oxygen element in the compound
× 100
Molar mass of the Na2 SO4
64.0 g
× 100
142.0 g
= 45.07 %
Page 2 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
3.
Some Basic Concepts Of Chemistry
Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1%
dioxygen by mass.
Solution:
Given : % of iron by mass = 69.9%
% of oxygen by mass = 30.1%
Atomic mass of iron = 55.85
Atomic mass of oxygen = 16.00
Formula used= Relative moles of atom in compound =
Relative moles of iron in iron oxide =
=
% by mass of atom
Atomic mass of atom
% by mass of iron
Atomic mass of iron
69.9
55.85
= 1.25
Relative moles of oxygen in iron oxide:
=
% of oxygen by mass
Atomic mass of oxygen
=
30.1
16.00
= 1.88
Simplest molar ratio of iron to oxygen:
= 1.25: 1.88
= 1: 1.5
= 2: 3
∴ The empirical formula of the iron oxide is Fe2 O3
4.
Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon is burnt in 16 g of dioxygen.
Page 3 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
Some Basic Concepts Of Chemistry
Solution:
The balanced reaction of combustion of carbon can be written as:
(i)
C
Initial mole
1
Mole
stoichiometry coefficient
Left mole
+
→
O2
−
1
1
=1
1
1.0
1
1−1=0
CO2
=1
-
1−1=0
1
As per stoichiometry of the balanced equation, 1 mole of carbon burns in 1 mole of
dioxygen (air) to Produced 1 mole of carbon dioxide.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
Given in question : dioxygen= 16 g ; mole of carbon = 1
moles
16
We know : Moles of oxygen = molecular weight = 32 = 0.5
C
Initial mole
Mole
stoichiometry coefficient
Left mole
+
→
O2
1
0.5
1
=1
1
0.5
1
1 − 0.5 = 0.5
CO2
0.5
= 0.5
0
0.5
0
According to the stoichiometry of the balanced equation
Hence, dioxygen will react with 0.5 moles of carbon to give 0.5 mole of carbon dioxide
(22 g). Hence, dioxygen is a limiting reactant.
(iii) C + O2 → CO2
Given : dioxygen (𝑂2 )= 16 g
; mole of carbon = 2
We know : molecular weigth of O2 = 32
𝑤𝑒𝑖𝑔ℎ𝑡
16
moles of dioxygen = 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 = 32 = 0.5
C
Initial mole
Mole
stoichiometry coefficient
+
→
O2
2
0.5
2
=2
1
0.5
1
= 0.5
CO2
0.5
0
Page 4 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
Left mole
Some Basic Concepts Of Chemistry
2 − 0.5 = 1.5
0
0.5
According to the question, only 0.5 𝑚𝑜𝑙𝑒 of dioxygen or is available. It is a limiting
reactant. Thus, 0.5 mole of dioxygen can combine with only 0.5 mole of carbon dioxide
Weight of carbon dioxide = moles × molecular weight of CO2 = 0.5 × 44 = 22 gram
5.
Calculate the mass of sodium acetate (CH3 COONa) required to make 500 mL of 0.375
molar aqueous solution. Molar mass of sodium acetate is 82.0245 gmol−1 .
Solution:
given : mass of sodium acetate (CH3 COONa) = ?
Molar mass of sodium acetate = 82.024 gmole‐1
Molarity of aqueous solution of sodium acetate = 0.375M
Volume of solution = 500 ml
we know ∶ ∴ A number of moles of sodium acetate in 500 mL. = Molarity × volume of
solution (litre)
=
0.375
× 500
1000
= 0.1875 mole
∴ Required mass of sodium acetate = moles × molecular weight
⇒ (82.0245gmo1−1 ) × (0.1875 mole)
⇒ 15.38 g
6.
Calculate the concentration of nitric acid in moles per liter in a sample which has a
density, 1.41gmL−1 and the mass percent of nitric acid in it being 69%.
Solution:
Given : density = 1.41gmL−1
Mass percent of nitric acid in the sample = 69%
Page 5 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
Some Basic Concepts Of Chemistry
We know, the mass percent of nitric acid in it being 69% = 100g of nitric acid contains
69g of nitric acid by mass.
