NCERT Exemplar Class 9 Science Solutions for Chapter 3 - Atoms And Molecules

NCERT Exemplar Solutions Class 9 Science Chapter 3 – Free PDF Download

NCERT Exemplar Solutions Class 9 Science Chapter 3 Atoms and Molecules are essential study materials for CBSE Class 9 students that play a pivotal role in providing a clear understanding of the concepts covered in the chapter. Students should follow the NCERT Exemplar Solutions provided here in order to score well in the Class 9 annual examination and for a better understanding of chemistry-related topics in future.

The NCERT Exemplar Class 9 Science Atoms and Molecules are prepared as per the latest CBSE Syllabus 2023-24 and have answers to the questions provided in the NCERT Exemplar book. This page has answers to 10 multiple-choice questions, 13 short answer questions and 25 long answer questions, which will help you get familiar with the questions that can be asked in the annual examination.

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Multiple-choice Questions

1. Which of the following correctly represents 360 g of water?

(i) 2 moles of H20

(ii) 20 moles of water

(iii) 6.022 × 1023 molecules of water

(iv) 1.2044×1025 molecules of water

Choose the option :

(a) (i)

(b) (i) and (iv)

(c) (ii) and (iii)

(d) (ii) and (iv)

Soln:

Answer is (d) (ii) and (iv)

Explanation:

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

Number of moles = 360g / 18g/mol

Number of moles = 20

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

Thus, option (d) is correct.

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

Soln:

Answer is (a) Atoms are not able to exist independently

3. The chemical symbol for nitrogen gas is

(a) Ni

(b) N₂

(c) N+

(d) N

Soln:

Answer is (b) N₂

Explanation:

The chemical formula of Nitrogen is N but Nitrogen exist in molecule of two ions hence chemical symbol of Nitrogen gas is written as N2.

4. The chemical symbol for sodium is

(a) So

(b) Sd

(c) NA

(d) Na

Soln:

Answer is (d) Na

Explanation:

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

Soln:

Answer is (c) 2 moles of CaCO3

Explanation:

The molecular mass of C₁₂H₂₂O₁₁ is = 12(12) + 22(1) + 11(16) gm = 342 gm

0.2 moles of it weigh = 0.2 x 342 gm = 68.4gm

The molar mass of CO₂ is = 12 + 2(16) gm = 44gm

2 moles of it weigh = 2 x 44 = 88gm

The molar mass of CaCO₃ weigh = 40 + 12 + 3(16) = 100gm

2 mole of it weigh = 2 x 100 gm = 200gm

Molar mass of H₂O is = 2(1) + 16 = 18gm

10 mole of it weigh = 10 x 18 gm = 180gm

6. Which of the following has maximum number of atoms?

(a) 18g of H2O

(b) 18g of O2

(c) 18g of CO2

(d) 18g of CH4

Soln:

Answer is (d) 18g of CH4

Explanation:

Number of atoms = Mass of substance × Number of atoms in the molecule/ Molar mass × NA

(a) 18 g of water =18 x3/18 ×NA = 3 NA

(b) 18 g of oxygen = 18 x2 /32 × NA = 1.12 NA

(c) 18 g of CO2 = 18 x3/44 × NA = 1.23 NA

(d) 18 g of CH4 =18 x5 /16 × NA = 5.63 NA

Note: NA = 6.023×10²³

7. Which of the following contains maximum number of molecules?

(a) 1g CO2

(b) 1g N2

(c) 1g H2

(d) 1g CH4

Soln:

Answer is (c) 1g H2

Note: NA = 6.023×10²³

Explanation:

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

8. Mass of one atom of oxygen is

a) 16/ 6.023 × 10²³g

b) 32/ 6.023 × 10²³g

c) 3/ 6.023 × 10²³g

(d) 8u

Soln:

Answer is (a) 16/ 6.023 × 10²³g

Explanation:

Mass of one atom of oxygen = Atomic mass/NA = 16/ 6.023 × 10²³g

Note: NA = 6.023×10²³

9. 3.42 g of sucrose are dissolved in 18g 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

