Science Chapter 3 Atoms And Molecules
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    NCERT Solution For Class 9 About 2.html

    Atoms And Molecules Here is the CBSE About 2.html Chapter 3 for Class 9 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 9 About 2.html Atoms And Molecules Chapter 3 NCERT Solutions for Class 9 About 2.html Atoms And Molecules Chapter 3 The following is a summary in Hindi and English for the academic year 2021-2022. You can save these solutions to your computer or use the Class 9 About 2.html.

    Question 1
    CBSEENSC9004969

    A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.

    Solution

    We have given,

    Mass of boron = 0.096g

    Mass of oxygen = 0.144g

    Mass of sample =0.24g

    Thus, percentage of boron by weight in the compound = 0.096 x 1000.24 =40%

    Thus, percentage of oxygen by weight in the compound =0.144 x 1000.24 =60%

     

    Question 2
    CBSEENSC9004970

    When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen ? Which law of chemical combination will govern your answer?

    Solution

    When 3.0 g carbon is burnt in 8.00 g oxygen, 11.00 g carbon dioxide is formed. It means all of carbon and oxygen are used up and carbon and oxygen are combined in the ratio of 3 : 8 to form carbon dioxide. Thus when there is 3 g carbon and 50.0 g oxygen, then also only 8 g oxygen will be used and 11.0 g carbon dioxide will be formed. The remaining oxygen is not used us. This indicates law of definite proportions which says that in compounds, the combining elements are present in definite proportions by mass.




    Question 3
    CBSEENSC9004971

    What are polyatomic ions? Give examples.

    Solution

    A group of atoms carrying a charge is known as a
    polyatomic ion. For example,  hydroxide ion (OH), carbonate ion (CO3–2).

    Question 4
    CBSEENSC9004972

    Write the chemical formulae of the following:

    (a) Magnesium chloride
    (b) Calcium oxide
    (c) Copper nitrate 
    (d) Aluminium chloride
    (e) Calcium carbonate

    Solution

    Chemical formulae of the following is:

    (a) Magnesium chloride – MgCl2

    (b) Calcium oxide – CaO

    (c) Copper nitrate – Cu(NO3)2

    (d) Aluminium chloride – AlCl3

    (e) Calcium carbonate – CaCO3

    Question 5
    CBSEENSC9004973

    Give the names of the elements in the following compounds.

    (a) Quick lime
    (b) Hydrogen bromide
    (c) Baking powder 
    (d) Potassium sulphate.

    Solution

    (a) Quick lime(CaO) – Calcium and oxygen

    (b) Hydrogen bromide(HBr) – Hydrogen and bromine

    (c) Baking powder(NaHCO3) – Sodium, Hydrogen, Carbon and Oxygen

    (d) Potassium sulphate (K2SO4)– Potassium, Sulphur and Oxygen

    Question 6
    CBSEENSC9004974

    Calculate the molar mass of the following substances:

    (a) Ethylene
    (b) Sulphur molecule, S8
    (c) Phosphorus molecule, P4
    (d) Hydrochloric acid, HCl
    (e) Nitric acid, HNO3.

    Solution

    (a) Ethylene, C2H2 = 2 x 12 + 2 x 1 = 26 u

    (b) Sulphur molecule, S8 = 8 x 32 = 256 u

    (c) Phosphorus molecule, P4 = 4 x 31 = 124 u

    (d) Hydrochloric acid, HCl = 1 x 1 + 1 x 35.5 = 36.5 u


    (e) Nitric acid, HNO3 = 1 x 1 + 1 x 14 + 3 x 16 = 63 u
    Question 7
    CBSEENSC9004975

    What is the mass of:

    (a) 1 mole of nitrogen atoms

    (b) 4 moles of aluminium atoms

    (c) 10 moles of sodium sulphite.

    Solution

    (a) 1 mole of nitrogen atoms = 14 u = 14 gm

    (b) 4 moles of aluminium atoms = 4 x 27 = 108 u = 108 gm

    (c) 1 mole of sodium sulphite, Na2 SO3 = (2 x 23 + 1 x 32 + 3 x 16) = 126 u

    10 moles of sodium sulphite = 126 x 10 = 1260 u = 1260 gm.

    Question 8
    CBSEENSC9004976

    Convert into mole

    (a) 12 gm of oxygen gas

    (b) 20 gm of water

    (c) 22 gm of carbon dioxide.

    Solution

    (a) Formula of oxygen is O2.
    1 mole of oxygen gas = 32 gm
     12 gm of oxygen gas = 12/32 = 0.375 mole

    (b) Formula of water is H2O. 
     18 gm of water = 1 mole
     20 gm of water =  20/18 = 1.1 mole

    (c) Formula of carbon dioxide CO2.
    44 gm of carbon dioxide = 1 mole
    22 gm of carbon dioxide = 0.5 mole

    Question 9
    CBSEENSC9004977

    What is the mass of?

    (a) 0.2 mole oxygen atoms.

    (b) 0.5 mole of water molecules.

    Solution

    (a) 1 mole of oxygen atoms = 16 gm

    0.2 moles of oxygen atoms = 16 x 0.2 = 3.2 gm

    (b) 1 mole of water molecules = 18 gm
    0.5 mole of water molecules = 0.5 x 18 =9.0 gm

    Question 10
    CBSEENSC9004978

    Calculate the number of molecules of sulphur (S8) present in 16 gm of solid sulphur.

    Solution
    1 mole of S8 = 32X8 = 256 gm
    1 mole of S8 contain  6.023 X 1023 molecules

    therefore, 256 gm S8 has = 6.023 X 1023 S8 molecules

    16 gm S8 has = (6.023 X 1023 X 16)/256
    = 3.76 X 1023 molecules


    Question 11
    CBSEENSC9004979

    Calculate the number of aluminium ions present in 0.56 gm of aluminium oxide.

    Solution

    1 mole of aluminium oxide, Al2O3 = 2 X 27+3 X 16 = 102 u = 102 gm

    So,       
    102 gm has Al2O3 has = 6.023 X 1023 Al2O3 molecules

              
       0.56 gm Al2O3 has  6.023 X1023 X 0.56
                                                   102
             
             = 3.01 X 1021 Al2O3 molecules
         
    1 molecules of Al2O3 gives =2Al3+ ions

    Hence 0.56 gm Al2O3 gives = 2 X 3.01 X 1021 Al3+ ions

      = 6.023 X 1021 aluminium ions

    Question 12
    CBSEENSC9004980

    Explain law of conservation of mass.

    Solution

    Law of consveration of mass state that mass can neither be created nor be destroy in chemical reaction. For example, take a solution of lead nitrate in water. Weigh it. Let it be x gm. Then add a given amount of solution (y gm) of sodium chloride. It will be found that a white precipitate is formed. Weigh the total solution. It will be found that total weight of the contents is (x + y) gm, i.e., equal to the total weight of the two solutions taken. Hence, it can be inferred that mass is neither created nor lost even during a chemical change. Similar results are obtained when any other chemical reaction has taken place. Only care to be taken is that no material is allowed to escape during the reaction.

    Question 13
    CBSEENSC9004981

    50 g of 10% lead nitrate is mixed with 50 g of 10% sodium chloride in a closed vessel. After the reaction has taken place, it was found that 6.83 g of lead chloride was precipitated. Besides, the reaction mixture contained 90 g water and sodium nitrate. Calculate the amount of sodium nitrate formed.

    Solution

    50 g of 10% lead nitrate means the solution contains 5 g lead nitrate and 45 g water. Similarly, 50 g of 10% sodium chloride means the solution contains 5 g sodium chloride and 45 g water.
    Thus total content before reaction = 5+5+90 = 100 g
    After reaction, amount of water = 90 g
    Amount of precipitate = 6.83 g
    Since according to law of conservation of mass, the total mass of reaction mixture = 100 g
    So, Amount of sodium nitrate = 100 - 90 - 6.83
                                           = 3.17 g

    Question 14
    CBSEENSC9004982

    Name and explain two important laws of chemical combination ?

    Solution

    (i) Law of constant proportion. According to law of constant composition or proportion, all pure samples of a compound contain the same elements combined together in the same proportion by mass. For example, a sample of water would always contain hydrogen and oxygen in the ratio of 1 : 8 by mass irrespective of the source of water. If we decompose 100 grams of pure water by passing electricity through it then we get 11 grams of hydrogen and 89 grams of oxygen showing that hydrogen and oxygen are combined together in water in the same constant proportion of 1 : 8 by weight.

    (ii) Law of multiple proportions. According to law of multiple proportions, when two elements combine to form two or more compounds, then the weights of one of these elements which combine with a fixed weight of the other element, bear a simple ratio to one another. For example, hydrogen and oxygen combine to form water (H2O) and hydrogen peroxide (H2O2) under different conditions. In water 2 grams of hydrogen combine with 16 grams of oxygen while in hydrogen peroxide, 2 grams of hydrogen combine with 32 grams of oxygen. Now the weights of oxygen which combine with a fixed weight of hydrogen (2 grams) in water and hydrogen peroxide respectively are 16 and 32 grams which are in simple ratio of 16 : 32 or 1 : 2.

    Question 15
    CBSEENSC9004983

    What are the postulates of Dalton’s atomic theory of matter?

    Solution

    According to Dalton’s atomic theory, all matter, whether an element, a compound or a mixture is composed of small particles called atoms.Each substances is made by extremely small particles called atoms.

    The postulates of this theory may be stated as follows:
    (i) All matter is made of very tiny particles called atoms.
    (ii) Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction.
    (iii) Atoms of a given element are identical in mass and chemical properties.
    (iv) Atoms of different elements have different masses and chemical properties.
    (v) Atoms combine in the ratio of small whole numbers to form compounds.
    (vi) The relative number and kinds of atoms are constant in a given compound.

    Question 16
    CBSEENSC9004984

    How will you explain the laws of constant proportion on the basis of Dalton’s atomic theory?

    Solution

    Explanation on the basis of Dalton’s atomic theory:
    According to Dalton’s atomic theory, each element consists of atoms which are similar and have same weights. Further atoms of one element combine with atoms of another to form compounds. for example that x atoms of element A combine with y atoms of element B and the compound formed is AxBy (Dalton’s atomic theory).
    If ‘a’ stands for the atomic mass of A and ‘b’ for that of B,then
    Percentage space of space straight A space in space the space compound space equals space fraction numerator ax space cross times space 100 over denominator ax space plus space by end fraction
Percentage space of space straight B space in space the space compound space equals space fraction numerator by space cross times space 100 over denominator ax space plus space by end fraction

    Now a and b (atomic masses of elements) are fixed, x and y are also fixed whole numbers according to atomic theory. Therefore per cent of A and B in the compound is also constant. This shows that the composition of various elements in a compound is also fixed. This is the law of constant proportion.

    Question 17
    CBSEENSC9004985

    Hydrogen and oxygen combine in two different ways to produce water and hydrogen peroxide. Which of Dalton’s law does this observation confirm? Explain.

    Solution

    In water 1 g of hydrogen combines with 8 g of oxygen. And in hydrogen peroxide, 2 g of hydrogen combines with 16 g of oxygen. The ratio of oxygen in water and hydrogen peroxide combining with the same amount of hydrogen i.e., 1 g is 8 : 16 or 1 : 2—a small integral ratio. This is law of multiple proportions which defines as under.

    When two elements combine to form more than one compound, the masses of one element that combine with a given mass of the other element in different compounds are in the ratio of small whole numbers.

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    Question 18
    CBSEENSC9004986

    State drawbacks of Dalton’s atomic theory of matter.

    Solution

    Some of the drawbacks of Dalton’s atomic theory of matter are:

    1. According to Dalton’s atomic theory, matter is indivisible, i.e., which cannot be divided. But now it is known that under special circumstances, atoms can be divided into still smaller particles called electrons, protons and neutrons.

    2. Dalton’s atomic theory fails to explain why substances like charcoal, graphite and diamond have different properties when all these substances are made up of the same type of atoms, called carbon atoms.

    3. Dalton’s atomic theory postulated that all the atoms of the same element have exactly the same mass. It is now known that isotopes are atoms of the same element but have different masses.

    4. It is now known that some atoms called isobars have same masses but these belong to different elements. Whereas Dalton’s atomic theory says that atoms of different atoms have different mass.

    Question 19
    CBSEENSC9004987

    Give an illustration based on atomic theory for the difference between an element, compound and mixture.

    Solution

    Characteristic

    Compound

    Element

    Define

    A compound contains atoms of different together in a fixed ratio.

    An element is a pure chemical substance made of the same type of atom

    Distinguishing features

    Compounds contain different elements in a fixed ratio arranged in a defined manner through chemical bonds.

    Elements are distinguished by their atomic number (number of protons in their nucleus).

    Ability to breakdown

    A compounds can be separated into the simpler substance by chemical method reaction only.

    A element cannot be broken down into the simpler substance by chemical reactions.

    Types

    The list of compounds is endless.

    There are about 117 elements that have observed, can be classified as metal, non-metal or metalloid.

    Representation

    A compound is represented by using formula.

    An element is represented by using symbols

    Examples

    Water (H2O), sodium chloride (NaCl), sodium bicarbonate (NaHCO3) etc.

    Iron (Fe), silver (Ag), gold (Au), nickel (Ni) etc.

    Question 20
    CBSEENSC9004988

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

    Solution

    According to Dalton’s atomic theory, matter is indivisible, i.e., which cannot be divided. But now it is known that under special circumstances, atoms can be divided into still smaller particles called electrons, protons and neutrons. This was the major drawback of Dalton's atomic theory of matter.

    Question 21
    CBSEENSC9004989

    A sample of ammonia contains 9 g hydrogen and 42 g nitrogen. Another sample contains 5 g hydrogen. Calculate the amount of nitrogen in the second sample.

    Solution

    The ratio of hydrogen and nitrogen in the first sample of ammonia is 9:42 or 3:14.
    According to ‘law of definite proportions' the second sample of ammonia should also contain hydrogen and nitrogen in the ratio 3:14.
    If space hydrogen space is space 5 straight g comma space nitrogen space is space fraction numerator 5 space cross times space 14 over denominator 3 end fraction space equals space 23.3 space straight g.

     

    Question 22
    CBSEENSC9004990

    Carbon and oxygen combine in two different ways to form carbon dioxide and carbon monoxide. In carbon monoxide, 12 g carbon combines with 16 g oxygen. In carbon dioxide 12 g carbon combines with 32 g oxygen. Which law of combination is illustrated by this experiment? Explain.

    Solution

    The above experiment illustrates law of multiple proportions. 16 g and 32 g of oxygen respectively are required to combine with a fixed weight, i.e., 12 g of carbon to form carbon monoxide and carbon dioxide. The ratio of oxygen which combines with a fixed weight of carbon is 16 : 32 or 1 : 2, a simple whole number. This is law of multiple proportions.

    Question 23
    CBSEENSC9004991

    In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.

    Solution

    Mass of reactants = Mass of sodium carbonate + Mass of ethanoic acid
                           = 5.3+6.0 = 11.3 g
    Mass of products = Mass of carbon dioxide + Mass of water + Mass of sodium ethanoate
                          = 2.2 + 0.9 +8.2 = 11.3 g
    Since the mass of products is equal to mass of reactants. Thus mass is neither creater nor lost during the given chemical change or the observation made is in agreement with the law of conservation of mass.

    Question 24
    CBSEENSC9004992

    Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?

    Solution

    We know that, 1 g of hydrogen reacts with oxygen =8g

    Therefore, 3g of hydrogen reacts with oxygen =8x 3 =24g

    Thus, 24g of oxygen gas would be required to react completely with 3 g of hydrogen gas.

    Question 25
    CBSEENSC9004993

    Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

    Solution

    Atoms are indivisible particles, which cannot be created nor destroyed in a chemical reaction. This is law which result of the law of conservation of mass.

    Question 26
    CBSEENSC9004994

    Which postulate of Dalton’s atomic theory can explain the law of definite proportions?

    Solution

    The relative number and kinds of atoms are constant in a given compound. This theory explain the law of definite proportions.

    Question 27
    CBSEENSC9004995

    What is an atom?

    Solution

    The smallest particle that retains properties of an element. Atom is Composed of electrons and a nucleus.

    Question 28
    CBSEENSC9004996

    In what form atoms of a solid exist?

    Solution

    Atoms of a solid exist in closely packed and compressed to each other. A solid element is a cluster of atoms. The properties of an element in the solid state is not of single atom of that element but due to cluster of atoms. For example, diamond and graphite though both are composed of carbon atoms but due to different arrangements of carbon atoms in these, they differ in their physical and chemical properties.

    Question 29
    CBSEENSC9004997

    What is the arrangement of atoms in elements ?

    Solution

    Atoms are very small atoms, they are smaller than anything that we can imagine or compare with. More than millions of atoms when stacked would make a layer barely as thichk as a sheet of paper. Nowdays, It is possible to take in depth photographs of the surfaces of the elements. The surface images of silicon with a scanning tunneling microscope (STM) shows that the silicon atoms are arranged in a regular pattern.




    Question 30
    CBSEENSC9004998

    How were the names of elements derived?

    Solution

    In the beginning, the names of elements were derived from the name of the place they were found for the first time or from some specific colours. For example, the name of copper was taken from Cyprus, a place where it was found. Gold was taken from the English word meaning yellow.

    Question 31
    CBSEENSC9004999

    What is the present accepted or IUPAC (International Union of Pure and Applied Chemistry) system of symbols of elements?

    Solution

    Many of the symbols are the first one or two letters of the element’s name in English. The first letter of a symbol is always written as a capital letter (upper case) and the second letter as a small letter (lower case). For example, Hydrogen is written as H and Aluminium is written as Al.

    Some symbols are formed from the first letter of the name and a letter, appearing later in the name. Examples : Chlorine, Cl and Zinc, Zn.

    Some other symbols are taken from the names of elements in Latin, German or Greek. For example, the symbol of iron is Fe from Latin name ferrum.

    Question 32
    CBSEENSC9005000

    What do you mean by symbols of elements?

    Solution

    In order to write the chemical reactions conveniently, each element is represented by a ‘letter’ or a group of two letters called symbol. Thus hydrogen is represented as ‘H’. Calcium is represented as ‘Ca’.

    Question 33
    CBSEENSC9005001

    Give a list of a few elements with their symbols.

    Solution

    Symbol of Common Elements

    Element

    Symbol

     

    Element

    Symbol

    Hydrogen

    H

     

    Iron

    Fe

    Carbon

    C

     

    Silicon

    Si

    Nitrogen

    N

     

    Gold

    Au

    Oxygen

    O

     

    Silver

    Ag

    Sulphur

    S

     

    Mercury

    Hg

    Phosphorus

    P

     

    Sodium

    Na

    Chlorine

    Cl

     

    Potassium

    K

    Calcium, Barium

    Ca,Ba

     

    Magnesium, Lead

    Mg, Pb

    Copper

    Cu

     

    Nickel

    Ni

    Manganese

    Mn

     

    Tin

    Sn

    Question 34
    CBSEENSC9005002

    Calculate  the relative molecular mass of water.