Molar mass of nitric acid (HNO3 ) = {1 + 14 + 3(16)} gmo1−1
= (1 + 14 + 48) gmo1−1
= 63 gmo1−1
∴ Number of moles in 69 g of HNO3
=
weight (g)
69g
=
molecular mass (gmo1−1 )
63 gmo1−1
= 1.095mol
Mass of solution
Volume of 100g nitric acid solution= density of solution
=
100 g
1.41gmL−1
=
100g
× 100
141gmL−1
= 70.92mL ≡ 70.92 × 10−3 L
Concentration of nitric acid=
=
(∴ 1 𝑚𝑙 = 10−3 𝑙𝑖𝑡𝑒𝑟) )
mole
litre
1.095 mo1e
70.92 × 10−3 L
= 15.44 mol⁄L
∴ Concentration of nitric acid = 15.44 mol⁄L
7.
How much copper can be obtained from 100 g of copper sulphate (CuSO4 )?
Solution:
Given : mass of CuSO4 = 100 g
We know : Molar mass of CuSO4 = (63.5) + (32.00) + 4(16.00)
= 63.5 + 32.00 + 64.00
= 159.5 g
159.5g of CuSO4 contains 63.5 g of copper.
Page 6 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
Some Basic Concepts Of Chemistry
63.5×100g
159.5
⇒ 100g of CuSO4 will contain copper =
= 39.81 g
∴ Amount of copper that can be obtained from 100 g CuSO4 = 39.81 g
8.
Determine the molecular formula of an oxide of iron, in which the mass percent of iron
and oxygen are 69.9 and 30.1, respectively.
Solution:
Given :Mass percent of iron (Fe) = 69.9%
Mass percent of oxygen (O) = 30.1%
We know : The atomic mass of iron = 55.85
The atomic mass of oxygen = 16
69.90
Number of moles of iron present in the oxide = 55.85 = 1.25
30.1
Number of moles of oxygen present in the oxide = 16.0 = 1.88
The ratio of iron to oxygen in the oxide,
= 1.25: 1.88
=
1.25 1.88
:
1.25 1.25
= 1: 1.5
= 2: 3
∴ The empirical formula of the oxide is Fe2 O3
Empirical formula mass of Fe2 O3 = [2(55.85) + 3(16.00)]g
Molar mass of Fe2 O3 = 159.69g
∴n=
Molar mass
159.69g
=
Empirical formula mass
159.7g
= 0.999
= 1(approx)
Molecular formula of a compound = (empirical formula of a compound) n.
Thus , molecular formula = ( Fe2 O3 ) × 1.
Hence, the molecular formula of the oxide is Fe2 O3 .
Page 7 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
9.
Some Basic Concepts Of Chemistry
Calculate the atomic mass (average) of chlorine using the following data:
% Natural
Abundance
Molar Mass
35
75.77
34.9689
37
24.23
36.9659
Cl
Cl
Solution:
given in question :
for
35
Cl : % Natural Abundance =75.77
Molar Mass=34.9689
37
Cl : % Natural Abundance = 24.23
Molar Mass = 36.9659
The average atomic mass of chlorine.
Fractional abundance Molar mass
Fractional abundance Molar mass
= [(
)(
)+(
)(
)]
35
35
of Cl
of Cl
of 37Cl
of 37Cl
75.77
24.23
= [{(
) (34.9689u)} + {(
) (36.9659u)}]
100
100
= 26.4959 u + 8.9568 u
= 35.4527u
∴ The average atomic mass of chlorine = 35.4527u
10.
In three moles of ethane (C2 H6 ) , calculate the following:
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atoms.
(iii) Number of molecules of ethane.
Solution:
(i) 1 mole of C2 H6 contains 2 moles of carbon atoms.
Page 8 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
Some Basic Concepts Of Chemistry
∴ Number of moles of carbon atoms in 3 moles of C2 H6
= 2 × 3 = 6 moles of carbon
(ii) 1 mole of C2 H6 contains 6 moles of hydrogen atoms.
∴ Number of moles of carbon atoms in 3 moles of C2 H6
= 3 × 6 = 18 moles of hydrogen atoms
(iii) 1 mole of C2 H6 contains= NA ×molecules of ethane= 6.023 × 1023 × molecules of
ethane.
∴ Number of molecules in 3 moles of C2 H6 = 3 × NA = 3 × 6.023 × 1023
= 18.069 × 1023 molecules of ethane
11.
