Halaman

    Social Items

NCERT Exemplar Class 9 Science Chapter 3 Atoms and Molecules Solutions

NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules covers all the important questions and answers as well as advanced level questions. It helps in learning about the Atom and its size, characteristics, modern symbols of atoms, Law of Conservation of mass, chemical symbols, valency, cation, anion, solubility of substances and molecules.

The NCERT Exemplar solutions for class 9 science is very important in the examination. NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like bonding, molecule of elements, molecules of compounds, ions and writing formula for the compound.


Chapter Name

Chapter 3 Atoms and Molecules

Book Title

NCERT Exemplar for Class 9 Science

Related Study

  • NCERT Solutions for Class 9 Science Chapter 3 Atoms and Molecules
  • Revision Notes for Class 9 Science Chapter 3 Atoms and Molecules
  • MCQ for Class 9 Science Chapter 3 Atoms and Molecules
  • Important Questions for Class 9 Science Chapter 3 Atoms and Molecules

Topics Covered

  • MCQ
  • Short Answers Questions
  • Long Answers Questions

NCERT Exemplar Solutions for Chapter 3 Atoms and Molecules Class 9 Science

Multiple Choice Questions

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

(iv) 1.2044 × 1025 molecules of water

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

Solution

(d) (ii) and (iv)

Number of moles = Mass of water/Molar mass of water

Number of moles = 360/120 = 20

Number of molecules = 20 ×6.022× 1023
= 1.2044 ×1025 molecules of water


2. Which of the following statements is not true about an atom?
(a) Atoms are not able to exist independently.
(b) Atoms are the basic units from which molecules and ions are formed.
(c) Atoms are always neutral in nature.
(d) Atoms aggregate in large numbers to form the matter that we can see, feel or touch.

Solution

(a) Atoms are not able to exist independently.


3. The chemical symbol for nitrogen gas is :
(a) Ni
(b) N2
(c) N+
(d) N

Solution

(b) N2

Chemical formula of Nitrogen is N but nitrogen exist as a molecule of two atoms. Therefore, the chemical symbol of nitrogen gas is N2.


4. The chemical symbol for sodium is :
(a) So
(b) Sd
(c) NA
(d) Na

Solution

(d) Na

Sodium word is derived from Latin word Natrium hence the chemical name of sodium is Na.


5. Which of the following would weigh the highest?
(a) 0.2 mole of sucrose (C12 H22 O11)
(b) 2 moles of CO2
(c) 2 moles of CaCO3
(d) 10 moles of H2O

Solution

(c) 2 moles of CaCO3

Weight of a sample in grant = Number of moles ×Molar mass

(a) 0.2 moles of C12H22O11= 0.2 ×342 = 68.4 g

(b) 2 moles of CO2is 2 × 44 = 88 g

(c) 2 moles of CaCO3= 2 ×100=200 g

(d) 10 moles of H2O= 10 ×18 = 180g


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

    Solution

    (d) 18 g of CH4


    7. Which of the following contains maximum number of molecules ?
    (a) 1 g CO2
    (b) 1 g N2
    (c) 1 g H2
    (d) 1 g CH4

      Solution

      (c) 1 g H2

      1 g of H2 = ½ ×NA
      = 0.5 NA
      = 0.5 × 6.022 × 1023
      = 3.011 × 1023


      8. Mass of one atom of oxygen is :
      (a) 16/(6.023 × 1023 g)
      (b) 32/(6.023 × 1023 g)
      (c) 1/(6.023 × 1023 g)
      (d) 8u

      Solution

      (a) 16/(6.023 ×1023 g)

      Mass of one atom of oxygen = Atomic mass/NA

      = 16/6.023× 1023 g


      9. 3.42 g of sucrose are dissolved in 18 g of water in a beaker. The number of oxygen atoms in the solution are :
      (a) 6.68 × 1023
      (b) 6.09 × 1022
      (c) 6.022 × 1023
      (d) 6.022 × 1021

        Solution

        (a) 6.68 × 1023

        1 mol of sucrose (C12H22O11) contains = 11× NA atoms of oxygen,

        Here, NA= 6.023×1023

        0.01 mol of sucrose (C12H22O11) contains = 0.01 × 11 × NA atoms of oxygen

        = 0.11× NA atoms of oxygen

        = 18 g/(1×2+ 16)gmol-1

        = 1mol

        1mol of water (H2O) contains 1× NA atom of oxygen

        Total number of oxygen atoms = Number of oxygen atoms from sucrose + Number of oxygen atoms from water

        = 0.11 NA + 1.0 NA = 1.11NA

        Number of oxygen atoms in solution = 1.11 × Avogadro’s number

        = 1.11 × 6.022 ×1023

        = 6.68 × 1023


        10. A change in the physical state can be brought about :
        (a) Only when energy is given to the system.
        (b) Only when energy is taken out from the system.
        (c) When energy is either given to, or taken out from the system.
        (d) Without any energy change.

        Solution

        (c) When energy is either given to, or taken out from the system.


        Short Answer Questions

        11. Which of the following represents a correct chemical formula? Name it.
        (a) CaCl
        (b) BiPO4
        (c) NaSO4
        (d) NaS

          Solution

          (a) CaCl = Wrong (valency of Ca = 2, Cl = 1)
          (b) BiPO4 = Correct (valency of Bi = 3, PO4 = 3)
          (c) NaSO4 = Wrong (valency of Na = 1, SO4 = 2)
          (d) NaS = Wrong (valency of Na = 1, Sulphide = 2)

          Bismuth phosphate is right because both ions are trivalent.


          12. Write the molecular formulae for the following compounds.

          Solution

          (a) Copper (II) bromide = CuBr2

          (b) Aluminium (III) nitrate = Al(NO3)3

          (c) Calcium (II) phosphate = Ca3(PO4)2

          (d) Iron (III) sulphide = Fe2S3

          (e) Mercury (II) chloride = HgCl2

          (f) Magnesium (II) acetate = Mg(CH3COO)2


          13. Write the molecular formulae of all the compounds that can be formed by the combination of given ions : Cu3+, Na+, Fe3+, Cl, SO34 PO43 .

