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NCERT Solution for Class 9 Science - Chapter 3, Atoms and Molecules, , Exercise-3.1, , Page: 32, , 1. In a reaction, 5.3g of sodium carbonate reacted with 6 g of acetic acid. The products were 2.2 g, of carbon dioxide, 0.9 g water and 8.2 g of sodium acetate. Show that these observations are in, agreement with the law of conservation of mass., Sodium carbonate + acetic acid → Sodium acetate + carbon dioxide + water, Solution:, Sodium carbonate + acetic acid → Sodium acetate + carbon dioxide + water, 5.3g, , 6g, , 8.2g, , 2.2g, , 0.9g, , As per the law of conservation of mass, the total mass of reactants must be equal to the total mass of, products, As per the above reaction, LHS = RHS, , i.e., 5.3g + 6g = 2.2g + 0.9 g + 8.2 g = 11.3 g, , Hence the observations are in agreement with the law of conservation of mass., , 2. Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass of oxygen, gas would be required to react completely with 3 g of hydrogen gas?, Solution:, We know hydrogen and water mix in the ratio 1: 8., For every 1g of hydrogen, it is 8g of oxygen., Therefore, for 3g of hydrogen, the quantity of oxygen = 3 x 8 = 24g, Hence, 24g of oxygen would be required for the complete reaction with 3g of hydrogen gas., , 3. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?, Solution:, The postulate of Dalton’s Atomic theory which is a result of the law of conservation of mass is,, “Atoms can neither be created nor destroyed”., , 4. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?, Solution:
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NCERT Solution for Class 9 Science - Chapter 3, Atoms and Molecules, The postulate of Dalton's atomic theory that can explain the law of definite proportions is - the, relative number and kinds of atoms are equal in given compounds., , Exercise-3.2, , Page: 35, , 5. Define the atomic mass unit?, Solution:, An atomic mass unit is a unit of mass used to express weights of atoms and molecules where one, atomic mass is equal to 1/12th the mass of one carbon-12 atom., 6. Why is it not possible to see an atom with naked eyes?, Solution:, Firstly, atoms are miniscule in nature, measured in nanometers. Secondly, except for atoms of noble, gasses, they do not exist independently. Hence, an atom cannot be visible to the naked eyes., , Exercise-3.3-3.4, 7. Write down the formulae of, (i) sodium oxide, (ii) aluminium chloride, (iii) sodium sulphide, (iv) magnesium hydroxide, Solution:, The following are the formulae:, (i) sodium oxide - Na2O, , Page: 39
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NCERT Solution for Class 9 Science - Chapter 3, Atoms and Molecules, (ii) aluminium chloride - AlCl3, (iii) sodium sulphide - Na2S, (iv) magnesium hydroxide - Mg (OH)2, 8. Write down the names of compounds represented by the following formulae:, (i) Al2(SO4)3, (ii) CaCl2, (iii) K2SO4, (iv) KNO3, (v) CaCO3., Solution:, Listed below are the names of the compounds for each of the following formulae, (i) Al2(SO4)3 - Aluminium sulphate, (ii) CaCl2 - Calcium chloride, (iii) K2SO4 - Potassium sulphate, (iv) KNO3 - Potassium nitrate, (v) CaCO3 - Calcium carbonate, , 9. What is meant by the term chemical formula?, Solution:, Chemical formula is the symbolic representation of a chemical compound. For example: The, chemical formula of hydrochloric acid is HCl., , 10. How many atoms are present in a
