Class 9 chemistry important questions with answers are provided here for Chapter 3 Atoms and Molecules. These important questions are based on the CBSE board curriculum and correspond to the most recent Class 9 chemistry syllabus. By practising these Class 9 important questions, students will be able to quickly review all of the ideas covered in the chapter and prepare for the Class 9 Annual examinations.
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Class 9 Chapter 3 Atoms and Molecules Important Questions with Answers
Short Answer Type Questions
Q1. Which of the following represents a correct chemical formula? Name it.
(a) CaCl
(b) BiPO4
(c) NaSO4
(d) NaS
Answer:
(b), BiPO4 represents the correct formulae of bismuth phosphate.
Q2. Write the molecular formulae for the following compounds
(a) Copper (I) bromide
(b) Aluminium (III) nitrate
(c) Calcium (II) phosphate
(d) Iron (III) sulphide
(e) Mercury (I) chloride
(f) Magnesium (I) acetate
Answer:
(a) The molecular formula of Copper (I) bromide is CuBr.
(b) The molecular formula of Aluminium (III) nitrate is Al(NO3)3.
(c) The molecular formula of Calcium (II) phosphate is Ca3(PO4)2.
(d) The molecular formula of Iron (III) sulphide is Fe2S3.
(e) The molecular formula of Mercury (I) chloride is HgCl2.
(f) The molecular formula of Magnesium (I) acetate is Mg(CH3COO)2.
Q3. Write the molecular formulae of all the compounds that can be formed by the combination of the following ions.
Cu2+, Na+, Fe3+, Cl–, SO42-, PO43-.
Answer:
The molecular formula of the compounds formed by the combination of Cu2+, Na+, Fe3+, Cl–, SO42-, and PO43- are CuCl2, CuSO4, NaCl, Na2SO4, FeCl3, and Fe2(SO4)3.
Q4. Write the cations and anions present (if any) in the following compounds
(a) CH3COONa
(b) NaCI
(c) H2
(d) NH4NO3
Answer:
(a) The cation and anion present in CH3COONa are Na+ and CH3COO–.
(b) The cation and anion present in NaCI are Na+ and Cl–.
(c) There is no cation and anion in H2.
(d) The cation and anion present in NH4NO3 are NH4+ and NH3–.
Q5. Give the formulae of the compounds formed from the following sets of elements
(a) Calcium and fluorine
(b) Hydrogen and sulphur
(c) Nitrogen and hydrogen
(d) Carbon and chlorine
(e) Sodium and oxygen
(f) Carbon and oxygen
Answer:
(a) The formulae of the compound formed by Calcium and fluorine is CaF2.
(b) The formulae of the compound formed by Hydrogen and sulphur is H2S.
(c) The formulae of the compound formed by Nitrogen and hydrogen is NH3.
(d) The formulae of the compound formed by Carbon and chlorine is CCl4.
(e) The formulae of the compound formed by Sodium and oxygen is Na2O.
(f) The formulae of the compound formed by Carbon and oxygen is CO2.
Q6. Which of the following symbols of elements are incorrect? Give their correct symbols
| S. No. | Element | Formula |
|---|---|---|
| 1. | Cobalt | CO |
| 2. | Carbon | c |
| 3. | Aluminium | AL |
| 4. | Helium | He |
| 5. | Sodium | So |
Answer:
The formula for cobalt, carbon, aluminium, and sodium is incorrect, while the formula for helium is correct.
The correct formulas are enlisted below.
| S. No. | Element | Correct Formula |
|---|---|---|
| 1. | Cobalt | Co |
| 2. | Carbon | C |
| 3. | Aluminium | Al |
| 4. | Helium | He |
| 5. | Sodium | Na |
Q7. 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
Answer:
(a) The chemical formula of ammonia is NH3, and its mass ratio is Mass of N: Mass of H = 14: 3.
(b) The chemical formula of Carbon monoxide is CO, and its mass ratio is Mass of C: Mass of O = 12: 16 = 3:4.
(c) The chemical formula of Hydrogen chloride is HCl, and its mass ratio is Mass of H: Mass of Cl = 1: 35.5.
(d) The chemical formula of Aluminium fluoride is AlF3, and its mass ratio is Mass of Al: Mass of F = 27: 19.
