ncert solution for class 9 science chapter 3 Atoms and Molecules

NCERT Solution for Class 9 Science Chapter 3 

Atoms and Molecules

Question 1: 

 In  a  reaction,  5.3  g  of  sodium  carbonate  reacted  with  6  g  of  ethanoic  acid.  The  products were  2.2 g of carbon dioxide,  0.9 g water and  8.2 g of sodium ethanoate.  Show that these observations are in agreement with the law of conservation of mass.  Sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water  

Answer:

In the given reaction, sodium carbonate reacts with ethanoic acid to produce sodium ethanoate, carbon dioxide, and water.

Mass of sodium carbonate = 5.3 g (Given)
Mass of ethanoic acid = 6 g (Given)
Mass of sodium ethanoate = 8.2 g (Given)
Mass of carbon dioxide = 2.2 g (Given)
Mass of water = 0.9 g (Given)

Now, total mass before the reaction = (5.3 + 6) g = 11.3 g

And, total mass after the reaction = (8.2 + 2.2 + 0.9) g = 11.3 g

∴Total mass before the reaction = Total mass after the reaction  Hence,  the  given  observations  are  in  agreement  with  the  law  of  conservation  of  mass.

Question 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?

Answer:

It is given that the ratio of hydrogen and oxygen by mass to form water is 1:8.

Then, the mass of oxygen gas required to react completely with 1 g of hydrogen gas is 8 g.

Therefore, the mass of oxygen gas required to react completely with 3 g of hydrogen gas is 8 × 3 g = 24 g.

Question 3:  

Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

Answer:

The postulate of Dalton’s atomic theory which is a result of the law of conservation of mass is:  Atoms  are  indivisible  particles,  which  can  neither  be  created  nor  destroyed  in  a chemical reaction.

Question 4:  

 Which   postulate   of   Dalton’s   atomic   theory   can   explain   the   law   of   definite proportions?

Answer:

The  postulate  of  Dalton’s  atomic  theory  which  can  explain  the  law  of  definite proportion is: The relative number and kind of atoms in a given compound remains constant.

Question 1:

Define atomic mass unit.

Answer:

Mass unit equal to exactly one-twelfth  the mass of one atom of carbon-12 is called one atomic mass unit. It is written as ‘u’.

Question 2:  Why is it not possible to see an atom with naked eyes?

Answer:

The size of an atom is so small that it is not possible to see it with naked eyes. Also, the atom of an element does not exist independently.

Question 1:

Write down the formulae of (i) sodium oxide (ii) aluminium chloride (iii) sodium suphide (iv) magnesium hydroxide

Answer:

(i) Sodium oxide →Na₂O

(ii) Aluminium chloride → AlCl₃

(iii) Sodium suphide → Na₂S

(iv) Magnesium hydroxide → Mg(OH)₂

Question 2:

Write down the names of compounds represented by the following formulae:

(i) Al₂(SO₄)₃ (ii) CaCl₂ (iii) K₂SO₄ (iv) KNO₃ (v) CaCO₃

Answer:

(i) Al₂(SO₄)₃ → Aluminium sulphate

(ii) CaCl₂ → Calcium chloride

(iii) K₂SO₄ → Potassium sulphate

(iv) KNO₃ → Potassium nitrate

(v) CaCO₃ → Calcium carbonate

Question 3:

What is meant by the term chemical formula?

Answer:

The  chemical  formula  of  a  compound  means  the  symbolic  representation  of  the composition of a compound. From the chemical formula of a compound, we can know the number and kinds of atoms of different elements that constitute the compound. For  example,  from  the  chemical  formula  CO₂  of  carbon  dioxide,  we  come  to  know that one carbon atom and two oxygen atoms are chemically bonded together to form one molecule of the compound, carbon dioxide.

Question 4:

How many atoms are present in a (i) H₂S molecule and (ii) PO₄-3 ion?

Answer:

(i)  In  an  H₂S  molecule,  three  atoms  are  present;  two  of  hydrogen  and  one  of sulphur.

(ii) In a PO₄₃₋ ion, five atoms are present; one of phosphorus and four of oxygen.

Question 1:

Calculate the molecular masses of H₂, O₂, Cl₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, CH₃OH.

 Answer:

Molecular mass of H₂ = 2 × Atomic mass of H = 2 × 1  = 2 u

Molecular mass of O₂ = 2 × Atomic mass of O = 2 × 16  = 32 u

Molecular mass of Cl₂ = 2 × Atomic mass of Cl = 2 × 35.5 = 71 u

Molecular mass of CO₂ = Atomic mass of C + 2 × Atomic mass of O = 12 + 2 × 16 = 44 u

Molecular mass of CH₄ = Atomic mass of C + 4 × Atomic mass of H = 12 + 4 × 1 = 16 u

Molecular mass of C₂H₆ = 2 × Atomic mass of C + 6 × Atomic mass of H = 2 × 12 + 6 × 1 = 30 u

Molecular mass of C₂H₄ = 2 × Atomic mass of C + 4 × Atomic mass of H = 2 × 12 + 4 × 1 = 28 u

Molecular mass of NH₃ = Atomic mass of N + 3 × Atomic mass of H = 14 + 3 × 1 = 17 u

Molecular  mass  of  CH₃OH  =  Atomic  mass  of  C  +  4  ×  Atomic  mass  of  H  +  Atomic mass of O  = 12 + 4 × 1 + 16  = 32 u

Question 2:

Calculate the formula unit masses of ZnO, Na₂O, K₂CO₃, given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.

