Your Progress — Chapter 7: Atoms and Molecules
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MCQI. Multiple Choice Questions1 mark each
Q.1 Which of the following has the smallest mass?
✓ Answer: (A)
Q.2 Which of the following is a triatomic molecule?
✓ Answer: (A)
Q.3 The volume occupied by 4.4 g of CO2 at S.T.P
✓ Answer: (A)
Q.4 Mass of 1 mole of Nitrogen atom is
✓ Answer: (A)
Q.5 Which of the following represents 1 amu?
✓ Answer: (A)
Q.6 Which of the following statement is incorrect?
✓ Answer: (A)
Q.7 The volume occupied by 1 mole of a diatomic gas at S.T.P is
✓ Answer: (A)
Q.8 In the nucleus of 20Ca40, there are
✓ Answer: (A)
Q.9 The gram molecular mass of oxygen molecule is
✓ Answer: (A)
Q.10 1 mole of any substance contains ____ molecules.
✓ Answer: (A)
Q.1 Atoms of different elements having same mass number, but different atomic numbers are called isobars.
Options not available — refer to textbook.
✓ Answer: (—)
Q.2 Atoms of different elements having same number of neutrons are called isotones.
Options not available — refer to textbook.
✓ Answer: (—)
Q.3 Atoms of one element can be transmuted into atoms of other element by artificial transmutation
Options not available — refer to textbook.
✓ Answer: (—)
Q.4 The sum of the numbers of protons and neutrons of an atom is called its mass number
Options not available — refer to textbook.
✓ Answer: (—)
Q.5 Relative atomic mass is otherwise known as standard atomic weight
Options not available — refer to textbook.
✓ Answer: (—)
Q.6 The average atomic mass of hydrogen is 1.0079 amu.
Options not available — refer to textbook.
✓ Answer: (—)
Q.7 If a molecule is made of similar kind of atoms, then it is called Homo atomic molecule.
Options not available — refer to textbook.
✓ Answer: (—)
Q.8 The number of atoms present in a molecule is called its atomicity
Options not available — refer to textbook.
✓ Answer: (—)
Q.9 One mole of any gas occupies 22400 ml at S.T.P 10 Atomicity of phosphorous is 4
Options not available — refer to textbook.
✓ Answer: (—)
MatchIII. Match the Following1 mark each
| Column A | Column B |
|---|---|
| 8 g of O2 | 0.25 moles |
| 4 g of H2 | 2 moles |
| 52 g of He | 13 moles |
| 112 g of N2 | 4 moles |
| 35.5 g of Cl2 | 0.5 moles |
T/FIV. True or False1 mark each
| # | Statement | Answer | Correction (if False) |
|---|---|---|---|
| 1 | Two elements sometimes can form more than one compound. | True | — |
| 2 | Noble gases are Diatomic | False | Noble gases are monoatomic. |
| 3 | The gram atomic mass of an element has no unit | False | The gram atomic mass of an element is exposed in grams. |
| 4 | 1 mole of Gold and Silver contain same number of atoms | True | — |
| 5 | Molar mass of CO2 is 42g. | False | Molar mass of CO2 is 44g. |
A&RV. Assertion & Reasoning2 marks each
Q.1
Assertion: Atomic mass of aluminium is 27
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✓ Answer
Reason: An atom of aluminium is 27 times heavier than 1/12th of the mass of the C - 12 atom.
Answer: i) A and R are correct, R explains the A.
Answer: i) A and R are correct, R explains the A.
Q.2
Assertion: The Relative Molecular Mass of Chlorine is 35.5 a.m.u.
▾
✓ Answer
Reason: The natural abundance of Chlorine isotopes are not equal.
Answer: i) A and R are correct, R explains the A.
Answer: i) A and R are correct, R explains the A.
ShortVI. Short Answer Questions2 marks each
Q.1
Define: Relative atomic mass.
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✓ Answer
The Relative Molecular Mass of a molecule is the ratio between the mass of one molecule of the substance to 1/12th mass of an atom of Carbon-12.
Q.2
Write the different types of isotopes of oxygen and its percentage abundance.
▾
✓ Answer
Oxygen exists as a mixture of three stable isotopes in nature.
Isotope: Mass (amu) : % abundance
8o16 : 15.9949 : 99.757
8O17 : 16.9991 : 0.038
8O18 : 17.9992 : 0.205
Isotope: Mass (amu) : % abundance
8o16 : 15.9949 : 99.757
8O17 : 16.9991 : 0.038
8O18 : 17.9992 : 0.205
Q.3
Define: Atomicity
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✓ Answer
The number of atoms present in the molecule is called as its atomicity.
Q.4
Give any two examples for heterodiatomic molecules.