What is the concentration of sugar (C12 H22 O11 ) in molL−1 if its 20 g are dissolved in
enough water to make a final volume up to 2L?
Solution:
Given mass of glucose = 20 g
We know : molar mass of sugar (C12 H22 O11 ) = [(12 × 12) + (1 × 22) + (11 × 16)] =
342
Formula : Molarity of a solution (M) =
Number of moles of solute
Volume of solution in Litres
By putting the values in formula we get
M=
Mass of sugar⁄molar mass of sugar
2L
=
20g⁄342
2L
=
0.0585mol
2L
= 0.02925molL−1
∴ Molar concentration of sugar = 0.02925 molL−1
Page 9 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
12.
Some Basic Concepts Of Chemistry
lf the density of methanol is 0.793 kgL−1 what is its volume needed for making 2.5 L of
its 0.25 M solution?
Solution:
Given:
density of methanol is = 0.793 kgL−1
volume of solution (V2 ) = 2.5 L
molarity of solution (M2 ) = 0.25 M
we know Molar mass of methanol (CH3 OH) = (1 × 12) + (4 × 1) + (1 × 16) =
32gmo1−1 = 0.032 kgmol−1
density of methanol is 0.793 kgL−1 means one litre of solution contain
0.793 kg of methanol =793 gm of methanol =
793
32
moles of methanol
i. e. molarity of methanol volume (𝑀1 )= 24.78molL−1
formula used : M1 V1 = M2 V2
(24.78molL−1 ) V1 = (2.5L)(0.25molL−1 )
V1 = 0.0252L
(1 liter= 1000 mililiter)
V1 = 25.22mL
13.
Pressure is determined as force per unit area of the surface. The Sl unit of pressure, pascal
is as shown below:
1Pa = 1Nm−2
If the mass of air at sea level is 1034 gcm−2 calculate the pressure in Pascal.
Solution:
Given : mass of air at sea level = 1034 gcm−2
The pressure is defined as force acting per unit area of the surface.
P=
=
F
A
1034g × 9.8ms−2
1kg
(100)2 cm2
×
×
cm2
1000g
1m2
Page 10 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
Some Basic Concepts Of Chemistry
= 1.01332 × 105 kgm−1 s−2
We know,
1N = 1kgms−2
Then,
1Pa = 1Nm−2 = 1kgm−2 s−2
1Pa = 1kgm−1 s −2
∴ Pressure = 1.01332 × 105 Pa
14.
What is the SI unit of mass? How is it defined?
Solution:
The SI unit of mass is a kilogram (kg) .
1 kilogram is defined as the mass equal to the mass of the international
prototype(Platinum-Iridium cylinder in Paris) of the kilogram.
15.
Match the following prefixes with their multiples:
Prefixes
(i)
(ii)
(iii)
(iv)
(v)
Micro
Deca
Mega
Giga
Femto
Multiples
106
109
10−6
10−15
10
Solution:
Prefixes
(i)
(ii)
(iii)
(iv)
(v)
Micro
Deca
Mega
Giga
Femto
Multiples
10−6
10
106
109
10−15
Page 11 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
16.
Some Basic Concepts Of Chemistry
What do you mean by significant figures?
Solution:
Significant figures are those meaningful digits that are known with certainty. They
indicate uncertainty in an experiment or calculated value. For example, if 25.8 mL is the
result of an experiment, then 25 is certain, while 8 is uncertain, and the total number of
significant figures are 3. Hence, significant figures are defined as the total number of
digits in a number including the last digit that represents the uncertainty of the result.
17.
A sample of drinking water was found to be severely contaminated with chloroform
CHCl3 supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by
mass).
(i) Express this in percent by mass.
(ii) Determine the molality of chloroform in the water sample.
Solution:
1 ppm = 1 part out of 1 million (106 ) parts.
(i)
Let assume Mass of chloroform CHCl3 =15
Mass of solution = 106
Formula :
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 × 100
Mass % of solute =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
∴ Mass percent of l5 ppm chloroform in water.
=
15
× 100
106
≈ 1.5 × 10−3 %
(ii) 100 g of the sample contains 1.5 × 10−3 g of CHCl3.