    Solution

    Atomic mass of hydrogen =1u,
    Atomic mass of oxygen =16u
    SO the molecular mass of water, which contains two atoms of hydrogen and one atom of oxygen is
    = 2 x1+1 x16 = 18u

    Question 35
    CBSEENSC9005003

    Calculate the molecular mass of HNO3.

    Solution

    The molecular mass of HNO3 = the atomic mass of H+ the atomic mass of N +3 x the atomic mass of O =1+14+48 =63u

    Question 36
    CBSEENSC9005004

    Write the names of symbols of five elements where the symbols are taken from their name in English.

    Solution

    i) Copper (Cu)
    ii)Calcium (Ca)
    iii)Oxygen (O)
    iv)Magnesium (Mg)
    v)Zinc (Zn).

    Question 37
    CBSEENSC9005005
    Question 38
    CBSEENSC9005006

    Write the symbols of the following chemical elements :

    Mercury, Potassium, Silver, Boron, Chlorine, Gold, Lead, Copper, Aluminium.

    Solution

    Mercury (Hg),
    Potassium (K),
    Silver (Ag),
    Boron (B),
    Chlorine (Cl),
    Gold (Au),
    Lead (Pb),
    Copper (Cu),
    Aluminium (Al).

    Question 39
    CBSEENSC9005007

    State the contribution of Dalton’s atomic theory to the mass of atom. What steps were taken to resolve this issue?

    Solution

    Dalton’s atomic theory states that each atom has a characteristic mass and all atoms of an element are identical. Since, the mass of an atom is very very small, it was a difficult task to determine the mass of an individual atom. Therefore, it was decided to measure the atomic mass of an atom. Atomic mass is the relative atomic mass of an atom as compoared to some particular atom. For example, carbon and oxygen combine to form carbon monoxide. It is observed that 3 g of carbon combines with 4 g of oxygen. In other words.
    carbon is 4/3 times its mass of oxygen, since,  oxygen combines with a large number of elements, oxygen atom was taken as standard and 1/16 of mass of oxygen was taken as one atomic mass unit (u).
    Now a days 1/12 of the mass of C-12 isotope of carbon is taken as the standard reference for measuring atomic masses.

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    Question 40
    CBSEENSC9005008

    Define and explain atomic mass unit.

    Solution

    The atomic mass unit (abbreviated as amu) is unit that is used to express the atomic masses of atoms and molecule at massed of compounds. One atomic mass unit is a mass unit equal to exactly one twelfth (1/12th) the mass of one atom of carbon-12. The atomic masses of all elements have been found relative to an atom is basically the one twelfth the mass of one atom of carbon-12 isotope.

    1/12th of carbon atom is basically the one twelfth the mass of one atom of carbon -12 isotope.

    Question 41
    CBSEENSC9005009

    How do atoms exist?

    Solution

    Individual atoms of most elements do not exist independently. These either aggregate to form elements or form molecules or ions which also aggregate in large numbers to form the matter that is available to us.

    Question 42
    CBSEENSC9005010

    Give atomic masses of some common elements.

    Solution

    Atomic masses of some common elements

    Element

    Symbol

    Mass (u)

    Element

    Symbol

    Mass(u)

    Aluminium

    Al

    27.0

    Magnesium

    Mg

    24.3

    Argon

    Ar

    39.9

    Manganese

    Mn

    54.9

    Arsenic

    As

    74.9

    Mercury

    Hg

    200.6

    Barium

    Ba

    137.3

    Neon

    Ne

    20.1

    Boron

    B

    10.8

    Nickel

    Ni

    58.7

    Bromine

    Br

    79.9

    Nitrogen

    N

    14.0

    Cesium

    Cs

    132.9

    Oxygen

    O

    16.0

    Calcium

    Ca

    40.1

    Phosphorus

    P

    31.0

    Carbon

    C

    12.0

    Platinum

    Pt

    195.1

    Chlorine

    Cl

    35.5

    Potassium

    K

    39.1

    Chromium

    Cr

    52.0

    Silicon

    Si

    28.1

    Cobalt

    Co

    58.9

    Silver

    Ag

    107.9

    Copper

    Cu

    63.5

    Sodium

    Na

    23.0

    Fluorine

    F

    19.0

    Sulphur

    S

    32.1

    Gold

    Au

    197.0

    Tin

    Sn

    118.7

    Helium

    He

    4.0

         
    Question 43
    CBSEENSC9005011

    Why is it not possible to see an atom with naked eyes?

    Solution

    The size of an atom is very very small. Further atoms of most elements do not exist independently. So it is impossible to see an atom with naked eyes.

    Question 44
    CBSEENSC9005012

    What are molecules? Give an account of the arrangement of the constituent atoms in the molecules.

    Solution

    A molecule is the smallest-particle of an element or of a compound which is stable under ordinary conditions and can exist freely and shows all the properties of that element or compound.
    A molecule may be made up of one, two or more atoms. Molecules with only one atom are called monoatomic, e.g., helium, argon, neon.
    Molecules with two atoms are diatomic. These may have similar or different atoms, e.g., chlorine (Cl2), oxygen (O2), hydrochloric acid gas (HCl). Similarly, there are molecules containing three atoms (H2O, CO2), four atoms (P4, NH3) and so on.
    Arrangement of atoms in molecules:
    The atoms in a molecule are chemically bonded in a definite pattern. This arrangement of atoms is given by the structural formula of the compound. For example, structural formula for water is H—O—H. This is a simple illustration for the structure of water. In fact water and all other molecules exhibit definite spatial arrangements depending on the nature of chemical bonding between atoms.
    Fig. 3.3  shows molecular model for water, ammonia and sulphur.


    Question 45
    CBSEENSC9005013

    Could you make a distinction between a molecule and a compound?

    Solution

    A molecule is usually formed when at least two atoms of the same or different kinds combine. If two or three atoms of the same kind of element combine, then it is referred to as molecule of an element or simply molecule. For example, molecule of oxygen (O2) is formed by the combination of two oxygen atoms. Molecule formed by the union of two or more atoms of different elements is called a molecule of a compound or simply compound. For example, a molecule or a compound of carbon dioxide is formed by the union of one carbon atom and two oxygen atoms. However, the noble gases consists of molecules of single atoms only. For example, helium exists as He.

    Question 46
    CBSEENSC9005014

    Comment on the statement. “A molecule is the smallest particle of a substance (element or compound) which has the properties of the substance and is stable.”

    Solution

    It is true that a molecule is the smallest particle of a substance which has the properties of that substance and is stable. In case of gases as example oxygen, a molecule (O2) is stable and gives properties of oxygen gas, whereas atom ‘O’ is not stable and does not represent oxygen gas. In case of noble gases such as helium, a molecule is of monoatomic helium atom (He) and represents helium gas and is also stable. In case of a compound, e.g., carbon dioxide, properties of carbon dioxide molecule are the properties of carbon dioxide gas. A molecule of carbon dioxide is stable.

    Question 47
    CBSEENSC9005015

    Here are molecules of different substances. Differentiate between these. (i) Chlorine molecule, (ii) Argon molecule, (iii) Hydrogen chloride molecule, (iv) Nitrogen molecule, (v) Neon molecule, (vi) Sulphur dioxide molecule, (vii) Hydrogen molecule.

    Solution

    (ii) Argon molecule and (v) Neon molecule are monoatomic molecules. Here molecule is formed of only single atom.

    (i) Chlorine molecule, (iv) Nitrogen molecule and (vii) Hydrogen molecule are diatomic molecules. These are formed by the union of two atoms of the same element.

    (iii) Hydrogen chloride and (vi) Sulphur dioxide are molecular compounds and are formed by the union of different kinds of atoms.

    Question 48
    CBSEENSC9005016

    Define atomicity of an element and give one example each of mono, dia and triatomic molecules.

    Solution

    The number of atoms combined together to form a stable molecule of an element is called its atomicity.

    For example helium (He), neon (Ne) is a monoatomic and its atomicity is 1. In same manner  oxygen (O2) diatomic, therefore its atomicity is 2.

     

    Question 49
    CBSEENSC9005017

    What is the basic unit of all material substances?

    Solution

    Since all matters are composed of tiny particles called atoms or molecules. Therefore, the basic unit of all material substance is atom and molecule. 

    Question 50
    CBSEENSC9005018

    Name some molecules containing more than four atoms. What are these called?

    Solution

    Molecules containing more than four atoms are called polyatomic molecules. Examples are : sulphur (S8), ethyl alcohol (C2H5OH), sugar (C12H22O11), a form of carbon called buckminsterfullerene (C60).

    Question 51
    CBSEENSC9005019

    What is the ultimate particle on which depends the properties of a material substance?

    Solution

    The properties of matter depend upon the properties of atoms or molecules of which they are composed.

    Question 52
    CBSEENSC9005020

    Which of the following statements is more correct? Explain.

    (i) Two atoms of hydrogen combine with one atom of oxygen to give one molecule of water.

    (ii) Two molecules of hydrogen combine with one molecule of oxygen to give two molecules of water.

    Solution

    The (ii) statement that two molecules of hydrogen combine with one molecule of oxygen to give two molecules of water is correct because oxygen and hydrogen exist as molecules and not as atoms.

    Question 53
    CBSEENSC9005021

    Explain the difference between 2Cl and Cl2.

    Solution

    2Cl indicates 2 atoms of chlorine and Cl2 indicates one molecule of chlorine.

    Question 54
    CBSEENSC9005022

    Name two molecules each having one atom, two atoms, three atoms and four atoms.

    Solution

    Helium (He) and Neon (Ne) are monoatomic molecules.

    Chlorine (Cl2) and Oxygen (O2) are diatomic molecules.

    Water (H2O) and carbon dioxide (CO2) are triatomic molecules.

    Ammonia (NH3) and hydrogen peroxide (H2O2) are tetra-atomic molecules.

    Question 55
    CBSEENSC9005023

    Give an account of existence of various elements and compounds.

    Solution

    Metals exist as atomic crystals and do not form molecules. Generally, metals and other elements like carbon and silicon do not have single molecular structure but consist of a very large indefinite number of atoms bonded together. A form of carbon bonded with sixty atoms is called buckminsterfullerence.

    Non-metals exist as molecules containing one or more atoms. For example, argon exists as Ar and nitrogen as N2.

    Compounds containing non-metals only exist as molecules. For example, compound of nitrogen and hydrogen is a ammonia molecule. Compound of carbon and oxygen is a carbon dioxide molecule.

    Compounds containing metal and non-metal exist as ions. For example, compound of sodium and chlorine consists of positively charged sodium ions and negatively charged chloride ions.

    Question 56
    CBSEENSC9005024

    If A and B are two kinds of atoms, then state whether A and B are metals or non-metals in the following representative compounds or materials:

    Solution

    It can defined by the bonding properties and valency of metal and non-metal.
    (a) A is a non-metal.
    (b) A is a metal or non-metal like carbon or silicon.
    (c) B is a non-metal.
    (d) A and B, both are non-metals.
    (e) A is a metal and B is a non-metal.
    (f) A and B are non-metals.
    (g) A is a metal and B is a non-metal.
    (h) B is a non-metal.

    Question 57
    CBSEENSC9005025

    Define atomicity.

    Solution

    The number of atoms constituting a molecule is knowns as its atomicity. For example in hydrogen molecule ( H2 ), its atomicity is 2 becuase it is made of 2 hydrogen atom. 

    Question 58
    CBSEENSC9005026

    How does an atom differ from a molecule?

    Solution

    An atom is the smallest particle of an element which may or may not have independent existence. On the other hand, ‘molecule’ is the smallest particle of the element or compound which is capable of independent existence. For example, helium (He) is an atom and can exist as such, whereas hydrogen atom (H) cannot exist as such but exists as a molecule, i.e., H2. A molecule may be made up of similar atoms (homo atomic molecule) or dissimilar atoms (hetero atomic molecule such as HCl).

    Question 59
    CBSEENSC9005027

    How many kinds of atoms are present in a molecule of copper carbonate (CuCO3) ?

    Solution

    Copper carbonate is a hetero atomic molecule. It contains three types of atoms, i.e., one atom of copper, one atom of carbon and three atoms of oxygen.

    Question 60
    CBSEENSC9005028

    Separate the following stable elements into atoms and molecules : Oxygen, Nitrogen, Argon, Sodium, Neon, Chlorine.

    Solution

    Atoms : Argon, Sodium, Neon.
    Molecules : Oxygen, Nitrogen, Chlorine.

    Question 61
    CBSEENSC9005029

    How many hydrogen and oxygen atoms are obtained when one molecule of water decomposes.

    2 straight H subscript 2 straight O space space space rightwards arrow space space space 2 straight H subscript 2 space plus space straight O subscript 2


    Solution

    When water molecule decomposes it gives hydrogen and oxygen. according to law of conversation of mass " mass neither be created nor be destroy but change into one form to other form. Thus, 2 molecules of water gives 2 molecules of hydrogen, i.e., 4 atoms of hydrogen and one molecule of oxygen, i.e., two atoms of oxygen. Hence, one molecule of water on decomposition gives 2 atoms of hydrogen and one atom of oxygen.

    Question 62
    CBSEENSC9005030

    What is the difference between:

    (i) an atom of hydrogen and a molecule of hydrogen ?

    (ii) a molecule of oxygen and a molecule of water ?

    Solution

    (i) An atom of hydrogen is the smallest particle of hydrogen element which does not have independent existence. A molecule of hydrogen has two hydrogen atoms bonded together and is the smallest particle of hydrogen element which has independent existence.

    (ii) A molecule of oxygen has two atoms of oxygen only. It is the smallest particle of oxygen element which can exist independently. A molecule of water is a compound obtained from two hydrogen atoms and one oxygen atom.

    Question 63
    CBSEENSC9005031

    What is an ion?

    Solution

    An ion is a charged particle. It can be either positively charged or negatively charged. A negatively charged ion is called an anion and a positively charged ion is called a cation. Ions may consist a single charged atom or a group of atoms that have a net-charge on them. Such a group of atoms carrying a net charge is known as a polyatomic ion. For example, Na+ is a single charged atom and is known as sodium cation. Cl is a single charged atom and is known as chloride anion. NO-3 is a group of atoms carrying net negative charge and is known polyatomic anion.

    Question 64
    CBSEENSC9005032

    What are ionic compounds?

    Solution

    An ionic compound is a compound formed by ions bonding together through electrostatic forces. Such compounds are formed from combination of metals with non-metals. For example, sodium chloride is an ionic compound. Its constituent particles are positively charged sodium ions (Na+) and negatively charged chloride ions (Cl).

    Question 65
    CBSEENSC9005033

    Name the ions in the following compounds:

    Sodium fluoride, potassium bromide, calcium oxide, silver sulphide.

    Solution

    Sodium fluoride — Sodium cation and fluoride anion

    Potassium bromide — Potassium cation and bromide anion.

    Calcium oxide — Calcium cation and oxide anion

    Silver sulphide — Silver cation and sulphide anion

    Question 66
    CBSEENSC9005034

    What do you understand by formula of a molecule?

    Solution

    Chemical formula: It is an expression which states the number and type of atoms present in a molecule of a substance. For example, there are 6C atoms and 14 H atoms in a hexane molecule, which has molecular formula C6H14.

    Question 67
    CBSEENSC9005035

    What is valency? What is its use?

    Solution

    Valency define as,  the combining capacity of an element. It can be used to find out how many atoms of an element will combine with the other element to form a chemical formula. For example, hydrogen has a valency of +1 and chlorine has a valency of –1, so one atom of hydrogen combines with one atom of chlorine to form hydrochloric acid. Further oxygen has a valency of –2, so one atom of oxygen combines with 2 atoms of hydrogen to form water molecule, H2O.


    Question 68
    CBSEENSC9005036

    Give symbols and valencies for the common ions as given below:

    Potassium, Barium, Aluminium, Calcium, Cobalt, Copper, Fluorine, Gold, Lead, Zinc, Iodine, Sulphide.

    Solution

    Name of ion

    Symbol

    Valency

    Potassium

    K+

    + 1

    Barium

    Ba2+

    + 2

    Aluminium

    Al3+

    + 3

    Calcium

    Ca2+

    + 2

    Cobalt

    Co2+

    + 2

    Copper

    Cu2+

    + 1, +2

    Fluoride

    F

    –1

    Gold

    Au+1,+2

    +1,+2

    Lead

    Pb+1,+2

    + 1, +2

    Zinc

    Zn+2

    + 2

    Iodide

    I–1

    –1

    Sulphide

    S–2

    –2

    Question 69
    CBSEENSC9005037

    Do all elements have a charged valency? Explain.

    Solution

    All elements do not form ions so their valencies do not have a charge. For example, carbon and silicon have a valency of 4 and nitrogen has a valency of 3. Further molecules containing only non-metals are formed without a charged valency. For example, in carbon tetrachloride (CCl4), carbon has a valency of 4 and chlorine has a valency of 1.

    Question 70
    CBSEENSC9005038

    Give the name and valency of the following polyatomic ions :

    Ammonium, Hydroxide, Nitrate, Carbonate, Sulphate, Sulphite, Bicarbonate, Phosphate.

    Solution

    Polyatomic ion

    Symbol

    Valency

    Ammonium

    (NH4)+1

    + 1

    Hydroxide

    (OH)

    –1

    Nitrate

    (NO3)–1

    – 1

    Carbonate

    (CO3)–2

    –2

    Sulphate

    (SO4)–2

    –2

    Sulphite

    (SO3)–2

    -2

    Bicarbonate

    (HCO3)–1

    –1

    Phosphate

    (PO4)–3

    -3

    Question 71
    CBSEENSC9005039

    How is a molecular formula of a compound written?

    Solution

    In order to write formula of a compound comprised of cations and anions, the valencies of the ion must be known. Let us consider a compound composed of cation A with valency x+ and anion B with valency y . Then the following steps may be followed to write a molecular formula.

    (i) Write down the symbols of the cation and anion side by side.

    A B

    (ii) Write their valencies at top corners as

    Ax+ By–

    (iii) Interchange between the ions their valencies and these are placed on the lower side of each radical or used as subscripts.

    Ay Bx

    (iv) If a radical is multi-atomic, use a small bracket around it.

    (v) Eliminate the common factor, if any, from the numbers used in subscripts.

    vi) The valencies or charges on the ion must balance.

    vii) When a compound consists of a metal and a non-metal, the name or symbol of the metal is written first. For example: calcium oxide (CaO), sodium chloride (NaCl), iron sulphide (FeS), copper oxide (CuO) etc., where oxygen, chlorine, sulphur are non-metals and are written on the right, whereas calcium, sodium, iron and copper are metals, and are written on the left.

    viii) In compounds formed with polyatomic
    ions, the ion is enclosed in a bracket
    before writing the number to indicate the
    ratio. Thus calcium phosphate, compound of calcium ion (valency 2+) and phosphate ion (valency 3) is written as Ca3(PO4)2.