Answer:

Formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O = 65 + 16 = 81 u

Formula unit mass of Na₂O = 2 × Atomic mass of Na + Atomic mass of O = 2 × 23 + 16 = 62 u

Formula  unit  mass  of  K₂CO₃  =  2  ×  Atomic  mass  of  K  +  Atomic  mass  of  C  +  3  × Atomic mass of O = 2 × 39 + 12 + 3 × 16  = 138 u

Question 1:

If one mole of carbon atoms weighs  12 gram, what is the mass  (in gram) of  1 atom of carbon?

Answer:

One mole of carbon atoms weighs 12 g (Given)

i.e., mass of 1 mole of carbon atoms = 12 g

Then, mass of    6.022 × 10 ²³ number of carbon atoms = 12 g

Therefore, mass of 1 atom of carbon

=1.99  × 10 ^-23

Question 2:

Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na = 23 u, Fe = 56 u)?

Answer:

Atomic mass of Na = 23 u (Given)

Then, gram atomic mass of Na = 23 g

Now, 23 g of Na contains =          6.022 × 10 ²³  number of atoms

Thus, 100 g of Na contains

       number of atoms

2.62 × 10 ²⁴ =       number of atoms

Again, atomic mass of Fe = 56 u(Given)

Then, gram atomic mass of Fe = 56 g

Now, 56 g of Fe contains = 6.022 × 10 ²³                  number of atoms

Thus, 100 g of Fe contains            number of atoms

number of atoms

Therefore,  100 grams of sodium contain more number of atoms than  100 grams of
iron.

Question 1:

A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.

Answer:

Mass of boron = 0.096 g (Given)
Mass of oxygen = 0.144 g (Given)
Mass of sample = 0.24 g (Given)

Thus, percentage of boron by weight in the compound = = 40%

And, 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:

Carbon + Oxygen             à Carbon dioxide

3 g of carbon reacts with 8 g of oxygen to produce 11 g of carbon dioxide.

If  3 g of carbon is burnt in  50 g of oxygen, then  3 g 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 11 g 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:

Question 4:

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

Answer:

(a) Magnesium chloride → MgCl₂

(b) Calcium oxide → CaO

(c) Copper nitrate → Cu (NO₃)₂

(d) Aluminium chloride → AlCl₃

(e) Calcium carbonate → CaCO₃

Question 5:

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

Answer:

 

Question 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

Answer:

(a) Molar mass of ethyne, C₂H₂ = 2 × 12 + 2 × 1 = 28 g

(b) Molar mass of sulphur molecule, S₈ = 8 × 32 = 256 g

(c) Molar mass of phosphorus molecule, P₄ = 4 × 31 = 124 g

(d) Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5 g

(e) Molar mass of nitric acid, HNO₃ = 1 + 14 + 3 × 16 = 63 g

Question 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)?

Answer:

(a) The mass of 1 mole of nitrogen atoms is 14 g.

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

(c) The mass of 10 moles of sodium sulphite (Na₂SO₃) is  10 × [2 × 23 + 32 + 3 × 16] g = 10 × 126 g = 1260 g

Question 8:

Convert into mole. (a) 12 g of oxygen gas (b) 20 g of water (c) 22 g of carbon dioxide

Answer:

Question 9:

What is the mass of: (a) 0.2 mole of oxygen atoms? (b) 0.5 mole of water molecules?

Answer:

(a) Mass of one mole of oxygen atoms = 16 g

Then, mass of 0.2 mole of oxygen atoms = 0.2 × 16g = 3.2 g

(b) Mass of one mole of water molecule = 18 g

Then, mass of 0.5 mole of water molecules = 0.5 × 18 g = 9 g

Question 10:

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

Answer:

1 mole of solid sulphur (S₈) = 8 × 32 g = 256 g

i.e., 256 g of solid sulphur contains = 6.022 × 10²³ molecules

Then, 16 g of solid sulphur contains =

= 3.76 × 10²² molecules (approx)

Question 11:

Calculate the number of aluminium ions present in 0.051 g 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 = 27 u)

Answer:

1 mole of aluminium oxide (Al₂O₃) = 2 × 27 + 3 × 16 = 102 g

i.e., 102 g of Al₂O₃ = 6.022 × 10²³ molecules of Al₂O₃

Then, 0.051 g of Al₂O₃ contains =

= 3.011 × 10²⁰ molecules of Al₂O₃

The number of aluminium ions  (Al³⁺) present in one molecule of aluminium oxide is  2.

Therefore,  the  number  of  aluminium ions  (Al³⁺)  present  in  3.011  ×  10²⁰  molecules (0.051 g ) of aluminium oxide (Al₂O₃) = 2 × 3.011 × 10²⁰  = 6.022 × 10²⁰