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Page No 128:

Question 1:

Write the full form of IUPAC.

Answer:

The full form of IUPAC is International Union of Pure and Applied Chemistry.

Page No 128:

Question 2:

Name the scientist who gave :
(a) law of conservation of mass.
(b) law of constant proportions.

Answer:

(a) Law of conservation of mass was proposed by Antoine Lavoisier.
(b) Law of constant proportions was proposed by Joseph Proust.

Page No 128:

Question 3:

Name the law of chemical combination :
(a) Which was given by Lavoisier.
(b) which was given by Proust.

Answer:

(a) The law of chemical combination proposed by Lavoisier is well known as the law of conservation of mass.
(b) The law of chemical combination proposed by Proust is well known as the law of constant proportions.



Page No 129:

Question 21:

(a) Name the element used as a standard for atomic mass scale.
(b) Which particular atom of the above element is used for this purpose ?
(c) What value has been given to the mass of this reference atom ?

Answer:

(a) Carbon is used as a standard in the atomic mass scale.
(b) Carbon-12 atom is particularly used as a standard in the atomic mass scale.
(c) Carbon-12 atom has been given a mass of exactly 12 units (C612 has six protons and six neutrons).

Page No 129:

Question 22:

Give one major drawback of Dalton's atomic theory of matter.

Answer:

Dalton's atomic theory states that atoms are indivisible; however, under special circumstances, atoms can be further divided into smaller particles called electrons, protons and neutrons. So, atoms are made of subatomic particles into which they can be broken or divided.

Page No 129:

Question 23:

Dalton's atomic theory says that atoms are indivisible. Is this statement still valid ? Give reasons for your answer.

Answer:

Dalton’s atomic theory states that atoms are indivisible. This statement is not valid as, atoms can be further divided into electrons, protons and neutrons. This is one of the major drawbacks of Dalton’s atomic theory of matter.

Page No 129:

Question 24:

Is it possible to see atoms these days ? Explain your answer.

Answer:

Yes, it is possible to see atoms these days, with the help of advanced technology. Electronic microscopes are extremely powerful and can magnify even the smallest particle. A Scanning Tunnelling Microscope (STM) can be used to produce highly magnified images of the surface of elements on a computer screen. Therefore, a highly sophisticated STM can be used to view atoms indirectly.

Page No 129:

Question 25:

What is meant by the symbol of an element ? Explain with examples.

Answer:

The symbol of an element is either the first letter or the first and another letter of the English or Latin name of the element.
For example: Oxygen is represented by the symbol O. Similarly Iron, whose Latin name is Ferrum, is represented by the symbol Fe.

Page No 129:

Question 26:

(a) Give two symbols which have been derived from the "English names" of the elements.
(b) Give two symbols which have been derived from the " Latin names" of the elements.

Answer:

(a) Two examples of symbols which have been derived from the “English names” of their respective elements: Sulphur – S and Aluminium – Al.

(b) Following are the two symbols which have been derived from the “Latin names” of their respective elements: Iron – Ferrum (Fe) and Copper (Cu)-Cuprum.

Page No 129:

Question 27:

Give the names and symbols of five familiar substances which you think are elements.

Answer:

The names of five familiar elements are:
• Oxygen:       O
• Iron:            Fe
• Gold:           Au
• Silver:         Ag
• Hydrogen:   H

Page No 129:

Question 28:

State the chemical symbols for the following elements :
Sodium, Potassium, Iron, Copper, Mercury, Silver.

Answer:

Following are the chemical symbols of the listed elements:
• Sodium:       Na
• Potassium:   K
• Iron:            Fe
• Copper:       Cu
• Mercury:     Hg
• Silver:          Ag

Page No 129:

Question 29:

Name the elements represented by the following symbols :
Hg, Pb, Au, Ag, Sn

Answer:

Following are the chemical names for the listed symbols of elements:
• Hg: Mercury
• Pb: Lead
• Au: Gold
• Ag: Silver
• Sn: Tin

Page No 129:

Question 30:

What is meant by atomicity ? Explain with two examples.

Answer:

Atomicity refers to the total number of atoms present in one molecule of an element.
For example, argon is a noble gas and exists in a free state. One molecule of argon comprises only one atom. Therefore, the atomicity of argon is one.
Similarly, a molecule of chlorine gas has two atoms and it exists as a Cl2 molecule. Therefore, the atomicity of chlorine is two.

Page No 129:

Question 31:

What is the atomicity of the following ?
(a) Oxygen
(b) Ozone
(c) Neon
(d) Sulphur
(e) Phosphorus
(f) Sodium

Answer:

The atomicity of the following:
(a) Oxygen:        2
(b) Ozone:          3
(c) Neon:            1
(d) Sulphur:       8
(e) Phosphorus: 4
(f) Sodium:         1

Page No 129:

Question 32:

What is meant by a chemical formula ? Write the formulae of one element and one compound.

Answer:

A chemical formula is an expression that denotes the number and types of atoms present in a molecule of a substance. It is a representation of the composition of a molecule, in terms of the elements present, using their symbols.

For example: O2 is the chemical formula of the element oxygen. Therefore, one molecule of oxygen comprises two atoms.

CO2 is the chemical formula of the compound carbon dioxide. It consists of one atom of carbon and two atoms of oxygen.

Page No 129:

Question 33:

Write the formulae of the following compounds. Also name the elements present in them.
(a) Water
(b) Ammonia
(c) Methane
(d) Sulphur dioxide
(e) Ethanol

Answer:

  Compound Formula Elements present
(a) Water H2O Hydrogen(H) and Oxygen(O)
(b) Ammonia NH3 Nitrogen(N) and Hydrogen(H)
(c) Methane CH4  Carbon(C) and Hydrogen(H)
(d) Sulphur dioxide SO2 Sulphur(S) and Oxygen(O)
(e) Ethanol C2H5OH Carbon(C), Hydrogen(H) and Oxygen(O)
                     
              
   
       

Page No 129:

Question 34:

Explain the difference between 2N and N2.

Answer:

N2 is the chemical formula of nitrogen gas. It shows that two atoms of nitrogen form chemical bond to give one molecule of nitrogen gas.
2N, on the other hand, represents two atoms of nitrogen, which are not chemically bonded.

Page No 129:

Question 4:

Name the scientist who gave atomic theory of matter.

Answer:

John Dalton proposed the atomic theory of matter.

Page No 129:

Question 5:

Which postulate of Dalton's atomic theory is the result of law of conservation of mass given by Lavoisier ?

Answer:

The postulate of Dalton's atomic theory, which is the result of the law of conservation of mass proposed by Lavoisier, states that, “Atoms can neither be created nor destroyed.”

Page No 129:

Question 6:

Which part of the Dalton's atomic theory came from the law of constant proportions given by Proust ?

Answer:

The postulate of Dalton's atomic theory which came from the law of constant proportions proposed by Proust, states that “The 'number' and 'kind' of atoms in a given compound is fixed.”

Page No 129:

Question 7:

Which ancient Indian philosopher suggested that all matter is composed of very small particle ? What name was given by him to these particles ?

Answer:

Maharishi Kanada was the ancient Indian philosopher who suggested that  matter  comprises of very small particles. He named these particles ‘parmanu’.

Page No 129:

Question 8:

Name any two laws of chemical combination.

Answer:

Two laws of chemical combination are:
  i.  Law of conservation of mass
  ii. Law of multiple proportions

Page No 129:

Question 9:

'If  100 grams of pure water taken from different sources is decomposed by passing electricity, 11 grams of hydrogen and 89 grams of oxygen are always obtained'. Which chemical law is illustrated by this statement ?

Answer:

According to the given information, water has been taken from different sources and still decomposes into same amount of oxygen and hydrogen; therefore, this is an example of the law of constant proportions.

Page No 129:

Question 10:

'If 100 grams of calcium carbonate (whether in the form of marble or chalk) are decomposed completely, then 56 grams of calcium oxide and 44 grams of carbon dioxide are obtained'. Which law of chemical combination is illustrated by this statement ?

Answer:

If 100 grams of calcium carbonate is decomposed in any form (marble or chalk), 56 grams of calcium oxide and 44 grams of carbon dioxide are obtained. This reactions illustrates the law of conservation of mass.

Page No 129:

Question 11:

What are the building blocks of matter ?

Answer:

Atoms and molecules are the building blocks of matter.

Page No 129:

Question 12:

How is the size of an atom indicated ?

Answer:

The size of an atom is indicated by its radius.

Page No 129:

Question 13:

Name the unit in which the radius of an atom is usually expressed.

Answer:

The radius of an atom is usually expressed in nanometre (nm).

Page No 129:

Question 14:

Write the relation between nanometre and metre.

Answer:

1 nanometre = 1/109 metre
or
1 nanometre = 10–9 metre

Page No 129:

Question 15:

The radius of an oxygen atom is 0.073 nm. What does the symbol 'nm' represent ?

Answer:

The symbol 'nm' in 0.073 nm (radius of an oxygen atom) represents the unit of atomic radius, i.e. nanometre.

Page No 129:

Question 16:

Why is it not possible to see an atom even with the most powerful microscope ?

Answer:

Atoms are the building blocks of nature. They are tiniest particles in nature and optical microscopes cannot magnify the image of such a tiny object, so that becomes visible to us. It is possible to watch an atom under an electron microscope.   

Page No 129:

Question 17:

State whether the following statement is true or false :
The symbol of element cobalt is CO.

Answer:

False. The symbol for the element Cobalt is Co, as the second letter of the symbol of an element has to be in the lower case.

Page No 129:

Question 18:

Define 'molecular mass' of a substance.

Answer:

The molecular mass of a substance is the relative mass of its one molecule as compared with a C612 atom. The mass of an atom of C612 is considered to be 12 a.m.u.

Page No 129:

Question 19:

What is meant by saying that 'the molecular mass of oxygen is 32' ?

Answer:

The statement, "the molecular mass of oxygen is 32", means that a molecule of oxygen is 32 times heavier than 1/12th of the mass of a C612 atom.

Page No 129:

Question 20:

Fill the following blanks with suitable words :
(a) In water, the proportion of oxygen and hydrogen is ...................by mass.
(b) In a chemical reaction, the sum of the masses of the reactants and the products remains unchanged. This is called ....................

Answer:

(a) In water, the ratio of oxygen to hydrogen is 8:1 by mass.
(b) In a chemical reaction, the sum of masses of the reactants and their products remain unchanged. This is called the law of conservation of mass.



Page No 130:

Question 35:

What do the following abbreviations stand for ?
(i) O
(ii) 2O
(iii) O2
(iv) 3O2

Answer:

(i) O stands for an atom of oxygen.
(ii) 2O refers to two atoms of oxygen which are not chemically bonded.
(iii) O2 refers to a molecule of oxygen in which atoms are chemically bonded to give a molecule of oxygen gas.
(iv) 3O2 refers to three molecules of oxygen.

Page No 130:

Question 36:

What do the symbols, H2, S and O4 mean in the formula H2SO4 ?

Answer:

In H2SO4, H2 represents two atoms of hydrogen; S represents an atom of sulphur; O4 represents 4 atoms of oxygen where all these atoms are chemically bonded to each other to form a molecule of sulphuric acid i.e., H2SO4.

Page No 130:

Question 37:

(a) In what form does oxygen gas occur in nature ?
(b) In what form do noble gases occur in nature ?

Answer:

(a) Naturally, oxygen  has diatomic molecules. Two oxygen atoms chemically combine with each other to give a molecule of O2.

(b) Noble gases are present in the earth’s atmosphere. They exist as monoatomic gases.

Page No 130:

Question 38:

What is the difference between 2H and H2 ?

Answer:

H2 is the chemical formula of hydrogen gas. It denotes that two atoms of hydrogen chemically bond to give a molecule of hydrogen gas.
2H denotes the presence of two atoms of hydrogen, which are not chemically bonded.

