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.
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 uย
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 uย 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