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✓ Answer
Examples of heterodiatomic HCl, CO.
Q.5
What is Molar volume of a gas?
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✓ Answer
(i) The volume occupied by one mole of any gas at S.T.P is called molar volume.
$(ii) One mole of any gas occupies = 22.4 litre or 22400 ml at S.T.P.$
$(ii) One mole of any gas occupies = 22.4 litre or 22400 ml at S.T.P.$
Q.6
Find the percentage of nitrogen in ammonia.
▾
✓ Answer
% of nitrogen in NH3
$= [ Mass of nitrogen / Molar mass of NH3 ] x 100$
$= (14/17) X 100 = 82%$
$= [ Mass of nitrogen / Molar mass of NH3 ] x 100$
$= (14/17) X 100 = 82%$
NumericalVII. Numerical Problems3 marks each
Q.1
Calculate the number of water molecule present in one drop of water which weighs 0.18 g.
▾
✓ Answer
$Mass of water = 0.18 g$
$No of moles = Mass / Molecular Mass$
$= 0.18 / 18$
$= 0.01 mole$
$Number of molecules = No of moles \times Avogadro number$
$= 0.01 \times 6.023 \times 1023$
$= 0.06023 \times 1023$
$= 6.023 \times 1021 molecules$
$No of moles = Mass / Molecular Mass$
$= 0.18 / 18$
$= 0.01 mole$
$Number of molecules = No of moles \times Avogadro number$
$= 0.01 \times 6.023 \times 1023$
$= 0.06023 \times 1023$
$= 6.023 \times 1021 molecules$
Q.2
N2 + 3 H2 \rightarrow 2 NH3
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✓ Answer
(The atomic mass of nitrogen is 14, and that of hydrogen is 1)
1 mole of nitrogen ( 28 g) +
3 moles of hydrogen ( 6 g) \rightarrow
2 moles of ammonia ( 34 g)
Answer: 1 mole of nitrogen ( 28 g) + 3 moles of hydrogen ( 6 g) \rightarrow 2 moles of ammonia ( 34 g)
Answer: 28, 6, 34.
1 mole of nitrogen ( 28 g) +
3 moles of hydrogen ( 6 g) \rightarrow
2 moles of ammonia ( 34 g)
Answer: 1 mole of nitrogen ( 28 g) + 3 moles of hydrogen ( 6 g) \rightarrow 2 moles of ammonia ( 34 g)
Answer: 28, 6, 34.
Q.3
Calculate the number of moles in $i) 27g of Al ii) 1.51 \times 1023 molecules of NH4Cl$
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✓ Answer
(i) 27 g of Al
$No of moles = Mass / Atomic Mass$
$27 / 27 = 1 mole$
(ii) 1.51 x 1023 molecules of NH4C1
$No of moles = Number of molecules / Avogadro number$
$= (1.51 \times 1023) / (6.023 \times 1023)$
$= 0.25 mole$
$No of moles = Mass / Atomic Mass$
$27 / 27 = 1 mole$
(ii) 1.51 x 1023 molecules of NH4C1
$No of moles = Number of molecules / Avogadro number$
$= (1.51 \times 1023) / (6.023 \times 1023)$
$= 0.25 mole$
Q.4
Give the salient features of “Modern atomic theory”.
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✓ Answer
‘The main postulates of modern atomic theory’ are as follows:
(i) An atom is no longer indivisible
(ii) Atoms of the same element may have different atomic mass. Example - isotopes l7Cl35 , l7Cl37.
(iii) Atoms of different elements may have same atomic masses. Example - Isobars 18Ar40, 20Ca40.
(iv) Atoms of one element can be transmuted into atoms of other elements.
(v) Atoms may not always combine in a simple whole number ratio. Eg: Glucose C6H12O6.
(vi) Atom is the smallest particle that take part in a chemical reaction.
$(vii) Mass of an atom can be converted into energy. E = mc2.$
(i) An atom is no longer indivisible
(ii) Atoms of the same element may have different atomic mass. Example - isotopes l7Cl35 , l7Cl37.
(iii) Atoms of different elements may have same atomic masses. Example - Isobars 18Ar40, 20Ca40.
(iv) Atoms of one element can be transmuted into atoms of other elements.
(v) Atoms may not always combine in a simple whole number ratio. Eg: Glucose C6H12O6.
(vi) Atom is the smallest particle that take part in a chemical reaction.
$(vii) Mass of an atom can be converted into energy. E = mc2.$
Q.5
Derive the relationship between Relative molecular mass and Vapour density.
▾
✓ Answer
Answer: Relative molecular mass :
(i) (Hydrogen scale): The Relative Molecular Mass of a gas or vapour is the ratio between the mass of one molecule of the gas or vapour to mass of one atom of Hydrogen.