Given : weight of CHCl3 = 1.5 × 10−3g
Weight of solvent =100 g
Page 12 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
Some Basic Concepts Of Chemistry
We know : Molar mass of CHCl3 = 12.00 + 1.00 + 3(35.5) = 119.5gmo1−1
1.5×10−2 g
𝑤𝑒𝑖𝑔ℎ𝑡(𝑔𝑚)
Moles of CHCl3 = 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 =
Formula : Molaity =
Molar mass of CHC13
1.5×10−2 g
= 119.5gmo1−1
weight of solute (gram)
Molar mass of solute
𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 (𝑘𝑔)
Put all the above values in formula we get
Molality of chloroform in water.
1.5 × 10−2 g
weight (gram)
1.5 × 10−2 g
119.5 gmo1−1
Molar mass of CHCl3
Molar mass of CHCl3
=
=
=
𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 (𝑘𝑔) 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 (𝑘𝑔)
1
= 0.0125 × 10−2
= 1.25 × 10−4 molal
18.
Express the following in the scientific notation:
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012
Solution:
(i) 0.0048 = 4.8 × 10−3
(ii) 234,000 = 2.34 × 105
(iii) 8008 = 8.008 × 103
(iv) 500.0 = 5.000 × 102
(v) 6.0012 = 6.0012
19.
How many significant figures are present in the following?
(i) 0.0025
(ii) 208
(iii) 5005
Page 13 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
Some Basic Concepts Of Chemistry
(iv) 126,000
(v) 500.0
(vi) 2.0034
Solution:
(i) There are 2 significant figures.
(ii) There are 3 significant figures.
(iii) There are 4 significant figures.
(vi) There are 3 significant figures.
(v) There are 4 significant figures.
(vi) There are 5 significant figures.
20.
Round up the following upto three significant figures:
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808
Solution:
(i) 34. 2
(ii) 10.4
(iii) 0.0460
(iv) 2810
21.
The following data are obtained when dinitrogen and dioxygen react together to form
different compounds:
(i)
(ii)
(iii)
(iv)
Mass of dinitrogen
Mass of dioxygen
14 g
14 g
28 g
28 g
16 g
32 g
32 g
80 g
Page 14 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
Some Basic Concepts Of Chemistry
(a) Which law of chemical combination is obeyed by the above experimental data? Give
its statement.
(b) Fill in the blanks in the following conversions:
(i) 1 km = ____________ mm = ____________ pm
(ii) 1 mg = ____________ kg = ____________ ng
(iii) 1 mL = ____________ L = ____________ dm3
Solution:
(a) If we fix the mass of dinitrogen at 28 g, then the masses of dioxygen that will combine
with the fixed mass of dinitrogen are 32 g, 64g, 32g, and 80 g.
(i)
(ii)
(iii)
(iv)
Mass of dinitrogen
14 g ×2=28
14 g ×2=28
Mass of dioxygen
16 g ×2=23
32 g ×2=62
28 g
28 g
32 g
80 G
Ration of oxygen
32
64
32
80
The masses of dioxygen bear a whole number ratio of 1: 2: 1: 5. Hence, the given
experimental data obeys the law of multiple proportions. The law states that if two elements
combine to form more than one compound, then the masses of one element that combines
with the fixed mass of another element are in the ration of small whole numbers.
(b) (i) 1 km = 1 km ×
1000m
100cm
10mm
×
×
1km
1m
1cm
∴ 1 km = 106 mm
1 km = 1 km ×
1000m
1pm
× −12
1km
10 m
∴ 1 km = 1015 pm
Hence, 1 km = 106 mm = 1015 pm
1g
1kg
(ii) 1 mg = 1 mg × 1000mg × 1000g
⇒ 1 mg = 10−6 kg
1 mg = 1 mg ×
1g
1ng
× −9
1000mg 10 g
Page 15 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
Some Basic Concepts Of Chemistry
⇒ 1 mg = 106 ng
∴ 1 mg = 10−6 kg = 106 ng
1L
(iii) 1mL = 1mL × 1000mL
⇒ 1mL = 10−3 L
1mL = 1cm3 = 1cm3
1dm × 1dm × 1dm
10cm × 10cm × 10cm
⇒ 1mL = 10−3 dm3
∴ 1mL = 10−3 L = 10−3 dm3
22.
If the speed of light is 3.0 × 108 ms−1 calculate the distance covered by light in 2.00 ns.
Solution:
Given : speed of light = 3.0 × 108 ms−1
distance covered by light =?
