# 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 par**t 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.

**NCERT solutions of class 9 science chapter3-Atoms and molecules**

You can download PDF-NCERT solutions of class 9 science chapter3-Atoms and molecules

**PDF-NCERT solutions of class 9 science chapter3-Atoms and molecules**

Click for online shopping

**Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc**

**ย Q1- In a reaction, 5.3 g of sodium –ย ย 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.ย **

Sodium carbonate +ethanoic acid โsodium 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

**ย ย Q2-ย ย 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

Xย =24

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

**Q3-ย ย Which postulate of Daltonโs atomic theory is the result of the law of conservation of mass?**

Ans-2 nd postulate of Daltonโs 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.

**ย Q4-ย ย Which postulate of Daltonโs atomic theory can explain the law of definite proportions?**

Ans- 6 th postulate of Daltonโs 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**

**ย ย ย ย ย ย ย ย ย ย (i)ย ย Sodium oxide**

**ย ย ย ย ย ย ย ย ย ย (ii)Aluminium chloride**

**ย ย ย ย ย ย ย ย ย ย (iii) Sodium sulphide**

**ย ย ย ย ย ย ย ย ย ย (iv) Magnesium hydroxide**

Ans-

(i) Sodium chloride โ NaCl

(ii)ย ย Aluminium chloride โ AlCl_{3}

(iii)ย ย Sodium sulphide โ Na_{2}S

(iv)ย ย Magnesium hydroxide โ Mg(OH)_{2}_{ }

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

**(i)ย ย Al _{2}(SO_{4})_{3}**

**(ii)ย CaCl _{2}**

**(iii)ย K _{2}SO_{4}**

**(iv)ย KNO _{3}**

**(v)ย CaCO _{3}**

Ans-

(i)ย ย Al_{2}(SO_{4})_{3}ย โ Aluminium sulphate

(ii)ย ย CaCl_{2}ย โ Calcium chloride

(iii)ย ย K_{2}SO_{4}ย โ Potassium sulphate

(iv)ย ย KNO_{3}ย โ Potassium nitrate

(v)ย ย CaCO_{3}ย โ 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 โ Al_{2}(SO_{4})_{3}_{ย }

**Q9-How many atoms are present in a**

**(i)ย ย H _{2}S molecule and**

**(ii)ย ย PO _{4}^{3-}ย ions**

Ans-

(i)In H_{2}S, Atoms of H =2, sulphur atom =1, so total atoms = 3

(ii)ย In PO_{4}^{3-}ย ions, phosphorus atom = 1, oxygen atoms= 4, total atoms = 5

**Q10-Calculate the molecular masses of H _{2}, O_{2}, Cl_{2}, CO_{2,ย }CH_{4}, C_{2}H_{6}, NH_{3}, CH_{3}OH .**

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

H_{2}ย โ 2 X 1 = 2 u

O_{2}ย โ 2 X 16 = 32 u

Cl_{2}ย โ 2 x 35 = 70 u

CO_{2}ย โ 12 + 2 x 16 = 12 +32 = 44 u

CH_{4}ย โ 12 + 4 x 1 = 12 + 4 = 16 u

C_{2}H_{4}ย โ 2 x 12 + 4 x 1 = 24 + 4 = 28 u

NH_{3}ย โ 14 + 3 x 1 = 14 +3 = 17 u

CH_{3}OH โ 12 + 3 x 1 + 16 + 1 = 32 u**ย **

**Q11-Calculate the formula unit masses of ZnO, Na _{2}O, K_{2}CO_{3}.(given atomic masses of**

**ย Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u and O = 16 u).**

Ans-ZnO โ 65 + 16 = 81 u

Na_{2}O โ 23 x 2 + 16 =62 u

K_{2}CO_{3} โ 39 x 2 + 12 + 16x 3 = 138 u**ย ย ย ย ย ย ย ย ย ย ย ย **

**ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย **

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

Ans- 1 mole of carbon atoms = 6.022 x 10^{23}ย atoms

6.022 x 10^{23}ย atoms weighs = 12 grams

So 1 atom of carbon will weigh =12/(6.022 x 10^{23})

