Samacheer Kalvi 10th Science Solutions Chapter 7 Pdf
Tamilnadu Board Samacheer Kalvi 10th Science Solutions Chapter 7: Tamilnadu State Board Solution Class 10 Science Chapter 7 – Atoms And Molecules.
Samacheer Kalvi 10th Science Solutions Chapter 7: Overview
Board |
Samacheer Kalvi |
Class |
10 |
Subject |
Science |
Chapter |
7 |
Chapter Name |
Atoms And Molecules |
Samacheer Kalvi 10th Science Solutions Chapter 7 Atoms And Molecules
I.) Choose the best answer.
1.) Which of the following has the smallest mass?
a.) 6.023 × 1023 atoms of He
b.) 1 atom of He
c.) 2 g of He
d.) 1 mole atoms of He
Ans:b.) 1 atom of He
Because, mass of one mole of He is 4g or 0.004kg
And 1 mole of He = 6.023*1023 atoms of He
Hence, mass of one atom of He = 0.004/ 6.023*1023 which is very small.
2.) Which of the following is a triatomic molecule?
a.) Glucose
b.) Helium
c.) Carbon dioxide
d.) Hydrogen
Ans:c.) Carbon dioxide
Because, CO2 has one carbon atom and two oxygen atoms, total 3 atoms that means triatomic.
3.) The volume occupied by 4.4 g of CO2 at S.T.P
a.) 22.4 litre
b) 2.24 litre
c.) 0.24 litre
d.) 0.1 litre
Ans: b.) 2.24 litre
Because, molar volume of CO2 = 22.4 litre
And mass of 1 mole of CO2 gas = 44g
Hence, 44g of CO2occupies 22.4 liter of volume
Thus, 4.4g of CO2 occupies the volume = 22.4/44*4.4 = 22.4/10 = 2.24 litre
4.) Mass of 1 mole of Nitrogen atom is
a.) 28 amu
b.) 14 amu
c.) 28 g
d.) 14 g
Ans:c.) 28 g
5.) Which of the following represents 1 amu?
a.) Mass of a C – 12 atom
b.) Mass of a hydrogen atom
c.) 1/12th of the mass of a C – 12 atom
d.) Mass of O – 16 atom
Ans:c.) 1/12th of the mass of a C – 12 atom
6.) Which of the following statement is incorrect?
a.) 12 gram of C – 12 contains Avogadro’s number of atoms.
b.) One mole of oxygen gas contains Avogadro’s number of molecules.
c.) One mole of hydrogen gas contains Avogadro’s number of atoms.
d.) One mole of electrons stands for 6.023× 1023 electrons.
Ans: a.) 12 gram of C – 12 contains Avogadro’s number of atoms.
7.) The volume occupied by 1 mole of a diatomic gas at S.T.P is
a.) 11.2 litre
b.) 5.6 litre
c.) 22.4 litre
d.) 44.8 litre
Ans: c.) 22.4 litre
8.) In the nucleus of 20Ca40, there are
a.) 20 protons and 40 neutrons
b.) 20 protons and 20 neutrons
c.) 20 protons and 40 electrons
d.) 40 protons and 20 electrons
Ans: b.) 20 protons and 20 neutrons
9.) The gram molecular mass of oxygen molecule is
a.) 16 g
b.) 18 g
c.) 32 g
d.) 17 g
Ans: b.) 18 g
10.) 1 mole of any substance contains ____ molecules.
a.) 6.023 × 1023
b.) 6.023 × 10-23
c.) 3.0115 × 1023
d.) 12.046 × 1023
Ans: a.) 6.023 × 1023
II.) Fill in the blanks
1.) Atoms of different elements having _______ mass number, but ________ atomic numbers are called isobars.
Ans:Atoms of different elements having samemass number, butdifferentatomic numbers are called isobars.
2.) Atoms of different elements having same number of ___________ are called isotones.
Ans:Atoms of different elements having same number of neutronsare called isotones.
3.) Atoms of one element can be transmuted into atoms of other element by ________
Ans:Atoms of one element can be transmuted into atoms of other element by artificial transmutation.
4.) The sum of the numbers of protons and neutrons of an atom is called its __________
Ans:The sum of the numbers of protons and neutrons of an atom is called its mass number.
5.) Relative atomic mass is otherwise known as __________
Ans: Relative atomic mass is otherwise known as standard atomic weight.
6.) The average atomic mass of hydrogen is ___________ amu.
Ans:The average atomic mass of hydrogen is 1.008amu.
7.) If a molecule is made of similar kind of atoms, then it is called ____________ atomic molecule. Ans: If a molecule is made of similar kind of atoms, then it is calledhomoatomic molecule.
8.) The number of atoms present in a molecule is called its ____________
Ans:The number of atoms present in a molecule is called itsatomicity.
9.) One mole of any gas occupies ________ ml at S.T.P
Ans: One mole of any gas occupies 22400 ml at S.T.P
10.) Atomicity of phosphorous is ___________
Ans:Atomicity of phosphorous is 4.