    The same method is used to write formula even for non-ionic substances.

    Question 72
    CBSEENSC9005040

    Write ‘step-wise’ the formulae for the following:
    (a) Zinc Sulphate (b) Ammonium carbonate
    (c) Methane (d) Potassium sulphide
    (e) Cupric chloride (f) Hydrogen sulphate.

    Solution

     

    (a) Zinc sulphate                                      (b) Ammonium carbonate



    Question 73
    CBSEENSC9005041

    Calcium pyrophosphate is represented by the formula Ca2P2O7. Write the formula of ferric pyrophosphate.

    Solution

    Calcium has a valency of +2. There are two Ca atoms. Thus Ca carries a total of +4 charges. Therefore, pyrophosphate has a valency of -4. Since ferric ion has a valency of +3, the formula of ferric pyrophosphate is Fe4(P2O7)3.

    Question 74
    CBSEENSC9005042

     

    Write the formula of aluminium chloride, aluminium sulphate, sodium chloride and sodium sulphate.


    Solution

    Aluminium chloride = AlCl3

    Aluminium sulphate = Al2(SO4)3

    Sodium chloride = NaCl

    Sodium sulphate = Na2SO4

    Question 75
    CBSEENSC9005043

    Valencies or charges of some ions are given below:

    Ions

    Valency/charge

    Aluminium ion

    3+

    Nitride ion

    3

    Magnesium ion

    2+

    Sulphate ion

    2

    Fluoride ion

    1

    Potassium ion

    1+

    Using the above information, write down the chemical formulae of the following:

    (i) Aluminium nitride (ii) Magnesium nitride

    (iii) Aluminium sulphate (iv) Potassium fluoride

    (v) Magnesium fluoride (vi) Potassium nitride.

    Solution

     

    Name of compound        

    Cation 

    Anion

    Chemical formula

     

    (i) Aluminium nitride       

    Al3+  

    N3-

    AlN


    (ii) Magnesium nitride      

    Mg2+   

    N3- 

    Mg3N2

    (iii) Aluminium sulphate   

    Al3+     

    SO42- 

    Al2(SO4)3

    (iv) Potassium fluoride     

    K+   

    F

    KF

    (v) Magnesium fluoride    

    Mg2+ 

    F- 

    MgF2

    (vi) Potassium nitride      

    K+     

    N3- 

    K3N

    Question 76
    CBSEENSC9005044

    If the calcium salt of a hypothetical anion Z has the molecular formula Ca3Z2. What is the valency of Z and what would be molecular formula of the aluminium salt of Z?

    Solution

    In a ionic compound the charge excange, thus Valency of Z = –3

    Valency of Al = +3

    Formula of aluminium salt of Z = AlZ.

    Question 77
    CBSEENSC9005045

    The formula of sulphuric acid is H2SO4 and the formula of metal chloride is MCl3. Write the formula of metal sulphate.

    Solution

    Valency of SO4–2 is –2 and that of metal is +3 [since it combines with 3 chloride ions (–1)].


    Metal sulphate = M2(SO4)3.

    Question 78
    CBSEENSC9005046

    Give the chemical name, chemical formulae for the following compounds.

    Baking soda, Washing soda, Blue vitriol, Green vitriol, Gypsum, Oil of vitriol or white vitriol, Soda ash, Marble chips, Lime water.

    Solution

    Common name

    Chemical name

    Chemical formula

    Baking soda

    Sodium bicarbonate

    NaHCO3

    Washing soda

    Sodium carbonate

    Na2CO3. 10H2O

    Blue vitriol

    Copper sulphate

    CuSO4 . 5H2O

    Green vitriol

    Ferrous sulphate

    FeSO4 . 7H2O

    Gypsum

    Calcium sulphate

    CaSO4.2H2O

    Oil of vitriol

    Sulphuric acid

    H2SO4

    Soda ash

    Sodium carbonate

    Na2CO3

    Marble chips

    Calcium carbonate

    CaCO3

    Lime water

    Calcium hydroxide

    Ca(OH)2

    Question 79
    CBSEENSC9005047

    What is molecular formula? Explain with example what information can be derived from a molecular formula?

    Solution

    Molecular formula of a substance is the symbolic representation of its molecule. It gives the number and kind of atoms united chemically. For example, water is given by H2O. H2SO4 represents one molecule of sulphuric acid.

    Significance of a formula 

    To signify the information derived from a formula let us consider carbon dioxide whose formula is CO2. It indicates :

    (i) The elements present in carbon dioxide are ‘carbon’ and ‘oxygen’.

    (ii) One molecule of carbon dioxide is given by CO2.

    (iii) On molecule of carbon dioxide has one atom of carbon and two atoms of oxygen.

    (iv) In carbon dioxide, carbon and oxygen are in a weight ratio of 12 : 32.

    (v) Mass of one molecule (relative weight or molecular weight) of carbon dioxide is 44 u, 12 u contributed from carbon atom and 32, from the two oxygen atoms.

    (vi) It shows that the valency of oxygen here is 2 and that of carbon is 4.

    (vii) Molecule of CO2 is not charged and is not an ion.

    Sponsor Area

    Question 80
    CBSEENSC9005048

    The chemical formula for calcium oxide is CaO and not Ca2O2. Comment.

    Solution

    The chemical formula of calcium oxide is CaO because writing formula of a compound, valencies are divided by highest common factor. Similarly formula for aluminium nitrate is AlN and not Al3N3.

    Question 82
    CBSEENSC9005050

    Write down the formulae of (i) aluminium oxide, (ii) aluminium chloride, (iii) hydrogen sulphide, (iv) calcium hydroxide, (v) ammonium sulphate, (vi) calcium bicarbonate, (vii) ammonium phosphate, (viii) silver nitrate.

    Solution

    (i) Aluminium oxide — Al2O3

    (ii) Aluminium chloride — AlCl3

    (iii) Hydrogen sulphide — H2S

    (iv) Calcium hydroxide — Ca(OH)2

    (v) Ammonium sulphate — (NH4)2SO4

    (vi) Calcium bicarbonate — Ca(HCO3)2

    (vii) Ammonium phosphate — (NH4)3PO4

    (viii) Silver nitrate — AgNO3

    Question 83
    CBSEENSC9005051

    Write down the name of the compounds represented by the following formulae : (i) Al2(SO)3, (ii) MgCl2, (iii) K2SO4, (iv) KNO3, (v) CaCO3, (vi) (NH4)2CO3, (vii) H2CO3, (viii) SO2

    Solution

    i) Al2(SO4)3 — Aluminium sulphate

    (ii) MgCl2 — Magnesium chloride

    (iii) K2SO4 — Potassium sulphate

    (iv) KNO3 — Potassium nitrate

    (v) CaCO3 — Calcium carbonate

    (vi) (NH4)2CO3 — Ammonium carbonate

    (vii) H2CO3 — Hydrogen carbonate

    (viii) SO2 — Sulphur dioxide

    Question 84
    CBSEENSC9005052

    Write down the formulae of

    (i) sodium oxide

    (ii) aluminium chloride

    (iii) sodium sulphate

    (iv) magnesium hydroxide.

    Solution

    The formulae of given compounds are:

    (i) sodium oxide — Na2O

    (ii) aluminium chloride — AlCl3

    (iii) sodium sulphate — Na2SO4

    (iv) magnesium hydroxide — Mg(OH)2

    Question 85
    CBSEENSC9005053

    Write down the names of compounds represented by the following formulae:

    (i) Al2(SO)3, (ii) CaCl2, (iii) K2SO4, (iv) KNO3, (v) CaCO3.

    Solution

    (i) Al2(SO4)3 —Aluminium sulphate

    (ii) CaCl2 — Calcium chloride

    (iii) K2SO4 — Potassium sulphate

    (iv) KNO3 — Potassium nitrate

    (v) CaCO3 — Calcium carbonate.

    Question 86
    CBSEENSC9005054

    Define ' Molecular Mass'

    Solution

    The molecular mass of a substance is the sum of the atomic masses of all the atoms in a molecule of substance. it is therefore the relative mass of a molecule expressed in atomic mass units.

    Question 87
    CBSEENSC9005055

    What is relative formula mass?

    Solution

    The formula mass of a substance is the sum of atomic masses of all atoms in a formula unit of a compound. Formula mass is used for substances whose constituent particles are ions. In ionic compound, one molecule may contain several positive and negative ions. One molecule of sodium chloride contains six Na+ and 6 C1 ions. But formula unit is NaC1. So formula mass is used in ionic compounds.

    Question 89
    CBSEENSC9005057

    Give an account of the ‘mole concept’.

    Solution

    A mole signifies a collection of 6.023 x 1023 atoms, molecules, ions, etc. Thus, 1 mole of oxygen atom refers to 6.023 x 1023 atoms of oxygen.
    A mole also refers to the mass in grams numerically equal to the molecular mass or atomic mass in atomic mass units. Thus, mole signifies both number and amount in grams. For example,
    One mole of hydrogen molecule refers to

    (i) 6.02323 x 1023 molecules of hydrogen.

    (ii) 2.016 g (equivalent to molecular mass in u i.e., 2.016 u) of hydrogen.

    Question 90
    CBSEENSC9005058

    What is the utility of the mole concept?

    Solution

    Utility of mole concept.

    (i) From the number of moles of a substance, we can calculate the number of elementary particles because the number of moles of a substance is directly proportional to the number of elementary particles.

    (ii) One mole of a gas occupies 22.4 litres at S.T.P. (273 K and 1 atm).

    (iii) One mole of any gas under the same conditions of temperature and pressure occupies the same volume.

    (iv) One mole is equal to molecular mass in grams which is equal to 6.023 x 1023 atoms, molecules, ions etc. Thus, we can calculate absolute masses of atoms and molecules.

    Question 91
    CBSEENSC9005059

    What is Avogadro number constant?

    Solution

    The number 6.023 x 1023 is referred to as Avogadro number constant and is denoted by symbol NA.

    Question 92
    CBSEENSC9005060

    Define the term gram atom. How is it related to the mole and Avogadro number?

    Solution

    Gram atom or Gram atomic mass. The atomic mass of an element expressed in grams is called gram atomic mass or gram atom.

    1 gram atom of any element contains 6.023 x 1023 (Avogadro number) atoms of the

    element. It is equal to 1 mole of atoms.

    1 gram atomic mass = 6.023 x 1023 atoms = 1 mole.

    (Avogadro number)

    The gram atomic mass of an element is equal to the masses of its 6.023 x 1023 atoms. Examples. One gram atom of hydrogen atom weighs one gram and contains 6.023 x 1023 hydrogen atoms. 1 mole of hydrogen atom also weighs 1 gram.

    Question 93
    CBSEENSC9005061

    Define the term:Gram molecular mass. How is it related to the mole and Avogadro number?

    Solution

    The molecular mass of a substance (compound) expressed in grams is called its gram molecular mass. Ionic compounds do not contain molecules (they consist of ions) and therefore, molecular masses of ionic compounds are equal to their formula masses.
    A gram molecular mass of a compound is equal to the mass of 6.023 x 1023 molecules or 1 mole of the substance.
    Examples;
    (i) A gram molecular mass of hydrogen molecules (molecular mass = 2)
    weighs 2 grams and contains 6.023 x 1023 hydrogen molecules.
    1 space gram space molecular space mass space of space hydrogen space space
space equals space 2 space grams space equals stack space 6.023 space cross times space 10 to the power of 23 with Avogadro space number below space equals space space space space 1 space mole space of stack space straight H subscript 2 with Hydrogen space molecule below
space space space space space space space space space space space

    (ii) A gram molecular mass of sodium chloride (molecular or formula mass = 58.5) weighs 58.5 grams and contains 6.023 x 1023 sodium chloride units.

    Question 94
    CBSEENSC9005062

    How do you account for the fact that gram atomic mass or gram molecular mass of any substance contains the same number of particles?

    Solution

    Atomic mass of any substance indicates the times it is heavier than 1/12th of C-12 atom. Now atomic mass of H atom is one u, it means that one C-12 atom is 12 times as heavy as one H atom or to say 1000 C-12 atoms are 12 times as heavy as 1000 H atoms or 6.023 x 1023 C-12 atoms (12 grams C-12 are found to contain 6.023 x 1023 atoms) are 12 times as heavy as 6.023 x 1023 H atoms. Since atomic mass of C-12 is 12 u and that of H atom is 1 u, it can be inferred that weight equal to atomic mass of C-12 or H atom contains the same number of atoms. Thus, it can be shown that gram atomic mass or gram molecular mass of all substances have the same number, i.e., 6.023 x 1023 particles.

    Question 96
    CBSEENSC9005064

    Define mole and Avogadro's number? Will the weight of 1 mole of sodium and 1 mole of oxygen be the same?

    Solution

    Mole:  Mole is the measurement in chemistry. It is used to express the amount of a chemical substance.
    One mole is defined as the amount of substance of a system which contains as many entities like, atoms, molecules, and ions as there are atoms in 12 grams of carbon - 12'.
    Avogadro number:  The number of the particles present in one mole of any substance is equal to 6.022x1023. This is called Avogadro's number or Avogadro's constant. 
    1 mole of sodium (6.023 x 1023 atoms) weighs 23 g (atomic mass). And 1 mole of oxygen (6.023 x 1023 molecules) weighs 32 g (molecular mass). Thus, particles present in 1 mole of oxygen and sodium is same such as (6.023 x 1023 atoms) but the weight could not be same.

    Question 97
    CBSEENSC9005065

    What  is an ion ?

    Solution

    The metal and non-metal contain charged species. The charged species are known as ion. It can be negtive or positive. A negatively charged ion is called anion and the postively charged ion is called cation. 

    Question 99
    CBSEENSC9005067

    How many moles of chlorine atoms are present in one mole of the following compounds:

    (i) Cl2O7 (ii) HCl (iii) BaCl2 (iv) AlCl3 (v) FeCl3 (vi) CCl4.

    Solution

    (i) 2 moles of chlorine atoms in one mole of Cl2O7

    (ii) 1 mole of chlorine atoms in one mole of HCl

    (iii) 2 moles of chlorine atoms in one mole of BaCl2

    (iv) 3 moles of chlorine atoms in one mole of AlCl3

    (v) 3 moles of chlorine atoms in one mole of FeCl3

    (vi) 4 moles of chlorine atoms in one mole of CCl4.

    Question 105
    CBSEENSC9005073
    Question 106
    CBSEENSC9005074

    What gives the number and kind of atoms in each molecule of a substance?

    Solution

    Molecular formula gives information of the number and kind of atoms in each molecule of a substance.

    Question 107
    CBSEENSC9005075

    What is the name given to the number 6.022 x 1023 ?

    Solution

    Avogadro Number.

    Question 108
    CBSEENSC9005076

    How are C12 and C14 related?

    Solution

    These are isotopes of each other.

    Question 109
    CBSEENSC9005077
    Question 110
    CBSEENSC9005078

    What is the weight of 0.5 mole of oxygen?

    Solution

    Mass = molar mass  x number of moles
    m= M x n =16 x0.5 = 8 g

    Tips: -

    16 gram.

    Question 111
    CBSEENSC9005079

    How many moles does 11 g of carbon dioxide make?

    Solution

    Molecular weight of carbon dioxide is 44g/mol.

    Such that CO2 contains 1 carbon and 2 oxygen atom. Thus,

    1 x12 +2x16 =44  (mass of carbon =12, mass of oxygen =16)

    Now by dividing the given grams (11g) by the molecular weight of CO2:

    11/44 =0.25 mol

    Question 112
    CBSEENSC9005080

    How many moles of oxygen atoms are present in 1 mole of KMnO4?

    Solution

    Four moles of oxygen atoms are present in 1 mole of KMnO4.

    Question 113
    CBSEENSC9005081

    Calculate the mass of 6.022 x1022 atoms of He.

     

    Solution

    6.022 x1022 atom of Helium weight 4g
    Therefore 6.022 x 1022 atoms of Helium will weighs

    4 x6.022 x1022/6.022 x1023 =0.4g

    Question 114
    CBSEENSC9005082

    What is the percentage of oxygen in water?

    Solution

    Total molecular mass of water =18 g
    mass of oxygen is  =16g
    Thus, the percentage of oxygen in water is,
    16/18 x100 =88.9%

    Question 117
    CBSEENSC9005085

    Name the smallest particle that has all the properties of an element.

    Solution

    Atom is smallest particles of an element. 

    Question 118
    CBSEENSC9005086

    Name the smallest particle that has all the properties of a compound.

    Solution

    Molecule is smallest particle of compound.

    Question 119
    CBSEENSC9005087

    Name the element used as a standard for atomic mass scale.

    Solution

    Carbon-12 atom.

    Sponsor Area

    Question 120
    CBSEENSC9005088
    Question 122
    CBSEENSC9005090

    What is gram molecular mass of CaCO3. (Atomic mass of Ca, C and O are 40, 12 and 16 amu respectively)?

    Solution

    Gram molecular mass of CaCO3 :
    Mass of Calcium =40
    Mass of carbon =12
    Mass of oxygen =16
    Thus,

    40 +12 +3 x16 =100grams.

    Question 123
    CBSEENSC9005091

    What is the ‘term’ called ‘the mass in grams’ of 1 mole of a substance?

    Solution

    Molar mass or gram molecular mass.

    Question 124
    CBSEENSC9005092

    How many mole atoms of sulphur are present in two moles of sulphuric acid, H2SO4?

     

    Solution

    In 2 moles of sulphuric acid (2H2SO4)  2 moles of sulphur is present.

    Question 125
    CBSEENSC9005093

    How many ions are present in one mole of ions?

    Solution

    6.022 x 1023 ions.

    Question 126
    CBSEENSC9005094

    Give the molecular mass and molar mass of C2H6.

    Solution

    mass of carbon atom =12 u
    mass of hydrogen atom =1 u
    therefore molecular mass of C2H6
    12 x2 +1x6 =30 u
    Hence, molecular mass of C2H6 = 30 u 

    Molar mass of C2H6 = 30 g mol–1

    Question 127
    CBSEENSC9005095

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

    Solution

    In one molar mass of a substance 6.022 x 1023 molecules are present.

    Question 128
    CBSEENSC9005096

    What is the approximate size of an atom?

    Solution

    The diameter of an atom ranges from about 0.1 to 0.5 nanometers (1 × 10-10 m to 5 × 10-10 m)

    Question 129
    CBSEENSC9005097
    Question 130
    CBSEENSC9005098

    Calculate the molecular masses of H2, O2, Cl2, CO2, CH4, CH6, C2H4, NH3, CH3 OH.