Soln:

Answer is (a) 6.68 × 10²³

Explanation:

1 mol of sucrose ( C12H22O11) contains = 11× NA atoms of oxygen, where NA = 6.023×10²³

0.01 mol of sucrose (C12 H22 O11) contains = 0.01 × 11 × NA atoms of oxygen

= 0.11× NA atoms of oxygen

= 18 g/(1×2+ 16)gmol-1

=18 g /18 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 ×10”²³ = 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

Soln:

Answer is (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

Soln:

Answer is (b) BiPO4, Its name is Bismuth Phosphate

Explanation:

Bismuth phosphate is right because Both ions are trivalent Bismuth phosphate(Bi3+- Trivalent anion. anion is an ion that is negatively charged).

12. Write the molecular formulae for the following compounds

(a) Copper (II) bromide

(b) Aluminium (III) nitrate

(c) Calcium (II) phosphate

(d) Iron (III) sulphide

(e) Mercury (II) chloride

(f) Magnesium (II) acetate

Soln:

Answers are

(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 following ions Cu²⁺, Na⁺, Fe³⁺, Cl^–  SO₄ ^-2, PO₄ ^-3

Soln:

Answers are

CuCl₂/ CuSO₄/ Cu₃(PO₄) ²

NaCl/ Na₂SO⁴/ Na₃PO⁴

FeCl₃/ Fe₂(SO₄) ³ / FePO₄

14. Write the cations and anions present (if any) in the following compounds

(a) CH₃COONa

b) NaCl

(c) H₂

(d) NH₄NO₃

Soln:

1. In CH₃COONa-CH₃COO is anion and Na is cation.
2. In NaCl-Cl anion Na is cation
3. In H₂both the ions are cations as they share electrovalent bond between them
4. In NH₄NO_3- NO₃ is anion NH₄ is cation

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

(a) Calcium and fluorine

(b) Hydrogen and sulphur

(c) Nitrogen and hydrogen

(d) Carbon and chlorine

(e) Sodium and oxygen

(f) Carbon and oxygen

Soln:

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

(b) Hydrogen and sulphur- H₂S- Hydrogen Sulphide

(c) Nitrogen and hydrogen- NH_3- Ammonia

(d) Carbon and chlorine – CCl_4- Carbon Tetrachloride

(e) Sodium and oxygen – Na₂O-Sodium Oxide

(f) Carbon and oxygen- CO₂ ; CO- Carbon-di-oxide; Carbon Monoxide

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

Soln:

Cobalt CO is wrong, correct symbol is Co

Carbon c is wrong, correct symbol is C

Aluminium AL is wrong, correct symbol is Al

Helium He is the right symbol

Sodium So is wrong, correct symbol 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. (You may use appendix-III).

(a) Ammonia

(b) Carbon monoxide

(c) Hydrogen chloride

(d) Aluminium fluoride

(e) Magnesium sulphide

Soln:

18. State the number of atoms present in each of the following chemical species

(a) CO3^-2

(b) PO4^-3

(c) P₂ O⁵

(d) CO

Soln:

(a) CO3^-2 – 1+3=4

(b) PO4^-3 – 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?

Soln:

Mass of 1 mole of a substance is equal to its relative atomic or molecular mass in grams.

Mass of one mole (Avogadro Number) of neutrons =1g

Mass of one neutron = 1/ Avogadro number(NA) g

Mass of one molecule of water = Molar mass / NA = 18/ NA g

The molar mass of water is 18.015 g/mol. This was calculated by multiplying the atomic weight of hydrogen (1.008) by two and adding the result to the weight for one oxygen (15.999)

Mass of one molecule of water = Molar mass / NA = 18/ NA g

Avogadro number(NA) =6.022 x 1023mol¯1

There are 8 neutrons in one atom of oxygen

Number of neutrons in oxygen= number of oxygen – Atomic number of oxygen

Oxygen’s atomic weight= 15.9994

increases with an increase in temperature.

