Class 9th Science Solution: Chapter 3 Atoms and Molecules

Question 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.

 

Answer 1:

 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  

  = 40%

  Thus, percentage of oxygen by weight in the compound  

  = 60 %

 

Question 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?

 

Answer 2:

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.

Question 3:

 What are polyatomic ions? Give examples? 

Answer 3:

A polyatomic ion is a group of atoms carrying a charge (positive or negative). For example, ammonium ion (𝑁𝐻4 +), hydroxide ion (OH−), carbonate ion (𝐶𝑂3 2−), sulphateion (𝑆𝑂4 2−).   

Question 4: Write the chemical formula of the following:             (a) Magnesium chloride             (b) Calcium oxide             (c) Copper nitrate             (d) Aluminium chloride             (e) Calcium carbonate 

Answer 4:

1. Magnesium chloride →MgCl2           

2. Calcium oxide →CaO          

3. Copper nitrate →Cu(NO3)2         

4. Aluminium chloride →AlCl3        

5. Calcium carbonate →CaCO3

Question 5:

Give the names of the elements present in the following compounds:

1. Quick lime

2. Hydrogen bromide

3. Baking powder

4. Potassium sulphate.

    

   

    

    

    

Question 6:

Calculate the molar mass of the following substances:

1. Ethyne, C₂H₂

2. Sulphur molecule, S₈

3. Phosphorus molecule, P₄ (atomic mass of phosphorus = 31)

4. Hydrochloric acid, HCl

5. Nitric acid, HNO₃

    

Answer 6:

1. Molar mass of ethyne, C₂H₂                     = 2 × 12 + 2 × 1 = 28g

2. Molar mass of sulphur molecule, S₈          = 8 × 32 = 256g

3. Molar mass of phosphorus molecule, P₄  = 4 × 31 = 124g

4. Molar mass of hydrochloric acid, HCl      = 1 + 35.5 = 36.5g

5. Molar mass of nitric acid, HNO₃              = 1 + 14 + 3 × 16 = 63g

    

Question 7:

What is the mass of  

1. 1 mole of nitrogen atoms?

2. 4 mole of aluminium atoms (Atomic mass of aluminium = 27)?

3. 10 moles of sodium sulphite (Na₂SO₃)?

Answer 7:

1. The mass of 1 mole of nitrogen atoms is 14g.

2. The mass of 4 moles of aluminium atoms is (4 × 27)g = 108g

3. The mass of 10 moles of sodium sulphite (Na₂SO₃) is  10 × [2 × 23 + 32 + 3 × 16]g 

   = 10 × 126g = 1260g

    

    

Question 8:

Convert into mole.

1. 12g of oxygen gas  

2. 12g of water            

3. 22g of carbon dioxide  

Answer 8:

1. 32 g of oxygen gas = 1 mole

              Then, 12g of oxygen gas = 12/32 mole = 0.375 mole

2. 18g of water = 1 mole

              Then, 20 g of water = 20/18 mole = 1.11 moles (approx.)

3. 44g of carbon dioxide = 1 mole

              Then, 22g of carbon dioxide = 22/44 mole = 0.5 mole

    

    

    

    

Question 9:

What is the mass of:

1. 0.2 mole of oxygen atoms?

2. 0.5 mole of water molecules?

Answer 9:

(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

Question 10:

Calculate the number of molecules of sulphur (S₈) present in 16g of solid sulphur.

Answer 10:

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 molecules

 = 3.76 × 10²² molecules (approx)  

 

 

 

Question 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)

Answer 11:

  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  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

 

Author : Rautish Kumar

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