(ii) Vapour Density : Vapour density is the ratio of the mass of a certain volume of a gas or vapour, to the mass of an equal volume of hydrogen, measured under the same conditions of temperature and pressure.
Vapour Density
$(VD) = Mass of given volume of gas or vapour at S.T.P / Mass of same volume of hydrogen$
According to Avogadro's law, equal volumes of all gases contain equal number of molecules.
$Thus, let the number of molecules in one volume = n, then$
(i) (Hydrogen scale): The Relative Molecular Mass of a gas or vapour is the ratio between the mass of one molecule of the gas or vapour to mass of one atom of Hydrogen.
(ii) Vapour Density : Vapour density is the ratio of the mass of a certain volume of a gas or vapour, to the mass of an equal volume of hydrogen, measured under the same conditions of temperature and pressure.
Vapour Density
$(VD) = Mass of given volume of gas or vapour at S.T.P / Mass of same volume of hydrogen$
According to Avogadro's law, equal volumes of all gases contain equal number of molecules.
$Thus, let the number of molecules in one volume = n, then$
A&RV. Assertion & Reasoning
D at S.T.P
$= Mass of ´n´molecules of gas or vapour at S.T.P / Mass of n molecules of hydrogen$
Cancelling 'n' which is common, we get
$V.D = Mass of 1 molecule of gas or vapour at S.T.P / Mass of 1 molecule of hydrogen$
However, since hydrogen is diatomic
$V.D = Mass of 1 molecule of a gas or vapour at S.T.P / Mass of 2 atoms of hydrogen$
$V.D = ( Mass of 1 molecule of a gas or vapour at S.T.P ) / ( 2 \times Mass of 1 atom of hydrogen)$
$V.D = Molecular Mass / 2$
$∴ 2 \times vapour density = Molecular Mass of a gas$
$Molecular Mass = 2 \times Vapour density.$
$= Mass of ´n´molecules of gas or vapour at S.T.P / Mass of n molecules of hydrogen$
Cancelling 'n' which is common, we get
$V.D = Mass of 1 molecule of gas or vapour at S.T.P / Mass of 1 molecule of hydrogen$
However, since hydrogen is diatomic
$V.D = Mass of 1 molecule of a gas or vapour at S.T.P / Mass of 2 atoms of hydrogen$
$V.D = ( Mass of 1 molecule of a gas or vapour at S.T.P ) / ( 2 \times Mass of 1 atom of hydrogen)$
$V.D = Molecular Mass / 2$
$∴ 2 \times vapour density = Molecular Mass of a gas$
$Molecular Mass = 2 \times Vapour density.$
HOTIX. Higher Order Thinking3 marks each
Q.1
Calcium carbonate is decomposed on heat-ing in the following reaction
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✓ Answer
CaCO3 \rightarrow CaO + CO2
i. How many moles of Calcium carbonate are involved in this reaction?
ii. Calculate the gram molecular mass of calcium carbonate involved in this reaction
iii. How many moles of CO2 are there in this equation?
CaCO3 \rightarrow CaO + CO2
$(i) No of moles of CaCO3 involved in this reaction = 1$
$(ii) Molar mass of calcium = 40$
$Molar mass of carbon = 12$
$Molar mass of oxygen = 16$
$Gram molecular mass of CaCO3 = Molar mass of calcium + Molar mass of carbon + Molar mass of oxygen x 3$
$= 40 + 12 + (16 x 3)$
$= 100 g / mol.$
$(iii) No of moles of CO2 in the reaction = 1$
i. How many moles of Calcium carbonate are involved in this reaction?
ii. Calculate the gram molecular mass of calcium carbonate involved in this reaction
iii. How many moles of CO2 are there in this equation?
CaCO3 \rightarrow CaO + CO2
$(i) No of moles of CaCO3 involved in this reaction = 1$
$(ii) Molar mass of calcium = 40$
$Molar mass of carbon = 12$
$Molar mass of oxygen = 16$
$Gram molecular mass of CaCO3 = Molar mass of calcium + Molar mass of carbon + Molar mass of oxygen x 3$
$= 40 + 12 + (16 x 3)$
$= 100 g / mol.$
$(iii) No of moles of CO2 in the reaction = 1$
NumericalVII. Numerical Problems3 marks each
Q.1
How many grams are there in the following?