          Solution

          CuCl2/CuSO4/Cu3(PO4)

          NaCl/Na2SO4/Na3PO4

          FeCl3/Fe(SO4)3/FePO4


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

          Solution

          Compounds

          Cations

          Anions

          CH3COONa

          Na+

          CH3COO

          NaCl

          Na+

          Cl

          H2

          Nil

          Nil

          NH4NO3

          NH4+

          NO3



          15. Give the formulae of the compounds formed from the following sets of elements.

            Solution

            (a) Calcium and fluoride -Calcium Fluoride (CaF2)

            (b) Hydrogen and sulphur-Hydrogen Sulphide(H2S)

            (c) Nitrogen and hydrogen-Ammonia(NH3)

            (d) Carbon and chlorine -Carbon Tetrachloride(CCl4)

            (e) Sodium and oxygen -Sodium Oxide(Na2O)

            (f) Carbon and oxygen-Carbon dioxide(CO2); Carbon Monoxide(CO)


            16. Which of the following symbols of elements are incorrect? Give their correct symbols

            (a) Cobalt CO
            (b) Carbon c
            (c) Aluminium AL
            (d) Helium He
            (e) Sodium So

            Solution

            (a) Incorrect, the correct symbol of cobalt is Co

            (b) Incorrect, the correct symbol of carbon is C

            (c) Incorrect, the correct symbol of aluminium is Al

            (d) Correct (He)

            (e) Incorrect, the correct symbol of sodium is Na


            17. Give the chemical formulae for the following compounds and compute the ratio by mass of the combining elements in each one of them.
            (a) Ammonia
            (b) Carbon monoxide
            (c) Hydrogen chloride
            (d) Aluminium fluoride
            (e) Magnesium sulphide

            Solution

            (a) NH3

            (b) CO

            (c) HCl

            (d) AlF3

            (e) MgS

            N : H × 3

            C : O

            H : Cl

            Al : F × 3

            Mg : S

            14 : 1 × 3

            12 : 16

            1 : 35.5

            27 : 19 × 3

            24 : 32

            14 : 3

            3 : 4

            1 : 35.5

            9 : 19

            3 : 4


            18. State the number of atoms present in each of the following chemical species.
            (a) CO2-3
            (b) PO3-4
            (c) P2O5
            (d) CO

            Solution

            (a)

            CO2-3

            1 + 3 = 4

            (b)

            PO3-4 

            1 + 4 = 5

            (c)

            P2O5

            2 + 5 = 7

            (d)

            CO

            1 + 1 = 2


            19. What is the fraction of the mass of water due to neutrons?

            Solution

            • Mass of 1 molecule of water = 18 amu
            • No. of proton in 2 atoms of H = 2 and no. of neutron = 0
            • No. of proton in 1 atom of O = 8 and no. of neutron = 8
            • Fraction of mass due to neutron in water 8/18 = 4/9


            20. Does the solubility of a substance change with temperature? Explain with the help of an example.

            Solution

            Yes, the solubility of a substance depends on its temperature. The solubility generally, increases with an increase in temperature.

            For example: one can dissolve more sugar in hot water than in cold water.


            21. Classify each of the following on the basis of their atomicity.

            (a) F2
            (b) NO2
            (c) N2O
            (d) C2H6
            (e) P4
            (f) H2O2
            (g) P4O10
            (h) O3
            (i) HCl
            (j) CH4
            (k) He
            (l) Ag
            Solution

            (a) Diatomic (2 atoms)

            (b) Triatomic (3 atoms)

            (c) Triatomic (3 atoms)

            (d) Polyatomic (8 atoms)

            (e) Polyatomic (4 atoms)

            (f) Polyatomic (4 atoms)

            (g) Polyatomic (14 atoms)

            (h) Triatomic (3 atoms)

            (i) Diatomic (2 atoms)

            (j) Polyatomic (5 atoms)

            (k) Monoatomic (1 atom)

            (l) Monoatomic(1 atom)


            22. You are provided with a fine white coloured powder which is either sugar or salt. How would you identify it without tasting?

            Solution

            We can identify the powder by heating. The powder will char if it is a sugar. Alternatively, the powder can be dissolved in water and checked for electrical conductivity. If it conducts, then the powder is salt.


            23. Calculate the number of moles of magnesium present in a magnesium ribbon weighing 12 g. Molar atomic mass of magnesium is 24gmol–1.

            Solution


            Long Answer Questions

            24. Verify by calculating that :
            (a) 5 moles of CO2 and 5 moles of H2O do not have the same mass.
            (b) 240 g of calcium and 240 g of magnesium elements have a mole ratio of 3 : 5.

              Solution

              (a) CO2 has molar mass = 44 g mol–1
              5 moles of CO2 have molar mass = 44 × 5 = 220 g
              H2O has molar mass = 18 g mol-1
              5 moles of H2O have mass = 18 × 5g= 90 g

              (b) Number of moles in 240 g Ca metal 240/40 = 6
              Number of moles in 240 g of Mg metal 24 = 240/24 = 10
              Ratio, 6 : 10 = 3 : 5


              25. Find the ratio by mass of the combining elements in the following compounds :
              (a) CaCO3
              (b) MgCl2
              (c) H2SO4
              (d) C2H3OH
              (e) NH3
              (f) Ca(OH)2

              Solution

              (a) CaCO3
              Ca : C : O × 3
              40 : 12 : 16 × 3
              40 : 12 : 48
              10 : 3 : 12

              (b) MgCl2
              Mg : Cl × 2
              24 : 35.5 × 2
              24 : 71

              (c) H2SO4
              H × 2 : S : O × 4
              1 × 2 : 32 : 16 × 4
              2 : 32 : 64
              1 : 16 : 32

              (d) C2H5OH
              C × 2 : H × 6 : O
              12 × 2 : 1 × 6 : 16
              24 : 6 : 16
              12 : 3 : 8

              (e) NH3
              N : H × 3
              14 : 1 × 3
              14 : 3

              (f) Ca(OH)2
              Ca : O × 2 : H × 2
              40: 16 × 2 : 1 × 2
              40 : 32 : 2


              26. Calcium chloride when dissolved in according to the following equation :
              CaCl2(aq) 
               Ca2+(aq) + 2Cl(aq)
              Calculate the number of ions obtained from CaCl2 when 222 g of it is dissolved in water.