    Solution

    Mass of hydrogen =1 u
    Molecular mass of  H2 = 1+1 =2 u

    Mass of oxygen=16u
    Molecular mass of O2 = 16+16 = 32 u

    Mass of Chlorine=35.5u
    Molecular mass of Cl2 = 35.5 + 35.5 = 71 u

    Mass of carbon =12u
    Mass of oxygen=16u
    Molecular mass of CO2 = 12 +32 = 44 u

    Mass of carbon = 12u
    Mass of hydrogen=1u
    Molecular mass of C2H6 = 12 X 2 + 1 X6 = 30 u

    Mass of carbon = 12u
    Mass of hydrogen=1u
    Molecular mass of C2H4 = 12 X 2 + 1 X 4 = 28 u

    Mass of hydrogen=1u
    Mass of carbon =12u
    Molecular mass of NH3  = 1 X 14 + 1 X 3 = 17 u

    Mass of carbon = 12u
    Mass of hydrogen=1u
    Mass of oxygen=16u
    Molecular mass of CH3OH = 1 X 12 + 1 X 3 + 1 X 16 +1 X 1 = 32 u

    Question 131
    CBSEENSC9005099

    Calculatte the formula unit masses of Zno, Na2O, K2CO3 given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u and O = 16 u.

    Solution

    We have given,
    Atomic masses of elements:  Zn = 65 u, Na = 23 u, K = 39u, C = 12 u and O = 16u

    Formula unit mass of ZnO = 65 + 16 = 81 u

    Formula unit mass of Na2O =23 X 2 +16 = 62 u

    Formula unit mass of K2CO3 = 39 X 2 + 12 X 1 + 16 X 3 = 138 u

    Question 132
    CBSEENSC9005100

    If one mole of carbon atoms weights is 12 grams, then what is the mass  ( in grams) of 1 atom of carbon?

    Solution

    1 mole of carbon atoms = 6.022 X 1023 atoms
    Now,
    6.022 X 1023 atoms of carbon weights = 12g
    So, One atom of carbon weights =12/6.022 X 1023 = 1.99 X 1023g.

    Question 133
    CBSEENSC9005101
    Question 134
    CBSEENSC9005102

    The unit of atomic mass is _______.

    Solution

    gram/mol

    Question 135
    CBSEENSC9005103
    Question 136
    CBSEENSC9005104

    Most elements are_____________.

    Solution

    solids

    Question 137
    CBSEENSC9005105
    Question 140
    CBSEENSC9005108
    Question 146
    CBSEENSC9005114
    Question 147
    CBSEENSC9005115
    Question 153
    CBSEENSC9006061

    How does the force of gravitation between two bodies change when the distance between them is reduced to half?

    Solution

    According to the law of gravitation, the force of attraction between two bodies is, 
                     space straight F space equals space straight G fraction numerator straight m subscript 1 straight m subscript 2 over denominator straight r squared end fraction
    When the distance is reduced to half,
                      straight r apostrophe space equals space straight r over 2 comma space so
    straight F apostrophe space equals space straight G fraction numerator straight m subscript 1 straight m subscript 2 over denominator straight r apostrophe squared end fraction space equals space straight G fraction numerator straight m subscript 1 straight m subscript 2 over denominator left parenthesis straight r divided by 2 right parenthesis squared end fraction space equals space 4 straight G fraction numerator straight m subscript 1 straight m subscript 2 over denominator straight r squared end fraction space equals space 4 straight F 
    Thus, graviational force becomes four times the original force. 

    Question 154
    CBSEENSC9006062

    Gravitational force acts on all objects in proportion to their masses. Why, then, does not a heavy object fall faster than a light object?

    Solution

    The gravitational force acting on the body is given by, 
        space straight F space equals space straight G Mm over straight r squared space equals space mg
    rightwards double arrow space space space straight g equals space space GM over straight r squared

    We can see that, F ∝ m but g does not depend on m.
    Hence, all bodies fall with the same rapidness when there is no air resistance. 

    Question 155
    CBSEENSC9006063

    What is the magnitude of the gravitational force between the earth and 1 kg object on its surface? (Mass of the Earth = 6 x 104 kg and radius of earth is 6.4 x 106 m)

    Solution

    Given,
    Mass of the body (m) = 1 kg
    Mass of the earth (M) =  6 x 1024kg
    Radius of the earth (R) = 6.4 x 106m
    Magnitude of the gravitational force between the earth and 1 kg body,   
                   straight F equals space straight G Mm over straight R squared

space space equals space fraction numerator 6.67 space cross times space 10 to the power of negative 11 end exponent cross times space 6 space cross times 10 to the power of 24 cross times 1 over denominator left parenthesis 6.4 space cross times space 10 to the power of 6 right parenthesis squared end fraction straight N

space space equals space fraction numerator 6.67 space cross times space 6 space cross times space 10 over denominator 6.4 space cross times space 6.4 end fraction space

space space equals space 9.77 space straight N space approximately equal to space 9.8 space straight N.

    Question 156
    CBSEENSC9006064

    The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon, with a force that is greater, or smaller, or the same as the force with which the moon attracts the earth ? Why?

    Solution

    According to Newton's third law, these two forces are equal and opposite. The earth attracts the moon with the same force as the force with which the moon attracts the earth.

    Question 157
    CBSEENSC9006065

    If the moon attracts the earth, why does the earth not move towards the moon?

    Solution

    According to Newton's third law, the moon attracts earth with a force equal to that with which the earth attracts the moon. But, the mass of Earth is much massive than that of the moon. So, the acceleration produced in the earth,

                             (a ∝1/m)
    is not noticeable. 

    Question 158
    CBSEENSC9006066

    What happens to the force between two bodies, if
    (i) the mass of one body is doubled?
    (ii) the distance between the bodies is doubled and tripled?
    (iii) the masses of both bodies are doubled?

    Solution

    Gravitational Force, F = G  fraction numerator straight m subscript 1 straight m subscript 2 over denominator straight r squared end fraction
    (i) The force gets doubled on increasing the mass of the body (m1 or m2).
    New force becomes, 
                          straight F apostrophe space equals straight G fraction numerator left parenthesis 2 straight m subscript 1 right parenthesis space straight m subscript 2 over denominator straight r squared end fraction space equals space 2 straight G fraction numerator straight m subscript 1 straight m subscript 2 over denominator straight r squared end fraction space equals space 2 straight F

    (ii)
    a) When distance between the bodies is doubled,

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    i.e. the force becomes one fourth of the original force. 
    b) When the distance between the two bodies is tripled,

           straight F apostrophe space equals space straight G fraction numerator straight m subscript 1 straight m subscript 2 over denominator left parenthesis 3 straight r right parenthesis squared end fraction space equals space 1 over 9 straight G fraction numerator straight m subscript 1 straight m subscript 2 over denominator straight r squared end fraction space equals space 1 over 9 straight F 
    i.e. the force becomes one ninth of the original force. 
    (iii) When the masses of both sids are doubled,
                     straight F apostrophe space equals space straight G fraction numerator left parenthesis 2 straight m subscript 1 right parenthesis thin space left parenthesis 2 straight m subscript 2 right parenthesis over denominator straight r squared end fraction space equals space 4 straight G fraction numerator straight m subscript 1 straight m subscript 2 over denominator straight r squared end fraction space equals space 4 straight F
    i.e, the force becomes four times the original force when masses of both bodies are doubled. 
                    
    Question 159
    CBSEENSC9006067

    What is the importance of universal law of gravitation?

    Solution

    The universal law of gravitation successfully explained many phenomena occurring in nature.

    Some of these phenomena are as follows:

    1. The force that binds us to the earth.

    2. The motion of the moon around the earth.

    3. The motion of planets around the sun.

    4. The tides due to the moon and the sun. 

    Question 160
    CBSEENSC9006068

    What is the acceleration of free fall?

    Solution
    The acceleration produced by a body when it falling solely under the effect of gravity is known as acceleration due to gravity.

    Near the surface of the earth, its value is 9.8 ms-2
    Question 161
    CBSEENSC9006069

    What do we call the gravitational force between the earth and an object?

    Solution

    Gravitational force between the earth and an object is called weight of the object. 

    Question 162
    CBSEENSC9006070

    Amit buys few grams of gold at poles as per the instruction of one of his friends. He hands over the same when he meets him at equator. Will the friend agree with the weight of gold bought? If not, why?

    Solution

    The friend of Amit will not agree with the weight of the gold bought. The value of 'g' at the equator is less than that at the poles. Hence, the few gram of gold at poles will measure less when taken to the equator. 

    Question 163
    CBSEENSC9006071

    Why will a sheet of paper fall slower than one that is crumpled into a ball?

    Solution

    The  sheet of paper crumpled into a ball will fall faster. The sheet of paper will experience a larger air resistance due to its larger surface area. Therefore, the sheet falls slowly. 

    Question 164
    CBSEENSC9006072

    Gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the earth. What is the weight in newtons of a 10 kg object on the moon and on the earth? What is its mass on each?

    Solution

    Mass of the object on the moon = 6 kg.
     
    Mass of the object on the earth  = 6 kg. 
    Weight of the object on the earth = mg = 10 x 9.8 = 98 N. 
    Weight of the object on the moon
    space equals space 1 over 6 cross times 98 space equals 16.3 space straight N. 
    Therefore, the weight of object on earth and moon is different. 

    Question 165
    CBSEENSC9006073

     A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate

    (i) the maximum height to which it rises,

    (ii) the total time it takes to return to the surface of the earth.

    Solution

    (i) Upward velocity is taken positive and acceleration due to gravity is taken negative, according tho the cartesian sign convention.
    So, 
    I n i t i a l space v e l o c i t y comma space straight u space equals space plus 49 space ms to the power of negative 1 end exponent

A c c e l e r a t i o n space d u e space t o space g r a v i t y comma space straight g space equals negative 9.8 space ms to the power of negative 2 end exponent

At space the space highest space point comma space straight v space equals space 0

U sin g space t h e space e q u a t i o n space o f space m o t i o n comma space
space space space space space space space space space space space straight v squared minus straight u squared space equals space 2 gs

therefore space space space 0 squared minus space 49 squared space equals space 2 left parenthesis negative 9.8 right parenthesis space cross times space straight s

Maximum space height comma space straight s space equals space fraction numerator 49 space cross times space 49 over denominator 2 space cross times space 9.8 end fraction space equals space 122.5 space straight m. 
         
    (ii) Let t be the time taken by the stone to reach the highest point. 
    According to the equation,
                       v = u + gt
      therefore space space space space space 0 space equals space space space 49 space minus space 9.8 space cross times space straight t
space space space space space space space space space space straight t space equals space fraction numerator 49 over denominator 9.8 end fraction space equals space 5 straight s
    Since time of ascent = time of descent 
    space Time space taken space by space the space stone space to space return space to space the space earth apostrophe straight s space surface space is comma space

space space space space space space space space space space space space space space space 2 space straight t space equals space 2 space cross times space 5 space equals space 10 space straight s.



     

     

    Question 166
    CBSEENSC9006074

    A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity.

    Solution

    Given, a stone is released from the top of a tower. 
    Initial velocity, u = 0 m/s
    Acceleration due to gravity, g = -9.8 m/s2 
    Distance covered by the ball during the course of it's motion = 19.6 m 
    Now, usign the equation of motion, 
                       v2 - u2 = 2ags
    We have,
         v2 - 02 = 2 * (-9.8) * (-19.6) 
    i.e.,  v2 - 02 = (19.6)2 
    i.e.,   v = - 19.6 m/s
    The negative sign indicates that the velocity is acting in the downward direction

    Question 167
    CBSEENSC9006075

    A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s2, find maximum height reached by the stone. What is the net displacement and total distance covered by the stone?

    Solution

    Given, a stone is thrown vertically upwards. 
    Initial velocity, u = 40 m/s
    Acceleration due to gravity, g = -10 m/s2
    When the stone reaches a maximum heigh, v = 0
    Now, using the equation of motion, 
                    v2 - u2 = 2 gs 
    We have, 
          02 - 402 = 2 (-10) h
    i.e.,  h = (40*40)/ 20 = 80 m
    Total distance covered  = h + h = 80 + 80 = 160 m
    Net displacement  = 80 - 80 = 0. 

    Question 169
    CBSEENSC9006077

    A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Find when and where the two stones meet?

    Solution

    Suppose the two stones meet at a height x from the ground, after time t from the start.

    For space the space downward space motion space of space stone space straight A colon

straight u space equals space 0 space straight m divided by straight s comma space straight g space equals space minus 10 space straight m divided by straight s squared comma space straight s space equals space left parenthesis negative 100 minus straight x right parenthesis space

As comma space straight s space equals space ut space plus space 1 half gt squared
therefore space minus left parenthesis 100 minus straight x right parenthesis space equals space 0 space minus space space 1 half cross times 10 cross times straight t squared space space space space... space left parenthesis 1 right parenthesis

For space upward space motion space of space stone space straight B colon

straight u space equals space plus 25 space straight m divided by straight s comma space straight g space equals space minus 10 straight m divided by straight s squared comma space straight s space equals space plus straight x space straight m

Using space the space formula comma space

space straight s space equals space ut space plus space 1 half gt squared
therefore space plus straight x space equals space 25 space straight t space minus space 1 half cross times 10 straight t squared space space space space space space space space space space space space space space space space space... thin space left parenthesis 2 right parenthesis

Now comma space subtracting space left parenthesis 1 right parenthesis space from space left parenthesis 2 right parenthesis comma space we space get

space space space space space space 100 space equals space 25 space straight t

rightwards double arrow space space straight t space equals space 4 space sec

From space equation space left parenthesis 2 right parenthesis comma space we space have

straight x space equals space 25 cross times 4 space minus 1 half cross times 10 cross times left parenthesis 4 right parenthesis squared
space space space equals 100 space minus space 80
space space space equals space 20 space straight m
    Hence, the two stones meet after 4 sec at a height of 20 m from the ground or 80 m from the top.

     

    Question 171
    CBSEENSC9006079

    In what direction does the buoyant force on an object immersed in a liquid act?

    Solution

    The upward force exerted by a liquid on an object is known as buoyant force. Buoyant force acts in the vertically upward direction, through the centre of gravity of the displaced fluid which is called centre of buoyancy. 

    Question 172
    CBSEENSC9006080

    Why does a block of plastic piece released under water come to the surface of water?

    Solution

    The upthrust or the buoyant force exerted by water brings the plastic block to the surface of water.

    Question 173
    CBSEENSC9006081

    The volume of 50 g substance is 20 cm3. If the density of water is 1 g cm-3, will the substance float or sink?

    Solution

    Mass of a substance = 50 g
    Volume of substance = 20 cm3
    Density space of space substance space equals space Mass over Volume space equals space fraction numerator 50 space straight g over denominator 20 space cm cubed end fraction space equals space 2.5 space straight g space cm to the power of negative 3 end exponent.
Density space of space water space equals space 1 space straight g space cm to the power of negative 3 end exponent.
    As the denisty of substance is greater than that of water, so the substance will sink in water.

    Question 174
    CBSEENSC9006082

    The volume of a 500 g sealed packet is 350 cm3. Will the packet float or sink in water if the density of water is 1 g cm-3? What will be the mass of the water displaced by this packet?

    Solution

    Mass of a packet  = 500 g
    Volume of packet = 350 cm3
    Density space of space packet space equals space Mass over Volume space equals space fraction numerator 500 space straight g over denominator 350 space cm cubed end fraction space equals space 1.43 space straight g space cm to the power of negative 3 end exponent

    The sealed packet will sink in water because its density (1.43 g cm-3) is more than that of water (1 g cm-3). 

    Question 175
    CBSEENSC9006083

    Distinguish between the terms gravitation and gravity, giving suitable examples.

    Solution

    Gravitation: Every single body in this universe attracts every other body with a force known as gravitational force. Gravitation is the force of attraction between any two bodies in the universe.

    Given are some of the examples on which gravitational force acts. 
    Attraction between the sun and the earth, attraction between a table and a chair lying in a room, attraction between the earth and a satellite revolving around it, etc.

    Gravity: Gravity is the attraction between the earth and any object lying on or near its surface. A body thrown up falls back on the surface of the earth due to earth's force of gravity. 
    Gravity is a special case of gravitation. 

    Question 176
    CBSEENSC9006084

    Give some examples of the gravitational force in nature.

    Solution

    The gravitational force plays an important role in nature. Below are mentioned some examples where gravitaional force comes into play:  

    1. All the planets revolve around the sun due to the gravitational force between the sun and the planets. Thus, the gravitational force is responsible for the existence of the solar system.

    Fig. Our solar system. Gravitational force holds together the sun and the planets.

    2. The motion of the moon around the earth is due to the gravitational force between the moon and the earth.

    3. Tides in oceans are formed due to the gravitational force between the moon and the water in oceans. 

    4. The earth holds the atmosphere (envelope of gases) near its surface due to its gravitational force.
    5. The gravitational force of the earth is responsible for rainfall and snowfall.

    Question 177
    CBSEENSC9006085

    What is centripetal force? From where does the moon get the centripetal force required for its motion around the earth?

    Solution

    Centripetal force:
    i) When a body moves along a circular path with a uniform speed, its direction of motion changes at every point.

    ii) The change in direction involves change in velocity or acceleration. The force that provides this acceleration and keeps the body moving along the circular path, acts towards the centre. This force is called centripetal (centre seeking) force.
    iii) Therefore, a force which is required to make a body move along a circular path with uniform speed is called centripetal force.
    iv) Centripetal force always acts along the radius and towards the centre of the circular path.
    v) The figure given below shows that it is the centripetal force which continuously deflects a particle from its straight line path to make it move along a circle. 

    Fig. Centripetal force.

    Example:
    The moon needs a centripetal force, for its circular motion around the earth. This centripetal force is provided by the gravitational attraction exerted by the earth on the moon. 

     

    Question 178
    CBSEENSC9006086

    State Kepler’s law of planetary motion.

    Solution

    Kepler's laws of planetary motion:
    John Kepler derived three laws, which govern the motion of planets. These laws are called Kepler's law and can be stated as follows:

    1. Law of orbits (first law): 
    Each planet revolves around the sun in an elliptical orbit with the sun S situated at one of the two foci, as shown in Fig. below. 

    Fig. Motion of a planet in an ellipse. 

    2. Laws of areas (second law): 
    The line joining the planet and the sun sweeps equal areas in equal intervals of time. Thus, if the time of travel from A to B is the same as that from C to D, then the shaded areas SAB and SCD are equal.

    3. Law of periods (third law): 
    The cube of the mean distance r of a planet from the sun is proportional to the square of its orbital period T.
    or space space space space space space space straight T squared space proportional to space space straight r cubed space space space space or space space space space straight T squared over straight r cubed space equals constant 

     

    Question 179
    CBSEENSC9006087

    How did Newton guess the inverse square rule obeyed by the gravitational force?