Time taken to cover the distance = 2.00 ns = 2.00 × 10−9 s
We know: velocity of light = 3.0 × 108 ms −1
Formula : Distance traveled by light = Speed of light × Time taken
= (3.0 × 108 ms−1 ) (2.00 × 10−9 s)
= 6.00 × 10−1 m
= 0.600 m
23.
In a reaction A + B2 → AB2
Identify the limiting reagent, if any, in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
Page 16 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
Some Basic Concepts Of Chemistry
Solution:
A reagent determines the extent of a reaction. It is the reactant which is the first to get
consumed during a reaction, thereby causing the reaction to stop and limiting the amount
of products formed.
(i) 300 atoms of A + 200 molecules of B
According to the stoichiometry of reaction, 1 atom of A reacts with 1 molecule of B.
A
+
B2
300 atoms = 300 NA
Initial moles
→
AB2
200 molecule =200 NA
−
Mole
stoichiometry coefficient
Left mole
300NA
=300NA
1
200NA
1
(300-200) NA =100NA
0
= 200NA
200NA
Thus, 200 NA moles of B2 will react with 300 NA moles of A, thereby leaving 100NA
moles of A unused. Hence, B2 is the limiting reagent.
(ii) 2 mol A + 3 mol B
According to the reaction, 1 mol of A reacts with 1 mol of 3.
Initial moles
Mole
stoichiometry coefficient
2
=2
1
Left mole
2-2=0
A +
B2
2
3
3
1
=3
→
AB2
−
-
3-2=1
2
Thus, 2 mol of A will react with only 2 mol of B. As a result, 1 mol of A will not be
consumed. Hence, A is the limiting reagent.
(iii) 100 atoms of A + 100 molecules of B
According to the given reaction, 1 atom of A combines with 1 molecule of B.
A
Initial moles
+
B2
100 atoms = 100 NA
→
AB2
100 molecule =100 NA
−
Mole
stoichiometry coefficient
100NA
=100NA
1
100NA
1
= 100NA
-
Page 17 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
Some Basic Concepts Of Chemistry
Left mole
0
100NA
0
Thus, all 100NA moles of A will combine with all 100NA moles of B. Hence, the mixture
is stoichiometric where no limiting reagent is present.
(ii)
5 mol A + 2.5 mol B
A +
B2
Initial moles
5
2.5
Mole
5
2.5
stoichiometry coefficient
1
1
Left mole
=5
5-2.5=2.5
→
AB2
−
= 2.5
2.5-2.5=0
2.5
1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of B will combine
with only 2.5 mol of A. As a result, 2.5 mol of A will be left as such. Hence, B is the
limiting reagent.
(iii)
2.5 mol A + 5 mol B2
A
Initial moles
+
2.5
B2
5
Mole
2.5
5
stoichiometry coefficient
1
1
Left mole
2.5-2.5=0
=2.5
→
AB2
−
=5
5-2.5=2.5
_
2.5
According to the reaction, 1 mol of atom A combines with 1 mol of molecule B. Thus,
2.5 mol of A will combine with only 2.5 mol of B and the remaining 2.5 mol of B will be
left as such. Hence, A is the limiting reagent.
24.
Dinitrogen and dihydrogen react with each other to produce ammonia according to the
following chemical reaction:
N2 (g) + H2 (g) → 2NH3 (g)
(i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts with
1.00 × 103 g of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?
Page 18 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
Some Basic Concepts Of Chemistry
Solution:
Given : mass of dinitrogen = 2.00 × 103 g
mass of dihydrogen = 1.00 × 103 g
we should know : 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 of N2 =28
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 of H2 =2
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 of NH3 = 34
𝑤𝑒𝑖𝑔ℎ𝑡(𝑔)
Moles of dinitrogen = 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 =
moles of dihydrogen = =
2.00×103 g
28
𝑤𝑒𝑖𝑔ℎ𝑡(𝑔)
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡
=
= 71.43
1.00×103 g
2
= 500
Balancing the given chemical equation,
N2 (g)
Initial moles
+ 3H2 (g)
→ 2NH3 (g)
71.43
500
Mole
71.43
500
stoichiometry coefficient
1
3
=71.43
= 166.66
−
_
From the equation, 1 mol of nitrogen reacts with 3 mol of dihydrogen to give 2 moles of
ammonia.
Mole
, the
stoichiometry coefficient
ratio is less for N2 Hence, N2 is the limiting reagent.