Therefore the mass is 16

Therefore number of neutrons= 16 – 8 = 8

Mass of one neutron = 1/ Avogadro number(NA) g

Mass of 8 neutrons = 8/ Avogadro number(NA) g

Fraction of mass of water due to neutrons = 8/18 g

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

Soln:

Solubility is the ability of a solute to get dissolved in 100g solvent. Solubility of a given solute to dissolve in specific solvent depends on the temperature. With an increase in temperature solubility of liquids and solids increase. In the same way solubility of gases decreases with increase in temperature.

Ex: Sugar dissolves faster 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

Soln:

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?

To examine if the fine white coloured powder is sugar pr salt we can conduct two experiments.

Soln:

Heating:

Upon heating sugar melts to liquid form because sucrose has a decomposition point and melting point at temperatures between 190 to 192 degrees Celsius. This will turn sugar to light brown colour. On heating further, sugar gets charred to black colour.

Salt has a melting point of 841 degrees Celsius and 1545.8 degrees Fahrenheit. If we don’t heat it to that point no change is observed.

Electrical conductivity:

If we dissolve the given substance in water we can check for electrical conductivity to check whether the substance is sugar or salt. If it is salt it conducts electricity. Because salt (NaCl) has positive sodium ions and negative chloride ions hence salt conducts electricity. But sugar doesn’t conduct electricity as sugar has only positive ions.

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

Soln:

Number of moles = weight atomic weight

= 12/24 = 0.5 moles

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 magnesium elements have a mole ratio of 3:5.

Soln:

a) Molar mass of CO2 = 12 + 2 x 16 = 12 + 32 = 44 g mol^-1

5 moles of CO2 have mass = 44 x 5 = 220 g

Similarly, molar mass of H2O = 2x 1 + 16 = 18 g mol ^-1

5 moles of H2O have mass =18 x 5 = 90 g

It is verified that 5 moles of CO₂ and 5 moles of H₂O are not same.

b) Number of moles = w/atomic weight

Atomic weight of Ca= 40 amu

Number of moles in 240g Ca metal 240/ 40 = 6

Number of moles in 240g of Mg metal 240/ 24 = 10

Atomic weight of Mg = 24amu

Ratio 6:10

25. Find the ratio by mass of the combining elements in the following compounds. (You may use Appendix-III) (a) CaCO3 (d) C2H5OH (b) MgCl2 (e) NH3 (c) H2SO4 (f) Ca(OH)2

Soln:

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 x 2 : S : O × 4

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

20 : 16 : 1

26. Calcium chloride when dissolved in water dissociates into its ions 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.

Soln:

1 mole of calcium chloride = 111g

Therefore 222g of CaCl2 is equivalent to 2 moles of CaCl2

Since 1 formula unit CaCl2 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.

Soln:

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

Mass of 1 mole of electron = 5.48002 / 100 g

Mass of one electron =  5.48002 /100 × 6.022 ×10²³

= 9.1 ×10²⁸ g

=  9.1×10^-31 kg

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.

Soln:

The molar mass of HgS = The molar mass of Hg + the molar mass of S

= 200.6 + 32 = 232.6 g mol^–1

1molecule of HgS contains 1 atom of Hg

232.6 g of HgS contains 200.6 g of Hg

Therefore, Mass of Hg in 225 g of HgS = 200.6 X 225 /232.6 = 194.04g

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?

Soln:

One mole of screws weigh = 2.475 ×10²⁴g

= 2.475×10²¹ kg

Mass of the Earth / Mass of 1 mole of screws = 5.98 ×10²⁴ kg 2.475×10²¹

= 2.4 ×10

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?

Soln:

We know,

1 mole = 6.022 x 10²³

The number of moles= Given number of particles / Avogadro Number

n= 2.58×10²⁴ /6.022×10²³

n= 4.28 mol

31. Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles of sodium atoms in another container of same weight.

(a) Whose container is heavier?

(b) Whose container has more number of atoms?