Page No 130:

Question 39:

What do the following denote ?
(i) N
(ii) 2N
(iii) N2
(iv) 2N2

Answer:

(i) N refers to an atom of nitrogen.
(ii) 2N represents two atoms of nitrogen which are not chemically bonded.
(iii) N2 refers to a molecule of nitrogen gas. Two atoms of nitrogen chemically bond to give a molecule of nitrogen gas.
(iv)2N2 refers to two molecules of nitrogen gas.

Page No 130:

Question 40:

What is the significance of the formula of a substance ?

Answer:

The significance of the chemical formula of any substance is as follows:
• A chemical formula of a substance (element or compound) represents its name.
• A chemical formula provides important information on the chemical makeup of any substance (element or compound). It names all the elements present in the molecule.
• A chemical formula also provides important information on the number of atoms of each constituent element of the molecule of the substance.
• A chemical formula represents one mole of molecules of the substance (element or compound), which is 6.022 × 1023 molecules of the substance.
• Since, a chemical formula represents one mole of molecules of a substance, it also represents the definite mass of that substance, which is expressed as gram molecular mass.

Page No 130:

Question 41:

What is the significance of the formula H2O ?

Answer:

The significance of the formula H2O is:

• It represents one molecule of water.
• It provides important information on the constituents of water i.e. hydrogen and oxygen, through the chemical formula.
• It also provides information on the number of atoms of each element present in one molecule of water.
• It refers to the presence of one mole of molecules of water; it refers to 6.022 × 1023 molecules of water.

Page No 130:

Question 42:

The molecular formula of glucose is C6H12O6. Calculate its molecular mass. (Atomic masses : C = 12 u ; H = 1 u ; O = 16 u)

Answer:

Given:
Atomic mass of C = 12 u
Atomic mass of H = 1 u
Atomic mass of O = 16 u
Therefore, molecular mass of C6H12O6 can be calculated as follows:
(6 × 12) + (12 × 1) + (6 ×16) = 180
Hence, molecular mass of glucose is 180 u.

Page No 130:

Question 43:

Calculate the molecular masses of the following :
(a) Hydrogen, H2
(b) Oxygen, O2
(c) Chlorine, Cl2
(d) Ammonia, NH3
(e) Carbon dioxide, CO2

(Atomic masses : H = 1 u ; O = 16 u ; Cl = 35.5 u ; N = 14 u ; C = 12 u)

Answer:

(a) Given:
Atomic mass of H = 1 u
Therefore, molecular mass of H2 can be calculated as follows:
1 × 2 = 2
Hence, molecular mass of H2 is 2 u.

(b) Given:
Atomic mass of O = 16 u
Therefore, molecular mass of O2 can be calculated as follows:
16 × 2 = 32
Hence, molecular mass of O2 is 32 u.

(c) Given:
Atomic mass of Cl = 35.5 u
Therefore, molecular mass of Cl2 can be calculated as follows:
35.5 × 2 = 71
Hence, molecular mass of Cl2 is 71 u.

(d) Given:
Atomic mass of N = 14 u
Atomic mass of H = 1 u
Therefore, molecular mass of NH3 can be calculated as follows:
(1 × 14) + (3 × 1) = 17
Hence, molecular mass of NH3 is 17 u.

(e) Given:
Atomic mass of C = 12 u
Atomic mass of O = 16 u
Therefore, molecular mass of CO2 can be calculated as follows:
(1 × 12) + (2 ×16) = 44
Hence, molecular mass of CO2 is 44 u.

Page No 130:

Question 44:

Calculate the molecular masses of the following compounds  :
(a) Methane, CH4
(b) Ethane, C2H6
(c) Ethene, C2H4
(d) Ethyne, C2H2

(Atomic masses : C = 12 u ; H = 1 u)

Answer:

(a) Given:
Atomic mass of C = 12 u
Atomic mass of H = 1 u
Therefore, molecular mass of CH4 (methane) can be calculated as follows:
(1 × 12) + (4 × 1) = 16
Hence, molecular mass of methane is 16 u.

(b) Given:
Atomic mass of C = 12 u
Atomic mass of H = 1 u
Therefore, molecular mass of C2H6 (ethane) can be calculated as follows:
(2 × 12) + (6 × 1) = 30
Hence, molecular mass of ethane is 30 u.

(c) Given:
Atomic mass of C = 12 u
Atomic mass of H = 1 u
Therefore, molecular mass of C2H4 (ethylene) can be calculated as follows:
(2 × 12) + (4 × 1) = 28
Hence, molecular mass of ethylene is 28 u.

(d) Given:
Atomic mass of C = 12 u
Atomic mass of H = 1 u
Therefore, molecular mass of C2H2 (acetylene) can be calculated as follows:
(2 × 12) + (2 × 1) = 26
Hence, molecular mass of acetylene is 26 u.

Page No 130:

Question 45:

Calculate the molecular masses of the following compounds :
(a) Methanol, CH3OH
(b) Ethanol, C2H5OH

Answer:

(a) Given:
Atomic mass of C = 12 u
Atomic mass of O = 16 u
Atomic mass of H = 1 u
Therefore, molecular mass of CH3OH can be calculated as follows:
(1 ×12) + (4 × 1) + (1 × 16) = 32
Hence, molecular mass of methanol is 32 u.

(b) Given:
Atomic mass of C = 12 u
Atomic mass of O = 16 u
Atomic mass of H = 1 u
Therefore, molecular mass of C2H5OH can be calculated as follows:
(2 × 12) + (6 × 1) + (1 ×16) = 46
Hence, molecular mass of ethanol is 46 u.

Page No 130:

Question 46:

Calculate the molecular mass of ethanoic acid, CH3COOH.
(Atomic masses : C = 12 u ; H = 1 u ; O = 16 u)

Answer:

Given:
Atomic mass of C = 12 u
Atomic mass of O = 16 u
Atomic mass of H = 1 u
Therefore, molecular mass of CH3COOH can be calculated as follows:
(2 × 12) + (4 × 1) + (2 ×16) = 60
Hence, molecular mass of acetic acid is 60 u.

Page No 130:

Question 47:

Calculate the molecular mass of nitric acid, HNO3. (Atomic masses : H = 1 u ; N = 14 u ; O = 16 u)

Answer:

Given:
Atomic mass of H = 1 u
Atomic mass of N = 14 u
Atomic mass of O = 16 u
Therefore, molecular mass of HNO3 can be calculated as follows:
(1 × 1) + (1 × 14) + (3 ×16) = 63
Hence, molecular mass of nitric acid is 63 u.

Page No 130:

Question 48:

Calculate the molecular mass of chloroform (CHCl3). (Atomic masses : C = 12 u ; H = 1 u ; Cl = 35.5 u)

Answer:

Given:
Atomic mass of C = 12 u
Atomic mass of H = 1 u
Atomic mass of Cl = 35.5 u
Therefore, molecular mass of CHCl3 can be calculated as follows:
(1 × 12) + (1 × 1) + (3 × 35.5) = 119.5
Hence, molecular mass of chloroform is 119.5 u.

Page No 130:

Question 49:

Calculate the molecular mass of hydrogen bromide (HBr). (Atomic masses : H = 1 u ; Br = 80 u)

Answer:

Given:
Atomic mass of H = 1 u
Atomic mass of Br = 80 u
Therefore, molecular mass of HBr can be calculated as follows:
(1 × 1) + (1 × 80) = 81
Hence, molecular mass of hydrogen bromide is 81 u.

Page No 130:

Question 50:

Calculate the molecular masses of the following compounds :
(a) Hydrogen sulphide, H2S
(b) Carbon disulphide, CS2

(Atomic masses : H = 1 u ; S = 32 u ; C = 12 u)

Answer:

(a) Given:
Atomic mass of H = 1 u
Atomic mass of S = 32 u
Therefore, molecular mass of H2S can be calculated as follows:
(2 × 1) + (1 × 32) = 34
Hence, molecular mass of hydrogen sulphide is 34 u.

(b) Given:
Atomic mass of C = 12 u
Atomic mass of S = 32 u
Therefore, molecular mass of CS2 can be calculated as follows:
(1 × 12) + (2 × 32) = 76
Hence, molecular mass of carbon disulphide is 76 u.

Page No 130:

Question 51:

State the law of conservation of mass. Give one example to illustrate this law.

Answer:

Law of conservation of mass states that the matter is neither created nor destroyed in a chemical reaction.
It was proposed by Antoine Lavoisier in 1774. It means that in a chemical reaction, the total mass of products (new substances that are formed in a chemical reaction) is equal to the total mass of reactants (substances which combine together in a chemical reaction).

Law of conservation of mass can be illustrated by the following chemical reaction where 100 grams of calcium carbonate decomposes to produce 56 grams of calcium oxide and 44 grams of carbon dioxide.


Calcium carbonateCalcium oxide+Carbon dioxide100 gm56 gm44 gm


Here,
Mass of reactant (calcium carbonate) = 100gm
Mass of products (calcium oxide & carbon dioxide) = 56gm + 44gm = 100gm
Since, the mass of reactant (100gm) is equal to the mass of the products (100gm), there is no change in mass in this chemical reaction. Therefore, this example illustrates the law of conservation of mass.

Page No 130:

Question 52:

State the law of constant proportions. Give one example to illustrate this law.

Answer:

Law of constant proportions states that a chemical compound always consists of the same elements combined together in the same proportion by mass. It was put forward by Joseph Proust in 1779
It means that a pure chemical compound is always made up of the same elements in the same mass percentage.

Law of constant proportions can be illustrated by the following example:

Carbon dioxide exists in nature in different forms and can be chemically produced in different ways. Yet, all the different samples of carbon dioxide will always contain the same elements i.e. carbon and oxygen in the same proportion by mass i.e. 3:8.

Page No 130:

Question 53:

(a) State the various postulates of Dalton's atomic theory of matter.
(b) Which postulate of Dalton's atomic theory can explain the law of conservation of mass ?
(c) Which postulate of Dalton's atomic theory can explain the law of constant proportions ?

Answer:

(a) Postulates of Dalton's atomic theory of matter :

• Matter comprises very small particles called atoms.
• Atoms of different elements differ in mass, size and chemical properties.
• Atoms can neither be created nor destroyed.
• Chemical reactions involve the combination, separation or rearrangement of atoms to form molecules or compounds.
• Atoms cannot be divided.
• A compound has a fixed number and kind of atoms.
• Atoms of the same element can combine in more than one ratio to form more than one compound.
• All the atoms in an element are identical in every aspect i.e. they have same mass, size and chemical properties.
• Like elements, there are many kinds of atoms.
• Atoms of different elements join together to form molecules of the compound in a chemical combination.

(b) Dalton’s third postulate explains the law of conservation of mass. It states that atoms can neither be created nor destroyed.

(c)  Dalton’s postulate explains the law of constant proportions. It states that elements consist of atoms which have a fixed mass and also the number and kind of atoms in each element of a compound is fixed.

Page No 130:

Question 54:

(a) What is the significance of the symbol of an element ? Explain with the help of an example.
(b) Explain the significance of the symbol H.

Answer:

(a) The symbol of an element is a short form used to represent individual elements or the atoms of the elements. Using symbols is advantageous, as using full names of the elements can be tedious. Symbols are commonly used to designate elements in a chemical reaction. Following are the significant points of using a symbol to designate an element:

• Symbol represents the name of the element. For example, the symbol O refers to the element oxygen.
Symbol represents one atom of the element. For example, the symbol O denotes one atom of the element oxygen.
• Using the symbol of an element refers to one mole of atoms of that element; therefore, the symbol of an element represents 6.022 × 1023 atoms. For example, the symbol O represents one mole of oxygen  i.e. 6.022 × 1023 atoms.
• Symbols of different elements represent a definite mass of those elements which is equal to gram atomic mass. For example, the symbol O refers to 1 gram atomic mass of oxygen.

(b) Significance of symbol H

• The symbol H refers to the element hydrogen.
• The symbol H refers to one atom of  hydrogen.
• The symbol H refers to one mole of hydrogen  i.e. 6.022 × 1023 atoms.
• The symbol H refers to 1 gram atomic mass of hydrogen.

Page No 130:

Question 55:

(a) What is an atom ? How do atoms usually exist ?
(b) What is a molecule ? Explain with an example.
(c) What is the difference between the molecule of an element and the molecule of a compound ? Give one example of each.