Hence, the moles of ammonia produced by 71.43 moles of N2 = 2 × 71.43
mass of ammonia produced by 71.43 moles of N2 = 2 × 71.43 ×
molecular weight of ammonia
= 2 × 71.43 × 17 g
= 2428.62g
(ii) N2 is the limiting reagent and H2 is the excess reagent. Hence, H2 will remain
unreacted.
(iii) moles of dihydrogen left unreacted = (500 − 71.43 × 3)moles
Mass of dihydrogen left unreacted = (500 − 71.43 × 3)2 𝑔= 285.71 ×2 g= 571.4g
25.
How are 0.50 mol Na2 CO3 and 0.50M Na2 CO3 different?
Page 19 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
Some Basic Concepts Of Chemistry
Solution:
Given: mols of Na2 CO3= 0.50
Molarity of Na2 CO3 = 0.50 M
We know : Molar mass of Na2 CO3 = (2 × 23) + 12.00 + (3 × 16)
= 106gmol−1
Now, 1 mole of Na2 CO3 means 106g of Na2 CO3
106g
∴ 0.5 mol of Na2 CO3 = 1mole × 0.5molNa2 CO3
= 53 g Na2 CO3
⇒ 0.50M of Na2 CO3 = 0.50
mol
Na2 CO3
L
Hence, 0.50 mol of Na2 CO3 is present in 1 L of water
or (0.5 × 106) = 53 g of Na2 CO3 is present in 1 L of water.
26.
If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many
volumes of water vapour would be produced?
Solution:
Reaction of dihydrogen with dioxygen can be written as:
2H2 (g) + O2 (g) → 2H2 O(g)
Mole ratio
2
1
2
∴ moles ∝ volume
According to stiochimerty of reaction two volumes of dihydrogen react with one volume
of dioxygen to produce two volumes of water vapour. Hence, ten volumes of dihydrogen
will react with five volumes of dioxygen to produce ten volumes of water vapour.
27.
Convert the following into basic units:
(i) 28.7 pm
(ii) 15.15 pm
(iii) 25365 mg
Page 20 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
Some Basic Concepts Of Chemistry
Solution:
(i) 28.7 pm:
1pm = 10−12 m
∴ 28.7pm = 28.7 × 10−12 m
= 2.87 × 10−11 m
(ii) 15.15 pm:
1 pm = 10−12 m
∴ 15.15 pm = 15.15 × 10−12 m
= 1.515 × 10−12 m
(iii) 25365 mg :
1 mg = 10−3 g
25365 mg = 2.5365 × 104 × 10−3 g
Since,
1 g = 10−3 kg
2.5365 × 101 g = 2.5365 × 10−1 × 10−3 kg
∴ 25365 mg = 2.5365 × 10−2 kg
28.
Which one of the following will have the largest number of atoms?
(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1g Cl2 (g)
Solution:
we know ∶ NA = 6.022 × 1023
Atomic mass of Au = 107
Atomic mass of Na = 23
Atomic mass of Li = 7
Molar mass of Cl2 molecule = Atomic mass of chlorine × 2 = 35.5 × 2 = 71gmol−1
Page 21 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
1
1g Au(s) = 197 mol of Au(s)
(i)
1
×
197
Some Basic Concepts Of Chemistry
NA =
6.022×1023
197
atoms of Au (s)
= 3.06 × 1021 atoms of Au(s)
(ii) 1g of Na(s) =
1
23
× NA =
1
23
6.022×1023
23
mol of Na(s)
atoms of Na(s)
= 0.262 × 1023 atoms of Na(s)
= 26.2 × 1021 atoms of Na(s)
1
7
(iii)1 g Li(s) = mol of Li(s)
1
= 7 × NA =
6.022×1033
7
atoms of Li(s)
= 0.86 × 1023 atoms of Li (s)
= 86.0 × 1021 atoms of Li (s)
1
(iv)1g Cl2 (g) = mol of Cl2 (g)
71
=
1
×
71
NA =
6.022×1033
71
atoms of Cl2 (g)
= 0.0848 × 1023 atoms of Cl2 (g)
= 8.48 × 1021 atoms of Cl2 (g)
Hence, 1 g of Li(s) will have the largest number of atoms.
29.
Calculate the molarity of a solution of ethanol in water, in which the mole fraction of
ethanol is 0.040 (assume the density of water to be one).