= 1.99 x 10^{-23}ย 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 N_{0}

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

given mass of Na = 100 g

N_{0}ย = ย 6.022 x 10^{23}

Number of Na atoms = (100/23) x6.022 x 10^{23}

= 2.62 x 10^{24}

On the same way number of iron atoms in 100 g of iron = (100/56) x 6.022 x 10^{23}

= 1.08 x 10^{24}

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

**ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย **

**Q14.ย ย 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.ย ย ย 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 ofย 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 โ 8 = 42 mg will be freed or unreacted, So hear law of definite proportion is followed.

**ย Q16 –ย ย ย What are polyatomic ions ? Give examples.**

Ans-Polyatomic are group of atoms that have a net charge on them as an exampleย (sulphate ion)SO_{4}^{2-}, (nitrate ion)NO_{3}^{–}ย etc.

**Q17-ย ย ย Write the chemical formulae of the following.**

**(a)ย Magnesium chloride**

**(b)Calcium oxide**

**ย (c) Copper nitrate**

**ย (d)Aluminium chloride**

**(e)Calcium carbonate**

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

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

**ย (a)Quick lime**

**(b)Hydrogen bromide**

**(c)ย Baking powder**

**(d)Potassium sulphate**

Ans-

(a)Quick lime โ CaO( Calcium, oxygen)

(b)ย Hydrogen bromide โ HBr( Hydrogen, Bromine)

(c)Baking powder โ NaHCO_{3}(Sodium, Hydrogen, Carbon, Oxygen)

(d)ย Potassium sulphate โ K_{2}SO_{4}( Potassium, Sulphor, Oxygen)

**Q18-Calculate the molar mass of the following substances.**

**(a)Ethyne, C _{2}H_{2}**

**(b)The sulphur molecule, S _{8}**

**(c)Phosphorus molecule, P _{4}ย (Atomic mass of phosphorus = 31)**

**(d)Hydrochloric acid, HCl**

**(e)Nitric acid, HNO _{3}**

_{ }Ans-

(a)ย ย Molecular mass of C_{2}H_{2}ย = 2 x 12 + 2 x 1 =26 u

So molar mass of C_{2}H_{2}ย = 26 g

(b)Molecular mass of sulphur molecule, S_{8}ย = 8 x 32 =256 u

Molar mass of S_{8}ย = 256 g

(c)molecular mass of P_{4}ย =4 x31 = 124 u

Molar mass of P_{4}ย =124 g

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

Molar mass of HCl = 36 g

(e)Molecular mass of HNO_{3}ย = 1 + 14 + 16 x 3 =63 u

Molar mass of HNO_{3ย }= 63 g

**ย **

**Q19-What is the mass of โ**

**(a)1 mole of nitrogen atoms?**

**(b)4 moles of aluminium atoms (Atomic mass of aluminium =27)**

**(c)10 moles of sodium sulphite(Na _{2}SO_{3})?**

Ans-

(a)Mass of 1 mole nitrogen atom

= its molar mass =14ย

**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, O_{2}= 16 x2 = 32 u

Molar mass of O_{2}ย = 32 g

No. of moles in 12 g of O_{2}ย = m/M

Given mass of O_{2}, m= 12 g

Molar mass of O_{2}, M =32 g

So, no. of moles of O_{2}ย =12/32 = 0.375

(b)ย ย Molecular mass of water, H_{2}O = 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 CO_{2}ย = 12 + 16 x 2 = 44 u

Molar mass of CO_{2}, M =44 g

Given mass of CO_{2}, m =22 g

No. of moles of CO_{2}ย = 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 โ