(e) The chemical formula of Magnesium sulphide is MgS, and its mass ratio is Mass of Mg: Mass of S = 24: 32 = 3: 4.
Q8. State the number of atoms present in each of the following chemical species
(a) CO32-
(b) PO33-
(c) P2O5
(d) CO
Answer:
(a) There are four atoms in CO32-.
(b) There are four atoms in PO33-.
(c) There are seven atoms in P2O5.
(d) There are two atoms in CO.
Q9. What is the fraction of the mass of water due to neutrons?
Answer:
The mass of one neutron = 1 amu.
The mass of one water molecule = 18 amu.
The oxygen atom has eight neutrons, while the hydrogen atom has 0 neutrons.
So the mass of neutrons in one water molecule is eight amu.
The fraction of mass of water due to neutrons = 8 / 18 = 4 / 9.
Q10. Does the solubility of a substance change with temperature? Explain with the help of an example.
Answer:
Yes, the solubility of a substance changes with temperature. The solubility generally increases with an increase in temperature.
Example: You can dissolve more sugar in hot water than in cold water.
Q11. Classify each of the following based on their atomicity.
(a) F2
(b) NO2
(c) N2O
(d) C2H6
(e) P4
(f) H2O2
(g) P4O10
(h) O3
(i) HCI
(i) CH4
(k) He
(l) Ag
Answer:
| Monoatomic | Diatomic | Triatmoc | Polyatomic |
|---|---|---|---|
| He | F2 | NO2 | C2H6 |
| Ag | HCl | N2O | P4 |
| O3 | H2O2 | ||
| P4O10 | |||
| CH4 |
Q12. You are provided with a fine white coloured powder, either sugar or salt. How would you identify it without tasting it?
Answer:
We can differentiate sugar and salt by
(a) Heating salts separately. Sugar will melt while salt will not.
(b) Dissolving them separately in water. The salt solution will conduct electricity due to Na+ ion and Cl– while the sugar solution will not conduct electricity. So, we can immediately tell the difference by testing a drop of the solution with an ohmmeter.
Q13. Calculate the number of moles of magnesium present in a magnesium ribbon weighing 12 g. The molar atomic mass of magnesium is 24g mol-1.
Answer:
Given
Mass of magnesium ribbon = 12 g
Molar mass of magnesium = 24 g
Number of moles = Mass / Molar Mass
Number of moles = 12 / 24
Number of moles = 0.5 moles.
Hence, there are half moles of magnesium in a 12 g magnesium ribbon.
Long Answer Type Questions
Q1. 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.
Answer:
(a) The number of moles of CO2 = 5 moles
Molar mass of CO2 = 44 g mol -1.
Hence, the mass of 5 moles of CO2 = the number of moles of CO2 X molar mass of CO2.
Hence, the mass of 5 moles of CO2 = 5 X 44.
Hence, the mass of 5 moles of CO2 = 220 g.
The number of moles of H2O = 5 moles
Molar mass of H2O = 18 gmol -1.
Hence, the mass of 5 moles of H2O = the number of moles of H2O X molar mass of H2O.
Hence, the mass of 5 moles of H2O = 5 X 18.
Hence, the mass of 5 moles of H2O = 90 g.
Thus, we can see the mass of 5 moles of CO2 and 5 moles of H2O is not the same.
(b) Mass of calcium = 240 g
Molar mass of calcium = 40
Number of moles = 240 / 40
Number of moles = 6
Mass of magnesium = 240 g
Molar mass of magnesium = 24 g
Number of moles = 240 / 24
Number of moles = 10
Mole ratio of calcium and magnesium = the number of moles of calcium/number of moles of magnesium
Mole ratio of calcium and magnesium = 6 / 10
Mole ratio of calcium and magnesium = 3 / 5
Hence, the mole ratio of 240 g of calcium and 240 g of magnesium is 3:5.
Q2. Find the ratio by mass of the combining elements in the following compounds. (You may use Appendix-III)
(a) CaCO3
(b) MgCl2
(c) H2SO4
(d) C2H5OH
(e) NH3
(f) Ca(OH)2
Answer:
(a) The ratio of the mass of CaCO3
Ca: C: O X 3
40: 12: (16 X 3)
40: 12: 48
10: 3: 12
Hence, the ratio of the mass of CaCO3 is 10: 3: 12.