Solution:
Given : mole fraction of ethanol = 0.040
We know : density of water = 1 gm/ml
Page 22 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
Some Basic Concepts Of Chemistry
the molecular mass of water (H2 O) =18
Number of moles of C H OH
2 5
Mole fraction of C2 H5 OH = Number of moles of solution
Number of moles of solution
= Number of moles of solute + Number of moles of solvent
Number of moles of solute = nC2 H5 OH
1000 g
Number of moles of solvent = nH2 O = 18 gmo𝑙−1 = 55.55mol
0.040 =
nC2 H5 OH
nC2 H5 OH + nH2 O
… (1)
Substituting the value of nH2 O in equation (1),
nC2 H5 OH
= 0.040
nC2 H5 OH + 55.55
nC2 H5 OH = 0.040 nC2 H5 OH + (0.040) × (55.55)
0.96nC2 H5 OH = 2.222mol
nC2 H5 OH =
2.222
0.96
mol
nC2 H5 OH ≡ 2.314mol
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 =
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝑙𝑖𝑡𝑒𝑟)
Molarity of solution =
2.314mo1
1L
= 2.314M
30.
What will be the mass of one 12C atom in g?
Solution:
1 mole of carbon atoms = 6.023 × 1023 atoms of carbon
= 12g of carbon
∴ Mass of one 12C = 12 × 1.67 × 10−24 g = 1.993 × 10−23 g
Page 23 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
31.
Some Basic Concepts Of Chemistry
How many significant figures should be present in the answer of the following
calculations?
(i)
0.02856×298.15×0.112
0.5785
(ii) 5 × 5.364
(iii) 0.0125 + 0.7864 + 0.0215
Solution:
(i)
0.02856×298.15×0.112
0.5785
Least precise number of calculation = 0.112
∴ Number of significant figures in the answer.
= Number of significant figures in the least precise number.
=3
(ii) 5 × 5.364
Least precise number of calculation = 5.364
∴ Number of significant figures in the answer = Number of significant figures in 5.364
=4
(iii) 0.0125 + 0.7864 + 0.0215
Since the least number of decimal places in each term is four, the number of significant
figures in the answer is also 4.
32.
Use the data given in the following table to calculate the molar mass of naturally
occuring argon isotopes:
Isotope
36
Ar
Ar
40
Ar
38
Isotopic molar mass
−1
35.96755 g mol
37.96272 g mol−1
39.9624 g mol−1
Abundance
0.337%
0.063%
99.600%
Page 24 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
Some Basic Concepts Of Chemistry
Solution:
Given :
for Isotope 36Ar ∶Isotopic molar mass
35.96755 g mol−1 , percentage Abundance 0.337%
for Isotope 38Ar ∶
Isotopic molar mass 37.96272 g mol−1 percentage Abundance0.063%
for Isotope
40
Ar ∶ Isotopic molar mass 39.9624 g mol−1 percentage Abundance99.600%
Average molar mass
=
molar mass of isotope 1×% abundance of isotope 1+ molar mass of isotope 2×% abundance of isotope 2+molar mass of isotope 3×% abundance of isotope3
100
Average molar mass of argon
= [ (35.96755 ×
0.337
0.063
90.60
) + (37.96272 ×
) + (39.9624 ×
)] gmo1−1
100
100
100
= [0.121 + 0.024 + 39.802] gmo1−1
= 39.947 g mol−1
33.
Calculate the number of atoms in each of the following
(i) 52 moles of Ar
(ii) 52 u of He
(iii) 52 g of He.
Solution:
We know : atomic mass of He=4 ; NA = 6.022 × 1023
(i) 1 mole of Ar = NA atoms of Ar = 6.022 × 1023 atoms of Ar
∴ 52 mol of Ar = 52 × 6.022 × 1023 atoms of Ar
= 3.131 × 1025 atoms of Ar
(ii) 1 atom of He = 4u of He
Or,
4u of He = 1 atom of He
1
1u of He = 4 atom of He
Page 25 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
52u of He =
52
𝑢
4
Some Basic Concepts Of Chemistry
atom of He
= 13u atoms of He
(ii)
I mole of He atom =4 g He
I mole of He atom = 𝑁𝐴 atoms of He = 6.022 × 1023 atoms of He
So 4 g of He = 6.022 × 1023 atoms of He
∴ 52 g of He =
6.022×1023 ×52
4
atoms of He
= 7.8286 × 1024 atoms of He
34.