              Solution

              1 mole of calcium chloride = 111 g
              ∴ 222 g of CaCl2 is equivalent to 2 moles of CaCl2

              Since 1 formula unit CaCl3 gives 3 ions, therefore, 1 mol of CaCl2 will give 3 moles of ions.
              2 moles of CaCl2 would give 3 × 2 = 6 moles of ions
              No. of ions = No. of moles of ions × Avogadro number
              = 6 × 6.022 × 1023
              = 36.132 × 1023 = 3.6132 × 1024 ions


              27. The difference in the mass of 100 moles each of sodium atoms and sodium ions is 5.48002 g. Compute the mass of an electron.

              Solution

              A sodium atom and ion, differ by one electron. For 100 moles each of sodium atoms and ions there would be a difference of 100 moles of electrons.
              Mass of 100 moles of electrons= 5.48002 g


              28. Cinnabar (HgS) is a prominent ore of mercury. How many grams of mercury are present in 225 g of pure HgS ? Molar mass of Hg and S are 200.6 g mol-1 and 32 g mol-1 respectively.

                Solution

                Molar mass of HgS = 200.6 + 32 = 323.6 g mol-1
                Mass of Hg in 232.6 g of HgS 200.6 g


                29. The mass of one steel screw is 4.11g. Find the mass of one mole of these steel screws. Compare this value with the mass of the Earth (5.98 × 1024kg). Which one of the two is heavier and by how many times?

                Solution

                One mole of screws weigh = 2.475 ×1024g
                = 2.475×1021 kg


                Mass of earth is 2.4×103 times the mass of screws.
                The earth is 2400 times heavier than one mole of screws.


                30. A sample of vitamin C is known to contain 2.58 × 1024 oxygen atoms. How many moles of oxygen atoms are present in the sample?

                Solution

                1 mole of oxygen atoms = 6.022 × 1023 atoms
                Number of moles of oxygen atoms

                = 4.28 moles of oxygen atoms.


                31. Raunak took 5 moles of carbon atoms in a container arid Krish also took 5 moles of a sodium atoms in another container of same weight.
                (a) Whose container is heavier ?
                (b) Whose countainer has more number of atoms ?

                Solution

                (a) Mass of sodium atoms carried by Krish
                = (5 × 23) g = 115 g
                While mass of carbon atom carried by Raunak
                = (5 × 12) g = 60 g
                Thus, Krish’s container is heavy.
                (b) As the two bags have the same number of moles of atoms, both the bags will have the same number of atoms.


                32. Fill in the missing data in the table :

                Species property

                Water

                CO2

                Na – Atom

                MgCl2

                No. of moles

                2

                ………

                ………

                0.5

                No. of particles

                …….

                3.011 × 1023

                ……..

                ……..

                Mass

                36 g

                ……..

                115 g

                ……

                Solution

                Species property

                Water

                CO2

                Na – Atom

                MgCl2

                No. of moles

                2

                0.5 mole

                5 moles

                0.5

                No. of particles

                2×6.022×1023

                3.011×1023

                5×6.022×1023

                0.5×6.022×1023×3

                 Mass

                36 g

                22 g

                115 g

                47.5 g


                33. The visible universe is estimated to contain 1022 stars. How many moles of stars are present in the visible universe?

                Solution

                Number of moles of stars = 0.0166


                34. What is the SI prefix for each of the following multiples and submultiples of a unit?

                (a)

                103

                (b)

                10-3

                (c)

                10-2

                (d)

                10-6

                (e)

                10-9

                (f)

                10-19

                Solution

                (a)

                kilo

                (b)

                deci

                (c)

                centi

                (d)

                micro

                (e)

                nano

                (f)

                pico


                35. Express each of the following in kilograms :
                (a) 5.84 × 10–3 mg
                (b) 58.34 g
                (c) 0.584 g
                (d) 5.873 × 10–21 g

                Solution

                (a) 5.84 × 10–9 kg
                (b) 5.834 × 10–2 kg
                (c) 5.84 × 10–4 kg
                (d) 5.873 × 10–24 kg


                36. Compute the difference in masses of 103moles each of magnesium atoms and magnesium ions. (Mass of an electron = 9.1×10–31 kg).

                Solution

                A Mg2+ion and Mg atom differ by two electrons.

                103moles of Mg2+and Mg atoms would differ by 103× 2 moles of electrons.

                Mass of 2 ×103moles of electrons = 2×103× 6.023×1023×9.1×10–31kg

                =2×6.022 × 9.1×10–5kg
                =109.6004 ×10–5kg
                =1.096 × 10–3kg


                37. Which has more number of atoms?
                100 g of N2 or 100 g of NH3

                Solution

                (i) 100 g of N2 = 100/28 moles
                Number of molecules = 100/28 × 6.022 × 1023
                Number of atoms = 2 × 100/28 × 6.022 × 1023
                = 43.01 × 1023
                (ii) 100 g of NH3 = 100/17 moles
                = 100/17 × 6.022 × 1023 molecules
                = 100/17 × 6.022 × 1023 × 4 atoms
                = 141.69 × 1023
                ∴ NH3 would have more atoms.


                38. Compute the number of ions present in 5.85 g of sodium chloride.

                Solution

                5.85 g of NaCl = 5.85/58.5 = 0.1 mole
                or 0.1 mole of NaCl particle.
                Each NaCl particle is equivalent to one Na+ one Cl- = 2 ions
                Total moles of ions = 0.1 × 2 = 0.2 mole
                No. of ions = 0.2 × 6.022 × 1023
                = 1.2042 × 1023 ions


                39. A gold sample contains 90% of gold and the rest copper. How many atoms of gold are present in one gram of this sample of gold?

