10th Science - Book Back Answers - Chemistry Unit 7 - English Medium Guides

 

SSLC / 10th - Science - Book Back Answers - Chemistry Unit 7 - English Medium

Tamil Nadu Board 10th Standard  Science - Chemistry Unit 7: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Unit 7 – Chemistry from the Tamil Nadu State Board 10th Standard  Science textbook. These detailed answers have been carefully prepared by our expert teachers at KalviTips.com.

    We have explained each answer in a simple, easy-to-understand format, highlighting important points step by step under the relevant subtopics. Students are advised to read and memorize these subtopics thoroughly. Once you understand the main concepts, you’ll be able to connect other related points with real-life examples and confidently present them in your tests and exams.

    By going through this material, you’ll gain a strong understanding of Chemistry Unit 7 along with the corresponding book back questions and answers (PDF format).

Question Types Covered:

• 1 Mark Questions: Choose the correct answer, Fill in the blanks, Identify the correct statement, Match the following 
• 2 Mark Questions: Answer briefly 
• 3, 4, and 5 Mark Questions: Answer in detail

All answers are presented in a clear and student-friendly manner, focusing on key points to help you score full marks.

All the best, Class 10 students! Prepare well and aim for top scores. Thank you!

Topic: Atoms and Molecules

I. Choose the correct 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
Answer Key:
(b) 1 atom of He

2. Which of the following is a triatomic molecule?
(a) Glucose
(b) Helium
(c) Carbon dioxide
(d) Hydrogen.
Answer Key:
(c) Carbon dioxide

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
Answer Key:
(b) 2.24 litre

4. Mass of 1 mole of Nitrogen atom is _____.
(a) 28 amu
(b) 14 amu
(c) 28 g
(d) 14 g.
Answer Key:
(b) 14 amu

5. Which of the following represents 1 amu?
(a) Mass of a C – 12 atom
(b) Mass of a hydrogen atom
(c) 1/12 th of the mass of a C – 12 atom
(d) Mass of O – 16 atom
Answer Key:
(c) 1/12 th of the mass of a C – 12 atom

6. Which of the following statement is incorrect?
(a) One 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.
Answer Key:
(a) One 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
Answer Key:
(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
Answer Key:
(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.
Answer Key:
(c) 32 g

10. 1 mole of any substance contains molecules.
(a) 6.023 × 10²³
(b) 6.023 × 10²³
(c) 3.0115 × 10²³
(d) 12.046 × 10²³
Answer Key:
(a) 6.023 × 10²³

II. Fill in the blanks

1. Atoms of different elements having ……… mass number, but ………. atomic numbers are called isobars.
2. Atoms of different elements having same number of ………. are called isotones.
3. Atoms of one element can be transmuted into atoms of other element by ………….
4. The sum of the numbers of protons and neutrons of an atom is called its …………
5. Relative atomic mass is otherwise known as …………
6. The average atomic mass of hydrogen is ……….. amu.
7. If a molecule is made of similar kind of atoms, then it is called ……….. atomic molecule.
8. The number of atoms present in a molecule is called its ………….
9. One mole of any gas occupies ………… ml at S.T.P
10. Atomicity of phosphorous is …………
Answer Key:
1. same, different
2. neutrons
3. artificial transmutation
4. mass number
5. standard atomic weight
6. 1.008
7. homo
8. atomicity
9. 22, 400
10. four

 

 

III. Match the following

Column I		Column II
A	8 g of O2	i	4 moles
B	4 g of H2	ii	0.25 moles
C	52 g of He	iii	2 moles
D	112 g of N2	iv	0.5 moles
E	35.5 g of Cl2	v	13 moles

Answer Key:

Column I		Column II
A	8 g of O2	ii	0.25 moles
B	4 g of H2	iii	2 moles
C	52 g of He	v	13 moles
D	112 g of N2	i	4 moles
E	35.5 g of Cl2	iv	0.5 moles

IV.  True or False (If false give the correct statement)

1. Two elements sometimes can form more than one compound.
2. Nobel gases are diatomic.
3. The gram atomic mass of an element has no unit.
4. 1 mole of Gold and Silver contain same number of atoms.
5. Molar mass of CO2 is 42 g.
Answer Key:
1. True
2. False – Noble gases are Monoatomic.
3. False – The unit of gram atomic mass of an element is gram.
4. True
5. False – Molar mass of CO2 is 44 g.

 

V.  Assertion & Reasoning

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: Atomic mass of aluminium is 27
Reason: An atom of aluminium is 27 times heavier than 1/12 th of the mass of the C-12 atom.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer Key:
(a) Assertion and Reason are correct, Reason explains the Assertion.

2. Assertion: The Relative Molecular Mass of Chlorine is 35.5 a.m.u.
Reason: The natural abundance of Chlorine isotopes are not equal.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer Key:
(c) Assertion is wrong, Reason is correct.