III. Match the following
Ans:
1) 8 g of O2:0.25 moles
2) 4 g of H2: 2 moles
3) 52 g of He: 13 moles
4) 112 g of N2: 4 moles
5) 5 g of Cl2: 0.5 moles
IV. True or False: (If false give the correct statement)
1) Two elements sometimes can form more than one compound.
Ans: True
2) Noble gases are Diatomic
Ans: False
Correct statement: Nobel gases are monoatomic.
3) The gram atomic mass of an element has no unit
Ans: False
Correct statement: The gram atomic mass of an element is expressed in the unit grams.
4) 1 mole of Gold and Silver contain same number of atoms
Ans: True
5) Molar mass of CO2 is 42g.
Ans: False
Correct statement: The molar mass of CO2 is 44 g.
V.) Assertion and Reason: Answer the following questions using the data given below:
i) A and R are correct, R explains the A.
ii) A is correct, R is wrong.
iii) A is wrong, R is correct.
iv) A and R are correct, R doesn’t explains A.
1.) Assertion: The Relative Atomic mass of aluminium is 27
Reason: An atom of aluminium is 27 times heavier than 1/12th of the mass of the C – 12 atom.
Ans:i) A and R are correct, R explains the A.
2.) Assertion: The Relative Molecular Mass of Chlorine is 35.5 a.m.u.
Reason: The natural abundance of Chlorine isotopes are not equal.
Ans:i) A and R are correct, R explains the A.
VI) Short answer questions
a) Define: Relative atomic mass.
Ans:
Relative atomic mass:
- Relative atomic mass of an element is defined as, it is the ratio of average mass of its isotopes and 1/12th part of the mass of carbon-12 atom.
- Relative atomic mass is also called as Standard atomic weight.
- It is denoted by Ar and given by,
- Relative Atomic Mass (Ar) = Average mass of the isotopes of the element/ 1/12th part of mass of the carbon – 12 atom.
2.) Write the different types of isotopes of oxygen and its percentage abundance.
Ans:
The following table shows the different types of isotopes of oxygen and its percentage abundance.
Isotopes of oxygen | Mass in amu | Percentage abundance |
8O16 | 15.9949 | 99.757 |
8O17 | 16.9991 | 0.038 |
8O18 | 17.9992 | 0.205 |
3) Define: Atomicity
Ans:
Atomicity:
The number of atoms present in the molecule is called as its atomicity.
For example:
- If atomicity is 1, then number of atoms present in a molecule is also one and it is called as monoatomic.
- If atomicity is 2, then number of atoms present in a molecule will be 2 and it is called as diatomic molecule.
- If atomicity is 3, then number of atoms present are 3 and the molecule is called as triatomic molecule.
- If the atomicity is greater than 3, then the no. of atoms present are also greater than 3 and the molecule is called as polyatomic molecule.
4) Give any two examples for heterodiatomic molecules.
Ans:
Heteroatomic molecule:
The molecule which consist of atoms of different elements is called heteroatomic molecule.
For example: H20, HCl are the heteroatomic molecules.
5.) What is Molar volume of a gas?
Ans:
Molar volume of a gas:
The gas of one mole occupies 22.4 litre or 22400 ml at standard temperature pressure, this volume of a gas is called as molar volume of a gas.
6) Find the percentage of nitrogen in ammonia.
Ans:
Molecular mass of NH3 = 14 + 3 = 17 g
Mass % of nitrogen = 14/17*100= 82.35 %
VII. Long answer questions
1) Calculate the number of water molecule present in one drop of water which weighs 0.18 g.
Ans:
The weight of the one mole of water is 18 g.
And 1 mole of water molecule having weight 18 g contains 6.023*1023 water molecules.
Hence, the number of molecules in 0.18 g of water are = (6.023*1023/ 180*0.18
= 6.023*1021 water molecules
Thus, the number of water molecules present in one drop of water which weighs 0.18g are 6.023*1021
2) N2 + 3 H2 → 2 NH3
(The atomic mass of nitrogen is 14, and that of hydrogen is 1) 1 mole of nitrogen (_______g) + 3 moles of hydrogen (_________ g) → 2 moles of ammonia (_________ g)
Ans:
Given that, N2 + 3 H2 → 2 NH3
1 mole of N2 = 2*14 = 28g
3 moles of H2 = 3* 2 = 6g
2 moles of NH3 = 34g
Thus, 1 mole of H2 (28g) + 3 moles of H2(6g) → 2 moles of NH3 (34g)
3) Calculate the number of moles in
Ans:
i) 27g of Al :
The number of moles is given by,
Number of moles = mass/ atomic mass
Hence, number of moles in 27 g of Al = 27/27 = 1 mole
ii) 51 × 1023 molecules of NH4Cl:
Also, number of moles = number of molecules/ Avogadro’s number
= 1.51*1023/ 6.023*1023
= 0.25 moles
4) Give the salient features of “Modern atomic theory”.
Ans:
The salient features of Modern atomic theory are as follows:
- An atom is no longer divisible, after the discovery of electron, proton and neutron.