                Solution

                One gram of gold sample will contain = 90/100 = 0.9 g of gold
                Number of moles of gold = Mass of gold/Atomic mass of gld
                = 0.9/197 = 0.0046
                One mole of gold contians Na atoms = 6.022 × 1023
                ∴ 0.0046 mole of gold will contain = 0.0046 × 6.022 × 1023
                = 2.77 × 1021 


                40. What are ionic and molecular compounds? Give examples.

                Solution

                While forming some compounds, atoms may gain or lose electrons, thereby forming electrically charged particles called ions. Compounds that are formed by the attraction of cations and anions are known as ionic compounds.
                Example: 2Na + Cl2→ 2Na+ Cl-→ 2NaCl (sodium chloride-common salt.)

                Compounds formed by the bonding of uncharged species are known as molecular compounds. The bonding is called covalent bonding. Molecular compounds are formed by sharing of electrons between the two atoms.
                Example: 2C + O2→ 2CO (Carbon monoxide)


                41. Compute the difference in masses of one mole each of aluminium atoms and one mole of its ions (mass of an electron is 9.1 × 10–28 g). Which one is heavier?

                Solution

                Mass of 1 mole of aluminium atom = the molar mass of aluminium = 27 g mol–1
                An aluminium atom needs to lose three electrons to become an ion, Al3+
                For one mole of Al3+ ion, three moles of electrons are to be lost.
                The mass of three moles of electrons
                = 3 × (9.1 × 10-28) × 6.022 × 1023 g
                = 27.3 × 6.022 × 10-5 g
                = 164.400 × 10-5 g = 0.00164 g
                Molar mass of Al3+ = (27 – 0.00164) g mol-1
                = 26.9984 g mol-1
                Difference = 27 – 26.9984 = 0.0016 g


                42. A silver ornament of mass ‘m’ gram is polished with gold equivalent to 1% of the mass of silver. Compute the ratio of the number of atoms of gold and silver in the ornament.

                Solution

                Mass of silver = mg
                Mass of gold = m/100 g
                Number of atoms of silver = [Mass/(Atomic mass)] × NA
                = m/108 × NA
                Number of atoms of gold = (m/100) × 197 × NA
                Ratio of number of atoms of gold to silver = An : Ag

                = 108 : 100 × 197
                = 108 : 19700 = 1 : 182.41


                43. A sample of ethane (C2H6) gas has the same mass as 1.5 ×1020 molecules of methane (CH4). How many C2Hmolecules does the sample of gas contain?

                Solution


                44. Fill in the blanks :
                (a) In a chemical reaction, the sum of the masses of the reactants and products remains unchanged.
                This is called _____.
                (b) A group of atoms carrying a fixed charge on them is called _____.
                (c) The formula unit mass of Ca3(PO4)2 is _____.
                (d) Formula of sodium carbonate is _____ and that of ammonium sulphate is _____.

                Solution

                (a) Law of conservation of mass
                (b) Polyatomic ion
                (c) (3 × atomic mass of Ca) + (2 × atomic mass of phosphorus) + (8 × atomic mass of oxygen) = 310
                (d) Na2CO3: (NH4)2SO4


                45. Complete the following crossword puzzle (Fig. 3.1) by using the name of the chemical elements. Use the data given in Table 3.2.

                Across

                Down

                The element used by Rutherford during his α–scattering experiment

                A white lustrous metal used for making ornaments and which tends to get tarnished black in the presence of moist air.

                An element which forms rust on exposure to moist air

                Both brass and bronze are alloys of the element

                A very reactive non–metal stored under water

                The metal which exists in the liquid state at room temperature

                Zinc metal when treated with dilute hydrochloric acid produces a gas of this element which when tested with burning splinter produces a pop sound.

                An element with symbol Pb

                Answer


                46. Write the formulae for the following and calculate the molecular mass for each one of them :
                (a) Caustic potash
                (b) Baking powder
                (c) Limestone
                (d) Caustic soda
                (e) Ethanol
                (f) Common salt

                  Solution

                  (a) KOH = (39 + 16 + 1) = 56 g mol–1
                  (b) NaHCO3 = 23 + 1 + 12 + (3 × 16) = 84 g mol–1
                  (c) CaCO3 = 40 + 12 + (3 × 16) = 100 g mol–1
                  (d) NaOH = 23 + 16 + 1 = 40 g mol–1
                  (e) C2H5OH = C2H6O = 2 × 12 + (6 × 1) + 16 = 46 g mol–1
                  (f) NaCl = 23 + 35.5 = 58.5 g mol–1 


                  47. In photosynthesis, 6 molecules of carbon dioxide combine with an equal number of water molecules through a complex series of reactions to give a molecule of glucose having a molecular formula C6H12O6. How many grams of water would be required to produce 18 g of glucose? Compute the volume of water so consumed, assuming the density of water to be 1 g cm–3.

                  Solution

                  6CO2 + 6H2O → C6H12O6 + 6O2
                  1 mole of glucose needs 6 moles of water.
                  180 g of glucose needs (6×18) g of water.
                  1 g of glucose will need 108/180 g of water.
                  18 g of glucose would need 108/180 × 18 g of water = 10.8 g
                  Volume of water used = Mass/Density
                  = 10.8 g/1 g cm-3
                  = 10.8 cm3

                  NCERT Exemplar Chapter 3 Atoms and Molecules Class 9 Science Solutions

                  NCERT Exemplar Class 9 Science Chapter 3 Atoms and Molecules Solutions

                  NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules covers all the important questions and answers as well as advanced level questions. It helps in learning about the Atom and its size, characteristics, modern symbols of atoms, Law of Conservation of mass, chemical symbols, valency, cation, anion, solubility of substances and molecules.

                  The NCERT Exemplar solutions for class 9 science is very important in the examination. NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like bonding, molecule of elements, molecules of compounds, ions and writing formula for the compound.