    Solution

    Kepler's third law of planetary motion was used by Newton to guess how the force of gravitation of the earth is weakened by distance. 
    Suppose a planet revolves around the sun in a circular orbit of radius r with speed v, then the force acting on the planet is given by,
                          straight F proportional to straight v squared over straight r               ...(1)
    If 'T' is the period of revolution of the planet around the sun, then its orbital velocity will be
    straight v space equals space fraction numerator D i s t a n c e over denominator T i m e end fraction
space space space equals space fraction numerator C i r c u m f e r e n c e over denominator P e r i o d space o f space r e v o l u t i o n end fraction
space space space equals space fraction numerator 2 pi r over denominator straight T end fraction
    rightwards double arrow space space straight v squared space equals space fraction numerator 4 straight pi squared straight r squared over denominator straight T squared end fraction 
    According to Kepler's third law,
                         straight T squared space proportional to space space straight r cubed
    therefore space space space space space space space space space straight v squared space proportional to space space fraction numerator 4 pi r cubed over denominator straight r cubed end fraction
rightwards double arrow space space space space space space space space straight v squared space proportional to space space 1 over straight r space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis

    This is inverse-square rule for the force of gravitation.
    According to the inverse-square law of gravitaion, the force of gravitation on a planet should decrease as the square of the distance of the planet from the sun.
    Newton further showed that the gravitational force is proportional to the product of the masses of the two objects.
    Thus, Newton formulated the universal law of gravitation.
    Question 180
    CBSEENSC9006088

    State Newton's universal law of gravitation. Hence define universal gravitational constant. Give the value of G.

    Solution

    Newton's universal law of gravitation
    The gravitational law states that every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
    The direction of the force is along the line joining the centres of two bodies.

    Fig. Law of gravitation.

    As shown in Fig. above, consider two bodies A and B of masses m1 and m2.
    Let r be the distance between their centres.
    If F is the force with which the two bodies attract each other, then according to the law of gravitation,
                                 straight F proportional to space straight m subscript 1 straight m subscript 2
                          and space space space straight F space proportional to space space 1 over straight r squared
    Combining the two factors, we get 
          straight F proportional to fraction numerator straight m subscript 1 straight m subscript 2 over denominator straight r squared end fraction
    or space space space straight F space equals space straight G fraction numerator straight m subscript 1 straight m subscript 2 over denominator straight r squared end fraction

    where G is a proportionality constant, called universal gravitational constant.
    The value of G is 6.754×1011 m3 kg1 s2

    Question 181
    CBSEENSC9006089

    Determine the units of G in SI system of measurement.

    Solution

    According to Newton's law of gravitation,
    Gravitational force is given by, 
    space space space space space space space space straight F space equals space straight G fraction numerator straight m subscript 1 straight m subscript 2 over denominator straight r squared end fraction space space
rightwards double arrow space space space straight G equals space fraction numerator F r squared over denominator straight m subscript 1 straight m subscript 2 end fraction 
    In SI system, force F is measured in Newton; distance r in metres and masses m1 and m2 in kilo gram.
    Therefore,
    SI unit of  straight G space equals space fraction numerator N m squared over denominator k g. space k g end fraction space equals N m squared space k g to the power of negative 2 end exponent.


     

    Question 182
    CBSEENSC9006090

    Name the scientist who first determined the value of G experimentally. What is the value of G accepted presently?

    Solution

    Henry Cavendish first determined the value of G experimentally in the year 1778, using a sensitive balance.

    The presently accepted value of G = 6.673 x 10–11 Nm2 kg-2.

    Question 183
    CBSEENSC9006091

    Why is G called ‘a universal gravitational constant’?

    Solution

    The value of G does not depend on the nature of the intervening medium. Also, G is constant for any object in the universe. Hence, is is known as 'universal gravitational constant'. 

    Question 184
    CBSEENSC9006092

    Do the gravitational forces obey Newton's third law of motion?

    Solution

    Yes, the gravitational forces obey Newton's third law of motion.
    Body A attracts body B with a force FBA and body B attracts body A with a force FBA. These two forces are equal and opposite, forming action and reaction pair. Hence gravitational forces obey Newton's third law of motion. 

    Question 185
    CBSEENSC9006093

    What is the importance of the universal law of gravitation in nature?

    Solution

    The universal law of gravitation successfully explained many phenomena occurring in nature.
    Some of these phenomena are as follows:

    1. The force that binds us to the earth.

    2. The motion of the moon around the earth.

    3. The motion of planets around the sun.

    4. The tides due to the moon and the sun.

    Question 186
    CBSEENSC9006094

    State the universal law of gravitation.

    Solution

    Universal law of gravitation states that every body in the universe attracts every other body with a force which is directly proportional to product of their masses and inversely proportional to the square of the distance between them.
    This force acts along the line joining the centres of the two bodies.

    Question 187
    CBSEENSC9006095

    Write the formula to find the magnitude of gravitational force between the earth and an object on the surface of the earth.

    Solution

    Consider, the earth as a sphere of mass M and radius R.
    If an object of mass m is placed on the surface of the earth, then the magnitude of the gravitational force between the body and the earth is, 
                                  straight F space equals space straight G Mm over straight R squared

    Question 190
    CBSEENSC9006098

    A sphere of mass 40 kg is attracted by a second sphere of mass 60 kg with a force equal to 4 x 10–5 N. If G = 6 x 10–11 Nm2/kg 2, calculate the distance between the two spheres.

    Solution

    Given,
    Mass of the first sphere, m1 = 40 kg
    Mass of the second sphere, m2 = 60 kg
    Forcing acting between the bodies, F = 4 x 10–5 N
    Gravitaional constant, G = 6 x 10–11 Nm2/kg2.
    Now, using the formula, 
    As space space space space space straight F space equals space straight G fraction numerator straight m subscript 1 straight m subscript 2 over denominator straight r squared end fraction
therefore space space space space straight r space equals space square root of fraction numerator Gm subscript 1 straight m subscript 2 over denominator straight F end fraction end root space equals space square root of fraction numerator 6 cross times 10 to the power of 11 cross times 40 cross times 60 over denominator 4 cross times 10 to the power of negative 5 end exponent end fraction end root space equals space square root of 36 space cross times space 10 to the power of negative 4 end exponent end root space equals space 6 space cross times space 10 to the power of negative 2 end exponent straight m
space space space space space space space equals space 6 space cross times space 10 to the power of negative 2 end exponent space cross times space 100 space cm space equals space 6 space cm. space Ans.


    Therefore, the distance between the two spheres is 6 cm. 

    Question 191
    CBSEENSC9006099

    Compare the gravitational forces exerted by the sun and the moon on the earth. Which exerts a greater force and by how many times?

    Solution

    Mass of the earth, M = 6 x 1024 kg
    Mass of the sun,   Ms = 2 x 1030 kg
    Mass of the moon, Mm = 7.3 x 1022 kg
    Distance of Earth from the sun, rs = 1.5 x 1011 m
    Distance of Earth from the moon, rm = 3.84 x 108 m
    Gravitational force exerted by the sun on the earth,
                            straight F subscript straight s space equals space straight G fraction numerator straight M subscript straight s straight M over denominator straight r subscript straight s superscript 2 end fraction 
    Gravitational space force space exerted space by space the space moon space on space the space earth comma
straight F subscript straight m space equals space straight G fraction numerator straight M subscript straight m straight M over denominator straight r subscript straight m superscript 2 end fraction
therefore space space space space straight F subscript straight s over straight F subscript straight m space equals space fraction numerator GM subscript straight S straight M over denominator straight r subscript straight s superscript 2 end fraction cross times fraction numerator straight r subscript straight m superscript 2 over denominator GM subscript straight m straight M end fraction space equals space straight M subscript straight s over straight M subscript straight m cross times open parentheses straight r subscript straight m over straight r subscript straight s close parentheses squared
space space space space space equals space fraction numerator 2 space cross times 10 to the power of 30 over denominator 7.3 space cross times space 10 to the power of 22 end fraction cross times open parentheses fraction numerator 3.84 space cross times space 10 to the power of 8 over denominator 1.5 space cross times space 10 to the power of 11 end fraction close parentheses space equals space 179.55

    Thus the sun exerts force, about 180 times the force exerted by the moon on the earth. 

     

    Question 192
    CBSEENSC9006100

    The mass of the earth is 6 x 1024 kg and that of the moon is 7.4 x 1022 kg. If the distance between the earth and the moon be 3.84 x 105 km, calculate the force exerted by the earth on the moon. The value of G = 6.7 x 10–11 Nm2 kg–2.

    Solution

    Given semicolon
space space The space mass space of space the space earth comma space straight M space equals space 6 space cross times space 10 to the power of 24 kg
space The space space mass space of space the space moon comma space straight m space equals space 7.4 space cross times space 10 to the power of 22 kg 
    The distance between the earth and the moon is, 
                        r = 3.84 x 105 m
                          = 3.84 x 105 x 1000 m
                          = 3.84 x 108 m
    The value of G = 6.7 x 10-11 Nm2 kg-2

    The force exerted by the earth on the moon is, 

    straight F space equals space straight G Mm over straight r squared
space equals fraction numerator 6.7 space cross times space 10 to the power of negative 11 end exponent cross times 6 space cross times space 10 to the power of 24 space cross times space 7.4 space cross times 10 to the power of 22 over denominator left parenthesis 3.84 space cross times space 10 to the power of 8 right parenthesis squared end fraction
equals 2.01 space cross times space 10 to the power of 20 straight N 

    Thus the force exerted by the earth on the moon is 2.01 x 1020 N. 

    Question 194
    CBSEENSC9006102

    What were the views of Aristotle regarding the rate of fall of bodies on the surface of the earth?

    Solution

    In the early days, it was observed that a piece of stone dropped from a certain height, reaches the Earth earlier than a light piece of paper dropped from the same height. This led the great greek philospher, Aristotle to conclude that,  a heavier body falls faster on the earth in comparison to a lighter body. So a body having mass twice as that of another body will fall twice fast than the lighter body.

    But, in 1590, Galileo disapproved this belief. He proved that all bodies irrespective of their mass and nature, fall at the same rate towards the earth. 

    Question 195
    CBSEENSC9006103

    How did Galileo prove that different bodies fall towards the earth at the same rate?

    Solution

    Galileo proved that all bodies fall at the same rate towards the earth. This experiment was done as mentioned below: 

    Galileo climbed to the top of the leaning tower of Pisa in the presence of a large gathering. He dropped spheres of different masses and materials simultaneously from the top. To a surprise of the gathering, all the spheres reached the earth's surface at the same time.
    So, he concluded that the acceleration of an object, falling freely towards the earth does not depend on the mass of the object.

    Question 196
    CBSEENSC9006104

    How did Robert Boyle prove experimentally that in vacuum, all bodies fall at the same rate?

    Solution

    Boyle's experiment: 

    Fig. Boyle's experiment of coin and paper (a) in vacuum, (b) in air.
    i) Robert Boyle took a long glass tube, a heavy coin and a piece of paper were placed inside it as shown in the Fig. below 
    ii) The ends of the tube were closed and air of the tube was removed by a vacuum pump.
    iii) When the tube was quickly inverted, it was observed that both the coin and the paper hit the bottom at the same time.
    iv) On repeating the experiment with air inside the tube, it was observed that the piece of paper falls slowly while the coin hits the bottom immediately.
    v) thus, Galileo's assertion was proved, that in vacuum all bodies irrespective of their mass fall towards the earth with the same acceleration.

    Question 197
    CBSEENSC9006105

    What do you mean by the term free fall? 

    Solution

    When the objects fall towards the Earth solely under the effect of gravitational force, the fall is known as free fall. The effect of air resistance is neglected in these cases. 

    E.g., The motion of a small heavy body in air, may be taken as a free fall as the resistance of air on it is very small. 

    Question 198
    CBSEENSC9006106

    Define acceleration due to gravity. Deduce an expression for it in terms of mass of the earth (M) and universal gravitational constant (G).

    Solution

    Acceleration due to gravity:
    Acceleration produced in the motion of a body falling under the force of gravity is called acceleration due to gravity.
    It is denoted by ‘g’.

    Relation between g and G
    Consider the earth as a sphere of mass M and radius R, as shown in Fig.
      
    Fig. To determine ‘g’.
    Consider a body of mass 'm' situated at distance 'r' from the centre of the earth.
    According to Newton's law of gravitation, the force of attraction between the earth and the body is given by, 
     
                               straight F space equals space GMm over straight r squared space space space space... left parenthesis 1 right parenthesis

    This force of gravity produces on acceleration ‘g’, called acceleration due to gravity in the body of mass m.
    Hence, from Newton's second law,
    F = Mass x Acceleration = mg            ...(2)
    From equations (1) and (2), we get 
        space space space space space space space space space space space space space space mg space equals GMm over straight r squared
or space space space space space space space space space space straight g space equals space GM over straight r squared space space space space space space space space space space... left parenthesis 3 right parenthesis
space 
    The above derived equation gives acceleration due to gravity at points far away from the earth.
    If body is located on the surface of the earth, then r = R, the radius of the earth.
    The above equation becomes, 

                  straight g subscript surface space equals GM over straight R squared

    This equation gives acceleration due to gravity at the surface of the earth. 

    The value of 'g' is almost constant near the surface of the earth because the radus of the earth does not change much over it's entire surface. 

    The value of ‘g’ on earth’s surface = 9.8 ms-2.

    Question 199
    CBSEENSC9006107

    Why do we take distance 'r' of the body from the centre of the earth for points outside or on the surface of the body?

    Solution

    We take distance 'r' of the body from the centre of the Earth because for gravitational force at points outside or on the surface, the spherical body behaves as if the entire mass of the body is concentrated at its centre. 

    Question 202
    CBSEENSC9006110

    What do you mean by acceleration due to gravity?

    Solution

    Whenever an object falls towards the Earth under the influence of gravitational force, an acceleration is involved. This is known as acceleration due to gravity. 

    Question 204
    CBSEENSC9006112

    A particle is thrown up vertically with a velocity of 50 m/s. What will be its velocity at the highest point of its journey? How high would the particle rise? What time would it take to reach the highest point?

    Solution

    According to the cartesian sign convention,
    Upward velocity is taken positive, and
    acceleration due to gravity is taken negative.
    So,
    Velocity with which the particle is thrown, u = +50 m/s
    Acceleration due to gravity, g = -10 m/s2

    At the highest point, v = 0. 
    Using the formula,
               v = u + gt
    rightwards double arrow       0 = 50 - 10 x t
    rightwards double arrow        space space straight t space equals space 50 over 10 space equals space 5 space straight s 
    Height space covered comma space straight s space equals space ut space plus space 1 half gt squared space
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 50 space cross times space 5 space minus space 1 half cross times 10 cross times left parenthesis 5 right parenthesis squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 250 minus space 125 space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 125 space straight m
space space

     

    Question 209
    CBSEENSC9006117

    A ball is dropped from the jumping board of a swimming pool, which is at a height of 20 m. A second ball is thrown from the same board after one second with initial velocity u. Both the balls hit the water together. What was the initial velocity with which the second ball was thrown ? Do they hit water with the same velocity ? Explain your answer.

    Solution

    (i) Given that, 
    s = -20 m, u = 0 m/s and g = -10 m/s2
    Now, according to the third equation of motion, 
    straight v squared space equals space straight u squared space plus space 2 gs
space space space space equals space 0 plus 2 cross times left parenthesis negative 10 right parenthesis cross times left parenthesis negative 20 right parenthesis space equals space 400
straight i. straight e. comma space straight v space equals space square root of 400 space equals space 20 space straight m divided by straight s

Downward space velocity space is space taken space as space negative. space

Therefore comma space

straight v space equals space minus space 20 space straight m divided by straight s 
    ii) Now, 
    Distance travelled by the ball, s = -40 m
    u = 0 and g = -10 m/s2 
    Now, using the third equation of motion, 
    straight v squared space equals space straight u squared space plus space 2 gs

space space space space equals space 0 plus 2 cross times left parenthesis negative 10 right parenthesis cross times left parenthesis negative 40 right parenthesis space equals space 800

straight i. straight e. comma space straight v space equals space square root of 800

space space space space space space space space space space space equals space plus-or-minus space square root of 800

space space space space space space space space space space space equals space minus 20 square root of 2 space straight m divided by straight s

Because space the space velocity space is space acting space in space the space downward space direction.

     

    Question 214
    CBSEENSC9006122

    Take data in Fig. 10.12 and find the value of ‘a’.




    Solution

    From A to B, 
    u = 0 m/s, s = - 4.9 m, t = 1 sec
    Now, using the 2nd equation of motion, 
    space space space space straight s space equals space ut space plus space 1 half at squared space
space minus space 4.9 space equals space 0 plus 1 half straight a cross times left parenthesis 1 right parenthesis squared space
rightwards double arrow space space space space straight a space equals space minus 9.8 space straight m divided by straight s squared space

From space straight B space to space straight C comma space

Let space straight v subscript straight B space be space the space velocity space of space the space body space at space straight B. space
Now comma space

straight v space equals space straight u space plus straight a space straight t

therefore space straight v subscript straight B space equals space 0 space minus space 9.8 space cross times 1 space
space space space space space space space space space space equals space space minus 9.8 space straight m divided by straight s

Again comma space using space the space equation comma
space space space space straight s space equals space ut space plus space 1 half at squared space
minus 14.7 space equals space left parenthesis negative 9.8 cross times 1 right parenthesis space plus space 1 half cross times straight a cross times left parenthesis 1 right parenthesis squared space
So comma space minus 4.9 space equals space straight a over 2
rightwards double arrow space space straight a space equals space 9.8 space straight m divided by straight s squared

From space straight C space to space straight D comma space

Let space straight v subscript straight c space be space the space velocity space of space the space body space at space straight C. space

Using space the space first space equation space of space motion comma space we space have

straight v space equals space straight u space plus space at

straight v subscript straight c space equals space minus 9.78 space minus 9.8 cross times 1 space equals space minus 19.6 space straight m divided by straight s

Using space the space formula comma space
space space space straight s space equals space ut space plus space 1 half at squared space
minus 24.5 space equals space left parenthesis negative 19.6 cross times 1 right parenthesis space plus space 1 half cross times straight a cross times left parenthesis 1 right parenthesis squared space
minus 4.9 space equals space straight a over 2 space
rightwards double arrow space straight a space equals space minus 9.8 space straight m divided by straight s squared

Thus comma space the space acceleration comma space throughout space is space 9.8 space straight m divided by straight s squared
in space the space downward space direction. space

    Question 215
    CBSEENSC9006123

    A particle is thrown up vertically with a velocity of 50 m/s. Make a table showing the distance covered by the particle, the velocity and acceleration at the end of each second of its journey. Use this data to draw distance-time, velocity-time and acceleration-time graphs.