Soln:

a) Mass of sodium atoms carried by Krish = (5 ×23) g = 115 g

The atomic weight of Na = 23

While the mass of carbon atom carried by Raunak = (5 ×12) g = 60g

b) Both container has an equal number of atoms

32. Fill in the missing data in the Table 3.1

Soln:

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

Sol :

Number of moles of stars = 10²² / 6.023 × 10²³

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^–1

(c) 10^–2

(d) 10^–6

(e) 10^–9

(f) 10^–12

Soln:

a) 10³ = 1000= kilo

(b) 10^–1 =1/10= 0.1= deci

(c) 10^–2 =1/100 = 0.01= centi

(d) 10^–6 = 0.000 001= micro

(e) 10^–9 =0.000 000 001 = nano

(f) 10^–12 =0.000 000 000 001 = pico

35. Express each of the following in kilograms

(a) 5.84×10-3 mg

(b) 58.34 g

(c) 0.584g

(d) 5.873×10-21g

Soln:

(a) 5.84 × 10^–3 mg = 5.84 ×10^{–9 kg}

(b) 58.34 g =5.834 ×10^–2 kg

(c) 0.584g =5.84 ×10^–4 kg

(d) 5.873×10^-21g=5.873 ×10^–24 kg

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

Soln:

Mg2+ ion and Mg atom differ by two electrons.

103 moles of Mg2+ and Mg atoms would differ by

10³ × 2 moles of electrons

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

2×6.022 × 9.1×10^–5kg

109.6004 ×10^–5 kg

1.096 × 10^–3kg

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

Soln:

No. of moles of atoms = weight / atomic weight.

For N₂

100 gms of N₂ = 100/2 x 14 moles = 100/28 moles

Number of molecules = 100 / 28 x 6.022 x 10²³

Molar mass of N₂ = 2 x molar mass of monoatomic  N

Molar mass of N₂ = 2 x 14.0067 = 28 moles.

Number of molecules = 100/28 x 6.022 x 10²³

No. of atoms = 2 x 100/28 x 6.022 x 10²³ = 43.01 x 10²³

For NH₃

100 gm of NH₃ = 100/17 moles

Number of molecules = 100/17 x 6.022 x 10²³ molecules

No. of atoms in NH₃ = (1 + 3) = 4 x 100/17 x 6.022 x 10²³ =

141.69 x 10²³ atoms.

Therefore, NH₃ has more atoms than N₂.

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

Soln:

58.5 g NaCl contains 6.023 x 10²³ molecules

therefore 58.5 g NaCl contains 12.046 x 10²³ ions.

Hence, 5.85 g NaCl contains 5.85 x 12.046 x 10²³/ 58.5

= 1.2046 x 10²³ ions

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

one gram of gold sample will contain 90/100 = 0.9g of gold

number of moles of gold = mass of gold / atomic mass of gold

= 0.9/19.7 = 0.0046

one mole of gold contains NA atoms = 6.022 × 10²³

Therefore, 0.0046 mole of gold will contain = 0.0046 × 6.022

= 2.77 × 10²¹

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

Soln:

While forming some compounds, atoms gain or lose electrons, and form electrically charged particles called ions. Compounds that are formed by the attraction of cations and anions are called as ionic compounds.

Ex : 2Na + Cl2 → 2Na+ Cl- → 2NaCl (sodium chloride- common salt.)

Sodium is a group 1 metal, thus forms a +1 charged cation. Chlorine is a non-metal, and has the ability to form a -1 charged anion.

Compounds formed due to bonding of uncharged ions are called as molecular compounds and the bonding between them is called covalent bonding. Molecular compounds are formed by sharing of electrons between the two atoms and the elements are held together by covalent bonds.

Ex: 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?

Soln:

Mass of one mole of Aluminium atom = {13 × mass of each electron + 13 × mass of each proton + 14 × mass of each neutron} × Avogadro’s constant.