Answer:

(a) Atoms are building blocks of all matter. They are the smallest particles of elements that can take part in a chemical reaction.

Atoms usually exist in combined form, in two ways:
• in the form of molecules
• in the form of ions

(b) Molecule is the smallest particle of a substance (element or compound), which has the properties of that substance. It can exist in a free state.

Example: Oxygen atoms combine and exist in the form of O3 molecules. Solid sulphur has eight sulphur atoms joined together, and exists in the form of S8 molecule.

(c)

Molecule of an element Molecule of a compound
1. It consists of two or more atoms of the same element chemically bonded together. 1. It consists of two or more atoms of different elements chemically bonded together.
2. Example: Bromine exists in nature as a diatomic element (Br2), wherein two bromine atoms are covalently bonded to give a molecule of bromine. 2. Example: A molecule of nitric acid, HNO3, comprises one atom of hydrogen, one atom of nitrogen and three atoms of oxygen chemically bonded together.

Page No 130:

Question 56:

(a) Define atomic mass unit. What is its symbol ?
(b) Define atomic mass of an element.
(c) What is meant by saying that 'the atomic mass of oxygen is 16' ?

Answer:

(a) Atomic mass unit can be defined as one twelfth (1/12) of the mass of an unbound atom of carbon-12. The atomic masses of all the elements are determined by comparing the mass of their atom with that of the carbon-12 atom.

Atomic mass unit = 1/12 the mass of a carbon-12 atom

(b) The symbol of atomic mass unit is “u”.

(c) Atomic mass of an element refers to the relative mass of its atom compared to the mass of a carbon-12 atom. (Mass of carbon-12 atom is taken to be 12 u).

(d) By saying that the atomic mass of oxygen is 16, we mean that the atom of oxygen is 16 times heavier compared to 1/12th the mass of a carbon-12 atom.

Page No 130:

Question 57:

The atomicities of ozone, sulphur, phosphorus and argon are respectively :
(a) 8, 3, 4 and 1
(b) 1, 3, 4 and 8
(c) 4, 1, 8 and 3
(d) 3, 8, 4 and 1

Answer:

(d)  3, 8, 4 and 1

Page No 130:

Question 58:

The symbol of a metal element which is used in making thermometers is :
(a) Ag
(b) Hg
(c) Mg
(d) Sg

Answer:

(b) Hg

The symbol of the metal used in making thermometers is Hg (Mercury).



Page No 131:

Question 59:

The Latin language name of an element is natrium. The English name of this element is :
(a) sodium
(b) potassium
(c) magnesium
(d) sulphur

Answer:

(a) Sodium

The English name of the element natrium is Sodium (Na).

Page No 131:

Question 60:

The atomic theory of matter was proposed by :
(a) John Kennedy
(b) Lavoisier
(c) Proust
(d) John Dalton

Answer:

(d) John Dalton

The atomic theory of matter was proposed by John Dalton.

Page No 131:

Question 61:

One of the following elements has an atomicity of 'one'. This element is :
(a) helium
(b) hydrogen
(c) sulphur
(d) ozone

Answer:

(a) Helium
The atomicity of Helium (He) is 1.

Page No 131:

Question 62:

The English name of an element is potassium, its Latin name will be :
(a) plumbum
(b) cuprum
(c) kalium
(d) natrium

Answer:

(c) kalium

The Latin name of the element potassium is kalium. Hence, the symbol for potassium is K.

Page No 131:

Question 63:

The law of conservation of mass was given by :
(a) Dalton
(b) Proust
(c) Lavoisier
(d) Berzelius

Answer:

(c) Antoine Lavoisier

The law of conservation of mass was proposed by Antoine Lavoisier.

Page No 131:

Question 64:

The element having atomicity 'four' is most likely to be :
(a) argon
(b) fluorine
(c) phosphorus
(d) francium

Answer:

(c) Phosphorus

Phosphorus has an atomicity of 4.

Page No 131:

Question 65:

If 1.4 g of calcium oxide is formed by the complete decomposition of calcium carbonate, then the amount of calcium carbonate taken and the amount of carbon dioxide formed will be respectively :
(a) 2.2 g and 1.1 g
(b) 1.1 g and 2.5 g
(c) 2.5 g and 1.1 g
(d) 5.0 g and 1.1 g

Answer:

(c) 2.5 gm and 1.1 gm

Molecular mass of CaCO3: 40 + 12 + (16 × 3) = 100g
Molecular mass of CaO: 40 + 16 = 56g
Molecular mass of CO2: 12 + (16 × 2) = 44g
Reaction for the decomposition of calcium carbonate:

CaCO3CaO+CO2100g56g44gX1.4gY

56 g of CaO is obtained when 100g of CaCO3 is burnt.
Therefore, to get 1.4g of CaO, the amount of CaCO3 to be burnt will be
100×1.456=2.5g
For CO2,
100g of CaCO3 yields 44g of CO2
Therefore, 2.5g of CaCO3 will yield;
44×2.5100=1.1 g

Page No 131:

Question 66:

The law of constant proportions was given by :
(a) Proust
(b) Lavoisier
(c) Dalton
(d) Berzelius

Answer:

(a) Proust

The law of constant proportion was proposed by Joseph Proust.

Page No 131:

Question 67:

Out of ozone, phosphorus, sulphur and krypton, the elements having the lowest and highest atomicities are respectively :
(a) sulphur and krypton
(b) krypton and ozone
(c) phosphorus and sulphur
(d) krypton and sulphur

Answer:

(d) krypton and sulphur

Out of ozone, phosphorus, sulphur and krypton, krypton has the lowest atomicity and sulphur  the highest .

Page No 131:

Question 68:

One nm is equal to :
(a) 10–9 mm
(b) 10–7 cm
(c) 10–9 cm
(d) 10–6 m

Answer:

(b) 10−7 cm

One nanometre is equal to 10−7 centimetre.

Page No 131:

Question 69:

The scientist who proposed the first letter (or first letter and another letter) of the Latin or English name of an element as its symbol, was :
(a) Dalton
(b) Proust
(c) Lavoisier
(d) Berzelius

Answer:

(d) Berzelius

Berzelius proposed the first letter (or first letter and another letter) of the Latin or English name of any element as its symbol.

Page No 131:

Question 70:

The atoms of which of the following pair of elements are most likely to exist in free state ?
(a) hydrogen and helium
(b) argon and carbon
(c) neon and nitrogen
(d) helium and neon

Answer:

(d) helium and neon

Atoms of helium and neon exist in a free state because they are inert gases and chemically inactive.

Page No 131:

Question 71:

Which of the following elements has the same molecular mass as its atomic mass ?
(a) nitrogen
(b) neon
(c) oxygen
(d) chlorine

Answer:

(b) Neon

Neon’s molecular mass is equivalent to its atomic mass.

Page No 131:

Question 72:

In water , the proportion of oxygen and hydrogen by mass is :
(a) 1 : 4
(b) 1 : 8
(c) 4 : 1
(d) 8 : 1

Answer:

(d) 8:1

In water, the ratio of oxygen to hydrogen is 8:1 by mass.

Page No 131:

Question 73:

In hydrogen peroxide (H2O2), the proportion of hydrogen and oxygen by mass is :
(a) 1 : 8
(b) 1 : 16
(c) 8 : 1
(d) 16 : 1

Answer:

(b) 1:16

In hydrogen peroxide, the ratio of hydrogen to oxygen is 1:16 by mass.

Page No 131:

Question 74:

The symbols of the elements cobalt, aluminium, helium and sodium respectively written by a student are as follows. Which symbol is the correct one ?
(a) CO
(b) AL
(c) He
(d) So

Answer:

(c)  He

He is the correct symbol of helium.

Page No 131:

Question 75:

Copper sulphate reacts with sodium hydroxide to form a blue precipitate of copper hydroxide and sodium sulphate. In an experiment, 15.95 g of copper sulphate reacted with 8.0 g of sodium hydroxide to form 9.75 g of copper hydroxide and 14.2 g of sodium sulphate. Which law of chemical combination is illustrated by this data ? Give reason for your choice.

Answer:

Chemical reaction:
CuSO4 + NaOH → Cu(OH)2 + NaSO4
15.94g      8.0g         9.75g        14.2g

Total mass of reactants = Total mass of products
 23.95g                                  23.95 g

The above reaction and data illustrate the law of conservation of mass which was proposed by Antoine Lavoisier.

The law states that the mass of elements remains constant over time i.e. mass can neither be created nor destroyed. In the above reaction, the mass of reactants (23.95 g) is exactly equal to the mass of their product (23.95 g). There is no change in the mass of the product after the reaction.

Page No 131:

Question 76:

Potassium chlorate decomposes, on heating, to form potassium chloride and oxygen. When 24.5 g of potassium chlorate is decomposed completely, then 14.9 g of potassium chloride is formed. Calculate the mass of oxygen formed. Which law of chemical combination have you used in solving this problem?

Answer:

Chemical reaction:

2KClO32KCl+3O224.5 g14.9 g

So, mass of oxygen formed during the reaction:
Mass of reactants = 24.5 g
Mass of products = mass of the potassium chloride + mass of oxygen
24.5 = 14.9 + mass of oxygen
Mass of oxygen = 24.5 – 14.9 = 9.6 g
We have used the law of conservation of mass, proposed by Antoine Lavoisier, to solve this problem.

Page No 131:

Question 77:

In an experiment, 4.90 g of copper oxide was obtained from 3.92 g of copper. In another experiment, 4.55 g of copper oxide gave, on reduction, 3.64 g of copper. Show with the help of calculations that these figures verify the law of constant proportions.

Answer:

To solve this problem, we need to work out the proportion of copper and oxygen in the two reactions.
1st chemical reaction:

4Cu+O22Cu2O3.92 gX g4.90 g

Therefore, mass of oxygen(X) = mass of copper oxide – mass of copper
Mass of oxygen = 4.9 – 3.92 = 0.98 g

In the above reaction, the ratio of mass of copper to the mass of oxygen is 3.92 : 0.98 = 4:1

2nd chemical reaction:
CuO + H2  → Cu + H2O
4.55g   Xg    3.64g  Yg

Molecular mass of copper oxide = 75.5g
Molecular mass of hydrogen = 2g
Molecular mass of copper = 63.5g
Molecular mass of water = 18g
Molecular mass of oxygen = 16g
Mass of hydrogen = 2/63.5 × 3.64 = 0.11g

Mass of water = 18/75.5 × 4.55 = 1.08g
Mass of oxygen in water = 16/18 × 1.08 = 0.96g

In the above reaction, the ratio of mass of copper to the mass of oxygen is 3.64 : 0.96 = 4:1
Ratio of copper to oxygen in both the reactions is 4:1. This illustrates and verifies the law of constant proportions.

Page No 131:

Question 78:

Magnesium and oxygen combine in the ratio of 3 : 2 by mass to form magnesium oxide. What mass of oxygen gas would be required to react completely with 24 g of magnesium ?

Answer:


Magnesium reacts with oxygen in the ratio of 3:2 to form magnesium oxide.

The above relation states that magnesium and oxygen combine in a fixed ratio of 3:2 by mass.
Therefore, 3g of magnesium react with 2g of oxygen to form magnesium oxide.
So, 1g of magnesium requires = 2/3g of oxygen = 0.66g of oxygen
24g of magnesium require = 24 × 0.66g of oxygen = 16g
16g of oxygen would be required to react completely with 24g of magnesium.



Page No 132:

Question 79:

When 5 g of calcium is burnt in 2 g of oxygen, then 7 g of calcium oxide is produced. What mass of calcium oxide will be produced when 5 g of calcium is burnt in 20 g of oxygen ? Which law of chemical combination will govern your answer ?

Answer:


5g of calcium react with 2g of oxygen to form 7g of calcium oxide.

This problem can be solved using the law of constant proportions.
We know that calcium and oxygen react in a fixed ratio of 5:2, by mass, to produce 7g of calcium oxide. When 5g of calcium react with 20g of oxygen, the same amount of calcium oxide will be formed.