                  Chapter Name

                  Chapter 3 Atoms and Molecules

                  Book Title

                  NCERT Exemplar for Class 9 Science

                  Related Study

                  • NCERT Solutions for Class 9 Science Chapter 3 Atoms and Molecules
                  • Revision Notes for Class 9 Science Chapter 3 Atoms and Molecules
                  • MCQ for Class 9 Science Chapter 3 Atoms and Molecules
                  • Important Questions for Class 9 Science Chapter 3 Atoms and Molecules

                  Topics Covered

                  • MCQ
                  • Short Answers Questions
                  • Long Answers Questions

                  NCERT Exemplar Solutions for Chapter 3 Atoms and Molecules Class 9 Science

                  Multiple Choice Questions

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

                  (iv) 1.2044 × 1025 molecules of water

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

                  Solution

                  (d) (ii) and (iv)

                  Number of moles = Mass of water/Molar mass of water

                  Number of moles = 360/120 = 20

                  Number of molecules = 20 ×6.022× 1023
                  = 1.2044 ×1025 molecules of water


                  2. Which of the following statements is not true about an atom?
                  (a) Atoms are not able to exist independently.
                  (b) Atoms are the basic units from which molecules and ions are formed.
                  (c) Atoms are always neutral in nature.
                  (d) Atoms aggregate in large numbers to form the matter that we can see, feel or touch.

                  Solution

                  (a) Atoms are not able to exist independently.


                  3. The chemical symbol for nitrogen gas is :
                  (a) Ni
                  (b) N2
                  (c) N+
                  (d) N

                  Solution

                  (b) N2

                  Chemical formula of Nitrogen is N but nitrogen exist as a molecule of two atoms. Therefore, the chemical symbol of nitrogen gas is N2.


                  4. The chemical symbol for sodium is :
                  (a) So
                  (b) Sd
                  (c) NA
                  (d) Na

                  Solution

                  (d) Na

                  Sodium word is derived from Latin word Natrium hence the chemical name of sodium is Na.


                  5. Which of the following would weigh the highest?
                  (a) 0.2 mole of sucrose (C12 H22 O11)
                  (b) 2 moles of CO2
                  (c) 2 moles of CaCO3
                  (d) 10 moles of H2O

                  Solution

                  (c) 2 moles of CaCO3

                  Weight of a sample in grant = Number of moles ×Molar mass

                  (a) 0.2 moles of C12H22O11= 0.2 ×342 = 68.4 g

                  (b) 2 moles of CO2is 2 × 44 = 88 g

                  (c) 2 moles of CaCO3= 2 ×100=200 g

                  (d) 10 moles of H2O= 10 ×18 = 180g


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

                    Solution

                    (d) 18 g of CH4


                    7. Which of the following contains maximum number of molecules ?
                    (a) 1 g CO2
                    (b) 1 g N2
                    (c) 1 g H2
                    (d) 1 g CH4

                      Solution

                      (c) 1 g H2

                      1 g of H2 = ½ ×NA
                      = 0.5 NA
                      = 0.5 × 6.022 × 1023
                      = 3.011 × 1023


                      8. Mass of one atom of oxygen is :
                      (a) 16/(6.023 × 1023 g)
                      (b) 32/(6.023 × 1023 g)
                      (c) 1/(6.023 × 1023 g)
                      (d) 8u

                      Solution

                      (a) 16/(6.023 ×1023 g)

                      Mass of one atom of oxygen = Atomic mass/NA

                      = 16/6.023× 1023 g


                      9. 3.42 g of sucrose are dissolved in 18 g of water in a beaker. The number of oxygen atoms in the solution are :
                      (a) 6.68 × 1023
                      (b) 6.09 × 1022
                      (c) 6.022 × 1023
                      (d) 6.022 × 1021

                        Solution

                        (a) 6.68 × 1023

                        1 mol of sucrose (C12H22O11) contains = 11× NA atoms of oxygen,

                        Here, NA= 6.023×1023

                        0.01 mol of sucrose (C12H22O11) contains = 0.01 × 11 × NA atoms of oxygen

                        = 0.11× NA atoms of oxygen

                        = 18 g/(1×2+ 16)gmol-1

                        = 1mol

                        1mol of water (H2O) contains 1× NA atom of oxygen

                        Total number of oxygen atoms = Number of oxygen atoms from sucrose + Number of oxygen atoms from water

                        = 0.11 NA + 1.0 NA = 1.11NA

                        Number of oxygen atoms in solution = 1.11 × Avogadro’s number

                        = 1.11 × 6.022 ×1023

                        = 6.68 × 1023


                        10. A change in the physical state can be brought about :
                        (a) Only when energy is given to the system.
                        (b) Only when energy is taken out from the system.
                        (c) When energy is either given to, or taken out from the system.
                        (d) Without any energy change.

                        Solution

                        (c) When energy is either given to, or taken out from the system.


                        Short Answer Questions

                        11. Which of the following represents a correct chemical formula? Name it.
                        (a) CaCl
                        (b) BiPO4
                        (c) NaSO4
                        (d) NaS

                          Solution

                          (a) CaCl = Wrong (valency of Ca = 2, Cl = 1)
                          (b) BiPO4 = Correct (valency of Bi = 3, PO4 = 3)
                          (c) NaSO4 = Wrong (valency of Na = 1, SO4 = 2)
                          (d) NaS = Wrong (valency of Na = 1, Sulphide = 2)

                          Bismuth phosphate is right because both ions are trivalent.


                          12. Write the molecular formulae for the following compounds.

                          Solution

                          (a) Copper (II) bromide = CuBr2

                          (b) Aluminium (III) nitrate = Al(NO3)3

                          (c) Calcium (II) phosphate = Ca3(PO4)2

                          (d) Iron (III) sulphide = Fe2S3

                          (e) Mercury (II) chloride = HgCl2

                          (f) Magnesium (II) acetate = Mg(CH3COO)2


                          13. Write the molecular formulae of all the compounds that can be formed by the combination of given ions : Cu3+, Na+, Fe3+, Cl, SO34 PO43 .