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, H_{2}O = 18 u

Molar mass, M of water = 18 g

No. of moles of H_{2}O = m/M

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

**Q22-Calculate the number of molecules of sulphur(S _{8}) present in 16 g of solid sulphur.**

Ans- Molecular mass of S_{8}ย = 8 x 32 =256 u

Molar mass ,M of S_{8}ย = 256 g

Given mass, m = 16 g

No. of molecules of sulphur = (m/M) x N_{0}

N_{0}ย = 6.022 x 10^{23}

No. of molecules of sulphur = (16/256) x 6.022 x 10^{23}

= 0.376 x 10^{23}

= 3.76 x 10^{22}

**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 ,Al_{2}O_{3}ย = 27 x2 + 16 x 3 =54 + 48 = 102 u

Molar mass of aluminium oxide ,M= 102 g

No. of molecules of ,Al_{2}O_{3}ย = (m/M) x6.022 x 10^{23}

Given mass , m = 0.051 g

No. of molecules of ,Al_{2}O_{3}ย = (0.051/102) x6.022 x 10^{23}

= 0.0030 x 10^{23}

=3.0 x 10^{20}

No. of ions in Al_{2}O_{3}ย = no. of atoms Al_{2}O_{3}ย = 2

So,total number of ions present in 0.051 g of Al_{2}O_{3}

= 3.0 x 2 x 10^{20}

^{ย }= 6 x 10^{20}

### Class 9 CBSE Science Notes

**Three Laws of Motion: Class 9 CBSE**

**Evoporation,Vapourization and Latent heat -Class 9 CBSE notes**

**What is an atom,molecule and atomicity of a substance?**

**How to determine Valency,net charge of an ion and Molecular formula of a substance.**

**Thrust and Pressure : Difference**

**The Complete Detail of Archimedes Principal**

**Average Speed and Average Velocity: Differences**

**How to evaluate recoil velocity of gun**

**If energy is conserved then why do we need to save it for future generations?**

**Molar mass,molecular mass and mole concept**

**What is second law of of motion ?**

**What is universal law of gravitational force**

**NCERT Solutions of Science and Maths for Class 9,10,11 and 12**

**NCERT Solutions for class 9 maths**

**NCERT Solutions for class 9 scienceย **

**CBSE Class 9-Question paper of science 2020 with solutions**

**CBSE Class 9-Sample paper of science**

**CBSE Class 9-Unsolved question paper of science 2019**

**NCERT Solutions for class 10 maths**

**CBSE Class 10-Question paper of maths 2021 with solutions**

**CBSE Class 10-Half yearly question paper of maths 2020 with solutions**

**CBSE Class 10 -Question paper of maths 2020 with solutions**

**CBSE Class 10-Question paper of maths 2019 with solutions**

**NCERT Solutions for Class 10 Science**

**Solutions of Class 10 Science Sample Paper and Question Papers for Term-1 and Term 2 2021-22 CBSE Board**

**Solution of Class 10 Science Question Paper Preboard 2021-22:Term 2 CBSE Board Exam**

**Solutions of Class 10 Science Sample Paper Term-1 2021-22 CBSE Board**

**Class 10 Science Sample Paper for Term 2 CBSE Board Exam 2021-22 with Solution**

**Solutions of Class 10 Science Question Paper Preboard Examination (First) 2021 -22 Class 10 Science**

**CBSE Class 10 – Question paper of science 2020 with solutions**

**CBSE class 10 -Sample paper of Science 2020**

**NCERT Solutions for class 11 maths**

Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | ย Chapter 16- Probability |

**CBSE Class 11-Question paper of maths 2015**

**CBSE Class 11 – Second unit test of maths 2021 with solutions**

**NCERT solutions for class 12 maths**

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 |

Chapter 5- Continuity and Differentiability | Chapter 13-Probability |

Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |

Chapter 7- Integrals | |

Chapter 8-Application of Integrals |

**Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2**

**Class 12 Maths Important Questions-Application of Integrals**

**Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22**

**Solutions of class 12ย maths question paper 2021 preboard exam CBSE**

**ย **