Science NCERT Solutions of the chapter3-Atoms and Molecules of chemistry part

The NCERT solutions of the chapter 3-Atoms and molecules are solutions of chemistry part of class 9 NCERT science textbook, these NCERT solutions are the initiations to get to know about the composition of matter, these class 9 science NCERT solutions of chapter 3-Atoms and molecules of chemistry part reveal the information how and why the elements or compounds react with each other and forms another compound. In chapter 3-Atoms and molecules are the particles that participate in chemical reactions? All NCERT solutions are solved by an expert teacher as per the CBSE norms.

 

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 Q1- In a reaction, 5.3 g of sodium –  carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon oxide. 0.9 g of 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

Ans-The mass of reactants = mass of Sodium carbonate + mass of ethanoic acid

= 5.3 g + 6 g = 11.3 g

Mass of products formed = mass of sodium ethanoate + mass of carbon dioxide + mass of water

= 2.2 g + 0.9 g + 8.2 g

=11.3 g

So, observations are in agreement with the law of conservation of mass because in this chemical reaction the observation follows the law of conservation of mass as mentioned below.

Mass of reactants = Mass of products

   Q2-  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?

Ans-According to the law of constant proportion all the elements in a compound are always in the fixed ratio,so let 3 g of hydrogen reacts with X grams of oxygen.

Therefore

(1/8 ) = 3/X

X  =24

So, 24 g oxygen will be required to react with 3 g of hydrogen in the formation of water.

Q3-  Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

Ans-2 nd postulate of Dalton’s atomic theory is the result of the law of conservation of mass because it states that atoms are indivisible particles, which can not be created or destroyed in a chemical reaction.

 Q4-  Which postulate of Dalton’s atomic theory can explain the law of definite proportions?

Ans- 6 th postulate of Dalton’s theory explains the law of definite proportions because it states that relative number and kinds of atoms are constant in a given compound.

Q5-Write down the formulae of

           (i)  Sodium oxide

           (ii)Aluminium chloride

           (iii) Sodium sulphide

           (iv) Magnesium hydroxide

Ans-

(i) Sodium chloride – NaCl

(ii)  Aluminium chloride – AlCl₃

(iii)  Sodium sulphide – Na₂S

(iv)  Magnesium hydroxide – Mg(OH)₂

Q6-Write down the names of the compounds represented by the following formulae;

(i)  Al₂(SO₄)₃

(ii) CaCl₂

(iii) K₂SO₄

(iv) KNO₃

(v) CaCO₃

Ans-

(i)  Al₂(SO₄)₃ – Aluminium sulphate

(ii)  CaCl₂ – Calcium chloride

(iii)  K₂SO₄ – Potassium sulphate

(iv)  KNO₃ – Potassium nitrate

(v)  CaCO₃ – Calcium carbonate

Q7-What is meant by the term chemical formula?

Ans-Chemical formula of a compound shows what is its composition, from the chemical formula we get the combining capacity of each atom which is known as valency. The

way of writing chemical formula is i.e aluminium sulphate.

Valancy of aluminium 3 means it has capacity to hold 3 ions of sulphate and sulphate

has the charge 2 has capacity to hold 2 aluminium atoms.

So its chemical formula will be – Al₂(SO₄)₃ 

Q9-How many atoms are present in a

(i)  H₂S molecule and

(ii)  PO₄^3- ions

Ans-

(i)In H₂S, Atoms of H =2, sulphur atom =1, so total atoms = 3

(ii) In PO₄^3- ions, phosphorus atom = 1, oxygen atoms= 4, total atoms = 5

Q10-Calculate the molecular masses of H₂, O₂, Cl₂, CO_2, CH₄, C₂H₆, NH₃, CH₃OH .

Ans-Molecular masses of the given substances are as follows.