                          Solution

                          CuCl2/CuSO4/Cu3(PO4)

                          NaCl/Na2SO4/Na3PO4

                          FeCl3/Fe(SO4)3/FePO4


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

                          Solution

                          Compounds

                          Cations

                          Anions

                          CH3COONa

                          Na+

                          CH3COO

                          NaCl

                          Na+

                          Cl

                          H2

                          Nil

                          Nil

                          NH4NO3

                          NH4+

                          NO3



                          15. Give the formulae of the compounds formed from the following sets of elements.

                            Solution

                            (a) Calcium and fluoride -Calcium Fluoride (CaF2)

                            (b) Hydrogen and sulphur-Hydrogen Sulphide(H2S)

                            (c) Nitrogen and hydrogen-Ammonia(NH3)

                            (d) Carbon and chlorine -Carbon Tetrachloride(CCl4)

                            (e) Sodium and oxygen -Sodium Oxide(Na2O)

                            (f) Carbon and oxygen-Carbon dioxide(CO2); Carbon Monoxide(CO)


                            16. Which of the following symbols of elements are incorrect? Give their correct symbols

                            (a) Cobalt CO
                            (b) Carbon c
                            (c) Aluminium AL
                            (d) Helium He
                            (e) Sodium So

                            Solution

                            (a) Incorrect, the correct symbol of cobalt is Co

                            (b) Incorrect, the correct symbol of carbon is C

                            (c) Incorrect, the correct symbol of aluminium is Al

                            (d) Correct (He)

                            (e) Incorrect, the correct symbol of sodium is Na


                            17. Give the chemical formulae for the following compounds and compute the ratio by mass of the combining elements in each one of them.
                            (a) Ammonia
                            (b) Carbon monoxide
                            (c) Hydrogen chloride
                            (d) Aluminium fluoride
                            (e) Magnesium sulphide

                            Solution

                            (a) NH3

                            (b) CO

                            (c) HCl

                            (d) AlF3

                            (e) MgS

                            N : H × 3

                            C : O

                            H : Cl

                            Al : F × 3

                            Mg : S

                            14 : 1 × 3

                            12 : 16

                            1 : 35.5

                            27 : 19 × 3

                            24 : 32

                            14 : 3

                            3 : 4

                            1 : 35.5

                            9 : 19

                            3 : 4


                            18. State the number of atoms present in each of the following chemical species.
                            (a) CO2-3
                            (b) PO3-4
                            (c) P2O5
                            (d) CO

                            Solution

                            (a)

                            CO2-3

                            1 + 3 = 4

                            (b)

                            PO3-4 

                            1 + 4 = 5

                            (c)

                            P2O5

                            2 + 5 = 7

                            (d)

                            CO

                            1 + 1 = 2


                            19. What is the fraction of the mass of water due to neutrons?

                            Solution

                            • Mass of 1 molecule of water = 18 amu
                            • No. of proton in 2 atoms of H = 2 and no. of neutron = 0
                            • No. of proton in 1 atom of O = 8 and no. of neutron = 8
                            • Fraction of mass due to neutron in water 8/18 = 4/9


                            20. Does the solubility of a substance change with temperature? Explain with the help of an example.

                            Solution

                            Yes, the solubility of a substance depends on its temperature. The solubility generally, increases with an increase in temperature.

                            For example: one can dissolve more sugar in hot water than in cold water.


                            21. Classify each of the following on the basis of their atomicity.

                            (a) F2
                            (b) NO2
                            (c) N2O
                            (d) C2H6
                            (e) P4
                            (f) H2O2
                            (g) P4O10
                            (h) O3
                            (i) HCl
                            (j) CH4
                            (k) He
                            (l) Ag
                            Solution

                            (a) Diatomic (2 atoms)

                            (b) Triatomic (3 atoms)

                            (c) Triatomic (3 atoms)

                            (d) Polyatomic (8 atoms)

                            (e) Polyatomic (4 atoms)

                            (f) Polyatomic (4 atoms)

                            (g) Polyatomic (14 atoms)

                            (h) Triatomic (3 atoms)

                            (i) Diatomic (2 atoms)

                            (j) Polyatomic (5 atoms)

                            (k) Monoatomic (1 atom)

                            (l) Monoatomic(1 atom)


                            22. You are provided with a fine white coloured powder which is either sugar or salt. How would you identify it without tasting?

                            Solution

                            We can identify the powder by heating. The powder will char if it is a sugar. Alternatively, the powder can be dissolved in water and checked for electrical conductivity. If it conducts, then the powder is salt.


                            23. Calculate the number of moles of magnesium present in a magnesium ribbon weighing 12 g. Molar atomic mass of magnesium is 24gmol–1.

                            Solution


                            Long Answer Questions

                            24. Verify by calculating that :
                            (a) 5 moles of CO2 and 5 moles of H2O do not have the same mass.
                            (b) 240 g of calcium and 240 g of magnesium elements have a mole ratio of 3 : 5.

                              Solution

                              (a) CO2 has molar mass = 44 g mol–1
                              5 moles of CO2 have molar mass = 44 × 5 = 220 g
                              H2O has molar mass = 18 g mol-1
                              5 moles of H2O have mass = 18 × 5g= 90 g

                              (b) Number of moles in 240 g Ca metal 240/40 = 6
                              Number of moles in 240 g of Mg metal 24 = 240/24 = 10
                              Ratio, 6 : 10 = 3 : 5


                              25. Find the ratio by mass of the combining elements in the following compounds :
                              (a) CaCO3
                              (b) MgCl2
                              (c) H2SO4
                              (d) C2H3OH
                              (e) NH3
                              (f) Ca(OH)2

                              Solution

                              (a) CaCO3
                              Ca : C : O × 3
                              40 : 12 : 16 × 3
                              40 : 12 : 48
                              10 : 3 : 12

                              (b) MgCl2
                              Mg : Cl × 2
                              24 : 35.5 × 2
                              24 : 71

                              (c) H2SO4
                              H × 2 : S : O × 4
                              1 × 2 : 32 : 16 × 4
                              2 : 32 : 64
                              1 : 16 : 32

                              (d) C2H5OH
                              C × 2 : H × 6 : O
                              12 × 2 : 1 × 6 : 16
                              24 : 6 : 16
                              12 : 3 : 8

                              (e) NH3
                              N : H × 3
                              14 : 1 × 3
                              14 : 3

                              (f) Ca(OH)2
                              Ca : O × 2 : H × 2
                              40: 16 × 2 : 1 × 2
                              40 : 32 : 2


                              26. Calcium chloride when dissolved in according to the following equation :
                              CaCl2(aq) 
                               Ca2+(aq) + 2Cl(aq)
                              Calculate the number of ions obtained from CaCl2 when 222 g of it is dissolved in water.