We know, if atoms convert into ions, only transfer of electrons takes place, in Al+3 ion, Aluminium atom loss three electrons,

So,

Mass of Al+3 ={10× mass of each electron +13 × mass of each proton + 14 × mass of each neutron} Avogadro’s constant

Now,

You see mass of aluminium atom is greater than aluminium ion by 3 electrons

Difference in mass =Mass of Aluminium atom -mass of aluminium ion

= 3 × mass of each electron x Avogadro’s constant .

= 3 × 9.1 × 10^-28x 6.023 x 10²³

=27.3 x 10^-28 x 6.023 x 10²³ g

=164.4 x 10^-5 g

= 1.644 x 10^-3 g

= 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.

Soln:

Mass of silver = m g

Mass of gold = m /100g

Number of atoms of silver =  Mass/ Atomic mass X NA

= m /108NA   Number of atoms of gold

= m/ 100 X197

Ratio of number of atoms of gold to silver =  Au : Ag

= m/ 100 X 197 X NA : m/ 108 NA

= 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 C2H6 molecules does the sample of gas contain?

Soln:

6.02 x 10²³ molecules of methane = 1 mole

Hence 1.5 x 10²⁰ molecules of methane = (1.5 x 10²⁰ x 1) ÷ (6.02 x 10²³) moles

= 2.49 x 10 ^-4 moles

Molar mass of Methane (CH₄) = 12 + 1×4 = 16 g

Mass of methane = molar mass x no. of moles = 16 x 2.49 x 10 ^-4 = 3.984 x 10^-3 g (This is the same mass as Ethane)

Ethane (C2H6) = 12×2 + 1×6 = 30

If 30 g of Ethane has 6.02 x 10²³ molecules

So 3.984 x 10 ^-3 g = (3.984 x 10^-3 x 6.02 x 10²³) ÷ 30

= 8 x 10¹⁹ molecules of Ethane

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 ————.

Soln:

Answers

a) Law of conservation of mass
b) Ions
c) 310

Explanation

3 × atomic mass of Ca+ 2 × atomic mass of phosphorus + 8 × atomic mass of oxygen) = 310

3 × 40 + 2 × 31 + 8 × 16 = 120 + 62 + 128 = 310

d) Na2 CO3 and (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.

Soln:

46. (a) In this crossword puzzle (Fig 3.2), names of 11 elements are hidden. Symbols of these are given below. Complete the puzzle. 1. Cl 7. He 2. H 8. F 3. Ar 9. Kr 4. O 10. Rn 5. Xe 11. Ne 6. N

Soln:

b) Six : Helium (He); Neon ( Ne); Argon (Ar); Krypton (Kr); Xenon (Xe); Radon (Rn).

47. 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

Soln:

The formulae for the following and calculate the molecular mass for each one of them.

48. 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.

Soln:

6CO₂ + 6 H₂ O Chlorophyll /Sunlight → C₆ H₁₂ O₆ + 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/ 1g cm-3

= 10.8 cm3

NCERT Exemplar for Class 9 Science Atoms and Molecules

Atoms and molecules are one of the most important chapters for CBSE Class 9 students. In this chapter, they will learn some important concepts of Chemistry like the law of conservation of mass, the law of constant proportions, atomic mass, what a molecule is and how to write a chemical formula. To practise higher-level questions from this chapter, students are advised to follow NCERT Exemplar Science Class 9 Chapter 3, Atoms and Molecules.

NCERT Exemplar for Class 9 Science Atoms and Molecules Important Concepts

1. Laws of Chemical Combination, (1a) Law of Conservation of Mass, (1b) Law of Constant Proportions
2. What Is an Atom?, (2a) What Are the Modern Day Symbols of Atoms of Different Elements?, (2b) Atomic Mass, (2c) How Do Atoms Exist
3. What Is a Molecule?, (3a) Molecules of Elements, (3b) Molecules of Compounds, (3c) What Is an Ion?
4. Writing Chemical Formulae, (4a) Formulae of Simple Compounds
5. Molecular Mass and Mole Concept, (5a) Molecular Mass Ex, (5b) Formula Unit Mass, (5c) Mole Concept

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