    Solution

    Displacement covered is s.
    Velocity attained is given by,
                 v = u - gt
    Acceleration, a = -g
    Given:
    Initial velocity, u = 50 ms-1
    Acceleration due to gravity, a = -10 ms-2

    Time

    Displacement

    Distance

    Velocity

    Acceleration

    0

    1

    2

    3

    4

    5

    6

    7

    8

    9

    10

    0 m

    45 m

    80 m

    105 m

    120 m

    125 m

    120 m

    105 m

    80 m

    45 m

    0

    0

    45 m

    80 m

    105 m

    120 m

    125 m

    130 m

    145 m

    170 m

    205 m

    250 m

    50 ms–1

    40 ms–1

    30 ms–1

    20 ms–1

    10 ms–1

    0

    -10 ms–1

    -20 ms–1

    -30 ms–1

    -40 ms–1

    -50 ms–1

    -10 m/s2

    -10 m/s2

    -10 m/s 2

    -10 m/s2

    -10 m/s 2

    -10 m/s2

    -10 m/s 2

    -10 m/s-2

    -10 m/s 2

    -10 m/s 2

    -10 m/s 2

    Fig. (i) Distance-time graph.

    Fig. (ii): Velocity-time graph


    Fig. (iii) Acceleration-time graph.

    Question 216
    CBSEENSC9006124

    A particle is dropped from a tower 180 m high. How long does it take to reach the ground ? What is its velocity when it touches the ground ?

    Solution

    Given that, 
    Distance from which the ball is dropped, s = -180 m (negative sign for downward displacement)
    Initial velocity, u = 0 m/s
    Acceleration due to gravity, g = - 10 m/s2
    Using the second equation of motion, 
    space space space space space space space space space space space space space space space space space space space space space straight s space equals space ut space plus space 1 half gt squared space
rightwards double arrow space space space space space space minus 180 space equals space 0 space minus space 1 half cross times 10 cross times straight t squared
rightwards double arrow space space space space space space space space space straight t squared space equals space fraction numerator 180 cross times 2 over denominator 10 end fraction space equals space 36 space
rightwards double arrow space space space space space space space space space space straight t space equals space 6 space seconds

Velocity space on space reaching space the space ground comma space

straight v space equals space straight u space plus space gt space

space space space equals space 0 space minus space 10 cross times 6

space space space equals space minus 60 space straight m divided by straight s
Negative space sign space indicates space that space the
space velocity space is space in space the space downward space direction. space 

     
    Question 217
    CBSEENSC9006125

    Define mass of a body. What is its SI unit ? Is it a scalar or vector quantity?

    Solution

    The amount of matter which is contained in abody is called it's mass. Mass gives a measure of inertia of the body.

    Greater the mass of a body, greater is its inertia.

     SI unit of mass is kilogram (kg).

    Mass of a body is a scalar quantity because it has only magnitude and no direction. 

    Question 218
    CBSEENSC9006126

    Can the mass of a body be zero?

    Solution

    No, the mass of a body can never be zero. There is always a matter within the minute particle. 

    Question 219
    CBSEENSC9006127

    Is the mass of an object a constant quantity?

    Solution

    Yes, the mass of an object does not vary from place to place. Whether the object is on Earth, moon or outer space, an object contains same quantity of matter. 

    Question 220
    CBSEENSC9006128

    Define weight of a body. Write an expression for it.

    Solution

    The force with which a body is attracted towards the centre of the earth is known as the weight of a body. 

    When a body falls towards the earth, the gravitational force of the earth produces an acceleration 'g' in the motion of the body.

    According to Newton's second law, the force of attraction on the body of mass m is,

    F = Mass x Acceleration due to gravity
      = mg

    This force is known as the weight of the body.

    So substituting weight(W) with F, we get

                              W = mg  

    The weight of a body varies from place to place, as 'g' varies. 

    Question 221
    CBSEENSC9006129

    Is the weight of a body a scalar or vector quantity ? Give the SI unit for weight.

    Solution
    Weight of a body is a vector quantity because it has both magnitude and direction. 

    SI unit of weight is Newton(N), wich is same as that of force.
    Question 222
    CBSEENSC9006130

    Define one kilogram-weight. How many newtons are there in 1 kg-wt?

    Solution

    The force of gravity acting on a body of mass 1 kg is called as one-kilogram weight. 

    1 kg-wt = 1 kg x 9.8 m/s2 

                 = 9.8 N.

    Question 223
    CBSEENSC9006131

    How does the weight of a body depend on its mass at a given place?

    Solution

    At a given place, the value of ‘g’ is constant. 

    Therefore, at a given place, the weight of a body is directly proportional to its mass,

    That is,

                               W ∝ m

    It is due to this reason that at a given place we can use the weight of a body as a measure of its mass. 

    Question 224
    CBSEENSC9006132

    What do you mean by weight of an object on the moon ? Why is the weight of an object on the moon is less than that on the earth?

    Solution

    The force with which the object is attracted towards the centre of the moon is the weight of an object on the moon.

    The mass and radius of the moon is less than that of the earth. Due to this the moon exerts lesser force of attraction on the object. Hence, the weight of an object on the moon is less than that on the earth.
    The gravitational force of the moon is about one sixth of that on the earth.

    Example. Suppose the weight of a certain object is 6 kg-wt on the earth. If the same object is weighed on the moon, then the weight will be just 1 kg-wt.

    Question 225
    CBSEENSC9006133

    Show that weight of an object on the moon is 1/6 th its weight on the earth. Given that the mass of the earth is 100 times the mass of the moon and its radius is 4 times that of the moon.

    Solution

    Let, 
    Mass of the object = m 
    Mass of the Earth = Me 
    Mass of the moon = Mm
    Radius of the Earth = Re
    Radius of the moon = Rm
    Me = 100 Mm  and Re = 4 Rm
    Weight of the object on earth is the force with which the earth attracts the object towards it. 
      straight W subscript straight e space equals space straight G space fraction numerator straight M subscript straight e straight m over denominator straight R subscript straight e squared end fraction
Weight space of space object space on space the space moon space is comma space

straight W subscript straight m space equals space straight G fraction numerator straight M subscript straight m. straight m over denominator straight R subscript straight m squared end fraction
So comma space ratio space of space object space of space weight space on space moon space to space weight space on space earth space is comma space

straight W subscript straight m over straight W subscript straight e space equals space fraction numerator GM subscript straight m. straight m over denominator straight R subscript straight m squared end fraction cross times fraction numerator straight R subscript straight e squared over denominator straight G space straight M subscript straight e straight m end fraction

rightwards double arrow space straight W subscript straight m over straight W subscript straight e space equals space straight M subscript straight m over straight M subscript straight e cross times open parentheses straight R subscript straight e over straight R subscript straight m close parentheses squared space

space space space space space space space space space space space space space space space equals space fraction numerator straight M subscript straight m over denominator 100 space straight M subscript straight m end fraction cross times open parentheses fraction numerator 4 straight R subscript straight m over denominator straight R subscript straight m end fraction close parentheses squared space

space space space space space space space space space space space space space space space equals space 16 over 100 space equals space 1 over 6

Thus comma space the space weight space of space an space object space on space the space moon space is space
about space one minus sixth space of space it apostrophe straight s space weight space on space the space earth.

    Question 226
    CBSEENSC9006134

    What are the differences between the mass of an object and its weight?

    Solution

    Differences between mass and weight:

    Mass

    Weight

    1. Mass is the quantity of matter contained in a body and is the measure of its inertia.

    2. Its value remains constant at all places.

    3. It is a scalar quantity.

    4. It is measured by a pan balance.

    5. Mass of a body is never zero.

    6. Its unit is kg.

    1. Weight of a body is the force with which a body is attracted towards the centre of the earth.

    2. Its value (W = mg) changes from place to place due to the change in the value of acceleration due to gravity ‘g’.

    3. It is a vector quantity.

    4. It is measured by a spring balance.

    5. Weight of a body is zero at the centre of the earth because there ‘g’ becomes zero.

    6. Its unit is newton or kg-wt.

    Question 227
    CBSEENSC9006135

    Why is the weight of an object on the moon 1/6th its weight on the earth?

    Solution

    Mass of the moon is 1/100 times and radius of moon is 1/4 times that of the earth. 

    Therefore, the gravitational attraction on the moon is about one sixth of that on the earth. Hence, the weight of an object on the moon is 1/6 th of the weight on the earth. 

    Question 228
    CBSEENSC9006136

    (i) Define centre of gravity of a body.
    (ii) What is the difference between centre of gravity and centre of mass?

    Solution

    (i) Centre of gravity: The point at which the resultant of all the parallel forces due to gravity experienced by various particles of the body acts or at which the whole weight of the body acts is called the centre of gravity. 
    If a force equal to the weight of the body is applied at this point vertically upwards, the body will remain in equilibrium. 

    Hence a body can be supported at its centre of gravity.


    Fig. (i): Centre of Gravity of Some Regular Bodies

    Fig.

    Body

    Centre of gravity

    (a)

    (b)

    (c)

    (d)

    (e)

    (f)

    Right cylinder

    Circular disc

    Triangular lamina

    A solid cone

    Rectangular lamina

    Circular ring

    Mid. point

    Centre of the disc

    The point of intersection of medians.

    Point on the axis at a distance h/4 from the base of the cone, h being its height.

    Point of intersection of diagonals

    Centre of the ring (outside the body)


    (ii) Centre of mass is a point at which whole of the mass of the body may be assumed to be concentrated to describe its motion as a particle.
    Question 229
    CBSEENSC9006137

    Discuss the various factors on which the value of g depends. 

    Solution

    Variation of acceleration due to gravity depends on a following factors:
    Value of g varies from place to place. 
    (i) Effect of shape of earth: Acceleration due to gravity on the earth's surface is given by,
    space space space space space space space straight g space equals space fraction numerator G M over denominator straight R squared end fraction space space
straight i. straight e. comma space space space straight g proportional to 1 over straight R squared.

    That is, earth is not a perfect sphere.
    The Earth is slightly compressed at the poles and bulges out at the equator. So equatorial radius Re is about 21 km greater than the polar radius Rp.
     
    As,  straight R subscript straight e space greater than space straight R subscript straight p space and space straight g space proportional to space 1 over straight R squared space space space space
therefore space space space straight g subscript straight e less than straight g subscript straight p.
    Thus, the value of g is maximum at the poles and minimum at the equator. 
    The weight of the body increases as we move from the equator to the poles. 

    (ii) Effect of altitude (Height):
     The value g at a height h from the earth's surface is given by, 
    space space space space space space space space straight g subscript straight h space equals space fraction numerator G M over denominator left parenthesis straight R plus straight h right parenthesis squared end fraction space
rightwards double arrow space space space straight g subscript straight h space proportional to space fraction numerator 1 over denominator left parenthesis straight R plus straight h right parenthesis squared end fraction

    Clearly, with increase in heigh h, acceleration due to gravity decreases.
    That is why the value of g is lesser at mountains than at plains.
    At a height equal to the radius of the earth, h = R = 6400 km.
    So,
    straight g subscript straight h space equals space fraction numerator G M over denominator left parenthesis straight R plus straight R right parenthesis squared end fraction space equals space 1 fourth space fraction numerator G M over denominator straight R squared end fraction space equals space straight g over 4

    Hence the weight of a body becomes one-fourth at distance 2R from the centre of the earth of its weight at the earth's surface.

    (iii) Effect of depth: 
    At a depth d, a body experiences force of gravity only due to the inner spherical portion of radius (R – d) of the earth.
    Acceleration due to gravity at depth d is, 
    straight g subscript straight d space equals space straight g open parentheses 1 minus straight d over straight R close parentheses

    Clearly, the acceleration due to gravity decreases with the increase of depth d.
    At the centre of the earth, d = R, so
    straight g subscript straight d space equals space straight g open parentheses 1 minus straight R over straight R close parentheses space equals space 0

    Hence the weight of a body at the centre of the earth is zero.

    Question 230
    CBSEENSC9006138

    What is the effect of the distribution of mass inside the earth on the value of g? What is its utility?

    Solution

    Acceleration due to gravity in any region of the Earth is the force acting on a unit mass. If in any region the value of g is found to be large, it indicates the presence of deposits of some heavy ore in that region. 

    Question 231
    CBSEENSC9006139

    Mass of an object is 10 kg. What is its weight on the earth? Take g = 9.8 m/s2.

    Solution

    Given,
    Mass of  m = 10 kg
    Acceleration due to gravity, g = 9.8 m/s2

    Weight, W = mg
                     = 10 x 9.8 N
                     = 98 N. 

    Question 232
    CBSEENSC9006140

    An object weighs 10 N when measured on the surface of the earth. What would be its weight when measured on the surface of the moon?

    Solution

    Weight of object on the earth, Wm = 10 N
    Weight of object on the moon = 1 over 6 cross times space Weight space of space object space on space the space earth
    That is, 
    Weight of the object on the moon,
    straight W subscript straight m space equals space 1 over 6 straight W subscript straight e space
space space space space space space equals space 1 over 6 cross times 10 space
space space space space space space equals 1.67 space straight N

    Question 233
    CBSEENSC9006141

    A man weighs 600 N on the earth, what is his mass? (Take g = 10 m/s2). If he was taken to the moon, his weight would be 100 N. What is his mass on the moon? What is the acceleration due to gravity on the moon?

    Solution

    Weights of man on earth, W = 600 N
    Acceleration due to gravity 'g' on earth, g = 10 m/s2
    therefore space Mass space of space man space on space earth space equals space space straight W over straight g space equals space 600 over 10 space equals space 60 space kg
    We know that,
    Mass remains constant everywhere.
    Therefore,
    Mass of man on moon, m = 60 kg
    Weight of man on moon, W = 100 N
    Acceleration due to gravity on moon is,
      g equals space straight W over straight m space equals 100 over 60 straight m divided by straight s squared space equals space 1.67 space straight m divided by straight s squared. 

     

    Question 237
    CBSEENSC9006145

    If your weight is 60 kg on earth, how far must you go from the centre of the earth so that you weigh 30 kg?

    Solution

    Weight of a body varies inversely as the square of the distance from the centre of the earth.
    Therefore,
          straight W subscript 2 over straight W subscript 1 space equals space fraction numerator straight R subscript 1 superscript 2 over denominator straight R subscript 2 superscript 2 end fraction 
    rightwards double arrow 30 over 60 space equals space fraction numerator straight R subscript straight e superscript 2 over denominator straight R subscript 2 superscript 2 end fraction          (Re = Radius of the earth)
    rightwards double arrow space space space space space 1 half space equals space fraction numerator straight R subscript straight e superscript 2 over denominator straight R subscript 2 superscript 2 end fraction space space space space
rightwards double arrow space space fraction numerator 1 over denominator square root of 2 end fraction space equals space straight R subscript straight e over straight R subscript 2
    rightwards double arrow space space space space space space space space straight R subscript 2 space equals space square root of 2 straight R subscript straight e
B u t comma

R a d i u s space o f space t h e space E a r t h comma space straight R subscript straight e space equals space 6.4 space cross times space 10 to the power of 6 straight m

therefore space space space space space space space straight R subscript 2 space equals space 1.414 space cross times space 6.4 space cross times space 10 to the power of 6 straight m space
space space space space space space space space space space space space space space space space space equals space 9.05 space cross times space 10 to the power of 6 straight m

    So, at a distance of 9.05 x 106m from the centre of the earth, 100 kg -wt will become 30 kg-wt.


         


     

     

    Question 238
    CBSEENSC9006146
    Question 239
    CBSEENSC9006147

    At what height from the surface of the earth, will the value of g be reduced by 36% from the value at the surface ? Radius of the earth = 6400 km.

    Solution

    Given,
    At height h, the value of g reduces by 36%.
    i.e., it becomes 64% of that at the surface.
    Then,
                straight g subscript straight h space equals space 64 percent sign space o f space straight g space equals space 64 over 100 straight g
    But space space space straight g subscript straight h space equals space straight g open parentheses fraction numerator straight R over denominator straight R plus straight h end fraction close parentheses squared
therefore space space space space space space space 64 over 100 straight g space equals space straight g open parentheses fraction numerator straight R over denominator straight R plus straight h end fraction close parentheses squared space space
rightwards double arrow space space space space space space space space space space 8 over 10 space equals space fraction numerator straight R over denominator straight R plus straight h end fraction

rightwards double arrow space space space straight h space equals space straight R over 4 space
space space space space space space space space space space space equals 6400 over 4 space
space space space space space space space space space space space equals space 1600 space km. space space

     

    Question 241
    CBSEENSC9006149

    Find the percentage change in the weight of a body when it is taken from the equator to poles. The radius of the earth at the poles is 6357 km, the radius at the equator is 6378 km.

    Solution

    A t space t h e space p o l e s comma space space straight g subscript straight p space equals space fraction numerator G M over denominator straight R subscript straight p superscript 2 end fraction
A t space t h e space e q u a t o r comma space straight g subscript straight e space equals space fraction numerator G M over denominator straight R subscript straight e superscript 2 end fraction

R a t i o space i s space g i v e n space b y comma

therefore space space straight g subscript straight p over straight g subscript straight e space equals space open parentheses straight R subscript straight e over straight R subscript straight p close parentheses squared space

space space space space space space space space space space space space space space space equals space open parentheses 6378 over 6357 close parentheses squared space

space space space space space space space space space space space space space space space equals left parenthesis 1.0033 right parenthesis squared space

space space space space space space space space space space space space space space space equals space 1.0066

i. e. comma space space straight g subscript straight p space equals space 1.0066 space straight g subscript straight e 
    The percentage increase in weight the body is taken from the equator to the poles,
                          space fraction numerator m g subscript straight p space minus space m g subscript straight e over denominator m g subscript straight e end fraction cross times 100 space equals space fraction numerator straight g subscript straight p space space minus space straight g subscript straight e over denominator straight g subscript straight e end fraction cross times 100
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 1.006 space straight g subscript straight e space minus space straight g subscript straight e over denominator straight g subscript straight e end fraction cross times space 100
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.0066 space cross times space 100 space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.66 space percent sign space space asymptotically equal to space space space 0.7 space percent sign.

    Question 242
    CBSEENSC9006150

    Define the term thrust. Give its SI unit

    Solution

    The force acting on an object perpendicular to the surface is called thrust.

    Thrust is a type of force. So, it's SI unit is Newton.


    Question 243
    CBSEENSC9006151

    Define the term pressure. Give its SI unit.

    Solution

    Pressure: The thrust acting per unit area of a surface is called pressure.
    Mathematically, 
    Pressure space space equals space Force over Area
    If a force F acts normally over an area A of a surface, then the pressure, 
              straight P space equals space straight F over straight A
    The SI unit of pressure is newton per square metre (Nm-2).
    It is also called pascal (Pa).

     

    Question 244
    CBSEENSC9006152

    Describe some practical applications from daily life which make use of the concept of pressure.