                              Solution

                              1 mole of calcium chloride = 111 g
                              ∴ 222 g of CaCl2 is equivalent to 2 moles of CaCl2

                              Since 1 formula unit CaCl3 gives 3 ions, therefore, 1 mol of CaCl2 will give 3 moles of ions.
                              2 moles of CaCl2 would give 3 × 2 = 6 moles of ions
                              No. of ions = No. of moles of ions × Avogadro number
                              = 6 × 6.022 × 1023
                              = 36.132 × 1023 = 3.6132 × 1024 ions


                              27. The difference in the mass of 100 moles each of sodium atoms and sodium ions is 5.48002 g. Compute the mass of an electron.

                              Solution

                              A sodium atom and ion, differ by one electron. For 100 moles each of sodium atoms and ions there would be a difference of 100 moles of electrons.
                              Mass of 100 moles of electrons= 5.48002 g


                              28. Cinnabar (HgS) is a prominent ore of mercury. How many grams of mercury are present in 225 g of pure HgS ? Molar mass of Hg and S are 200.6 g mol-1 and 32 g mol-1 respectively.

                                Solution

                                Molar mass of HgS = 200.6 + 32 = 323.6 g mol-1
                                Mass of Hg in 232.6 g of HgS 200.6 g


                                29. The mass of one steel screw is 4.11g. Find the mass of one mole of these steel screws. Compare this value with the mass of the Earth (5.98 × 1024kg). Which one of the two is heavier and by how many times?

                                Solution

                                One mole of screws weigh = 2.475 ×1024g
                                = 2.475×1021 kg


                                Mass of earth is 2.4×103 times the mass of screws.
                                The earth is 2400 times heavier than one mole of screws.


                                30. A sample of vitamin C is known to contain 2.58 × 1024 oxygen atoms. How many moles of oxygen atoms are present in the sample?

                                Solution

                                1 mole of oxygen atoms = 6.022 × 1023 atoms
                                Number of moles of oxygen atoms

                                = 4.28 moles of oxygen atoms.


                                31. Raunak took 5 moles of carbon atoms in a container arid Krish also took 5 moles of a sodium atoms in another container of same weight.
                                (a) Whose container is heavier ?
                                (b) Whose countainer has more number of atoms ?

                                Solution

                                (a) Mass of sodium atoms carried by Krish
                                = (5 × 23) g = 115 g
                                While mass of carbon atom carried by Raunak
                                = (5 × 12) g = 60 g
                                Thus, Krish’s container is heavy.
                                (b) As the two bags have the same number of moles of atoms, both the bags will have the same number of atoms.


                                32. Fill in the missing data in the table :

                                Species property

                                Water

                                CO2

                                Na – Atom

                                MgCl2

                                No. of moles

                                2

                                ………

                                ………

                                0.5

                                No. of particles

                                …….

                                3.011 × 1023

                                ……..

                                ……..

                                Mass

                                36 g

                                ……..

                                115 g

                                ……

                                Solution

                                Species property

                                Water

                                CO2

                                Na – Atom

                                MgCl2

                                No. of moles

                                2

                                0.5 mole

                                5 moles

                                0.5

                                No. of particles

                                2×6.022×1023

                                3.011×1023

                                5×6.022×1023

                                0.5×6.022×1023×3

                                 Mass

                                36 g

                                22 g

                                115 g

                                47.5 g


                                33. The visible universe is estimated to contain 1022 stars. How many moles of stars are present in the visible universe?

                                Solution

                                Number of moles of stars = 0.0166


                                34. What is the SI prefix for each of the following multiples and submultiples of a unit?

                                (a)

                                103

                                (b)

                                10-3

                                (c)

                                10-2

                                (d)

                                10-6

                                (e)

                                10-9

                                (f)

                                10-19

                                Solution

                                (a)

                                kilo

                                (b)

                                deci

                                (c)

                                centi

                                (d)

                                micro

                                (e)

                                nano

                                (f)

                                pico


                                35. Express each of the following in kilograms :
                                (a) 5.84 × 10–3 mg
                                (b) 58.34 g
                                (c) 0.584 g
                                (d) 5.873 × 10–21 g

                                Solution

                                (a) 5.84 × 10–9 kg
                                (b) 5.834 × 10–2 kg
                                (c) 5.84 × 10–4 kg
                                (d) 5.873 × 10–24 kg


                                36. Compute the difference in masses of 103moles each of magnesium atoms and magnesium ions. (Mass of an electron = 9.1×10–31 kg).

                                Solution

                                A Mg2+ion and Mg atom differ by two electrons.

                                103moles of Mg2+and Mg atoms would differ by 103× 2 moles of electrons.

                                Mass of 2 ×103moles of electrons = 2×103× 6.023×1023×9.1×10–31kg

                                =2×6.022 × 9.1×10–5kg
                                =109.6004 ×10–5kg
                                =1.096 × 10–3kg


                                37. Which has more number of atoms?
                                100 g of N2 or 100 g of NH3

                                Solution

                                (i) 100 g of N2 = 100/28 moles
                                Number of molecules = 100/28 × 6.022 × 1023
                                Number of atoms = 2 × 100/28 × 6.022 × 1023
                                = 43.01 × 1023
                                (ii) 100 g of NH3 = 100/17 moles
                                = 100/17 × 6.022 × 1023 molecules
                                = 100/17 × 6.022 × 1023 × 4 atoms
                                = 141.69 × 1023
                                ∴ NH3 would have more atoms.