Therefore, 7g of calcium oxide will be formed when 5g of calcium will react with 20g of oxygen. Calcium and oxygen will react in a ratio of 5:2. The remaining oxygen will remain unreacted.
The law of constant proportions governs our answer.

Page No 132:

Question 80:

A liquid compound X of molecular mass 18 u can be obtained from a number of natural sources. All the animals and plants need liquid X for their survival. When an electric current is passed through 200 grams of pure liquid X under suitable conditions, then 178 grams of gas Y and 22 grams of gas Z are produced. Gas Y is produced at the positive electrode whereas gas Z is obtained at the negative electrode. moreover, gas Y supports combustion whereas gas Z burns itself causing explosions.

(a) Name (i) liquid X (ii) gas Y, and (iii) gas Z.
(b) What is the ratio of the mass of element Z to the mass of element Y in the liquid X ?
(c) Which law of chemical combination is illustrated by this example ?
(d) Name two sources of liquid X.
(e) State an important use of Y in our life


 

Answer:

(a) (i) Liquid X is water (H2O).
     (ii) Gas Y is oxygen.
    (iii) Gas Z is hydrogen.

(b) The ratio of element Z to element Y by mass = 22:178
Therefore, ratio of Z : Y by mass = 1:8

(c) The law of constant proportions is illustrated by this example.

(d) Liquid X has already been stated as water. Oceans and rivers are the important sources of water on earth.

(e) Y, oxygen, is essential for our survival. We breathe in oxygen during respiration.

Page No 132:

Question 81:

One of the forms of a naturally occurring solid compound P is usually used for making the floors of houses. On adding a few drops of dilute hydrochloric acid to P, brisk effervescence are produced. When 50 g of reactant P was heated strongly, than 22 g of a gas Q and 28 g of a solid R were produced as products. Gas Q is the same which produced brisk effervescence on adding dilute HCl to P. Gas Q is said to cause global warming whereas solid R is used for white-washing.

(a) What is (i) solid P (ii) gas Q, and (iii) solid R.
(b) What is the total mass of Q and R obtained from 50 g of P ?
(c) How does the total mass of Q and R formed compare with the mass of P taken ?
(d) What conclusion do you get from the comparison of masses of products and reactant ?
(e) Which law of chemical combination is illustrated by the example given in this problem ?

Answer:

(a) (i) Solid P is calcium carbonate (CaCO3).
     (ii) Gas Q is carbon dioxide (CO2).
    (iii) Solid R is calcium oxide (CaO).

(b) When 50g of reactant P is heated, Q and R are produced. According to the law of conservation of mass, same amount of Q and R will be produced. Therefore, 50g of Q and R will be produced on heating 50g of P.

(c)  According to the law of conservation of mass, same amount of Q and R will be produced depending upon  the amount of P by mass.

(d) When we compare the mass of reactants to that of their product, it will be seen that the total mass of the reactants is equal to the total mass of their product.

(e) In the example, the law of conservation of mass proposed by Antoine Lavoisier is illustrated.



Page No 150:

Question 1:

What do we call those particles which have more or less electrons than the normal atoms ?

Answer:

The particles which have more or less electrons than the normal atoms are called ions.

Page No 150:

Question 2:

What do we call those particles which have :
(a) more electrons than the normal atoms ?
(b) less electrons than the normal atoms ?

Answer:

(a) Particles which have more electrons than the normal atoms are called anions.

(b) Particles which have fewer electrons than the normal atoms are called cations.

Page No 150:

Question 3:

Define 'formula mass' of a compound.

Answer:

Formula mass of a compound is the relative mass of its 'formula unit' compared to the mass of a carbon-12 atom (which has a mass of exactly 12 units).



Page No 151:

Question 11:

Name the elements water is made of. What are the valencies of these elements ? Work out the chemical formula for water.

Answer:

Water (H2O) is made of Hydrogen and Oxygen.
Valency of Hydrogen is 1, and valency of Oxygen is 2.
Valency of one Hydrogen atom = 1
Valency of one Oxygen atom = 2
Therefore, to balance the valency of Oxygen atom, 2 atoms of Hydrogen atoms are required.
Hence, 2 atoms of Hydrogen will combine with 1 atom of Oxygen to form one molecule of water with the chemical formula H2O.

Page No 151:

Question 12:

If the valency of hydrogen is 1 and that of nitrogen is 3, work out the formula for ammonia.

Answer:


Valency of hydrogen = 1
Valency of nitrogen = 3
Ammonia consists of nitrogen and hydrogen.

Elements:NHValencies:31

Formula for ammonia can be worked out by crossing the valencies of the constituent elements.

Therefore, formula of ammonia is NH3.

Page No 151:

Question 13:

Work out the formula for sulphur dioxide. (Valencies : S = 4; O = 2)

Answer:

Valency of Sulphur = 4
Valency of Oxygen = 2

Elements:SOValencies:42

Formula for sulphur dioxide can be worked out by crossing the valencies of the constituent elements.
Thus, formula of sulphur dioxide is S2O4. The formula has a common factor of 2; therefore, dividing the formula by the common factor gives us the simplest chemical formula: SO2.

Page No 151:

Question 14:

If the valency of carbon is 4 and that of sulphur is 2, work out the formula of the compound formed by the combination of carbon with sulphur. What is the name of this compound ?

Answer:


Valency of carbon  =  4
Valency of sulphur = 2


Elements:CSValencies:42

Formula for the compound containing carbon and sulphur can be worked out by crossing the valencies of the constituent elements.
Thus, formula of the compound (formed by the combination of carbon and sulphur) is C2S4. The formula has a common factor of 2; therefore, dividing the formula by the common factor gives us the simplest chemical formula, i.e., CS2.
This compound is called carbon disulphide.

Page No 151:

Question 15:

An element X has a valency of 4 whereas  another element Y has a valency of 1. What will be the formula of the compound formed between X and Y ?

Answer:

Valency of element X = 4
Valency of element Y = 1

Elements:XYValencies:41

Formula for the compound formed between X and Y can be worked out by crossing  the valencies of the constituent elements.

Therefore, formula of the compound is XY4.

Page No 151:

Question 16:

An element B shows valencies of 4 and 6. Write the formulae of its two oxides.

Answer:

Case i

Valency of B = 4
Valency of O = 2

Elements:BOValencies:42

Formula for the oxide formed by the element B (valency 4) can be worked out by crossing over the valencies of the constituent elements.
Therefore, formula of the oxide of element B is B2O4. The formula has a common factor of 2; hence, dividing the formula by the common factor gives us the simplest chemical formula: BO2.

Case ii

Valency of element B = 6
Valency of O = 2

Elements:BOValencies:62

Formula for the oxide formed by the element B (valency 6) can be worked out by crossing over the valencies of the constituent elements.
Therefore, formula of the oxide of element B is B2O6. The formula has a common factor of 2; therefore, dividing the formula by the common factor gives us the simplest chemical formula: BO3.

Page No 151:

Question 17:

An element X of valency 3 combines with another element Y of valency 2. What will be the formula of the compound formed ?

Answer:


Valency of element X = 3
Valency of element Y = 2

Elements:XYValencies:32

Formula for the compound formed between X and Y can be worked out by crossing the valencies of the constituent elements.
Therefore, formula of the compound is X2Y3.

Page No 151:

Question 18:

Work out the formula for magnesium hydrogen carbonate.

Answer:

Valency of Magnesium = 2
Valency of hydrogen carbonate ion = −1

Elements/Ions:MgHCO3-Valencies:+2-1

Formula for Magnesium hydrogen carbonate can be worked out by


Therefore, formula for Magnesium hydrogen carbonate is: Mg(HCO3)2.

Page No 151:

Question 19:

An element X has a valency of 2. Write the simplest formula for :
(a) bromide of the element
(b) oxide of the element

Answer:

(a)

Valency of X = 2
Valency of Br = 1

Formula for bromide of element X can be worked out by crossing the valencies of the constituent elements.
Therefore, formula for bromide of X is: XBr2.

(b)

Valency of X = 2
Valency of Oxygen = 2

Formula for oxide of element X can be worked out by crossing the valencies of the constituent elements.
Therefore, formula for oxide of X is XO.

Page No 151:

Question 20:

Work out the formulae for the following compounds :
(a) Sodium oxide
(b) Calcium carbonate

Answer:

(a) Valency of Sodium = 1
Valency of Oxygen = 2



Formula for sodium oxide can be worked out by crossing the valencies of the constituent elements.
Therefore, formula for sodium oxide is: Na2O.

(b) Valency of Calcium = 2
Valency of Carbonate ion = −2



Formula for calcium carbonate can be worked out by crossing  the valencies of the constituent elements/ions. +2 positive charge of calcium neutralizes the −2 negative charge of carbonate ion.
Therefore, formula for calcium carbonate is: CaCO3.

Page No 151:

Question 21:

Calculate the formula masses of the following compounds :
(i) Sodium oxide, Na2O
(ii) Aluminium oxide, Al2O3
(Given : Atomic masses : Na = 23 u ; O = 16 u ; Al = 27 u)

Answer:

(i) Formula mass of Na2O:

Formula mass of Na2O = Mass of two Na atoms + Mass of one O atom

= 2 × 23 + 1 × 16
= 46 + 16
= 62
Therefore, the formula mass of Na2O is 62 u.

(ii) Atomic mass of aluminium = 27 u

Atomic mass of oxygen = 16 u
Formula mass of Al2O3 can be calculated as follows:
Formula mass of Al2O3 = Mass of two Al atoms + Mass of three O atoms
= 2 × 27 + 3 × 16            
= 54 + 48
= 102
Therefore, the formula mass of Al2O3 is 102 u.

Page No 151:

Question 22:

Name the following compounds. Also write he symbols/formulae of the ions present in them :
(a) CuSO4
(b) (NH4)2SO4
(c) Na2O
(d) Na2CO3
(e) CaCl2

Answer:

(a) Copper sulphate (Copper ion – Cu2+ and Sulphate ion – SO42)
(b) Ammonium sulphate (Ammonium ion −NH4+ and Sulphate ion −SO42)
(c) Sodium oxide (Sodium ion −Na+ and Oxygen ion −O2)
(d) Sodium carbonate (Sodium ion −Na+ and Carbonate ion −CO32)
(e) Calcium chloride (Calcium ion −Ca2+ and Chloride ion −Cl)

Page No 151:

Question 23:

Write the cations and anions present, if any, in the following :
(a) CH3COONa
(b) NaCl
(c) H2
(d) NH4NO3

Answer:


(a)CH3COONa:Cation:Na+Anion:CH3COO-
(b) NaCl:Cation:Na+Anion:Cl-
(c) H2 is a neutral molecule; therefore, it will not have a cation or an anion.
(d)NH4NO3:Cation:NH4+Anion:NO3-

Page No 151:

Question 24:

Give the formulae of the compounds formed from the following sets of elements :
(a) calcium and fluorine
(b) hydrogen and sulphur
(c) nitrogen and hydrogen
(d) carbon and chlorine
(e) sodium and oxygen
(f) carbon and oxygen

Answer:

(a) Compound formed with calcium and fluorine will have the chemical formula CaF2. (valency of Ca is 2 and that of F is 1)
(b) Compound formed with hydrogen and sulphur will have the chemical formula H2S. (valency of S is 2 and that of H is 1)
(c) Compound formed with nitrogen and hydrogen will have the chemical formula NH3. (valency of N is 3 and that of H is 1)
(d) Compound formed with carbon and chlorine will have the chemical formula CCl4. (valency of C is 4 and that of Cl is 1)
(e) Compound formed with sodium and oxygen will have the chemical formula Na2O. (valency of Na is 1 and that of O is 2)
(f) Compound formed with carbon and oxygen will have the chemical formula CO2. (valency of C is 4 and that of O is 2, divided by the common factor of 2)

Page No 151:

Question 25:

What are (i) ionic compounds, and (ii) molecular compounds ? Give two examples of each type of compounds.