H₂ – 2 X 1 = 2 u

O₂ – 2 X 16 = 32 u

Cl₂ – 2 x 35 = 70 u

CO₂ – 12 + 2 x 16 = 12 +32 = 44 u

CH₄ – 12 + 4 x 1 = 12 + 4 = 16 u

C₂H₄ – 2 x 12 + 4 x 1 = 24 + 4 = 28 u

NH₃ – 14 + 3 x 1 = 14 +3 = 17 u

CH₃OH – 12 + 3 x 1 + 16 + 1 = 32 u 

Q11-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).

Ans-ZnO – 65 + 16 = 81 u

Na₂O – 23 x 2 + 16 =62 u

K₂CO₃ – 39 x 2 + 12 + 16x 3 = 138 u                       

                            

Q12-If one mole of carbon atoms weighs 12 grams. What is the mass (in grams) of 1 atom of carbon ? 

Ans- 1 mole of carbon atoms = 6.022 x 10²³ atoms

6.022 x 10²³ atoms weighs = 12 grams

So 1 atom of carbon will weigh =12/(6.022 x 10²³)

= 1.99 x 10^-23 gram

Q13-Which has more number of atoms. 1OO grams of sodium or 100 grams of iron (given atomic mass of Na = 23 u, Fe = 56 u) ?

Ans-

Number of Na atoms

= (given mass of Na/molar mass of Na)x N₀

atomic mass of Na = 23 u, so molar mass of Na= 23 g

given mass of Na = 100 g

N₀ =  6.022 x 10²³

Number of Na atoms = (100/23) x6.022 x 10²³

= 2.62 x 10²⁴

On the same way number of iron atoms in 100 g of iron = (100/56) x 6.022 x 10²³

= 1.08 x 10²⁴

In 100 g of iron and sodium number of Na atoms are more than the number of Fe atoms. 

                            

Q14.  A 0.24 g sample of compound 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.

Ans

% of boron in the compound

= (quantity of boron )/(mass of sample)x 100

=( 0.096/0.24) x 100

= 40 %

Similarly % of oxygen = (0.144/0.24) x 100

= 60 %

Q15.   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 combination will govern your answer ?

Ans.If 3.0 g of carbon is burnt in 8.00 g oxygen as a result of  11 g of carbon dioxide is formed. According to law of constant proportion, all elements are in fixed proportion in a compound.So 3.00 g of carbon will react only with 8 g of oxygen and rest of the oxygen 50 – 8 = 42 mg will be freed or unreacted, So hear law of definite proportion is followed.

 Q16 –   What are polyatomic ions ? Give examples.

Ans-Polyatomic are group of atoms that have a net charge on them as an example  (sulphate ion)SO₄^2-, (nitrate ion)NO₃^– etc.

Q17-   Write the chemical formulae of the following.

(a) Magnesium chloride

(b)Calcium oxide

 (c) Copper nitrate

 (d)Aluminium chloride

(e)Calcium carbonate

Ans- Valency of Mg, Ca , Cu  and Al are 2, 2, (1,2) and 3 respectively.

Q18-Give the names of the elements present in the following compounds.

 (a)Quick lime

(b)Hydrogen bromide

(c) Baking powder

(d)Potassium sulphate

Ans-

(a)Quick lime – CaO( Calcium, oxygen)

(b) Hydrogen bromide – HBr( Hydrogen, Bromine)

(c)Baking powder – NaHCO₃(Sodium, Hydrogen, Carbon, Oxygen)

(d) Potassium sulphate – K₂SO₄( Potassium, Sulphor, Oxygen)

Q18-Calculate the molar mass of the following substances.

