ATOMS AND MOLECULES : NCERT Exercise Questions

Q.1     A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g if boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Sol.     Mass of boron = 0.096g(Given)
           Mass of oxygen = 0.144g (Given)
           Mass of sample = 0.24g (Given)
           Thus, percentage of boron by weight in the compound = 0.0960.24×100%
           = 40%
           Thus, percentage of oxygen by weight in the compound = 0.1440.24×100%
          = 60 %

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Q.2     When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen ? Which law of chemical combinations will govern your answer ?
Sol.      Carbon + Oxygen → Carbon dioxide 3g of carbon reacts with 8 g of oxygen to produce 11g of carbon dioxide. If 3g of carbon is burnt in 50g of oxygen, then 3g of carbon will react with 8 g of oxygen. The remaining 42 g of oxygen will be left un-reactive. In this case also, only 11g of carbon dioxide will be formed.  The above answer is governed by the law of constant proportions.

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Q.3     What are polyatomic ions? Give examples?
Sol.     A polyatomic ion is a group of atoms carrying a charge (positive or negative).For example, ammonium ion (NH+4), hydroxide ion (OH−), carbonate ion (CO2−3),sulphateion (SO2−4).

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Q.4      Write the chemical formula of the following:
            (a) Magnesium chloride
            (b) Calcium oxide
            (c) Copper nitrate
            (d) Aluminium chloride
            (e) Calcium carbonate
Sol.      (a) Magnesium chloride →MgCl2
            (b) Calcium oxide →CaO
            (c) Copper nitrate →Cu(NO3)2
            (d) Aluminium chloride →AlCl3
            (e) Calcium carbonate →CaCO3

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Q.5     Give the names of the elements present in the following compounds:
            (a) Quick lime
            (b) Hydrogen bromide
            (c) Baking powder
            (d) Potassium sulphate
Sol. .

Compound	Chemical formula	Elements present
Quick lime	Cao	Calcium,oxygen
Hydrogen bromide	HBr	Hydrogen,bromine
Baking powder	NaHCO3	Sodium,hydrogen, carbon, oxygen
Potassium sulphate	k2SO4	Potassium,sulphur,oxygen

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Q.6     Calculate the molar mass of the following substances:
           (a) Ethyne, C2H2
           (b) Sulphur molecule, S8
           (c) Phosphorus molecule, P4 (atomic mass of phosphorus = 31)
           (d) Hydrochloric acid, HCl
           (e) Nitric acid, HNO3
Sol.     (a) Molar mass of ethyne, C2H2 = 2 × 12 + 2 × 1 = 28g
           (b) Molar mass of sulphur molecule, S8 = 8 × 32 = 256g
           (c) Molar mass of phosphorus molecule,P4 = 4 × 31 = 124g
           (d) Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5g
           (e) Molar mass of nitric acid, HNO3 = 1 + 14 + 3 × 16 = 63g

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Q.7     What is the mass of --
           (a) 1 mole of nitrogen atoms?
           (b) 4 mole of aluminium atoms (Atomic mass of aluminium = 27)?
           (c) 10 moles of sodium sulphite (Na2SO3) ?
Sol.     (a) The mass of 1 mole of nitrogen atoms is 14g.
           (b) The mass of 4 moles of aluminium atoms is (4 × 27)g = 108g
           (c) The mass of 10 moles of sodium sulphite (Na2SO3) is
           10 × [2 × 23 + 32 + 3 × 16]g = 10 × 126g = 1260g

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Q.8     Convert into mole.
           (a) 12g of oxygen gas 
           (b) 12g of water
           (c) 22g of carbon dioxide
Sol.     (a) 32 g of oxygen gas = 1 mole
           Then, 12g of oxygen gas = 1232 mole = 0.375 mole
           (b) 18g of water = 1 mole
           Then, 20 g of water = 2018 mole = 1.11 moles (approx)
           (c) 44g of carbon dioxide = 1 mole
           Then, 22g of carbon dioxide = 2244 mole = 0.5 mole

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Q.9     What is the mass of :
           (a) 0.2 mole of oxygen atoms?
           (b) 0.5 mole of water molecules?
Sol.     (a) Mass of one mole of oxygen atoms = 16g
           Then, mass of 0.2 mole of oxygen atoms = 0.2 × 16g = 3.2g
           (b) Mass of one mole of water molecule = 18g
           Then, mass of 0.5 mole of water molecules = 0.5 × 18g = 9g

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Q.10     Calculate the number of molecules of sulphur (S8) present in 16g of solid sulphur.
Sol.        1 mole of solid sulphur (S8) = 8 × 32g = 256g
              i.e., 256g of solid sulphur contains = 6.022 × 1023 molecules
              Then, 16g of solid sulpur contains 6.022×1023256×16 molecules
              = 3.76 × 1022 molecules (approx) 

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Q.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)
Sol.        1 mole of aluminium oxide (Al2O3) = 2 × 27 + 3 × 16 = 102g
              i.e., 102g of Al2O3 = 6.022 × 1023 molecules of Al2O3
              Then, 0.051 g of  Al2O3 contains = 6.022×1023102×0.051 molecules
              = 3.011 ×  1020  molecules of  Al2O3
              The number of aluminium ions (Al3+) present in one  molecules of aluminium oxide is 2.
              Therefore, The number of aluminium ions  (Al3+) present in
              3.11 × 1020 molecules (0.051g) of aluminium oxide (Al2O3) = 2 × 3.011 × 1020
              = 6.022 × 1020