Answer:

Ionic Compounds: Ionic compounds that consist of oppositely-charged ions held together by strong electrostatic forces. In an ionic compound, positively-charged ions, cations, and negatively-charged ions, anions, are chemically bonded by ionic bonds.
Examples of ionic compounds: NaCl – It comprises positively-charged sodium ions chemically bonded to the negatively-charged chloride ions.
Similarly, CuSO4  comprises positively-charged copper ions chemically bonded to the negatively-charged sulphate ions.
Molecular compounds: Molecular compounds consist of molecules of two or more elements that are chemically bonded. Molecular compounds are made up of electrically-neutral molecules and not electrically-charged ions.
Examples of molecular compounds: SO2 – It consists of electrically-neutral sulphur chemically bonded to two atoms of oxygen.
Similarly, NH3 consists of electrically-neutral nitrogen atom chemically bonded to three atoms of hydrogen.

Page No 151:

Question 4:

What do we call those particles which are formed :
(a) by the gain of electrons by atoms ?
(b) by the loss of electrons by atoms ?

Answer:

(a) Particles formed by the gain of electrons are called anions.
(b) Particles formed by the loss of electrons are called cations.

Page No 151:

Question 5:

State whether the following statements are true or false :
(a) A sodium ion has positive charge because it has more protons than a neutral atom.
(b) A chloride ion has negative charge because it has more electrons than a neutral atom.

Answer:

(a) False. A sodium ion carries a positive charge because it has less number of electrons than a neutral atom.

(b) True. A chloride ion carries a negative charge because it has more electrons than a neutral atom.

Page No 151:

Question 6:

Write down the formulae for the following compounds :
(a) Calcium oxide
(b) Magnesium hydroxide

Answer:

(a) Calcium oxide: CaO
(b) Magnesium hydroxide: Mg(OH)2

Page No 151:

Question 7:

An element Z has a valency of 3. What is the formula of oxide of Z ?

Answer:

If Z has a valency of 3, then the formula of its oxide will be Z2O3.

Page No 151:

Question 8:

What is the name of a particle which contains 10 electrons, 11 protons and 12 neutrons ?

Answer:

The particle that contains 10 electrons, 11 protons and 12 neutrons is a positively-charged sodium ion (Na+).

Page No 151:

Question 9:

Name the particle which has 18 electrons, 18 neutrons and 17 protons in it.

Answer:

The particle which has 18 electrons, 18 neutrons and 17 protons is a negatively-charged Chloride ion (Cl).

Page No 151:

Question 10:

Fill in the following blanks with suitable words :
(a) The particle which is formed by gain of electrons by an atom is called ...................
(b) The particle which is formed by the loss of electrons by an atom is called ...................
(c) The particle which is formed by the loss or gain of electrons by an atom is called .....................
(d) A potassium ion has positive charge because it contains less .................... than ...................
(e) A sulphide ion has negative charge because it contains less ....................... than ...................

Answer:

(a) Particle formed by the gain of electrons by its atom is called anion.
(b) Particle formed by the loss of electrons by its atom is called cation.
(c) Particle formed by losing or gaining electrons by its atom is called ion.
(d) A potassium ion carries a positive charge because it contains less number of electrons than protons.
(e) A sulphide ion carries a negative charge because it contains less number of protons than electrons.



Page No 152:

Question 26:

(a) What is an ion ? How is an ion formed ? Explain with the help of two examples of different ions.
(b) The valencies (or charges) of some of the ions are given below :
 

Ion Valency (Charge) Ion Valency (Charge)
Sodium ion 1+ Bromide ion 1–
Ammonium ion 1+ Hydroxide ion 1–
Calcium ion 2+ Sulphate ion 2–
Lead ion 2+ Phosphate ion 3–

Using this information, write down the formulae of the following compounds :
(i) Sodium phosphate
(ii) Ammonium sulphate
(iii) Calcium hydroxide
(iv) Lead bromide

Answer:

(a) An ion is an electrically-charged atom or molecule in which the total number of electrons is not equal to the total number of protons.
An ion is formed when a neutral atom loses or gains electrons. This loss or gain of electrons imparts a charge (positive or negative) to the neutral atom making it an ion.
There are two different types of ions:
If a neutral atom loses an electron, an overall positive charge is imparted to the atom and it becomes a positively-charged ion. A positively charged ion is called a cation. In a cation, the total number of electrons is less than the total number of protons as compared to the neutral atom. For example, sodium readily loses an electron to become a positively-charged sodium ion (Na+).
If a neutral atom gains an electron, an overall negative charge is imparted to the atom and it becomes a negatively-charged ion. A negatively-charged ion is called an anion. It has  more no of electrons than the  number of protons compared to the neutral atom. For example, chlorine readily gains an electron to become a negatively-charged chloride ion (Cl-).

(b) Formula of the listed compounds are as follows:
Sodium phosphate –      Na3PO4
Ammonium sulphate –   (NH4)2SO4
Calcium hydroxide –     Ca(OH)2
Lead bromide –            PbBr2

Page No 152:

Question 27:

(a) What is the difference between a cation and an anion ? Explain with examples.
(b) The valencies (or charges) of some of the ions are given below :
 

Ion     Valency (Charge) Ion    Valency (Charge)
Sodium ion 1+ Nitrate ion 1–
Copper ion 2+ Sulphide ion 2–

Using this information, write down the formulae of :
(i) Sodium sulphide
(ii) Copper nitrate

Answer:

Cation Anion
1. Cation is a positively charged ion. 1. Anion is a negatively charged ion.
2. When a neutral atom loses one or more electrons, it develops an overall positive charge and becomes a cation. 2.  When a neutral atom gains one or more electrons, it develops an overall negative charge and becomes an anion.
3. A cation has fewer electrons than the neutral (normal) atom of that element. 3. An anion has more electrons than the neutral (normal) atom of that element.
4. Example: Sodium atom readily loses an electron to form a positively charged sodium ion.
Na → Na+ + e
4. Example: Chlorine atom accepts an electron to become a negatively charged chloride ion.
Cl + e → Cl
Similarly, calcium atom loses two electrons to become a positively charged calcium ion. Thus, the cation formed has two units of positive charge.
 Ca → Ca2++ 2e
Oxygen atom readily accepts two electrons to become a negatively charged oxygen ion. Thus, the oxygen anion formed has two units of negative charge.
O + 2e− → O2

(b) Formula of the listed compounds can be derived by crossing the individual valencies over the ions/elements. Therefore, the formula for sodium sulphide and copper nitrate are:
(i) Na2S
(ii) Cu(NO3)2

Page No 152:

Question 28:

Explain the formation of (i) sodium ion, and (ii) chloride ion, from their respective atoms giving the number of protons and number of electrons in each one of them. What is the reason for positive charge on a sodium ion and a negative charge on a chloride ion ?

Answer:

Formation of Sodium ion:
A neutral sodium atom consists of 11 protons and 11 electrons. The protons and electrons are equal in number, thus keeping the overall charge of the electrically neutral sodium atom zero.
A sodium ion is formed when the neutral atom loses an electron.

  Sodium atom (Na) Sodium ion (Na+)
Electrons 11 (−charge) 10 (−charge)
Protons 11 (+ charge) 11 (+charge)
Overall charge Zero  +1

It is clear that a sodium ion has one electron less than the electrically neutral sodium atom. In a sodium ion, the number of protons is more than the number of electrons, which gives the ion an overall positive charge of one unit.
A proton has an overall positive charge and an electron has an overall negative charge. In a neutral atom, these charges cancel out each other keeping the neutral atom electrically uncharged. However, when a neutral atom loses an electron, the number of protons in the ion is more than the number of electrons. The extra proton gives the ion an overall positive charge (since a proton is positively charged).

(ii) Formation of chloride ion:
A neutral chlorine atom consists of 17 protons and 17 electrons. The positive charge of protons and the negative charge of electrons cancel  each other, thus keeping the overall charge on the electrically neutral chlorine atom zero.
A chloride ion is formed when the neutral chlorine atom accepts an electron.
  Chlorine atom (Cl) Chloride ion (Cl)
Electrons 17(−charge) 18(−charge)
Protons 17(+charge)  17(+charge)
Overall charge Zero −1

It is clear that a chloride ion has one electron more than the electrically neutral chlorine atom. In a chloride ion, the number of protons is less than the number of electrons, which gives the ion an overall negative charge of one unit.
A proton has an overall positive charge and an electron has an overall negative charge. In a neutral atom, these charges cancel each other keeping the neutral atom electrically uncharged. However, when a neutral atom accepts an electron, the number of protons in the ion becomes less than the number of electrons. The extra electron gives the ion an overall negative charge (since an electron is negatively charged).

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Question 29:

(a) Write the symbols/formulae of two simple ions and two compound ions (or polyatomic ions).
(b) An element Y has a valency of 4. Write the formula for its :
(i) chloride
(ii) oxide
(iii) sulphate
(iv) carbonate
(v) nitrate

Answer:

(a) Example of a simple ion: Iodine atom forms an iodide ion by accepting an electron, I-. Similarly, magnesium loses two electrons to form a magnesium ion, Mg2+.
Example of a compound ion: Carbonate ion, CO32-, is made up of two types of atoms i.e. carbon and oxygen (chemically bonded). Similarly, ammonium ion, NH4+, is made up of two types of atoms i.e. nitrogen and hydrogen (chemically bonded).

(b) Formula for the listed compounds are as follows:
(i) Crossing the valencies, chloride for the element Y will have the chemical formula YCl4.


(ii) Crossing over the valencies, oxide for the element Y will have the chemical formula YO2.


(iii) Crossing the valencies, sulphate for the element Y will have the chemical formula Y(SO4)2.


(iv) Crossing the valencies, carbonate for the element Y will have the chemical formula Y(CO3)2.


(v)Crossing  the valencies, nitrate for the element Y will have the chemical formula Y(NO3)4.

Page No 152:

Question 30:

(a) Define 'formula unit' of an ionic compound. What is the formula unit of (i) sodium chloride, and (ii) magnesium chloride ?
(b) Calculate he formula masses of the following compounds :
(i) Calcium chloride
(ii) Sodium carbonate
(Given : Atomic masses : Ca = 40 u ; Cl = 35.5 u ; Na = 23 u ; C = 12 u ; O = 16 u)

Answer:

(a) Formula unit of an ionic compound refers to the empirical formula of an ionic or covalent molecule. It is the lowest whole number ratio of ions that make up the electrically neutral ionic compound. Formula unit represents the smallest unit of a compound.
The formula unit of sodium chloride is NaCl.
The formula unit of magnesium chloride is MgCl2.

(b) (i) Formula mass of calcium chloride = mass of one Ca atom + mass of two Cl atoms = 40 + 2 × 35.5 = 111
Thus, the formula mass of calcium chloride is 111 u.
(ii) Formula mass of sodium carbonate = mass of two Na atoms + mass of one C atom + mass    of three O atoms = 2 × 23 + 12 + 3 × 16 = 106
Thus, the formula unit of sodium carbonate is 106 u.

Page No 152:

Question 31:

The atomic number of an element X is 13. What will be the number of electrons in its ion X3+ ?
(a) 11
(b) 15
(c) 16
(d) 10

Answer:

(d) 10
If the atomic number of X is 13, the number of electrons in its X3+ ion would be 13 – 3 = 10.

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Question 32:

Which of the following represents a correct chemical formula ?
(a) CaCl
(b) Na3N
(c) NaSO4
(d) NaS

Answer:

(b) Na3N
The correct chemical formula among the listed chemicals is of Na3N.

Page No 152:

Question 33:

If the number of electrons in an ion Z3– is 10, the atomic number of element Z will be :
(a) 7
(b) 5
(c) 10
(d) 8

Answer:

(a) 7
If the number of electrons in an ion Z3– is 10, the atomic number of element Z will be 7 because the ion is formed by gaining three electrons.

Page No 152:

Question 34:

The anion of an element has :
(a) more electrons than the normal atom
(b) less electrons than the normal atom
(c) more protons than the normal atom
(d) same number of electrons as normal atom

Answer:

(a) more electrons than the normal atom

The anion of an element has more electrons than the normal atom because it is formed by gaining electrons.