(a)Ethyne, C₂H₂

(b)The sulphur molecule, S₈

(c)Phosphorus molecule, P₄ (Atomic mass of phosphorus = 31)

(d)Hydrochloric acid, HCl

(e)Nitric acid, HNO₃

Ans-

(a)  Molecular mass of C₂H₂ = 2 x 12 + 2 x 1 =26 u

So molar mass of C₂H₂ = 26 g

(b)Molecular mass of sulphur molecule, S₈ = 8 x 32 =256 u

Molar mass of S₈ = 256 g

(c)molecular mass of P₄ =4 x31 = 124 u

Molar mass of P₄ =124 g

(d)Molecular mass of HCl = 1 + 35 = 36 u

Molar mass of HCl = 36 g

(e)Molecular mass of HNO₃ = 1 + 14 + 16 x 3 =63 u

Molar mass of HNO₃ = 63 g

 

Q19-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)?

Ans-

(a)Mass of 1 mole nitrogen atom

= its molar mass =14 

Q20- Convert into mole.

(a)12 g of oxygen gas

(b)20 g of water

(c)22 g of carbon dioxide

Ans-

(a)Molecular mass of oxygen gas, O₂= 16 x2 = 32 u

Molar mass of O₂ = 32 g

No. of moles in 12 g of O₂ = m/M

Given mass of O₂, m= 12 g

Molar mass of O₂, M =32 g

So, no. of moles of O₂ =12/32 = 0.375

(b)  Molecular mass of water, H₂O = 2×1 +16 =18 u

Molar mass of water, M = 18 g

Given mass , m= 20 g

No. of moles of water =m/M = 20/18 = 1.11

(C) Molecular mass of CO₂ = 12 + 16 x 2 = 44 u

Molar mass of CO₂, M =44 g

Given mass of CO₂, m =22 g

No. of moles of CO₂ = m/M = 22/44 = 0.5

Q21-What is the mass of :

(a)0.2 moles of oxygen atoms ?

(b)0.5 moles of water molecules ?

Ans –

Atomic mass of oxygen = 16 u

So, molar mass (M )of ,O = 16 g

Given mass of O =m

No. of moles of O =m/M

0.2 = m/16

0.2 x 16 = m, m= 3.2 g

(b)Molecular mass of water, H₂O = 18 u

Molar mass, M of water = 18 g

No. of moles of H₂O = m/M

So, given mass, m = 0.5 x 18 = 9 g

Q22-Calculate the number of molecules of sulphur(S₈) present in 16 g of solid sulphur.

Ans- Molecular mass of S₈ = 8 x 32 =256 u

Molar mass ,M of S₈ = 256 g

Given mass, m = 16 g

No. of molecules of sulphur = (m/M) x N₀

N₀ = 6.022 x 10²³

No. of molecules of sulphur = (16/256) x 6.022 x 10²³

= 0.376 x 10²³

= 3.76 x 10²²

Q23-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)

Ans-

Molecular mass of aluminium oxide ,Al₂O₃ = 27 x2 + 16 x 3 =54 + 48 = 102 u

Molar mass of aluminium oxide ,M= 102 g

No. of molecules of ,Al₂O₃ = (m/M) x6.022 x 10²³

Given mass , m = 0.051 g

No. of molecules of ,Al₂O₃ = (0.051/102) x6.022 x 10²³

= 0.0030 x 10²³

=3.0 x 10²⁰

No. of ions in Al₂O₃ = no. of atoms Al₂O₃ = 2

So,total number of ions present in 0.051 g of Al₂O₃

= 3.0 x 2 x 10²⁰

 = 6 x 10²⁰

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Chapter 1- Number System	Chapter 9-Areas of parallelogram and triangles
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3- Coordinate Geometry	Chapter 11-Construction
Chapter 4- Linear equations in two variables	Chapter 12-Heron’s Formula
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Chapter 7-Triangles	Chapter 15-Probability
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Chapter 5-Complex numbers	Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities	Chapter 14-Mathematical Reasoning
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Chapter 1-Relations and Functions	Chapter 9-Differential Equations
Chapter 2-Inverse Trigonometric Functions	Chapter 10-Vector Algebra
Chapter 3-Matrices	Chapter 11 – Three Dimensional Geometry
Chapter 4-Determinants	Chapter 12-Linear Programming
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