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NCERT Solution for Class 9 Science - Chapter 3, Atoms and Molecules, (i) H2S molecule and, (ii) PO43- ion?, Solution:, The number of atoms present are as follows:, (i) H2S molecule has 2 atoms of hydrogen and 1 atom of sulphur hence 3 atoms in totality., (ii) PO43- ion has 1 atom of phosphorus and 4 atoms of oxygen hence 5 atoms in totality., , Exercise-3.5.1-3.5.2, , Page: 40, , 11. Calculate the molecular masses of H2 , O2 , Cl2, CO2, CH4, C2H6, C2H4, NH3, CH3OH., Solution:, The following are the molecular masses:, The molecular mass of H2 – 2 x atoms atomic mass of H = 2 x 1u = 2u, The molecular mass of O2 – 2 x atoms atomic mass of O = 2 x 16u = 32u, The molecular mass of Cl2 – 2 x atoms atomic mass of Cl = 2 x 35.5u = 71u, The molecular mass of CO2 - atomic mass of C + 2 x atomic mass of O = 12 + ( 2x16)u = 44u, The molecular mass of CH4 - atomic mass of C + 4 x atomic mass of H = 12 + ( 4 x 1)u = 16u, The molecular mass of C2H6- 2 x atomic mass of C + 6 x atomic mass of H = (2 x 12) +, (6 x 1)u=24+6=30u, The molecular mass of C2H4- 2 x atomic mass of C + 4 x atomic mass of H = (2x 12) +, (4 x 1)u=24+4=28u, The molecular mass of NH3- atomic mass of N + 3 x atomic mass of H = (14 +3 x 1)u= 17u, The molecular mass of CH3OH - atomic mass of C + 3x atomic mass of H + atomic mass of O +, atomic mass of H = (12 + 3x1+16+1)u=(12+3+17)u = 32u, 12. Calculate the formula unit masses of ZnO, Na2O, K2CO3, given atomic masses of Zn = 65u,
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NCERT Solution for Class 9 Science - Chapter 3, Atoms and Molecules, Na = 23 u, K=39u, C = 12u, and O=16u., Solution:, Given:, Atomic mass of Zn = 65u, Atomic mass of Na = 23u, Atomic mass of K = 39u, Atomic mass of C = 12u, Atomic mass of O = 16u, The formula unit mass of ZnO= Atomic mass of Zn + Atomic mass of O = 65u+16u = 81u, The formula unit mass of Na2O = 2 x Atomic mass of Na + Atomic mass of O = (2 x 23)u, +16u = 46u + 16u = 62u, The formula unit mass of K2CO3 = 2 x Atomic mass of K + Atomic mass of C + 3 x, Atomic mass of O = (2 x 39)u + 12u + (3 x 16)u = 78u + 12u + 48u = 138u, , Exercise-3.5.3, , Page: 42, , 13. If one mole of carbon atoms weighs 12grams, what is the mass (in grams) of 1 atom of carbon?, Solution:, Given: 1 mole of carbon weighs 12g, 1 mole of carbon atoms = 6.022 X 1023, Molecular mass of carbon atoms = 12g = an atom of carbon mass, Hence, mass of 1 carbon atom = 12 / 6.022 X 1023 = 1.99 X 10-23g, , 14. Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic, mass of Na = 23u, Fe = 56 u)?, Solution:, Given: Atomic mass of Na=23u, Atomic mass of Fe= 56u
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NCERT Solution for Class 9 Science - Chapter 3, Atoms and Molecules, To calculate the number of atoms in 100g of sodium:, 23g of Na contains = 6.022 X 1023 atoms, 1g of Na contains = 6.022 X 1023 atoms / 23, 100g of Na contains = 6.022 X 1023 atoms X 100 / 23, = 2.6182 X 1024 atoms, To calculate the number of atoms in 100g of sodium:, 56g of Fe contains = 6.022 X 1023 atoms, 1g of Fe contains = 6.022 X 1023 atoms / 56, 100g of Fe contains = 6.022 X 1023 atoms X 100 / 56, = 1.075 X 1024 atoms, Hence, through comparison, it is evident that 100g of Na has more atoms., , Exercise, , Page: 43, , 1. A 0.24g sample of compound of oxygen and boron was found by analysis to contain 0.096g of, boron and 0.144g of oxygen. Calculate the percentage composition of the compound by weight., Solution:, Given: Mass of the sample compound = 0.24g, mass of boron = 0.096g, mass of oxygen = 0.144g, To calculate percentage composition of the compound:, Percentage of boron = mass of boron / mass of the compound x 100, = 0.096g / 0.24g x 100 = 40%, Percentage of oxygen = 100 - percentage of boron, = 100 - 40 = 60%