- Atoms of the same elements may have different mass which leads to discovery of isotopes 17Cl35 and 17Cl37.
- Atoms of the same elements may have same masses which leads to discovery of isobars 18Ar40 and 20Ca40.
- Atom is no longer indestructible which leads to discovery of artificial transmutation.
- Atoms may not always combine in a simple whole number ratio.
For example:
In Glucose C6H12O6, C: H: O = 6:12:6 or 1:2:1
Andin Sucrose C12H22O11, C: H: O = 12:22:11
- Atom is the smallest particle which takes part in the chemical reaction.
The mass of an atom is converted into energy also.
5) Derive the relationship between Relative molecular mass and Vapour density.
Ans:
Relative molecular mass:
It is the ratio of mass of one molecule of gas or vapor to the mass of one atom of hydrogen.
Vapour density:
Vapour density is the ratio of mass of certain volume of a gas or vapour to the mass of an equal volume of hydrogen at S.T.P.
V.D. = mass of certain volume of a gas or vapour at S.T.P. / mass of an equal volume of hydrogen
Let, the no. of molecules in one volume = n,
Then according to Avogadro’s law,
V.D. at STP = mass of n molecules of gas or vapour at STP/ mass of n molecules of hydrogen
We get, V.D. = mass of 1 molecules of gas or vapour at STP/ mass of 1 molecules of hydrogen
But, hydrogen is diatomic.
Hence, V.D. = mass of 1 molecules of gas or vapour at STP/ mass of 2 atoms of hydrogen
By comparing the formula of vapour density with relative molecular mass we write as,
V.D. = mass of 1 moles of gas or vapour at STP/ 2* mass of one atom of hydrogen——-A
Hence,
Relative molecular mass = mass of 1 molecule of gas or vapour at STP/ mass of 1 atoms of hydrogen ———————-B
Thus, by putting equation B in equation A, we get,
V.D. = relative molecular mass/ 2
Thus, relative molecular mass = 2* vapour density
This is the relation between relative molecular mass and vapour density.
VIII.) HOT question
1) Calcium carbonate is decomposed on heating in the following reaction CaCO3 → CaO + CO2
Ans:
i) How many moles of Calcium carbonate are involved in this reaction?:
CaCO3 → CaO + CO2
In above reaction, one mole of calcium carbonate is involved.
ii) Calculate the gram molecular mass of calcium carbonate involved in this reaction iii. How many moles of CO2 are there in this equation?
The gram molecular mass of CaCO3= mass of Ca + mass of C + mass of 3 O-atoms
= 40 + 12 + 3*16
The gram molecular mass of CaCO3 = 100g
IX.) Solve the following problems
1) How many grams are there in the following?
Ans:
1) 2 moles of hydrogen molecule, H2:
2 moles of H2 molecule = 2*2* mass of one H-atom
= 2*2*1= 4g
2 moles of H2 molecule = 4g
2) 3 moles of chlorine molecule, Cl2:
3 moles of Cl2 molecule = 3*2* mass of one Cl-atom
= 3*2*35.6 = 3*71
3 moles of Cl2 molecule = 213g
iii) 5 moles of sulphur molecule, S8:
3 moles of S8 molecule = 5*8* mass of one S-atom
= 5*8* 32= 5*256
3 moles of S8 molecule= 1280g
iv) 4 moles of phosphorous molecule, P4:
4 moles of P4 molecule = 4*4* mass of one P-atom
= 4*4*31 = 4*124 =496g
4 moles of P4 molecule= 496g
2.) Calculate the % of each element in calcium carbonate. (Atomic mass: C-12, O-16,
Ca -40)
Ans:
The molar mass of CaCO3 = 40+ 12 + 3*16 = 100g
Thus, % of Ca = (40/100)*100 = 40%
% of C = (12/100)*100 = 12%
% of O = (48/100)*100 = 48%
3) Calculate the % of oxygen in Al2(SO4)3. (Atomic mass: Al-27, O-16, S -32)
Ans:
The molar mass of Al2(SO4)3 = 2*27 + 3*32 + 12*16 = 54 + 96 + 192 = 342g
Hence, % of oxygen = (192/ 342)*100 = 56.14%
4.) Calculate the % relative abundance of B -10 and B -11, if its average atomic mass is 10.804 amu.
Ans:
Given that, the average atomic mass of boron = 10.804 amu
Let the a is the % relative abundance of B-10
And b is the % relative abundance of B-11.
Then, a + b = 1
Hence, b = 1 – a
But, relative abundance = a*10 + (1-a)*11 = 10.804 amu
Hence, 10a + 11 – 11a = 10.804 amu
11 – a = 10.804 amu
-a = 10.804 – 11
Thus, a = 0.196
But, a = % relative abundance of B-10 = 0.196*100 = 19.6%
And
Thus, b = % relative abundance of B- 11 = 100 – 19.6 = 80.4%
% relative abundance of B-10= 19.6%
% relative abundance of B- 11 = 80.4%