A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in
oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of
10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate
(i) empirical formula,
(ii) molar mass of the gas, and
(iii) molecular formula.
Solution:
Given: a mass of carbon dioxide in welding fuel gas =3.38 g
Mass of Water welding fuel gas =0.690 g
The volume of welding fuel gas = 10 litre
Weight od welding fuel gas=11.6 g
(i)
We know 1 mole (44g) of CO2 contains 12g of carbon.
12g
∴ 3.38g of CO2 will contain carbon = 44g × 3.38g
= 0.9217g
we know 18 g of water(H2 O) contains 2 g of hydrogen.
2g
∴ 0.690 g of water will contain hydrogen = 18g × 0.690
= 0.0767g
Page 26 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
Some Basic Concepts Of Chemistry
Since carbon and hydrogen are the only constituents of the compound, the total mass of
the compound is:
= 0.9217g + 0.0767g
= 0.9984g
weight of atom
Formula : Percent of atom in the compound = weight of sample × 100
0.9217g
∴ Percent of C in the compound = 0.9984g × 100
= 92.32%
Percent of H in the compound =
0.0767g
×
0.9984g
100
= 7.68%
𝑤𝑒𝑖𝑔ℎ𝑡
Formula : 𝑚𝑜𝑙𝑒𝑠 = 𝑎𝑡𝑜𝑚𝑖𝑐 𝑤𝑒𝑖𝑔ℎ𝑡
92.32
12
Moles of carbon in the compound =
= 7.69
Moles of hydrogen in the compound =
7.68
1
= 7.68
∴ Ratio of carbon to hydrogen in the compound = 7.69: 7.68
= 1: 1
Hence, the empirical formula of the gas is CH.
(ii) Given,
Weight of 10.0L of the gas (at S.T.P) = 11.6g
∴ Weight of 22.4 L of gas at STP =
11.6 g
×
10 L
22.4L
= 25.984g
≈ 26g
Hence the molar mass of the gas is 26 g.
(iii) Empirical formula mass of CH = 12 + 1 = 13 g
n=
Molar mass of gas
Empirical formula mass of gas
Page 27 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
=
Some Basic Concepts Of Chemistry
26g
13g
n=2
formula : Molecular formula of gas= ( emperical formula of gas)× 𝑛 n
∴ Molecular formula of gas = (CH)n
= C2 H2
35.
Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the
reaction,
CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + CO2 (g) + H2 O(l).
What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?
Solution:
given : volume of HCl solution = 25 mL
molarity of HCl solution = 0.75 M
we know molecular formula of CaCO3 = 100
formula : moles = molarity × volume (litre)
moles of HCl =
0.75 M ×25 mL
1000
= 0.01875
From the given chemical equation,
CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + CO2 (g) + H2 O(l)
According to stoichiometry of reaction :
∴ 2 mol of HCl react with = 1 mol of CaCO3
1
2
0.01875 mol HCl of react with = × 0.01875 mol of CaCO3
(∴ weight = moles × molecular weight )
1
Amount of CaCO3 that will react = 2 × 0.01875 × 100
= 0.9375 g
Page 28 of 29
Practice more
www.embibe.com
Class-XI-CBSE-Chemistry
36.
Some Basic Concepts Of Chemistry
Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2 ) with
aqueous hydrochloric acid according to the reaction
4HCl(aq) + MnO2 (s) → 2H2 O(l) + MnCl2 (aq) + Cl2 (g)
How many grams of HCl react with 5.0 g of manganese dioxide?
Solution:
given : weight of manganese dioxide= 5 g
we know molecular mass of MnO2 = [55 + 2 × 16 = 87g]
molecular mass of HCl = [35.5 + 1 = 36.5]
weight
5
moles of MnO2 = molecular weight = 87 = 0.057 mole
according to stoichiometry
1 mole MnO2 reacts completely with 4 mol of HCl.
4
1
∴ 0.057 mole of MnO2 will react with HCl = × 0.057 =
0.2298 mole of HCl formed
weight of HCl = moles molecular × moles of HCl = 36.5 × 0.2298 =8.4g
Hence, 8.4g of HCl will react completely with 5.0g of manganese dioxide.
Page 29 of 29
Practice more
www.embibe.com