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Question 35:

A particle X has 17 protons, 18 neutrons and 18 electrons. This particle is most likely to be :
(a) a cation
(b) an anion
(c) a molecule
(d) a compound

Answer:

(b) anion
X, most likely, is an anion because the number of electrons is more than the number of protons.

Page No 152:

Question 36:

An element which can exhibit valencies of 2, 4, and 6 can be :
(a) copper
(b) iron
(c) mercury
(d) sulphur

Answer:

(d) Sulphur
Sulphur exhibits the valencies of 2, 4 and 6.



Page No 153:

Question 37:

The atomic number of an element E is 16. The number of electrons in its ion E2– will be :
(a) 16
(b) 18
(c) 15
(d) 14

Answer:

(b) 18
Element with an atomic number 16 will have 18 electrons in its E2- ion because anions are formed by gaining electrons.

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Question 38:

The cation of an element has :
(a) the same number of electrons as its neutral atom
(b) more electrons than a neutral atom
(c) less protons than a neutral atom
(d) less electrons than a neutral atom

Answer:

(d) number of electrons less than the neutral atom 

The cation of an element has less electrons than a neutral atom.

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Question 39:

Two elements X and Y have valencies of 5 and 3, and 3 and 2, respectively. The elements X and Y are most likely to be respectively :
(a) copper and sulphur
(b) sulphur and iron
(c) phosphorus and nitrogen
(d) nitrogen and iron

Answer:

(d) nitrogen and iron
Elements X and Y, most likely, are nitrogen and iron.

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Question 40:

The number of electrons in an ion Y2+ is 10. The atomic number of element Y is most likely to be :
(a) 8
(b) 12
(c) 10
(d) 14

Answer:

(b) 12
The atomic number of element Y would be 12.

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Question 41:

A particle P has 18 electrons, 20 neutrons and 19 protons. This particle must be :
(a) a molecule
(b) a binary compound
(c) an anion
(d) a cation

Answer:

(d) cation
Particle P must be a cation.

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Question 42:

An ionic compound will be formed by the combination of one of the following pairs of elements. This pair of elements is :
(a) chlorine and calcium
(b) calcium and sodium
(c) sulphur and carbon
(d) chlorine and chlorine

Answer:

(a) Chlorine and calcium
Chlorine and calcium can form an ionic compound CaCl2.

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Question 43:

Molecular compounds are usually formed by the combination between :
(a) a metal and a non-metal
(b) two different non-metals
(c) two different metals
(d) any two gaseous elements

Answer:

(b) two different non-metals
Molecular compounds are formed by the combination of two different non-metals.

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Question 44:

The formula of a compound is X3Y. The valencies of elements X and Y will be respectively :
(a) 1 and 3
(b) 3 and 1
(c) 2 and 3
(d) 3 and 2

Answer:

(a) 1 and 3

Valencies of element X and Y will be 1 and 3, respectively.

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Question 45:

The formula of the sulphate of an element X is X2(SO4)3. The formula of nitride of element X will be :
(a) X2N
(b) XN2
(c) XN
(d) X2N3

Answer:

(c) XN
Formula of nitride of the element X will be XN.

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Question 46:

An element A forms an oxide A2O5.
(a) What is the valency of element A ?
(b) What will be the formula of chloride of A ?

Answer:

(a) Element A forms an oxide A2O5. Crossing the valencies, we can see that the valency of O (oxide) is −2 and that of element A is 5.

(b) Formula of chloride of A:
Element/IonACl-Valency5-1
Formula of the chloride of element A can be worked out by crossing over the valencies. Thus, the formula is ACl5.

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Question 47:

An element X forms the following compounds with hydrogen, carbon and oxygen :

H2X, CX2, XO2, XO3

State the three valencies of element X which are illustrated by these compounds.

Answer:

H2X:
Element/IonHXValency12
To stabilize the compound with hydrogen, two atoms of X are required. Therefore, the valency of X in this compound will be two.

CX2:
Element/IonCXValency42
To stabilize the compound with carbon, two atoms of X are required. Thus, the valency of X in this compound is two.

XO2:
Element/IonXOValency4-2
To stabilize this compound with oxygen, four atoms of X are required. Thus, the valency of X in this compound is four.

XO3:
Element/IonXOValency6-2
To stabilize this compound with oxygen, six atoms of X are required. Thus, the valency of X in this compound  is six.

Page No 153:

Question 48:

If the aluminium salt of an anion X is Al2X3, what is the valency of X ? What will be the formula of the magnesium salt of X ?

Answer:

Valency of X in the aluminium salt Al2X3 can be worked out as follows (we know that valency of Al is 3):

ElementAlXValency32

Thus, it is clear that the valency of X is two.

Formula for the magnesium salt of X can be worked out as follows:
ElementMgXValency22
Crossing the valencies gives us the formula of the magnesium salt of X i.e. MgX.

Page No 153:

Question 49:

The formula of carbonate of a metal M is M2CO3.
(a) What will be the formula of its iodide ?
(b) What will be the formula of its nitride ?
(c) What will be the formula of its phosphate ?

Answer:

Let us first work out the valency of the metal M (we know that carbonate ion has a valency of −2):
Element/IonMCO32-Valency12
It is clear that M has a valency of 1.

(a) Formula of iodide of M:
ElementMIValency11
Crossing  the valencies, formula of the iodide of M is MI.

(b) Formula of nitride of M:
Element/IonMN3-Valency1-3
Crossing the valencies, formula of the nitride of M is M3N.

(c) Formula of phosphate of M:
Element/IonMPO43-Valency1-3
Crossing the valencies, formula of the phosphate of M is M3PO4.

Page No 153:

Question 50:

The atom of an element X contains 17 protons, 17 electrons and 18 neutrons whereas the atom of an element Y contains 11 protons, 11 electrons and 12 neutrons.

(a) What type of ion will be formed by an atom of element X ? Write the symbol of ion formed.
(b) What will be the number of (i) protons (ii) electrons, and (iii) neutrons, in the ion formed from X ?
(c) What type of ion will be formed by an atom of element Y ? Write the symbol of ion formed.
(d) What will be the number of (i) protons (ii) electrons, and (iii) neutrons, in the ion formed from Y ?
(e) What is the atomic mass of (i) X, and (ii) Y ?
(f) What could the elements X and Y be ?

Answer:

(a) Element X consists of 17 protons and electrons and 18 neutrons. Element X will readily accept an electron to form a negatively charged anion.
Symbol of the negatively charged X is X.

(b) In the ion formed by X:
  (i)  Number of protons = 17
  (ii) Number of electrons = 18
  (iii) Number of neutrons = 18

(c) Element Y consists of 11 protons and electrons and 12 neutrons. Element Y will readily lose an electron to form a positively charged cation.
Symbol of the positively charged Y is Y+.

(d) In the ion formed by Y:
(i) Number of protons = 11
(ii) Number of electrons = 10
(iii) Number of neutrons = 12

(e) (i) Atomic mass of X = number of protons of X + number of neutrons of X = 17 + 18 = 35 u

(ii) Atomic mass of Y = number of protons of Y + number of neutrons of Y = 11 + 12 = 23 u

(f) Atomic number of X is 17; therefore, element X is chlorine.
Atomic number of Y is 11; therefore, element Y is sodium.



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Question 12:

(a) How many atoms are there in exactly 12 g of carbon -12 element ? (C = 12 u)
(b) What name is given to this number ?
(c) What name is given to the amount of substance containing this number of atoms ?

Answer:

(a) 6.022 × 1023 atoms are present in 12g of carbon-12 element.
(b) The name given to this number is Avogadro’s number.
(c) A substance containing 6.022 × 1023 number of atoms is called 1 mole of that substance.

Page No 172:

Question 13:

Calculate the mass of 12.044 × 1025 molecules of oxygen (O2).

Answer:

12.044 × 1025 molecules of oxygen
Mass of 6.022 × 1023 molecules of oxygen = 32g
Mass of one molecule of oxygen = 32 / 6.022 × 1023 = 5.3 × 1023g
Mass of 12.044 × 1025 molecules of oxygen = 5.3 × 10−23 × 12.044 × 1025 g = 6400g or 6.4kg
Hence, the mass of 12.044 × 1025 molecules of oxygen is 6.4kg.

Page No 172:

Question 14:

What is the number of molecules in 1.5 moles of ammonia ?

Answer:

1.5 moles of ammonia
Number of molecules in 1 mole of ammonia = 6.022 × 1023
Number of molecules in 1.5 moles of ammonia = 1.5 × 6.022 × 1023
Hence, number of molecules in 1.5 moles of ammonia is 9.033 × 1023.

Page No 172:

Question 15:

How many moles of calcium carbonate (CaCO3) are present in 10 g of the substance ? (Ca = 40 u ; C = 12 u ; O = 16 u)

Answer:

Mass of calcium carbonate = 10g
Molecular mass of calcium carbonate = (1 × 40) + (1 × 12) + (3 × 16) = 100g
Number of moles in 100g of calcium carbonate = 1 mole
Therefore, number of moles in 10g of calcium carbonate is 0.1.

Page No 172:

Question 16:

How many moles of O2 are there in 1.20 × 1022 oxygen molecules ?

Answer:

1.20 × 1022 molecules of oxygen

One mole of oxygen contains 6.022 × 1023 molecules.
1.20 × 1022 molecules of oxygen contain = 1.20 × 1022 / 6.022 × 1023 = 0.01099 moles.
Hence, 1.20 × 1022 molecules of oxygen contain 0.0199 moles of oxygen.

Page No 172:

Question 17:

If one mole of nitrogen molecules weighs 28 g, calculate the mass of one molecule of nitrogen in grams.

Answer:

Mass of one mole of nitrogen = 28g
We know that, 1 mole of nitrogen contains 6.022 × 1023 molecules.
Mass of 6.022 × 1023 nitrogen molecules = 28g
Mass of one molecule of nitrogen = 28 / 6.022 × 1023 g = 4.64 × 10-23 grams.
Hence, one molecule of nitrogen weighs 4.64 × 10-23 grams.

Page No 172:

Question 18:

How many moles are there in 34.5 g of sodium ? (Atomic mass of Na = 23 u)

Answer:

Mass of sodium = 34.5g
We know that 1 mole of sodium weighs 23g.
Therefore, 34.5g of sodium  = 34.5 / 23 = 1.5 moles
Hence, 34.5g of sodium contain 1.5 moles of sodium.

Page No 172:

Question 19:

What is the number of zinc atoms in a piece of zinc weighing 10 g ? (Atomic mass of Zn = 65 u)

Answer:

Mass of zinc = 10g
gram atomic mass of zinc = 65g
We know that one mole of zinc contains 6.022 × 1023 atoms of zinc.
65g of zinc contain = 6.022 × 1023 atoms
Therefore, 10g of zinc contain = 10 × 6.022 × 1023/65 = 9.264 × 1022 atoms

Hence, 10 grams of zinc contain 9.264 × 1022 atoms.

Page No 172:

Question 20:

Calculate the mass of 3.011 × 1024 atoms of carbon.

Answer:

Number of carbon atoms = 3.011 × 1024
We know that 12g (atomic mass of carbon) of carbon contains 6.022 × 1023 carbon atoms.
Mass of 1 atom of carbon = 12/6.022 × 1023 = 1.99 × 10-23
Hence, mass of 3.011 × 1024 carbon atoms = 3.011 × 1024 ×1.99 × 10-23 = 59.9g or 60g (approximately)

Page No 172:

Question 21:

If 16 g of oxygen contains 1 mole of oxygen atoms, calculate the mass of one atom of oxygen.

Answer:

Given:
Mass of 1 mole of oxygen = 16g
We know that 1 mole of oxygen contains 6.022 × 1023 oxygen atoms.
Mass of 6.022 × 1023 oxygen atoms = 16g
Mass of 1 oxygen atom = 16/6.022 × 1023
Hence, 1 atom of oxygen molecule weighs 2.65 × 10-23 grams.

Page No 172:

Question 1:

What is a group of 6.022 × 1023 particles known as ?

Answer:

A group of 6.022 × 1023 particles is known as one mole.