(b) The ratio of the mass of MgCl2
Mg: Cl X 2
24: 35.5 X 2
24: 71.
Hence, the ratio of the mass of MgCl2 is 24:71.
(c) The ratio of the mass of H2SO4
H X 2: S: O X 4
1 X 2: 32: 64
2: 32: 64
1: 16: 32.
Hence, the ratio of the mass of H2SO4 is 1: 16: 32.
(d) The ratio of the mass of C2H5OH
C X 2: H X 6: O
12 X 2: 1 X 6: 16
24: 6: 16
12: 3: 8.
Hence, the ratio of the mass of C2H5OH is 12: 3: 8.
(e) The ratio of the mass of NH3
N: H X 3
14: 1 X 3
14: 3.
Hence, the ratio of the mass of NH3 is 14: 3.
(f) The ratio of the mass of Ca(OH)2
Ca: O X 2: H X 2
40: 16 X 2: 1 X 2
40: 32: 2
20: 16: 1
Hence, the ratio of the mass of Ca(OH)2 is 20: 16: 1.
Q3. When dissolved in water, calcium chloride dissociates into its ions according to the following equation.
CaCl2 (aq) → Ca2+ (aq) + 2 Cl– (aq)
Calculate the number of ions obtained from CaCl2 when 222 g of it is dissolved in water.
Answer:
1 mole of CaCl2 gives 111 g of CaCl2
∴ 222 g of CaCl2 is equal to 2 moles of CaCl2.
Since one formula unit, CaCl2 gives three ions.
Therefore, 1 mol of CaCl2 will give 3 moles of ions.
And, 2 moles of CaCl2 would give 3 X 2 = 6 moles of ions.
The number of ions = the number of moles of ions × Avogadro number.
The number of ions = 6 X 6.022 X 1023.
The number of ions = 36.132 X 1023.
The number of ions = 3.6132 X 1024 ions.
Q4. The difference in the mass of 100 moles of sodium atoms and sodium ions is 5.48002 g. Compute the mass of an electron.
Answer:
Na → Na+ + e−
A sodium atom and ion differs 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 X 6.022 X 1023)
Mass of one electron = 9.1 X 10−28 g
Mass of one electron = 9.1 X 10−31 kg
Hence, mass of one electron is equal to 9.1 X 10−31 kg.
Q5. Cinnabar (HgS) is a prominent ore of mercury. How many grams of mercury are present in 225 g of pure HgS? The molar mass of Hg and S is 200.6 g mol-1 and 32 g mol-1, respectively.
Answer:
Molar mass of HgS = molar mass of Hg + molar mass of S
Molar mass of Hg = 200.6 g mol–1
Molar mass of S = 32 g mol–1
Molar mass of HgS = (200.6 + 32) g mol–1
Molar mass of HgS = 232.6 g mol–1
1 molecule of HgS contains one atom of Hg
232.6 g of HgS contains 200.6 g of Hg
225 g of HgS contains (200.6 X 225) / 232.6 of Hg
225 g of HgS contains 45135 / 232.6 of Hg
225 g of HgS contains 194.04g of Hg
Hence, 225 g of HgS contains 194.04g of Hg.
Q6. 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 × 1024 kg). Which one of the two is heavier, and by how many times?
Answer:
Mass of one steel screw= 4.11 g.
Thus, the mass of one mole of screw = 4.11 X NA = 4.11 X 6.022 X 1023.
One mole of screws weighs 2.475 X 1024 g = 2.475 X 1021 kg
Ratio of mass of earth to the mass of screw = Mass of the earth / Mass of 1 mole of screws
Ratio of mass of earth to the mass of screw = 5.98 X 1024 kg / 2.75 X 1021 kg
Ratio of mass of earth to the mass of screw = 2.4 X 103
Hence, the mass of the earth is 2.4 X 103 times more than the mass of screw.
Or the earth is 2400 times heavier than a mole of the screw.
Q7. A sample of vitamin C is known to contain 2.58 x1024 oxygen atoms. How many moles of oxygen atoms are present in the sample?
Answer:
Given
The number of oxygen atoms in the given sample = 2.58 X 1024.