    Solution

    Practical applications based on the concept of pressure are as given below: 

    1. A sharp knife cuts better than a blunt one. The area of a sharp edge is much less than the area of blunt edge. Therefore, the effective force per unit area is more for the sharp edge than the blunt edge. Hence, a sharp edge cuts better. 

    2. Railway tracks are laid on large sized wooden sleepers. Therefore, the force due to the weight of the train is on a larger area. Hence, this reduces the pressure considerably. This prevents sinking of the ground under the weight of the train. 

    3. Pins and nails are made to have pointed ends. Their pointed ends have very small area. When a force is applied over the head of a pin or a nail, it transmits a large pressure (force/area) on the surface and hence easily penetrates through the surface. 

    Question 245
    CBSEENSC9006153

    When you stand on loose sand, your feet go deep into the sand. Now, lie down on the sand. You will find that your body will not go deep in the sand. Give reason for the difference in the two cases.

    Solution

    The force exerted on the sand is equal to the weight of our body in both the cases. 

    When we stand on loose sand, our weight acts on a smaller area of the sand. Hence, a large pressure (= force/area) is exerted onto sand and our feet penetrate deeper into the sand.

    When we lie down, our weight acts on a larger surface area of the sand. Hence, a smaller pressure is exerted onto sand and our feet stay on the ground only. 

    Question 246
    CBSEENSC9006154

    A force of 40 N is applied on a nail, whose tip has an area of cross-section of 0.001 cm2. Calculate the pressure on the tip.

    Solution

    Given,
    Force, F = 40 N
    Area of cross-section, A = 0.001 cm2
                                       = 0.001 x 10-4 m2
    So, Pressure is given by, 
    straight P space equals space straight F over straight A equals fraction numerator 40 over denominator 0.001 space cross times space 10 to the power of negative 4 end exponent end fraction space equals space 4 space cross times space 10 to the power of 8 space straight N divided by straight m squared.

    Question 249
    CBSEENSC9006157

    A block of wood is kept on a table top. The mass of wooden block is 5 kg and its dimensions are 40 cm x 20 cm x 10 cm. Find the pressure exerted by the wooden block on the table top if it is made to lie on the table top with its sides of dimensions (i) 20 cm x 10 cm and (ii) 40 cm x 20 cm.

    Solution

    Mass of the wooden block  = 5 kg
    Dimensions of the block = 40 cm x 20 cm x 10 cm
    Force exerted by the block on the table top,
     F = Weight of block = mg = 5 x 9.8 = 49 N
    (i) Let us consider the side of dimension 20 cm x 10 cm
    Therefore, 
    Area = Length x Breadth  = 20 cm x 10 cm
            = 0.20 x 0.10 m = 0.02 m2
    That is, 
    Pressure space space equals space Thrust over Area
space space space space space space space space space space space space space space space space space equals space fraction numerator Force space over denominator Area end fraction space equals space fraction numerator 49 space straight N over denominator 0.02 space straight m squared end fraction space equals space 2450 space Nm to the power of negative 2 end exponent.
    (ii) Consider the block lies on its side 40 cm x  20 cm,
    Area = Length x Breadth
            = 40 cm x 20 cm
            = 0.40 m x 0.20 m = 0.08 m2

    When the block lies on its side 40 cm x 20 cm, it exerts the same force F.
    therefore space space space Pressure space equals space Thrust over Area
space space space space space space space space space space space space space space space space space space space space space space space equals space Force over Area space equals fraction numerator 49 space straight N over denominator 0.08 space straight m squared end fraction space equals space 612.5 space Nm to the power of negative 2 end exponent.

     

    Question 250
    CBSEENSC9006158

    What is difference between the pressure exerted by a solid and a fluid?

    Solution

    A solid exerts pressure on a surface due to its weight. 
    Fluids also have weight. Hence, the fluids also exert pressure on the walls of the container in which they are kept. A fluid exerts pressure in all the directions. 

    Question 251
    CBSEENSC9006159

    What do you understand by buoyancy and centre of buoyancy? 

    Solution

    The upward force acting on a body immersed in a fluid is called upthrust or force of buoyancy and the phenomenon is called buoyancy. 
    When a body is immersed partially or wholly in a fluid, it displaces fluid. The displaced fluid tends to regain its original condition. Durig this time, it exerts an upward force on the body which is called as Buoyancy. 
    For example, a cork taken inside water comes to the surface after experiencing an upward thrust. 

    The buoyancy acts through the centre of gravity of the displaced fluid which is called centre of buoyancy.

    Question 252
    CBSEENSC9006160

    Why is it difficult to hold a school bag having a strap made of a thin and strong string?

    Solution

    The thin string exerts a large pressure on our hand due to the smaller area covered. Hence, we feel the weight of the bag and feel uncomfortable. 

    Question 253
    CBSEENSC9006161

    What do you mean by buoyancy?

    Solution

    The upward force acting on a body immersed in a fluid is called upthrust or force of buoyancy. The phenomenon is called buoyancy. 

    The magnitude of buoyant force depends upon the density of the fluid. 

    Question 254
    CBSEENSC9006162

    When does an object float or sink when placed on the surface of water?

    Solution

    (i) An object sinks in water if its density is greater than that of water. Upthrust of water on an object is less than the weight of the object. Hence, it sinks.

    (ii) An object floats on water when its density is less than that of water. When the upthrust of water on the object is greater than the weight of the object, the object floats.

    Question 255
    CBSEENSC9006163

    State Archimedes' principle.

    Solution

    Archimede's principle states that when a body is immersed fully or partially in a fluid, it experiences an upward force that is equal to the weight of the fluid displaced by it.

    Question 256
    CBSEENSC9006164

    Briefly explain how can Archimedes' principle be verified experimentally?

                                         Or

    Describe a simple activity to show that a body loses weight when immersed in a fluid.

    Solution

    Archimedes principle can be explained experimentally. 
    Look at the fig. below.

    Fig. 10.22. Verification of Archimedes' principle.

    i) Take an overflow can and fill it with water.
    ii) Hang a piece of stone on spring balance.
    iii) Weigh the piece of stone hung from the spring balance.
    iv) Take a beaker and weigh it.
    v) Place the beaker below the spout of the overflow can.
    vi) Immerse the stone in the overflow can and collect the water which overflows and read the weight of stone when it is immersed in water.
    vii) We find that, the weight of stone in water is less than the weight of stone in air.
    viii) Now weigh the beaker with the water.
    ix) It is observed that the loss in weight of the stone is equal to the weight of the water collected in the beaker, i.e., the weight of the water displaced.
    Hence, Archimedes principle is proved. 

     

    Question 257
    CBSEENSC9006165

    State Archimedes' principle. Write two applications of this principle.

    Solution

    Archimedes' principle states that when a body is immersed fully or partially in a fluid, it experiences an upward thrust equal to the weight of the fluid displaced by it. 

    The applications of Archimedes’ principle are:

    (i) Archimedes' principle is used in designing ships and submarines. 

    (ii) Lactometers based on Archimedes' principle are used to measure purity of a sample of milk.

    (iii) Hydrometers used to measure density of liquids are based on Archimedes' principle.

    Question 258
    CBSEENSC9006166

    You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?

    Solution

    The weighing machine reads slightly less than the actual value. This is because of the upthrust of air acting on our body. Hence, our actual weight is more than 42 kg. 

    Question 259
    CBSEENSC9006167

    You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality one is heavier than other. Can you say which one is heavier and why?

    Solution

    In reality, cotton bag is heavier than the iron bar. The cotton bag experiences larger upthrust of air than on the iron bar. So, the weighing machine indicates a smaller mass for cotton bag than its actual mass. 

    Question 260
    CBSEENSC9006168

    Define the terms density and relative density. Give their units.

    Solution

    Density of a substance is defined as its mass per unit volume. 

    Mathematically, we can write, 

                     Density = MassVolume

    The SI unit of density is kilogram per cubic metre (kg m–3).

    Relative density of a substance is the ratio of the density of the substance to the density of water.

    Therefore, 

    Relative density = Density of a substancedensity of water


    Since the relative density is a ratio of two similar quantities, it has no units.

    Question 261
    CBSEENSC9006169

    State the laws of floatation of a body in a liquid.

    Solution

    The laws of floatation are: 

    i) A body will sink in a liquid if its density is greater than the density of the liquid.

    (ii) A body will float in a liquid if its density is less than the density of the liquid.

    (iii) A body will be in equilibrium at any submerged depth, if the density of the body is equal to the density of the liquid. 

    Question 262
    CBSEENSC9006170

    Density of gold is 19300 kg/m3. What will be the mass of a gold brick with dimensions of 5 cm x 10 cm x 20 cm?

    Solution
    Density of gold = 19300 kg/m3

    Dimension of the brick = 5 cm x 10 cm x 20 cm

    Therefore, 

    Volume = l×b×h

                = 5 x 10 x 20 cm3

                = 5 x 10 x 10-6 m3

                = 10-3 m3

    So, Mass = Volume x Density

                  = 10-3 x 19300

                  = 19.3 kg
    Question 263
    CBSEENSC9006171

    The density of gold is 19300 kg/m3 and that of water is 1000 kg/m3. What is the relative density of gold?

    Solution

    Given,
    Density of gold = 19300 kg/m3
    Density of water = 1000 kg/m3
    Therefore, 

    Relative density of gold = Density of gold
                                           Density of water
                                     
                                        = 19300 kg/m3
                                          
    1000 kg/m3
                                      
                                         = 19.3

    Question 264
    CBSEENSC9006172

    Relative density of silver is 10.8. The density of water is 103 kg m–3. What is the density of silver in S.I. unit?

    Solution

    Given,

    Relative density of silver = 10.8
    Density of water = 103 kg/m3

    Now, using the formula of relative density, 

    Relative density = Density of silver
                                Density of water

    So,

    Density of silver = R. D. x Density of water

                             = 10.8 c 102 kg m-3

    The density of silver is 10.8 x 103 kg m-3 

    Question 265
    CBSEENSC9006173

    What is difference between gravitation and gravity?

    Solution
    Gravitation gravity
    1. Gravitation is the force of attraction between any two bodies in general. 1. Gravity refers to the force of attraction between any body and the earth.
    2. The value of G never varies and is constant for all bodies at all places.  2. Value of 'g' is different at both equator and poles. 
    3. G = 6.67 × 10-11 m3 kg-1s-2 3. g = 9.8 m/s2
    Question 266
    CBSEENSC9006174

    State the Universal law of gravitation.

    Solution

    Newton's law of gravitation states that every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them.

    The force is along the line joining the centres of two objects. 

    Question 267
    CBSEENSC9006175

    We cannot move finger without disturbing stars. Why? 

    Solution

    There exists a gravitaional force between all objects in the universe. When we move our fingers, force of attraction changes when the distance between two bodies changes. 
    Therefore, the position of the stars also changes. 

    Question 268
    CBSEENSC9006176

    How the gravitational force between two objects depends on the distance between them? What happens to the gravitational force between two objects when the distance between them is doubled?

    Solution

    Gravitational force F between two objects is inversely proportional to the square of the distance between them.
    That is,
                                straight F proportional to 1 over straight r squared

    When the distance between two objects is doubled, the force becomes one-fourth of its original value.

    Question 269
    CBSEENSC9006177

    What is the value of the universal gravitational constant?

    Solution

    The value of universal gravitational constant is 6.67 x 10–11 Nm2/kg2

    Question 270
    CBSEENSC9006178

    Does the value of ‘G’ depend on the medium present between the two bodies?

    Solution

    No, the value of ‘G’ is independent on the medium present between the two bodies.

    Question 271
    CBSEENSC9006179

    The value of G on the surface of the earth is 6.67 x 10–11 Nm2/kg2. What is its value on the moon?

    Solution

    G is a universal constant.
    So, its value on the moon = 6.67 x 10–11 Nm2/kg2.

    Question 272
    CBSEENSC9006180

    What does the small value of G indicate?

    Solution

    The small value of 'G' indicates that the force of gravitational attraction between two ordinary sized objects is a very weak force.

    Question 273
    CBSEENSC9006181

    The value of G is very small but all the objects near the surface of the earth fall towards it. Why?

    Solution

    Mass of the Earth is very large. Therefore, the force of gravitation is sufficiently large to make all objects near its surface to fall towards it.

    Question 274
    CBSEENSC9006182

    In which direction does the force of gravitational attraction act?

    Solution

    The force of gravitational attraction acts along the line joining the centres of mass of the two bodies.

    Question 275
    CBSEENSC9006183

    Does ‘G’ change with increasing height above the earth’s surface?

    Solution

    Value of G remains constant. Therefore, the value of G does not change with increasing or decreasing height. 

    Question 276
    CBSEENSC9006184
    Question 277
    CBSEENSC9006185

    Does the gravitational force between two bodies change, if a material body is placed between them?

    Solution
    No, the gravitaional force will not change if a material body is placed between the bodies.

    Tips: -

    Question 278
    CBSEENSC9006186

    Does the gravitational force between two bodies depend upon the presence of other bodies in the surroundings?

    Solution

    No, gravitaional force does not depend upon the presence of other bodies in the surroundings. It remanins constant. 

    Question 279
    CBSEENSC9006187

    Do all bodies fall with the same acceleration in the absence of air resistance?

    Solution

    Yes, all bodies fall with the same acceleration in the absence of air resistance and this acceleration is called 'acceleration due to gravity'. 

    Question 280
    CBSEENSC9006188

    Define acceleration due to gravity.

    Solution

    An object which is falling freely under the effect of gravity gets accelerated. This acceleration is called as 'acceleration due to gravity'. 

    Question 282
    CBSEENSC9006190

    What are the values of g and G at the centre of the earth?

    Solution

    At the centre of the earth,

    g = zero
    G = 6.67 x 10–11 Nm2/kg2

    Question 283
    CBSEENSC9006191

    Which is scalar and which is vector amongst g and G?

    Solution

    Acceleration due to gravity(g) is acting vertically downwards. Therefore, g is a vector quantity. But, 'G' is a scalar quantity. 

    Question 284
    CBSEENSC9006192

    Write down the formula for acceleration due to gravity. 

    Solution
    Acceleration due to gravity is given by, 

                               g = GMR2

    where, 
    G is the universal gravitaional constant, 
    M is the mass of the Earth, and
    R is the radius of the Earth. 

     
     
    Question 285
    CBSEENSC9006193

    Explain why a 10 kg stone will fall with the same acceleration as 1 kg stone?

    Solution

    since the acceleration due to gravity is given by,

                             g = GMR2 

    where, M is the mass of the Earth. 

    Therefore, 'g' does not depend on the mass of the body. So both the stones will fall with the same acceleration. 

    Question 286
    CBSEENSC9006194

    Name the factors on which the value of g depends.

    Solution

    The value of g depends on,

    (i) the shape of the earth,
    (ii) latitude, 
    (iii) altitude, and
    (iv) depth.

    Question 287
    CBSEENSC9006195

    What is the effect of altitude (height) on acceleration due to gravity?

    Solution

    'Acceleration due to gravity' decreases with altitude. 

    Question 288
    CBSEENSC9006196

    What is the effect of latitude and shape of earth on the value of g?

    Solution

    The value of 'g' is minimum at the equator and gradually increases from equator to poles.

    Question 289
    CBSEENSC9006197

    What is the effect of rotation of the earth on the acceleration due to gravity?

    Solution

    The value of 'g' decreases due to rotation of the earth. This effect is zero at poles and maximum at the equator. 

    Question 290
    CBSEENSC9006198

    A man can jump six times as high on the moon as that on the earth. Why?

    Solution

    A man can jump six times as high on the moon than on the earth because the value of 'g' on the surface of the moon is about one-sixth of its value on the surface of the earth. 

    Question 291
    CBSEENSC9006199

    Define centre of mass of a body.

    Solution

    Centre of mass of a body is the point where the whole mass of the body is assumed to be concentrated. 

    Question 292
    CBSEENSC9006200

    On which factors does the position of centre of mass depend?

    Solution

    The position of centre of mass depends on,

    (i) shape of the body and
    (ii) distribution of mass in the body.

    Question 293
    CBSEENSC9006201

    Where does the centre of mass of a triangular lamina lie?

    Solution

    The centre of mass of a triangular lamina is at the centroid of the triangular lamina. 

    Question 294
    CBSEENSC9006202

    Where does the centre of mass of a cone lie?

    Solution

    The centre of mass of a cone lies on the line joining the apex to the centre of the base at a distance equal to 1/4 of the length of this line from the base.

    Question 295
    CBSEENSC9006203

    Does the centre of mass of a body necessarily lie inside the body?

    Solution

    No, it is not necessary that centre of mass of a body lies inside the body.

    For example, the centre of mass of a ring lies in its hollow portion. 

    Question 296
    CBSEENSC9006204

    What is the mass of a body at a height above the earth's surface of one earth radius when its mass on earth's surface is m kg?

    Solution

    Given that mass of the body on the surface of the Earth = m kg

    Mass of the object is constant everywhere. 
    So, mass of the body at a height 'h' above the Earth's surface = m kg

    Question 297
    CBSEENSC9006205

    Define weight of a body.

    Solution

    Weight of a body is the force with which a body is attracted towards the centre of the earth. 

    Question 298
    CBSEENSC9006206

    What is the relation between mass and weight of a body?

    Solution

    The relation between mass and weight of a body is given by, 
     
                                  W = mg

    Question 299
    CBSEENSC9006207

    Is weight a vector or a scalar quantity?

    Solution

    Weight is a vector quantity because weight always acts in the direction of force.

    Question 300
    CBSEENSC9006208

    What is the SI unit of weight?

    Solution

    Weight has the same unit as that of force. 

     SI unit of weight is Newton. 

    Question 301
    CBSEENSC9006209

    One kg-wt is equal to how many Newtons?

    Solution

    One kg of weight is equal to 9.8 Newtons.

    Question 302
    CBSEENSC9006210

    A body has a weight of 10 kg on the surface of the earth. What will be its weight when taken to the centre of the earth?

    Solution

    At the center of the earth, acceleration due to gravity = 0. 

    Weight is dependent on the value of 'g'. 

    So, weight at the centre of earth = 0

    Question 303
    CBSEENSC9006211

    Where on the earth surface is the weight of a body (i) maximum, and (ii) minimum?

    Solution

    The value of acceleration due to gravity influence the weight of a body.
    (i) The value of g is maximum at the poles. Therefore, weight of the body is maximum on the poles. 

    (ii) The value of 'g' is  minimum at the equator. Therefore, weight of the body is minimum.

    Question 304
    CBSEENSC9006212

    A body of weight 600 N rests on the floor of a lift. If the lift begins to fall freely under gravity, what is the force with which the body presses on the floor?

    Solution

    The weight of the body during a free fall is zero. So, the force with which the body presses on the floor is also zero. 

    Question 305
    CBSEENSC9006213

    Name one factor on which the value of ‘g’ depends.

    Solution

    The value of ‘g’ depends on the distance from the centre of the earth. 