Page No 172:

Question 2:

What name is given to the amount of substance containing 6.022 × 1023 particles(atoms, molecules or ions) of a substance ?

Answer:

The name given to the amount of substance containing 6.022 × 1023 particles is 1 mole.

Page No 172:

Question 3:

What is the numerical value of Avogadro number ?

Answer:

The numerical value of Avogadro’s number is 6.022 × 1023.

Page No 172:

Question 4:

How many atoms are present in one gram atomic mass of a substance ?

Answer:

6.022 × 1023 (avogadro's number) atoms are present in one gram atomic mass of a substance.

Page No 172:

Question 5:

How many molecules are present in one gram molecular mass of a substance ?

Answer:

6.022 × 1023 molecules are present in one gram molecular mass of a substance.

Page No 172:

Question 6:

What name is given to the number of 6.022 × 1023 ?

Answer:

The name given to the number 6.022 × 1023 is Avogadro’s number.

Page No 172:

Question 7:

Convert 12 g of oxygen gas into moles.

Answer:

Gram molecular mass of O2 = 32 g
mass of oxygen gas given = 12 g
Therefore, 32 g of O2 = 1 mole of O2
So, 12 g of O21232 moles of O2
Hence, 12 g of oxygen contains 0.375 moles of oxygen.

Page No 172:

Question 8:

How many moles are 3.6 g of water ?

Answer:

Given  mass of water = 3.6g
Gram molecular mass of water = 18g
Therefore, 18g of water = 1 mole of water
So, 3.6g of water = 3.618 moles of water
Hence, 3.6g of water contain 0.2 moles of water.

Page No 172:

Question 9:

What is the mass of 0.2 mole of oxygen atoms ?

Answer:

Given mass of oxygen =  0.2 moles
Gram atomic mass of oxygen atom = 16g
Therefore, 16g of oxygen atoms = 1 mole of oxygen atoms
So, 0.2 moles of oxygen atoms = 16 × 0.2g of oxygen atoms
Hence, 0.2 moles of oxygen atoms weigh 3.2g.

Page No 172:

Question 10:

Find the mass of 2 moles of nitrogen atoms.

Answer:

Given moles on N atom = 2 moles
Gram atomic mass of nitrogen atoms = 14g
Therefore, 14g of nitrogen atoms = 1 mole of nitrogen atoms
So, 2 moles of nitrogen atoms = 14 × 2g of nitrogen atoms
Hence, 2 moles of nitrogen atoms weigh 28g.

Page No 172:

Question 11:

Fill in the following blanks :
(a) 1 mole contains ...................... atoms, molecules or ions of a substance.
(b) A mole represents an ..................... number of particles of a substance.
(c) 60 g of carbon element are ................... moles of carbon atoms.
(d) 0.5 mole of calcium element has a mass of ......................
(e) 64 g of oxygen gas contains ....................... moles of oxygen atoms.

Answer:

(a) 1 mole contains 6.022 × 1023 atoms, molecules or ions of a substance.
(b) A mole represents the Avogadro number of particles of a substance.
(c) 60g of carbon element are 5 moles of carbon atoms.
(d) 0.5 moles of calcium element have a mass of 20g.
(e) 64g of oxygen contain 4 moles of oxygen atoms.



Page No 173:

Question 22:

How many atoms are there in 0.25 mole of hydrogen ?

Answer:

0.25 moles of hydrogen
Number of hydrogen atoms in 1 mole = 6.022 × 1023
Therefore, number of atoms in 0.25 moles = 6.022 × 1023 × 0.25 = 1.50 × 1023 atoms
Hence, number of hydrogen atoms in 0.25 moles is 1.50 × 1023

Page No 173:

Question 23:

Calculate the number of moles in 12.044 × 1025 atoms of phosphorus.

Answer:

12.044 × 1025 atoms of phosphorus
6.022 × 1023 atoms of phosphorus = 1 mole
12.044 × 1025 atoms of phosphorus = 12.044 × 1025/6.022 × 1023 = 199.9 moles or 200 moles
Hence, 12.044 × 1025 atoms of phosphorus have 200 moles.

Page No 173:

Question 24:

Calculate the number of molecules present in a drop of chloroform (CHCl3) weighing 0.0239 g.
(Atomic mass : C = 12 u ; H = 1 u ; Cl = 35.5 u)

Answer:

Mass of chloroform = 0.0239g
Molar mass of chloroform = (1 × 12) + (1 × 1) + (3 × 35.5) = 119.5g
Number of molecules in 119.5g of chloroform = 6.022 × 1023
Number of molecules in 1g of chloroform = 6.022 × 1023/119.5
Number of molecules in 0.0239g of chloroform = 0.0239 × 6.022 × 1023/119.5     
Hence, a drop of chloroform weighing 0.0239g has 1.2 × 1020 molecules.

Page No 173:

Question 25:

What is the mass of 5 moles of sodium carbonate (Na2CO3) ?
(Atomic masses : Na = 23 u ; C = 12 u ; O = 16 u)

Answer:

Number of moles of sodium carbonate = 5 moles
Molar mass of sodium carbonate = (2 × 23) + (1 × 12) + (3 × 16) = 106g
We know that 1 mole of sodium carbonate weighs 106g.
Mass of 5 moles of sodium carbonate = 5 × 106 = 530g
Hence, 5 moles of sodium carbonate weigh 530 grams.

Page No 173:

Question 26:

Calculate the number of molecules in 4 g of oxygen.

Answer:

Mass of oxygen = 4g
We know that 32g of oxygen contain 6.022 × 1023 molecules.
Number of molecules in 32g of oxygen = 6.022 × 1023
Number of molecules in 4g of oxygen = 4 × 6.022 × 1023/32
Hence, 4g of oxygen have 7.52 × 1022 molecules.

Page No 173:

Question 27:

How many moles are represented by 100 g of glucose, C6H12O6 ? (C = 12 u, H = 1 u, O = 16 u)

Answer:

Mass of glucose = 100g
Molar mass of glucose    = (6 × 12) + (12 × 1) + (6 × 16) = 180g
1 mole of glucose = 180g
Or, 180g of glucose = 1 mole of glucose
Therefore, 100g of glucose = 100 / 180 moles = 0.55 moles
Hence, 100g of glucose contain 0.55 moles.

Page No 173:

Question 28:

Calculate the mass in grams of 0.17 mole of hydrogen sulphide, H2S.
(Atomic masses : H = 1 u, S = 32 u)

Answer:

Moles of hydrogen sulphide = 0.17
Molar mass of hydrogen sulphide = (2 × 1) + (1 × 32)g = 34g                    
1 mole of hydrogen sulphide weighs = 34g
Therefore, mass of 0.17 moles of hydrogen sulphide = 34 × 0.17 = 5.78g
Hence, 0.17 moles of hydrogen sulphide weigh 5.78 grams.

Page No 173:

Question 29:

Show by means of calculations that 5 moles of CO2 and 5 moles of H2O do not have the same mass. How much is the difference in their masses ?

Answer:

The mass of 5 moles of carbon dioxide and 5 moles of water is not the same. This has been proved below:
Mass of 5 moles of carbon dioxide:
Molar mass of carbon dioxide = (1 × 12) + (2 × 16) = 44g
We know that 44g of carbon dioxide represent 1 mole.
Therefore, 5 moles of carbon dioxide weigh = 5 × 44 = 220g
Mass of 5 moles of water:
Molar mass of water = (2 × 1) + (1 × 16) = 18g
We know that 18g of water represent 1 mole.
Therefore, 5 moles of water weigh = 5 × 18 = 90g
Hence, it is proved from the above calculation that the mass of 5 moles of carbon dioxide and water is not the same. 5 moles of carbon dioxide weigh 130g more than 5 moles of water.

Page No 173:

Question 30:

Calculate the mole ratio of 240 g of calcium and 240 g of magnesium. (Ca = 40 u ; Mg = 24 u)

Answer:

Mass of calcium = 240g
Molar mass of calcium = 40g
Number of moles in 40g of calcium = 1 mole
Therefore, number of moles in 240g of calcium = 240 / 40 = 6 moles
Mass of magnesium = 240g
Molar mass of magnesium = 24g
Number of moles in 24g of magnesium = 1 mole
Therefore, number of moles in 240g of magnesium = 240 / 24 = 10 moles
Therefore, mole ratio of 240g of calcium to 240g of magnesium is 3:5.

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Question 31:

(a) Define mole. What are the two things that a mole represents.
(b) What weight of each element is present in 1.5 moles of sodium sulphite, Na2SO3?

(Atomic masses : Na = 23 u ; S = 32 u ; O = 16 u)

Answer:

(a) Mole can be defined as, the amount of a pure substance, containing the same number of atoms, molecules or ions as present in 12g of carbon. This refers to the value of 6.022 × 1023 atoms, molecules or ions of a substance.
A mole represents Avogadro number of particles of a substance.

(b)
1.5 moles of sodium sulphite
Molar mass of sodium sulphite = (2 × 23) + (1 × 32) + (3 × 16) = 126g    
Therefore, 1 mole of sodium sulphite weighs 126g.
Now, 1 mole of sodium sulphite contains 2 moles of sodium.
So, mass of sodium in 1 mole of sodium sulphite = 2 × 23 = 46g
Therefore, mass of sodium in 1.5 moles of sodium sulphite is = 1.5 × 46 = 69g

1 mole of sodium sulphite contains 1 mole of sulphur.
So, mass of sulphur in 1 mole of sodium sulphite = 1 × 32 = 32g
Therefore, mass of sulphur in 1.5 moles of sodium sulphite is = 1.5 × 32 = 48g

1 mole of sodium sulphite contains 3 moles of oxygen.
So, mass of oxygen in 1 mole of sodium sulphite = 3 × 16 = 48g
Therefore, mass of oxygen in 1.5 moles of sodium sulphite is = 1.5 × 48 = 72g
Hence, the mass of sodium, sulphur and oxygen in sodium sulphite is 69g, 48g and 72g, respectively.

Page No 173:

Question 32:

(a) What is meant by 'a mole of carbon atoms' ?
(b) Which has more atoms, 50 g of aluminium or 50 g of iron ? Illustrate your answer with the help of calculations.
(Atomic masses : Al = 27 u ; Fe = 25 u)

Answer:

(a) A mole can be defined as the amount of a pure substance containing the same number of atoms, molecules or ions as present in exactly 12g of carbon. This refers to the value of 6.022 × 1023 atoms, molecules or ions of a substance.
Therefore, a mole of carbon atoms refers to a group of 6.022 × 1023 carbon atoms.

(b)
Mass of aluminium = 50g
We know that 27g of aluminium contain 6.022 × 1023 aluminium atoms
Number of aluminium atoms in 27g  aluminium = 6.022 × 1023
Number of aluminium atoms in 50g aluminium = 50 × 6.022 × 1023/ 27
Hence, 50g aluminium has11.15 × 1023 atoms.

Mass of iron = 50g
We know that 56g  iron contain 6.022 × 1023 atoms
Number of atoms in 56g iron = 6.022 × 1023
Number of atoms in 50g  iron = 50 × 6.022 × 1023 / 56
Hence, 50g iron comprises 5.376 × 1023 atoms.
It is clear that more number of atoms is present in 50g  aluminium than 50g  iron.

Page No 173:

Question 33:

(a) Define gram atomic mass of a substance. How much is the gram atomic mass of oxygen ?
(b) How many moles of oxygen atoms are present in one mole of the following compounds ?

(i) Al2O3
(ii) CO2
(iii) Cl2O7
(iv) H2SO4
(v) Al2(SO4)3

Answer:

(a) Gram atomic mass refers to the mass of an element , in one mole of atoms of that substance. It is equal to the gram atomic weight of that element.
Gram atomic mass of oxygen is 16 grams.

(b) Number of moles of oxygen in the listed compounds are as follows:
• Al2O3:  3 moles of oxygen are present in 1 mole of Al2O3.
• CO2:  2 moles of oxygen are present in 1 mole of CO2.
• Cl2O7: 7 moles of oxygen are present in 1 mole of Cl2O7.
• H2SO4:  4 moles of oxygen are present in 1 mole of H2SO4.
• Al2(SO4)3: 12 moles of oxygen are present in 1 mole of Al2(SO4)3 (4 × 3 moles).