We know that 1 mole contains 6.022 X 1023 oxygen atoms
So,
2.58 X 1024 oxygen atoms = 2.58 X 1024 / 6.022 X 1023
2.58 X 1024 oxygen atoms = 4.28 moles.
Q8. Raunak took 5 moles of carbon atoms in a container, and Krish took 5 moles of sodium atoms in another container of the same weight. (a) Whose container is heavier? (b) Whose container has more number of atoms?
Answer:
As both containers are of the same mass, the mass of atoms will decide which will be heavier.
Mass of 5 moles of sodium atoms in Krish’s container = (5 × 23) g = 115 g.
Mass of 5 moles of carbon atom in Raunak’s container =(5 × 12) g = 60 g.
Hence, Krish’s container is heavier.
Q9. Fill in the missing data in Table 3.1
Table 3.1
| Species | H2O | CO2 | Na Atom | MgCl2 |
|---|---|---|---|---|
| Property | ||||
| No. of moles | 2 | – | – | 0.5 |
| No. of particles | – | 3.011 X 1023 | – | – |
| Mass | 36 g | – | 115 g | – |
Answer:
| Species | H2O | CO2 | Na Atom | MgCl2 |
|---|---|---|---|---|
| Property | ||||
| No. of moles | 2 | 0.5 | 5 | 0.5 |
| No. of particles | 12.044 X 1024 | 3.011 X 1023 | 3.011 X 1023 | 3.011 X 1023 |
| Mass | 36 g | 22 g | 115 g | 47.5 g |
Q10. The visible universe is estimated to contain 1022 stars. How many moles of stars are present in the visible universe?
Answer:
1 mole of stars = 6.023 X 1023.
Hence, the number of moles of stars = 1022 / NA = 1022 / 6.023 X 1023 = 0.0166 moles.
Q11. 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
Answer:
(a) 103: Kilo
(b) 10-1: Deci
(c) 10-2: Centi
(d) 10-6: Micro
(e) 10-9: Nano
(f) 10-12: Pico
Q12. Express each of the following in kilograms
(a) 5.84 X 10-3 mg
(b) 58.34 g
(c) 0.584 g
(d) 5.873 X 10-21 g
Answer:
(a) 5.84 X 10–3 mg = 5.84 X 10–9 kg
(b) 58.34 g = 5.834 X 10–2 kg
(c) 0.584 g = 5.84 X 10–4 kg
(d) 5.873 X 10-21 g = 5.873 X 10–24 kg
Q13. Compute the difference in masses of 103 moles each of magnesium atoms and magnesium ions. (Mass of an electron = 9.1 x 10-31 kg)
Answer:
Mg → Mg+2 + 2 e−
A Mg2+ ion, and Mg atom differs by two electrons.
103 moles of Mg2+ and Mg atoms would differ by 103 X 2 moles of electrons.
Mass of 2 X 103 moles of electrons = 2 X 103 × 6.023 × 1023 × 9.1 × 10−31 kg
Mass of 2 X 103 moles of electrons = 2 X 6.022 X 9.1 X 10−5 kg
Mass of 2 X 103 moles of electrons = 109.6004 X 10−5 kg
Mass of 2 X 103 moles of electrons = 1.096 X 10−3 kg
Q14. Which has more number of atoms?
100g of N2 or 100 g of NH3.
Answer:
(i) 100 g N2 contains = 100 / 28 moles
Number of molecules = (100 X 6.022×1023) / 28
Number of atoms = (100 X 6.022×1023 X 2) / 28 = 43.01 X 1023
(ii) 100g NH3 contains = 100 / 17 moles
100g NH3 contains = (100 X 6.022 X 1023) / 17 molecules
100g NH3 contains = (100 X 6.022 X 1023 X 4) / 17 atoms
100g NH3 contains = 141.69 X 1023
Hence, NH3 would have more atoms than N2.
Q15. Compute the number of ions present in 5.85 g of sodium chloride.
Answer:
5.85 g of NaCl contains 5.85 / 58.5 moles of NaCl particle
5.85 g of NaCl contains 0.1 moles of NaCl particle.
Each NaCl particle contains one Na+ and one Cl− ion.