    Question 306
    CBSEENSC9006214

    The mass of a body is 9.8 kg on the earth. What would be its mass on the moon where the value of g is approximately 1.63 m/s2.

    Solution

    Mass of the body is constant everywhere. 

    Therefore, mass of body will remain same on the moon i.e. 9.8 kg. 

    Question 307
    CBSEENSC9006215

    How does the mass of a body change when it is taken into deep space?

    Solution

    Mass of the body remains constant. So, mass of a body will not change when taken into deep space. 

    Question 308
    CBSEENSC9006216

    Two balls are released simultaneously from a certain height, one allowed to fall freely and other thrown with some horizontal velocity. Will they hit the ground together? 

    Solution

    When two balls are released simultaneously in a free fall, the balls will hit the ground at the same rate. Free fall is independent of the mass of the body.

    Question 309
    CBSEENSC9006217

    What is the SI unit of pressure?

    Solution

    SI unit of pressure is Nm–2 or Pascal (Pa).

    Question 310
    CBSEENSC9006218

    What is the SI unit of density?

    Solution

    SI unit of density is kg m-3.

    Question 311
    CBSEENSC9006219

    Why does a mug full of water feel lighter inside water?

    Solution

    A mug full of water feels lighter inside water because of upthrust exerted by water on the mug is more than the weight of the mug. 

    Question 312
    CBSEENSC9006220

    How does the gravitational force change between two objects when:

    (i) the distance between them is reduced to half,

    (ii) the mass of each is reduced to half. Justify your answer in each case.

    Solution

    As per Newton's law of gravitation, the force of attraction between two bodies is given by, 
                           straight F space equals space straight G fraction numerator straight m subscript 1 straight m subscript 2 over denominator straight R squared end fraction
    (i)  Here, distance between the bodies is reduced to half.
    i.e.,                   straight R apostrophe space equals space straight R over 2 comma
    Therefore comma space

Force comma space straight F apostrophe space equals space straight G fraction numerator straight m subscript 1 straight m subscript 2 over denominator straight R apostrophe squared end fraction space
space space space space space space space space space space space space space space space space space equals space straight G fraction numerator straight m subscript 1 straight m subscript 2 over denominator open parentheses begin display style straight R over 2 end style close parentheses squared end fraction space equals space fraction numerator 4 G m subscript 1 straight m subscript 2 over denominator straight R squared end fraction space equals 4 straight F
    i.e, gravitational force increases four times when distance is reduced to half. 
    (ii) Given, mass of each body is reduced to half.
    That is, 
     straight m subscript 1 apostrophe space equals space straight m subscript 1 over straight m subscript 2 comma space space straight m subscript 2 apostrophe space equals space straight m subscript 2 over 2 
    therefore space space space space space straight F apostrophe space equals space straight G fraction numerator straight m subscript 1 apostrophe straight m subscript 2 apostrophe over denominator straight R squared end fraction
space space space space space space space space space space space space space space equals space fraction numerator straight G open parentheses begin display style straight m subscript 1 over 2 end style close parentheses space open parentheses begin display style straight m subscript 2 over 2 end style close parentheses over denominator straight R squared end fraction space
space space space space space space space space space space space space space space equals 1 fourth fraction numerator G m subscript 1 straight m subscript 2 over denominator straight R squared end fraction space equals 1 fourth straight F

    Hence, the gravitational force becomes one-fourth of the original value. 

     

    Question 313
    CBSEENSC9006221

    Why is the gravitational force between ordinary sized objects not noticeable?

    Solution

    Gravitational force between ordinary sized objects is hardly observable. This is because of the very small value of force of gravitation between two ordinary sized objects is very small and cannot be easily detected. 

    Question 314
    CBSEENSC9006222

    Explain why a small piece of stone is not attracted towards another big piece of stone on the earth's surface?

    Solution

    Because of very small value of G, the force of attraction between any two such ordinary sized bodies is so small that it cannot produce motion in them.

    Question 315
    CBSEENSC9006223

    According to Newton's third law of motion, action and reaction are equal and opposite, then why do we see a stone falling to the earth and not the earth rising towards the stone?

    Solution

    Any object which is falling towards the Earth exerts an equal and opposite force towards the earth. But, we see a stone falling towards the earth and not the movement of the Earth towards the stone. This is because of the large mass of the Earth as compared to that of the stone. Hence, the acceleration produced in the stone is very large as compared to that of the earth. 
    So the stone readily falls towards the earth while the earth does not rise at all. 

    Question 316
    CBSEENSC9006224

    What is the effect of gravitational force between two objects of comparable masses? Give an example.

    Solution

    When two masses are comparable, the force of gravitation between two objects produces comparable accelerations in them. Therefore, the two objects move towards each other. 

    For example, binary stars, Sirius and its companion move around each other because their masses are comparable. 



    Question 317
    CBSEENSC9006225

    Give some points of differences between g and G.

    Solution

    Acceleration due to gravity ‘g’

    Universal gravitational constant ‘G’

    1. It is the acceleration acquired by a body due to earth's gravitational pull on it.

    2. ‘g’ is not a universal constant. It is different at different places on the surface of the earth. Its value varies from one celestial body to another.



    3. It is a vector quantity.

    1. It is numerically equal to the force of attraction between two masses of 1 kg each separated by a distance of 1 m.

    2. ‘G’ is a universal constant i.e., its value is same viz., 6.67 x 10–11 N-m2 kg–2 everywhere in the universe.

    3. It is a scalar quantity.

    Question 319
    CBSEENSC9006227

    Under what conditions our weight becomes zero?

    Solution

    Our weight becomes zero when we fall freely under gravity. 
    Also, at the centre of the Earth, weight becomes zero because acceleration due to gravity is zero at the centre of the Earth. 

    Question 320
    CBSEENSC9006228

    Why do we feel giddy while moving on a merry-go-round?

    Solution

    While going up, our weight increases and it decreases when awe are going down. Therefore, we feel giddy in a merry-go-round. 

    Question 321
    CBSEENSC9006229

    An astronaut feels weightlessness in a spacecraft. Give reason.

    Solution

    An astronaut is in a state of free fall in a spacecraft. A body falling freely does not experience any force on his body. Hence, a state of weightlessness is felt for an astronaut in a spacecraft.

    Question 322
    CBSEENSC9006230

    Why is the value of g more at the poles than at the equator?

    Solution

    Acceleration due to gravity is given by, 
                            g =  GM over straight R squared
    We can see that value of 'g' is inversely proportional to the radius, R. 
    And, the value of R at the equator is greater than at the poles. Hence 'g' at poles is greater than g at the equator. 

    Question 323
    CBSEENSC9006231

    Why does a body weigh more at the poles than at the equator?

    Solution

    Acceleration due to gravity is greater at poles than that at the equator. 

    That is, g> ge

     mgp > mge

    So, 
    Weight at poles > Weight at equator.

    Question 324
    CBSEENSC9006232

    Where does a body weigh more ? At the sea level or on the mountains?

    Solution

    A body weigh more at the sea-level. 'g' decreases with height. 

    Question 325
    CBSEENSC9006233

    The weight of a body is less inside the earth than on the surface. Why? 

    Solution

    The value of attracting mass M decreases when we go inside the Earth. Therefore, the value of g decreases. So, the weight of the body(mg) is less inside the Earth than on the surface. 

    Question 326
    CBSEENSC9006234

    If the earth stops rotating about its axis, what will be the effect on the value of g? Will the effect be same at all places?

    Solution

    If the Earth stops rotating about it's axis, value of 'g' will increase at all places except at the poles. The variation will be different at different places. Value of 'g' will be maximum at the equator. 
     

    Question 327
    CBSEENSC9006235

    Which force is responsible for the motion of the planets around the sun?

    Solution

    Gravitational force of attraction of the sun keeps the planets moving in their orbits, just as the moon revolves round the earth due to gravitational attraction of the earth.

    Fig. Orbital motion of a Planet around the Sun.

     

    Question 328
    CBSEENSC9006236

    A satellite does not need any fuel to circle around the earth. Why?

    Solution

    The gravitational force between satellite and earth provides the centripetal force required by the satellite to move in a circular orbit. Hence no fuel is required for its orbital motion.

    Question 329
    CBSEENSC9006237

    What is difference between centre of gravity and centre of mass?

    Solution

    Centre of mass: The point at which whole of the mass of the body is concentrated to describe its motion as a particle.

    Centre of gravity: The point at which the entire weight of a body may be thought of as centered, so that, if supported at this point the body would balance perfectly. 

    Question 330
    CBSEENSC9006238

    Why does a projectile fired along the horizontal does not follow a straight line path?

    Solution

    A projectile which is fired horizontally is being constantly acted upon by acceleration due to gravity, acting vertically downwards. Hence, it does not follow a straight line path.

    Question 331
    CBSEENSC9006239

    You must have seen two types of balances. A grocer has a balance with two platforms, or pans, and a needle moving on a scale. Some junk dealers (kabadi walas) may be using a spring balance to weigh used newspapers. Suppose the two balances give the same measure for a given body on the earth. Now, you take the balances to moon. Would the two balances give the same measure ? Explain your answer.

    Solution

    No, the two balance would measure different on the moon.
    The pan balance measures inertial mass while the spring balance measures gravitational mass or weight of the body. Both masses are equivalent when measured on the earth. But, on the moon the value of 'g' becomes one-sixth of that on the earth. So, spring balance will measure one-sixth of the reading on the pan balance. 

    Question 332
    CBSEENSC9006240

    You buy a bag of sugar of weight W at a place on the equator. You take this to Antarctica. Would its weight be the same there ? If not, would it increase or decrease? 

    Solution

    The value of g is less at equator than at poles. Weight of a body is dependent on the value of 'g'. And, Antarctica  lies in the polar region of the earth.
    Therefore, weight of the bag of sugar in Antarctica would increase.


    Question 333
    CBSEENSC9006241

    In what sense does the moon fall towards the earth? Why does it not actually fall on the earth's surface?

    Solution

    Moon is orbiting around the Earth in a circular orbit. The centripetal force which is provided by the force of attraction of the Earth will enable the moon to go in a circular orbit. So, at each point of its orbit, the moon falls towards the earth instead of going straight. 


    Question 334
    CBSEENSC9006242

    The earth attracts an apple. Does the apple also attract the earth ? If it does, why does the earth not move towards the apple?

    Solution

    The Earth attracts an apple and so does the apple attracts the earth with an equal and opposite force. Mass of the Earth is extremely massive as compared to that of the moon. So, the acceleration produced is very small as compared to that in the apple. Hence, the motion of the earth towards the apple is not noticeable. 

    Question 335
    CBSEENSC9006243

    If the force of gravity somehow vanishes today, why would we be sent flying in space?

    Solution

    In the absence of force of gravity, the centripetal force required to keep us rotating along with the earth would not be available. Then, we would be flying off along the tangent into the space. 

    Question 336
    CBSEENSC9006244

    Suppose the mass of the earth somehow increases by 10% without any change in its size. What would happen to your weight? Suppose the radius of the earth becomes twice of its present radius without any change in its mass, what will happen to your weight?

    Solution

    Weight  = Force of gravity
    That space is comma space
space space space space space space straight W space equals space m g space equals space fraction numerator G M m over denominator straight R squared end fraction 
    Clearly, when the mass M of the earth increases by 10%, our weight also increase by 10% because weight is directly proportional to mass. 
    Also, weight is inversely proportional to the square of radius. Therefore, when radius R of the earth becomes twice of its present radius, our weight become 1/4  of its present value.    

    Question 337
    CBSEENSC9006245

    Suggest a method for calculating the mass of the moon.

    Solution

    The acceleration produced on the surface of moon is given by, 
                         straight a space equals space fraction numerator GM subscript straight m over denominator straight R subscript straight m superscript 2 end fraction
    therefore space space space space space Mass space of space the space moon comma space straight M subscript straight m space equals space fraction numerator aR subscript straight m superscript 2 over denominator straight G end fraction

    Thus knowing acceleration a, radius Rm of the moon and G, the mass of the moon can be determined.

    Question 338
    CBSEENSC9006246

    A ball moving on a table reaches the edge and falls. Sketch the path it will follow while falling.

    Solution

    The ball carries a horizontal velocity and a vertical downward acceleration due to gravity.
    Under the combined effect of these two motions, the ball moves along a parabolic trajectory as shown in the fig. below. 

     

    Question 339
    CBSEENSC9006247

    A sharp knife cuts better than a blunt one. Why?

    Solution

    The surface area of sharp edged knife is much less than the blunt edge. For the same total force applied, the effective force per unit area (or pressure) is more for the sharp edge than for the blunt edge. So sharp edge cuts better. 

    Question 340
    CBSEENSC9006248

    Define the terms thrust and pressure. State their SI units.

    Solution

    The total force acting on a body perpendicular to its surface is called thrust.

    SI unit of thrust is newton (N).

    Thrust acting per unit on the surface area is called pressure.


    Pressure = Thrust
                       Area

    The SI unit of pressure is Nm–2 or pascal (Pa).

    Question 341
    CBSEENSC9006249

    Define the terms (i) thrust (ii) pressure and (iii) relative density.

    Solution

    (i) Thrust: The total force acting on a body perpendicular to its surface is called thrust.

    (ii) Pressure: Thrust acting per unit on the surface area is called pressure. 

    (iii) Relative density: The relative density of a substance is the ratio of the density of the substance to the density of water.

    Relative density = Density of substance
                                     Density if water at

    Since, relative density is a pure ratio, it has no units. 

    Question 344
    CBSEENSC9006252

    The relation between ‘g’ and ‘G’ is

    • g = GMR2
    • G R2
        R2
    • g = G MR2
    •   G
      MR2

    Solution

    A.

    g = GMR2

    The relation between ‘g’ and ‘G’ is

    Question 345
    CBSEENSC9006253

    The value of universal gravitational constant in SI system is

     

    • 6.67 x 10–3

    • 6.67 x 10–6

       

    • 6.67 x 10–11

    • 6.67 x 10–5.

    Solution

    C.

    6.67 x 10–11

    6.67 x 10–3

    Tips: -

    6.67 x 10–6  

     

    Question 356
    CBSEENSC9006264
    Question 369
    CBSEENSC9006277
    Question 374
    CBSEENSC9006886

    Hydrogen and oxygen combine in the ratio 1:8 by mass to form water. What mass of oxygen gas would be required to react completely with 3g of hydrogen gas?

    Solution

    It is given that the ration of hydrogen and oxygen by mass to form water is 1:8. Then, the mass of oxygen gas required to react completely with 1g of hydrogen gas is 8g. Therefore, the mass of oxygen gas required to react completely with 3g of hydrogen gas is = 8 x 3g=24g.

    Question 375
    CBSEENSC9006887

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

    Solution

    The postulate of Dalton’s atomic theory which is result of the law of conservation of mass is “Atoms are indivisible particles, which can neither be created nor be destroyed in a chemical reaction.

    Question 376
    CBSEENSC9006888

    Which postulate of Dalton's atomic theory can explain the law of definite proportions?

    Solution

    The postulate of Dalton’s atomic theory which can explain the law of definite proportion is “The relative number and kind of atoms in a given compound remains constant”.

    Question 377
    CBSEENSC9006889

    Write down the formulae of:
    (i) sodium oxide

    (ii) aluminium chloride

    (iii) sodium sulphide

    (iv) magnesium hydroxide.

    Solution

    (i) Sodium oxide =  NaO
    (ii) Aluminum chloride =AlCl3
    (iii) Sodium sulphide = Na2S
    (iv)  Magnesium hydroxide = Mg(OH)2

    Question 378
    CBSEENSC9006890

    Write down the names of compounds represented by the following formulae:
    (i) Al2(SO4)3

    (ii) CaCl2

    (iii) K2SO4

    (iv) KNO3

    (v) CaCO3.

    Solution

    (i) Al(SO4)3→ Aluminium sulphate
    (ii) CaCl2→ Calcium chloride
    (iii) K2SO4→ Potassium sulphate
    (iv) KNO3→ Potassium Nitrate
    (v) CaCO3→ Calcium carbonate

    Question 379
    CBSEENSC9006891

    What is meant by the term chemical formula?

    Solution

    The chemical formula of a compound means the symbolic representation of the composition of a compound. From the chemical formula of a compound, we can know the number and kinds of atoms of different elements that constitute the compound. For example, from the chemical formula CO2 of carbon dioxide, we come to know that one carbon atom and two oxygen atoms are chemically bonded together to form one molecule of the compound, carbon dioxide.

    Question 380
    CBSEENSC9006892

    How many atoms are present in a

      left parenthesis straight i right parenthesis space straight H subscript 2 straight S space molecule space and
left parenthesis ii right parenthesis space PO subscript 4 superscript 3 minus end superscript space ion ?

    Solution

    (i) In an H2S molecule, three atoms are present; two of hydrogen and one of sulphur.
    (ii) In an  PO subscript 4 superscript 3 minus end superscript space ion five atoms are present; one of phosphorus and four of oxygen.

    Question 381
    CBSEENSC9006893

    Calculate the molecular masses of H2, O2, Cl2, CO2, CH4, C2H6, C2H4, NH3, CH3OH.

    Solution

    Molecular mass of H2 = 2 × Atomic mass of H = 2 × 1 = 2u
    Molecular mass of O2 = 2 × Atomic mass of O = 2 × 16 = 32u
    Molecular mass of Cl2 = 2 × Atomic mass of Cl = 2 × 35.5 = 71 u
    Molecular mass of CO2 = Atomic mass of C + 2 × Atomic mass of O = 12 + 2 × 16 = 44 u
    Molecular mass of CH4 = Atomic mass of C + 4 × Atomic mass of H = 12 + 4 × 1 = 16 u
    Molecular mass of C2H6 = 2 × Atomic mass of C + 6 × Atomic mass of H = 2 × 12 + 6 × 1 = 30u
    Molecular mass of C2H4 = 2 × Atomic mass of C + 4 × Atomic mass of H = 2 × 12 + 4 × 1 = 28u
    Molecular mass of NH3 = Atomic mass of N + 3 × Atomic mass of H = 14 + 3 × 1 =17 u
    Molecular mass of CH3OH Atomic mass of C+4 ×Atomic mass of H+Atomic mass of O = 12 + 4 × 1 + 16 = 32 u.

    Question 382
    CBSEENSC9006894

    Calculate the formula unit masses of ZnO,  Na subscript 2 straight O comma space space straight K subscript 2 CO subscript 3, given atomic masses of Zn = 65 u. Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.

    Solution

    Formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O = 65 + 16 = 81 u
    Formula unit mass of Na2O = 2 × Atomic mass of Na + Atomic mass of O = 2 × 23 + 16 = 62u
    Formula unit mass of K2CO3 = 2 × Atomic mass of K + Atomic mass of C + 3 × Atomic mass of O = 2 × 39 + 12 + 3 × 16 = 138u.

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