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✓ Answer
i. 2 moles of hydrogen molecule, H2
ii. 3 moles of chlorine molecule, Cl2
iii. 5 moles of sulphur molecule, S8
iv. 4 moles of phosphorous molecule, P4
(i) 2 moles of hydrogen molecule
$Molecular mass of H2 = 2$
$Mass = no of moles x molecular mass$
$= 2 x 2 = 4$
(ii) 3 moles of Chlorine molecule
$Molecular mass of Cl2 = 70.9$
$Mass = no of moles x molecular mass$
$= 3 x 70.9$
$= 212.7 g$
(iii) 5 moles of sulphur molecule S8
Molecular mass of sulphur molecule
$= 32 x 8 = 256$
$Mass = no of moles x molecular mass$
$= 5 x 256 = 1280 g$
(iv) 4 molecules of P4 molecule
$Molecular mass of P4 molecule = 31 x 4 = 124$
$Mass = no of moles x molecular mass$
$= 4 x 124 = 496 g$
ii. 3 moles of chlorine molecule, Cl2
iii. 5 moles of sulphur molecule, S8
iv. 4 moles of phosphorous molecule, P4
(i) 2 moles of hydrogen molecule
$Molecular mass of H2 = 2$
$Mass = no of moles x molecular mass$
$= 2 x 2 = 4$
(ii) 3 moles of Chlorine molecule
$Molecular mass of Cl2 = 70.9$
$Mass = no of moles x molecular mass$
$= 3 x 70.9$
$= 212.7 g$
(iii) 5 moles of sulphur molecule S8
Molecular mass of sulphur molecule
$= 32 x 8 = 256$
$Mass = no of moles x molecular mass$
$= 5 x 256 = 1280 g$
(iv) 4 molecules of P4 molecule
$Molecular mass of P4 molecule = 31 x 4 = 124$
$Mass = no of moles x molecular mass$
$= 4 x 124 = 496 g$
Q.2
Calculate the % of each element in calci-um carbonate. (Atomic mass: C-12, O-16, Ca -40)
▾
✓ Answer
$Molar mass of CaCO3 = 40 + 12 + (3 x 16)$
$= 100$
$% of C in CaCO3 = [ Mass of carbon / Molar Mass of CaCO3 ] x 100$
$= [12/100] x 100 =12%$
$% of O in CaCO3 = [ Mass of oxygen / Molar Mass of CaCO3 ] x 100$
$= [16x3 / 100] x 100 =48%$
$% of Ca in CaCO, = [ Mass of calcium / Molar Mass of CaCO3 ] x 100$
$= [40/100] x 100 =40%$
$= 100$
$% of C in CaCO3 = [ Mass of carbon / Molar Mass of CaCO3 ] x 100$
$= [12/100] x 100 =12%$
$% of O in CaCO3 = [ Mass of oxygen / Molar Mass of CaCO3 ] x 100$
$= [16x3 / 100] x 100 =48%$
$% of Ca in CaCO, = [ Mass of calcium / Molar Mass of CaCO3 ] x 100$
$= [40/100] x 100 =40%$
Q.3
Calculate the % of oxygen in Al2(SO4)3. (Atomic mass: Al-12, O-16, S -32)
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✓ Answer
Molar mass of Al2 (SO4)3
$= (27 x 2) + (32 x 3) + (16 x 12)$
$= 342$
% of Oxygen in Al2 (SO4)3
$= [ Mass of oxygen in compound / Molar mass of Al2 (SO4)3 ] x 100$
$= [192/342] x 100$
$= 56.1 %$
$= (27 x 2) + (32 x 3) + (16 x 12)$
$= 342$
% of Oxygen in Al2 (SO4)3
$= [ Mass of oxygen in compound / Molar mass of Al2 (SO4)3 ] x 100$
$= [192/342] x 100$
$= 56.1 %$
Q.4
Calculate the % relative abundance of B -10 and B -11, if its average atomic mass is 10.804 amu.
▾
✓ Answer
Let,
$% Abundance of B10 = x$
$% Abundance of B-11 = y$
$x+y = 100$
$x = 100-y$
$Atomic mass unit of B10 = 10.01294$
$Atomic mass unit of B11 = 11.009305$
$amu 10.804 = [10.01294 x +11.009305 y] / 100$
$Substitute x= 100 - y$
$10.804 = [10.01294 (100 - y) +11.009305 y] / 100$
$1080.4 = 1001.294 + 0.996365 y$
$79.106 = 0.996365 y$
$y = 79.364%$
$x = 20.636%$
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$% Abundance of B10 = x$
$% Abundance of B-11 = y$
$x+y = 100$
$x = 100-y$
$Atomic mass unit of B10 = 10.01294$
$Atomic mass unit of B11 = 11.009305$
$amu 10.804 = [10.01294 x +11.009305 y] / 100$
$Substitute x= 100 - y$
$10.804 = [10.01294 (100 - y) +11.009305 y] / 100$
$1080.4 = 1001.294 + 0.996365 y$
$79.106 = 0.996365 y$
$y = 79.364%$
$x = 20.636%$
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