Total moles of ions = 0.1 X 2 = 0.2 moles
No. of ions = 0.2 X 6.022 X 1023 = 1.2042 X 1023 ions.
Q16. 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?
Answer:
As the sample is 90% pure.
So one gram of sample will have 90 / 100 = 0.9 gm of Gold.
The number of moles of Gold = Mass of Gold / Atomic mass of gold.
The number of moles of Gold = 0.9 / 197.
The number of moles of Gold = 0.0046.
According to the mole concept,
Gold’s one mole contains NA atoms = 6.022 X 1023 atoms.
Hence, 0.0046 moles of Gold will contain = 0.0046 X 6.022 X 1023.
Hence, 0.0046 moles of Gold will contain = 2.77 X 1021 atoms.
Q17. What are ionic and molecular compounds? Give examples.
Answer:
Ionic compounds are those compounds that contain charged species. The charged species are called ions. An ion is a charged particle and can be positively or negatively charged. A positively charged ion is a cation, while a negatively charged ion is an anion. The transfer of electrons forms ionic compounds. Example: sodium chloride and calcium oxide.
Molecular compounds are formed by sharing electrons through covalent bonds. Example: water, ammonia, carbon dioxide.
Q18. Compute the difference in masses of one mole of aluminium atoms and one mole of its ions. (Mass of an electron is 9.1 × 10-28 g). Which one is heavier?
Answer:
Mass of 1 mole of aluminium atom = Molar mass of aluminium = 27 g mole−1
An aluminium atom loses three electrons to form an Al3+ ion.
For one mole of Al3+ ion, three moles of electrons will be lost.
Hence, the mass of three moles of electrons = 3 X (9.1 X 10−28) X 6.022 X 1023 g.
Mass of three moles of electrons = 27.3 X 10−28 X 6.022 X 1023 g.
Mass of three moles of electrons = 164.400 X 10−5 g
Mass of three moles of electrons = 0.00164 g
Hence, the molar mass of Al+3 = (27 − 0.00164) g mol− = 26.998 g mol−
Difference in mass = 27 – 26.9984 = 0.0016 g
Hence, aluminium atom is 0.0016 g heavier than aluminium ions.
Q19. 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.
Answer:
Mass of silver = m g
Mass of gold = m / 100 g
Number of atoms of Ag = m / (108 × Na)
Number of atoms of Au = m / (100 × 197 × Na)
Au: Ag ≡ m / (100 × 197 × Na) : m / (108 × Na)
= 108: 100 X 197
= 108: 19700
Q20. The ethane (C2H6) gas sample has the same mass as 1.5 x1020 molecules of methane (CH4). How many C2H6 molecules does the sample of gas contain?
Answer:
Mass of 1 molecule of CH4 = 16 / Na g
Mass of 1.5 X 1020 molecules of methane = (1.5 X 1020 X 16) / Na
Mass of x molecules of C2H6 is = (1.5 X 1020 X 16) / Na
We know that mass of 1 molecule of C2H6 = 30 / Na g
Hence, the number of molecules of ethane (x) = (1.5 X 1020 X 16 X Na) / (30 X Na)
Number of molecules of ethane (x) = 0.8 X 1020.
Q21. 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 Cag(PO4)2 is _____
(d) Formula of sodium carbonate is ______, and that of ammonium sulphate is ______
Answer:
(a) In a chemical reaction, the sum of the masses of the reactants and products remains unchanged. This is called the Law of conservation of mass.
(b) A group of atoms carrying a fixed charge on them is called a polyatomic ion.
(c) The formula unit mass of Ca3(PO4)2 is 310 g.
(d) Formula of sodium carbonate is Na2CO3, and that of ammonium sulphate is (NH4)2SO4.
Q22. Complete the following crossword puzzle (Fig. 3.1) by using the name of the chemical elements. Use the data given in Table 3.2.