                                38. Compute the number of ions present in 5.85 g of sodium chloride.

                                Solution

                                5.85 g of NaCl = 5.85/58.5 = 0.1 mole
                                or 0.1 mole of NaCl particle.
                                Each NaCl particle is equivalent to one Na+ one Cl- = 2 ions
                                Total moles of ions = 0.1 × 2 = 0.2 mole
                                No. of ions = 0.2 × 6.022 × 1023
                                = 1.2042 × 1023 ions


                                39. A gold sample contains 90% of gold and the rest copper. How many atoms of gold are present in one gram of this sample of gold?

                                Solution

                                One gram of gold sample will contain = 90/100 = 0.9 g of gold
                                Number of moles of gold = Mass of gold/Atomic mass of gld
                                = 0.9/197 = 0.0046
                                One mole of gold contians Na atoms = 6.022 × 1023
                                ∴ 0.0046 mole of gold will contain = 0.0046 × 6.022 × 1023
                                = 2.77 × 1021 


                                40. What are ionic and molecular compounds? Give examples.

                                Solution

                                While forming some compounds, atoms may gain or lose electrons, thereby forming electrically charged particles called ions. Compounds that are formed by the attraction of cations and anions are known as ionic compounds.
                                Example: 2Na + Cl2→ 2Na+ Cl-→ 2NaCl (sodium chloride-common salt.)

                                Compounds formed by the bonding of uncharged species are known as molecular compounds. The bonding is called covalent bonding. Molecular compounds are formed by sharing of electrons between the two atoms.
                                Example: 2C + O2→ 2CO (Carbon monoxide)


                                41. Compute the difference in masses of one mole each of aluminium atoms and one mole of its ions (mass of an electron is 9.1 × 10–28 g). Which one is heavier?

                                Solution

                                Mass of 1 mole of aluminium atom = the molar mass of aluminium = 27 g mol–1
                                An aluminium atom needs to lose three electrons to become an ion, Al3+
                                For one mole of Al3+ ion, three moles of electrons are to be lost.
                                The mass of three moles of electrons
                                = 3 × (9.1 × 10-28) × 6.022 × 1023 g
                                = 27.3 × 6.022 × 10-5 g
                                = 164.400 × 10-5 g = 0.00164 g
                                Molar mass of Al3+ = (27 – 0.00164) g mol-1
                                = 26.9984 g mol-1
                                Difference = 27 – 26.9984 = 0.0016 g


                                42. A silver ornament of mass ‘m’ gram is polished with gold equivalent to 1% of the mass of silver. Compute the ratio of the number of atoms of gold and silver in the ornament.

                                Solution

                                Mass of silver = mg
                                Mass of gold = m/100 g
                                Number of atoms of silver = [Mass/(Atomic mass)] × NA
                                = m/108 × NA
                                Number of atoms of gold = (m/100) × 197 × NA
                                Ratio of number of atoms of gold to silver = An : Ag

                                = 108 : 100 × 197
                                = 108 : 19700 = 1 : 182.41


                                43. A sample of ethane (C2H6) gas has the same mass as 1.5 ×1020 molecules of methane (CH4). How many C2Hmolecules does the sample of gas contain?

                                Solution


                                44. Fill in the blanks :
                                (a) In a chemical reaction, the sum of the masses of the reactants and products remains unchanged.
                                This is called _____.
                                (b) A group of atoms carrying a fixed charge on them is called _____.
                                (c) The formula unit mass of Ca3(PO4)2 is _____.
                                (d) Formula of sodium carbonate is _____ and that of ammonium sulphate is _____.

                                Solution

                                (a) Law of conservation of mass
                                (b) Polyatomic ion
                                (c) (3 × atomic mass of Ca) + (2 × atomic mass of phosphorus) + (8 × atomic mass of oxygen) = 310
                                (d) Na2CO3: (NH4)2SO4


                                45. Complete the following crossword puzzle (Fig. 3.1) by using the name of the chemical elements. Use the data given in Table 3.2.

                                Across

                                Down

                                The element used by Rutherford during his α–scattering experiment

                                A white lustrous metal used for making ornaments and which tends to get tarnished black in the presence of moist air.

                                An element which forms rust on exposure to moist air

                                Both brass and bronze are alloys of the element

                                A very reactive non–metal stored under water

                                The metal which exists in the liquid state at room temperature

                                Zinc metal when treated with dilute hydrochloric acid produces a gas of this element which when tested with burning splinter produces a pop sound.

                                An element with symbol Pb

                                Answer


                                46. Write the formulae for the following and calculate the molecular mass for each one of them :
                                (a) Caustic potash
                                (b) Baking powder
                                (c) Limestone
                                (d) Caustic soda
                                (e) Ethanol
                                (f) Common salt

                                  Solution

                                  (a) KOH = (39 + 16 + 1) = 56 g mol–1
                                  (b) NaHCO3 = 23 + 1 + 12 + (3 × 16) = 84 g mol–1
                                  (c) CaCO3 = 40 + 12 + (3 × 16) = 100 g mol–1
                                  (d) NaOH = 23 + 16 + 1 = 40 g mol–1
                                  (e) C2H5OH = C2H6O = 2 × 12 + (6 × 1) + 16 = 46 g mol–1
                                  (f) NaCl = 23 + 35.5 = 58.5 g mol–1 


                                  47. In photosynthesis, 6 molecules of carbon dioxide combine with an equal number of water molecules through a complex series of reactions to give a molecule of glucose having a molecular formula C6H12O6. How many grams of water would be required to produce 18 g of glucose? Compute the volume of water so consumed, assuming the density of water to be 1 g cm–3.

                                  Solution

                                  6CO2 + 6H2O → C6H12O6 + 6O2
                                  1 mole of glucose needs 6 moles of water.
                                  180 g of glucose needs (6×18) g of water.
                                  1 g of glucose will need 108/180 g of water.
                                  18 g of glucose would need 108/180 × 18 g of water = 10.8 g
                                  Volume of water used = Mass/Density
                                  = 10.8 g/1 g cm-3
                                  = 10.8 cm3

                                  No comments