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NCERT Solution for Class 9 Science - Chapter 3, Atoms and Molecules, 2. When 3.0g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced. What, mass of carbon dioxide will be formed when 3.00g of carbon is burnt in 50.00 g of oxygen?, Which law of chemical combination will govern your answer?, Solution:, 11.00g of carbon dioxide is formed when 3.00g carbon is burnt in 8.00g of oxygen., Carbon and oxygen are combined in the ratio 3:8 to give carbon dioxide using up all the carbon and, oxygen, Hence, for 3g of carbon and 50g of oxygen, 8g of oxygen is used and 11g of carbon is formed, the, left oxygen is unused i.e., 50-3=47g of oxygen is unused., This depicts the law of definite proportions - The combining elements in compounds are present in, definite proportions by mass., , 3. What are polyatomic ions? Give examples., Solution:, Polyatomic ions are ions that contain more than one atom but they behave as a single unit, Example: CO32- , H2PO4-, , 4. Write the chemical formula of the following., (a) Magnesium chloride, (b) Calcium oxide, (c) Copper nitrate, (d) Aluminium chloride, (e) Calcium carbonate, Solution:
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NCERT Solution for Class 9 Science - Chapter 3, Atoms and Molecules, The following are the chemical formula of the above-mentioned list:, (a) Magnesium chloride - MgCl2, (b) Calcium oxide - CaO, (c) Copper nitrate - Cu(NO3)2, (d) Aluminium chloride - AlCl3, (e) Calcium carbonate - CaCO3, , 5. Give the names of the elements present in the following compounds., (a) Quick lime, (b) Hydrogen bromide, (c) Baking powder, (d) Potassium sulphate., Solution:, The following are the names of the elements present in the following compounds:, (a) Quick lime - Calcium and oxygen (CaO), (b) Hydrogen bromide - Hydrogen and bromine (HBr), (c) Baking powder - Sodium, Carbon, Hydrogen, Oxygen (NaHCO3), (d) Potassium sulphate - Sulphur, Oxygen, Potassium (K2SO4), , 6. Calculate the molar mass of the following substances., (a) Ethyne, C2H2, (b) Sulphur molecule, S8
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NCERT Solution for Class 9 Science - Chapter 3, Atoms and Molecules, (c) Phosphorus molecule, P4 (Atomic mass of phosphorus =31), (d) Hydrochloric acid, HCl, (e) Nitric acid, HNO3, Solution:, Listed below is the molar mass of the following substances:, (a) Molar mass of Ethyne C2H2= 2 x Mass of C+2 x Mass of H = (2x12)+(2x1)=24+2=26g, (b) Molar mass of Sulphur molecule S8 = 8 x Mass of S = 8 x 32 = 256g, (c) Molar mass of Phosphorus molecule, P4 = 4 x Mass of P = 4 x 31 = 124g, (d) Molar mass of Hydrochloric acid, HCl = Mass of H+ Mass of Cl = 1+35.5 = 36.5g, (e) Molar mass of Nitric acid, HNO3 =Mass of H+ Mass of Nitrogen + 3 x Mass of O = 1 + 14+, 3x16 = 63g, , 7. What is the mass of (a) 1 mole of nitrogen atoms?, (b) 4 moles of aluminium atoms((Atomic mass of aluminium =27)?, (c) 10 moles of sodium sulphite (Na2SO3)?, Solution:, The mass of the above mentioned list is as follows:, (a) Atomic mass of nitrogen atoms = 14u, Mass of 1 mole of nitrogen atoms= Atomic mass of nitrogen atoms, Therefore, mass of 1 mole of nitrogen atom is 14g, (b) Atomic mass of aluminium =27u, Mass of 1 mole of aluminium atoms = 27g