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Question 34:

(a) Define gram molecular mass of a substance. How much is the gram molecular mass of oxygen ?
(b) If sulphur exists as S8 molecules, calculate the number of moles in 100 g of sulphur. (S = 32 u)

Answer:

(a) Gram molecular mass refers to the mass of a molecules in one mole of molecules of that substance. It is equal to the gram molecular weight of that element.
Oxygen exists as a diatomic molecule (O2); therefore, gram molecular mass of oxygen is 32 grams.

(b)
Mass of sulphur = 100g (sulphur exists as S8 molecules)
molecular mass of sulphur = (8 × 32) = 256g
Number of moles in 256g of sulphur  = 1 mole
Therefore, number of moles in 100g of sulphur = 100/256 = 0.39 moles
Hence, 100g of sulphur contains 0.39 moles of sulphur.

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Question 35:

(a) What is meant by the 'molar mass' of a substance ? State the unit in which molar mass is usually expressed.
(b) Calculate the molar masses of the following substances. Write the results with proper units.

(i) Ozone molecule, O3
(ii) Ethanoic acid, CH3COOH

Answer:

(a) Molar mass of a substance (molecular compound) is the mass in grams, numerically equivalent to the sum of atomic masses of individual atoms in the molecular formula of that substance.
Molar mass of a substance is usually expressed as grams/mole.

(b)

(i) Molar mass of ozone (O3)
Molar mass of ozone = 16 + 16 +16 = 48 g/mol

(ii) Molar mass of acetic acid (CH3COOH)
Molar mass of acetic acid = (2 × 12) + (4 × 1) + (2 × 16) = 60 g/mol

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Question 36:

Which of the following pair of elements represents a mole ratio of 1 : 1 ?
(a) 10 g of calcium and 12 g of magnesium
(b) 12 g of magnesium and 6 g of carbon
(c) 12 g of carbon and 20 g of calcium
(d) 20 g of sodium and 20 g of calcium

Answer:

(b) 12 g of magnesium and 6 g of carbon
      12 g of magnesium and 6 g of carbon will have a mole ratio of 1 : 1.

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Question 37:

Which of the following correctly represents 360 g of water ?
(i) 2 moles of H2O
(ii) 20 moles of water
(iii) 6.022 × 1023 molecules of water
(iv)1.2044 × 1025 molecules of water

(a) (i)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv)

Answer:

(d) (ii) and (iv)
20 moles of water and 1.2044×1025 molecules of water correctly represent 360 gms of water.

 

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Question 38:

If 32 g of sulphur has x atoms, then the number of atoms in 32 g of oxygen will be :
(a) x2
(b) 2x
(c) x
(d) 4x

Answer:

(b) 2x
The number of oxygen atoms in 32 g of oxygen will be 2x.

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Question 39:

A student wants to have 3.011 × 1023 atoms each of magnesium and carbon elements. For this purpose, he will have to weigh :
(a) 24 g of magnesium and 6 g of carbon
(b) 12 g of carbon and 24 g of magnesium
(c) 20 g of magnesium and 10 g of carbon
(d) 12 g of magnesium and 6 g of carbon

Answer:

(d) 12g of magnesium and 6g of carbon
The student will have to weigh 12 g of Magnesium and 6 g of Carbon.

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Question 40:

The ratio of moles of atoms in 12 g of magnesium and 16 g of sulphur will be :
(a) 3 : 4
(b) 4 : 3
(c) 1 : 1
(d) 1 : 2

 

Answer:

(c) 1:1
The ratio of moles of atoms in 12 g of Magnesium and 16 g of Sulphur will be 1:1.

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Question 41:

If 12 gram of carbon has x atoms, then the number of atoms in 12 grams of magnesium will be :
(a) x
(b) 2x
(c) x2
(d) 1.5x

Answer:

(c) The number of atoms in 12g of magnesium will be x/2.



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Question 42:

Which of the following has the maximum number of atoms ?
(a) 18 g of H2O
(b) 18 g of O2
(c) 18 g of CO2
(d) 18 g of CH4

Answer:

(d) 18g of CH4 has  maximum number of atoms.

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Question 43:

If 1 gram of sulphur dioxide contains x molecules, how many molecules will be present in 1 gram of oxygen ?
(S = 32 u ; O = 16 u)

Answer:

Given:
Mass of sulphur dioxide = 1g
We know that equal moles of all substances contain equal number of molecules. (1 mole of all substances contains 6.022 × 1023 molecules).
Therefore, the ratio of molecules in sulphur dioxide and oxygen will be same as the ratio of their moles.
Number of moles in 1g of sulphur dioxide:
1 mole of SO2 = (1 × 32) + (2 × 16) = 64g
64g of SO2 = 1 mole of SO2
Therefore, 1g of SO2 = 1/64 moles of SO2
Now, 1/64 moles of SO2 contain x molecules (given). Since equal number of molecules are present in equal moles of all substances, 1/64 moles of oxygen will also contain x number of molecules.
Number of moles in 1g of oxygen:
1 mole of O2 = 2 × 16 = 32g
32g of O2 = 1 mole of O2
Therefore, 1g of O2 = 1/32 moles of O2
We have already stated that 1/64 moles of oxygen contain x molecules. Therefore,
1/32 mole of oxygen will contain = x × 64 / 32 = 2x
Hence, 1 gm of oxygen consists of 2x number of molecules.

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Question 44:

The mass of one molecule of a substance is 4.65 × 10–23 g. What is its molecular mass ? What could this substance be ?

Answer:

Given:
Mass of one molecule of a substance = 4.65 × 10-23
We know that the mass of a molecule of a substance is equal to the mass of 1 mole of that substance. 1 mole of a substance consists of 6.022 × 1023 molecules; therefore, the mass of a molecule of a substance is equal to the mass of 6.022 × 1023 molecules.
Therefore, to calculate the molecular mass of the substance, we need to calculate the mass of 6.022 × 1023 molecules of that substance.
Mass of 1 molecule of the substance = 4.65 × 10-23
Mass of 6.022 × 1023 molecules of the substance = 6.022 × 1023 × 4.65 × 10-23 = 28g
Hence, molecular mass of the substance is 28 u.
Nitrogen has a molecular mass of 28 u; therefore, the substance is nitrogen.

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Question 45:

Which contains more molecules, 10 g of sulphur dioxide (SO2) or 10 g of oxygen (O2) ?
(Atomic masses : S = 32 u ; O = 16 u)

Answer:

Given:
Mass of sulphur dioxide = 10g
Molar mass of sulphur dioxide = (1 × 32) + (2 × 16) = 64g
We know that 64g of sulphur dioxide contain 6.022 × 1023 molecules.
Number of molecules in 64g  sulphur dioxide = 6.022 × 1023
Number of molecules in 10g  sulphur dioxide  = 10 × 6.022 × 1023/ 64
Hence, the number of molecules in 10g sulphur dioxide is 9.4× 1022.
Given:
Mass of oxygen = 10g
Molar mass of oxygen molecule = (2 × 16) = 32g
Number of molecules in 32g oxygen = 6.022 × 1023
Number of molecules in 10g oxygen = 10 × 6.022 × 1023/ 32
Hence, the number of molecules in 10g  oxygen is 18.8 × 1022
It is clear that the number of atoms in 10g  oxygen is greater than that in 10g  sulphur dioxide.

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Question 46:

What weight of oxygen gas will contain the same number of molecules as 56 g of nitrogen gas ? (O = 16 u ; N = 14 u)

Answer:

Mass of nitrogen gas = 56g
We know that equal moles of all substances contain equal number of molecules. (1 mole of all substances contains 6.022 × 1023 molecules).
The first step will be to convert 56g  nitrogen gas into moles.
1 mole  nitrogen gas N2 = 28g
28g  nitrogen = 1 mole
56g  nitrogen = 2 moles
Since, equal moles of all substances contain equal number of molecules, 2 moles of nitrogen will have the same number of molecules as 2 moles of oxygen.
The next step will be to find the mass of 2 moles of oxygen in grams.
1 mole of oxygen = 32g
2 moles of oxygen = 32 × 2 = 64g
Hence, 64g  oxygen will contain the same number of molecules as 56g  nitrogen gas.

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Question 47:

What mass of nitrogen, N2, will contain the same number of molecules as 1.8 g of water, H2O ? (Atomic masses : N = 14 u ; H = 1 u ; O = 16 u)

Answer:

Mass of water = 1.8g
We know that equal moles of all substance contain equal number of molecules (1 mole of all substances contains 6.022 × 1023 molecules).
The first step will be to convert 1.8g  water into moles.
1 mole of water = (2 × 1) + (1 × 16) = 18g
18g water = 1 mole
1.8g  water = 0.1 moles
Since, equal moles of all substances contain equal number of molecules, 0.1 moles of water will have the same number of molecules as present in 0.1 moles of nitrogen.
The next step will be to find the mass of 0.1 moles of nitrogen in grams.
1 mole of nitrogen = 28g
0.1 moles of nitrogen     = 28 × 0.1 = 2.8g
Hence, 2.8g  nitrogen will contain the same number of molecules as present in 1.8g  water.

Page No 174:

Question 48:

If one gram of sulphur contains x atoms, calculate the number of atoms in one gram of oxygen element, (Atomic masses : S = 32 u ; O = 16 u)

Answer:

Given:
1g  sulphur contains x atoms
We know that equal moles of all substances contain equal number of molecules (1 mole of all substances contains 6.022 × 1023 molecules).
Therefore, the ratio of molecules in sulphur and oxygen will be the same as their ratio of moles.
Therefore, 1g of S = 1/32 moles of S
Now, 1/32 moles of S contain x molecules (given). Since equal number of molecules are present in equal moles of all substances, 1/32 moles of oxygen will also contain x number of molecules.
Number of moles in 1g of oxygen:
1 mole of O = 16g
16g  O = 1 mole of O
Therefore, 1g  O2 = 1/16 moles of O2
We have already stated that 1/32 moles of oxygen contain x molecules; therefore,
1/16 mole of oxygen will contain = x × 32 / 16 = 2x molecules
Hence, 1 gram  oxygen consists of 2x the number of sulphur molecules.

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Question 49:

How many grams of magnesium will have the same number of atoms as 6 grams of carbon ? (Mg = 24 u ; C = 12 u)

Answer:

Mass of carbon = 6g
We know that equal moles of all substances contain equal number of atoms (1 mole of all substances contains 6.022 × 1023 atoms).
The first step will be to convert 6g  carbon into moles.
1 mole of carbon = 12g
12g  carbon = 1 mole
Therefore, 6g carbon = 0.5 moles
Since, equal moles of all substances contain equal number of atoms, 0.5 moles of carbon will have the same number of atoms as present in 0.5 moles of magnesium.
The next step will be to find out the mass of 0.5 moles of magnesium in grams.
1 mole of magnesium = 24g
0.5 moles of magnesium = 24 × 0.5 = 12g
Hence, 12g magnesium will contain the same number of atoms as present in 6g  carbon.

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Question 50:

The mass of one atom of an element X is 2.0 × 10–23 g.
(i) Calculate the atomic mass of element X.
(ii) What could element X be ?

Answer:

Given:
Mass of an atom of element X = 2.0 × 10-23g
We know that the mass of an atom of a substance is equal to the mass of 1 mole of that substance. 1 mole of a substance consists of 6.022 × 1023 atoms; therefore, the mass of an atom of a substance is equal to the mass of 6.022 × 1023 atoms.
Therefore, to calculate the atomic mass of element X, we need to calculate the mass of 6.022 × 1023 atoms of X.
Mass of 1 atom of X = 2.0 × 10-23
Mass of 6.022 × 1023 atoms of X = 6.022 × 1023 × 2.0 × 10-23 = 12 u
Hence, the atomic mass of element X is 12 u.
Carbon has an atomic mass of 12 u; therefore,  element X is carbon.



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