Table 3.2
| Across | Down |
|---|---|
| 2. The element used by Rutherford during his alpha scattering experiment | 1. A lustrous white metal used for making ornaments tends to get tarnished black in the presence of moist air |
| 3. An element which forms rust on exposure to moist air | 4. Both brass and bronze are alloys of element |
| 5. A very reactive non-metal stored underwater | 6. The metal exists in the liquid state at room temperature |
| 7. When treated with
dilute hydrochloric acid, zinc metal produces gas of this element which, when tested with a burning splinter. |
8. An element with the symbol Pb |
Answer:
| Across | Down |
|---|---|
| 2. Gold | 1. Silver |
| 3. Iron | 4. Copper |
| 5. Phosphorous | 6. Mercury |
| 7. Hydrogen | 8. Lead |
Q23. (a) In this crossword puzzle (Fig 3.2), the names of 11 elements are hidden. Symbols of these are given below. Complete the puzzle.
1. CI
2. H
3. Ar
4. O
5. Xe
6. N
7. He
8. F
9. Kr
10. Rn
11. Ne
(b) Identify the total number of inert gases, their names and symbols from this crossword puzzle.
Answer:
(a) 1. CI: Chlorine
2. H: Hydrogen
3. Ar: Argon
4. O: Oxygen
5. Xe: Xenon
6. N: Nitrogen
7. He: Helium
8. F: Fluorine
9. Kr: Krypton
10. Rn: Radon
11. Ne: Neon
(b) There are five inert gases in this crossword puzzle. Their names and symbols are Argon (Ar), Xenon (Xe), Helium (He), Krypton (Kr) and Radon (Rn).
Q24. Write the formulae for the following and calculate the molecular mass for each of them.
(a) Caustic potash
(b) Baking powder
(c) Limestone
(d) Caustic soda
(e) Ethanol
(f) Common salt
Answer:
(a) The formulae of Caustic potash is KOH.
The molecular mass of Caustic potash is 39 + 16 + 1 = 56 u.
(b) The formulae of Baking powder is NaHCO3.
The molecular mass of Baking powder is 23 + 1 + 12 + 3 X 16 = 84 u.
(c) The formulae of Limestone is CaCO3.
The molecular mass of Limestone is 40 + 12 + 3 X 16 = 100 u.
(d) The formulae of Caustic soda is NaOH.
The molecular mass of Caustic soda is 23 + 16 + 1 = 40 u.
(e) The formulae of Ethanol is C2H5OH.
The molecular mass of 2 × 2 + 5 X 1 + 16 + 1 = 46 u.
(f) The formulae of Common salt is NaCl.
The molecular mass of Common salt is 23 + 35.5 = 58.5 u.
Q25. In photosynthesis, six 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 produce 18 g of glucose? Compute the volume of water consumed, assuming the density of water to be 1g cm-3.
Answer:
6 CO2 + 6 H2O → C6H12O6 + 6 O2
1 mole of glucose requires 6 moles of water.
180 g of glucose requires 108 g of water.
1 g of glucose will require 108 / 180 g of water.
18 g of glucose would need (108 X 18) / 180 g of water
18 g of glucose would need = 10.8 g of water
Volume of water used = Mass / Density = 10.8 / 1 = 10.8 cm3.
CBSE Class 9 Science Chapter 3 Extra Questions
Q1. What is the compound Al2(SO4)3, and give the ions present in it?
Answer:
The given compound is aluminium sulphate.
It contains two ions: Al3+ and SO42−.
Q2. What is an atomic mass unit’? How is it linked with relative atomic mass?
Answer:
Atomic mass unit is defined as the 1 / 12 of the mass of carbon – 12 atom C – 12.
The relative atomic mass of an element is the number of times one atom of the element is heavier than the 1 / 12 times the mass of an atom of carbon-12.
Q3. What is Avogadro’s number? Why is it also known as Avogadro’s constant?
Answer:
Avogadro’s number is the number of particles (atoms, molecules or ions) present in one mole of any substance. It is known as Avogadro’s constant because its value is fixed (6.022 x 1023) irrespective of the nature of the particles.
Q4. Which are the six postulates of Dalton’s atomic theory?
Answer:
Postulates of Dalton’s atomic theory
1. The matter is composed of indivisible particles called atoms.
2. Atoms of the same element are similar in shape and mass but differ from the atoms of different elements.
3. Atoms can neither be created nor be destroyed.
4. Atoms of different elements combine in fixed, simple, whole-number ratios to form a compound.
5. Atoms of the same element can combine in more than one ratio to form two or more compounds.
6. The atom is the tiniest unit of matter that can take part in a chemical reaction.
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