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NCERT Solution for Class 9 Science - Chapter 3, Atoms and Molecules, 1 mole of aluminium atoms = 27g, 4 moles of aluminium atoms = 4 x 27 = 108g, (c) Mass of 1 mole of sodium sulphite Na2SO3 = Molecular mass of sodium sulphite = 2 x Mass, of Na + Mass of S + 3 x Mass of O = (2 x 23) + 32 +(3x 16) = 46+32+48 = 126g, Therefore, mass of 10 moles of Na2SO3 = 10 x 126 = 1260g, 8. Convert into mole., (a) 12g of oxygen gas, (b) 20g of water, (c) 22g of carbon dioxide, Solution:, Conversion of the above-mentioned molecules into moles is as follows:, (a) Given: Mass of oxygen gas=12g, Molar mass of oxygen gas = 2 Mass of Oxygen = 2 x 16 = 32g, Number of moles = Mass given / molar mass of oxygen gas = 12/32 = 0.375 moles, (b) Given: Mass of water = 20g, Molar mass of water = 2 x Mass of Hydrogen + Mass of Oxygen = 2 x 1 + 16 = 18g, Number of moles = Mass given / molar mass of water, = 20/18 = 1.11 moles, (c) Given: Mass of carbon dioxide = 22g, Molar mass of carbon dioxide = Mass of C + 2 x Mass of Oxygen = 12 + 2x 16 = 12+32=44g, Number of moles = Mass given/ molar mass of carbon dioxide = 22/44 = 0.5 moles, , 9. What is the mass of:, (a) 0.2 mole of oxygen atoms?
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NCERT Solution for Class 9 Science - Chapter 3, Atoms and Molecules, (b) 0.5 mole of water molecules?, Solution:, The mass is as follows:, (a) Mass of 1 mole of oxygen atoms = 16u, hence it weighs 16g, Mass of 0.2 moles of oxygen atoms = 0.2 x 16 = 3.2u, (b) Mass of 1 mole of water molecules = 18u, hence it weighs 18g, Mass of 0.5 moles of water molecules = 0.5 x 18 = 9u, 10. Calculate the number of molecules of sulphur (S8) present in 16g of solid sulphur., , Solution:, To calculate molecular mass of sulphur:, Molecular mass of Sulphur (S8) = 8xMass of Sulphur = 8x32 = 256g, Mass given = 16g, Number of moles = mass given/ molar mass of sulphur, = 16/256 = 0.0625 moles, To calculate the number of molecules of sulphur in 16g of solid sulphur:, Number of molecules = Number of moles x Avogadro number, = 0.0625 x 6.022 x 10²³ molecules, = 3.763 x 1022 molecules, , 11. Calculate the number of aluminium ions present in 0.051g of aluminium oxide., (Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al =, 27u)
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NCERT Solution for Class 9 Science - Chapter 3, Atoms and Molecules, Solution:, To calculate the number of aluminium ions in 0.051g of aluminium oxide:, 1 mole of aluminium oxide = 6.022 x 1023 molecules of aluminium oxide, 1 mole of aluminium oxide (Al2O3) = 2 x Mass of aluminium + 3 x Mass of Oxygen, = (2x 27) + (3 x16) = 54 +48 = 102g, 1 mole of aluminium oxide = 102g = 6.022 x 1023 molecules of aluminium oxide, Therefore, 0.051g of aluminium oxide has = 0.051 x 6.022 x 1023 / 102, = 3.011 x 1020 molecules of aluminium oxide, One molecule of aluminium oxide has 2 aluminium ions, hence number of aluminium ions, present in 0.051g of aluminium oxide = 2 x 3.011x 1020 molecules of aluminium oxide, = 6.022 x 1020