 

 

VI. Short answer questions

1. Define: Relative atomic mass.
Answer Key:
Relative atomic mass of an element is the ratio between the average mass of its isotopes to $\frac{\mathbf{1}}{{\mathbf{12}}^{\mathbf{t}\mathbf{h}}}$ part of the mass of a carbon-12 atom. It is denoted as Ar.
[OR]

${\mathbf{A}}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{A}\mathbf{v}\mathbf{e}\mathbf{r}\mathbf{a}\mathbf{g}\mathbf{e}\mathbf{ }\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{i}\mathbf{s}\mathbf{o}\mathbf{t}\mathbf{o}\mathbf{p}\mathbf{e}\mathbf{s}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{e}\mathbf{l}\mathbf{e}\mathbf{m}\mathbf{e}\mathbf{n}\mathbf{t}}{\frac{\mathbf{1}}{{\mathbf{12}}^{\mathbf{t}\mathbf{h}}}\mathit{o}\mathit{f}\mathbf{ }\mathit{t}\mathit{h}\mathit{e}\mathbf{ }\mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{o}\mathit{n}\mathit{e}\mathbf{ }\mathit{c}\mathit{a}\mathit{r}\mathit{b}\mathit{o}\mathit{n}\mathbf{ }\mathbf{-}\mathbf{12}\mathit{a}\mathit{t}\mathit{o}\mathit{m}}$
 
2. Write the different types of isotopes of oxygen and its percentage abundance.
Answer Key:
Oxygen has three stable isotopes. They are

	Mass	%abudance
	15.9949	99.757
	16.9991	0.038
	17.9992	0.205

 

3. Define Atomicity.
Answer Key:
The number of atoms present in the molecule is called its ‘Atomicity’.

4. Give any two examples for heteroatomic molecules.
Answer Key:
HI, HCl, CO, HBr, HF.
 
5. What is Molar volume of a gas?
Answer Key:
One mole of any gas occupies 22.4 litres.
(or)
22400 ml at S.T.R This volume is called as molar volume.

6. Find the percentage of nitrogen in ammonia.
Answer Key:
Molar mass of NH3 = 1(14) + 3(1) = 17 g
$\mathrm{Mass\; of\; \%\; of\; Nitrogen}=\frac{\mathrm{Mass} \mathrm{of} \mathrm{Nitrogen} \mathrm{in} \mathrm{the} \mathrm{compound}}{\mathrm{Molar} \mathrm{mass} \mathrm{of} \mathrm{hte} \mathrm{compound}}\times 100$
$\frac{\mathrm{=14}}{17}\times \mathrm{100=\; 82.35\%}$

$\mathrm{ }$

VII. Long answer questions

1. Calculate the number of water molecule present in one drop of water which weighs 0.18 g.
Answer Key:
The molecular mass of water (H₂O) is 18.
18 g of water molecule = 1 mole.
0. 18 g of water = $\frac{1}{18}\times 0.18$ = 0.01 mole.
1 mole of water (Avogadro’s number) contains 6.023 × 10²³ water molecules.
0. 01 mole of water contain $\frac{6.023\times {10}^{23}}{1}\times 0.01$ = 6.023 ×  10²¹ molecules.

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)
Answer Key:
1 mole of nitrogen (28 g) +
3 moles of hydrogen (6 g) →
2 moles of ammonia (34 g)

3. Calculate the number of moles in
(i) 27 g of Al;
(ii) 1.51 × 10²³ molecules of NH4Cl.
Answer Key:
(i) 27 g of Al
Given mass atomic mass = $\frac{\mathrm{Given} \mathrm{Mass} }{\mathrm{Atomic} \mathrm{Mass} }=\frac{27}{27}$
= 1 mole
(ii) 1.51 x 1023 molecules of NH4Cl
Number of moles
$=\frac{\mathrm{Number} \mathrm{of} \mathrm{molecules} \mathrm{given}}{6.023\times {10}^{23}}$

$=\frac{1.51\times {10}^{23}}{6.023\times {10}^{23}}=0.25 \mathrm{moles}$

$\mathrm{ }$
4. Give the salient features of “Modern atomic theory”.
Answer Key:
The salient features of “Modem atomic theory” are,
1. An atom is no longer indivisible.
2. Atoms of the same element may have different atomic mass.
3. Atoms of different elements may have the same atomic masses.
4. Atoms of one element can be transmuted into atoms of other elements. In other words, an atom is no longer indestructible.
5. Atoms may not always combine in a simple whole-number ratio.
6. Atom is the smallest particle that takes part in a chemical reaction.
7. The mass of an atom can be converted into energy [E = mc²].
 
5.Derive the relationship between Relative molecular mass and Vapour density.
Answer Key:
Relative molecular mass : 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.
Vapour density : Vapour density is the ratio of the mass of 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.

According to Avogadro’s law equal volumes of all gases contain equal number of molecules.
Let the number of molecules in one volume = n, then

When cancelling ‘n’ which is common at STP, we get
$\mathbf{V}\mathbf{.}\mathbf{D}\mathbf{=}\frac{\mathit{M}\mathit{a}\mathit{s}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathbf{1}\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{c}\mathit{u}\mathit{l}\mathit{e}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{g}\mathit{a}\mathit{s}\mathbf{ }\mathit{o}\mathit{r}\mathbf{ }\mathit{v}\mathit{a}\mathit{p}\mathit{o}\mathit{u}\mathit{r}\mathbf{ }\mathit{a}\mathit{t}\mathbf{ }\mathit{S}\mathit{T}\mathit{P}}{\mathit{S}\mathit{T}\mathit{P}\mathit{M}\mathit{a}\mathit{s}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathbf{1}\mathbf{ }\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{c}\mathit{u}\mathit{l}\mathit{e}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{h}\mathit{y}\mathit{d}\mathit{r}\mathit{o}\mathit{g}\mathit{e}\mathit{n}}$
Since hydrogen is diatomic,

2 × Vapour density = Relative Molecular mass of a gas
[OR]
Relative Molecular Mass = 2 × Vapour density

 

VIII. HOT question

1.Calcium carbonate is decomposed on heating in the following reaction
CaCO₃ → CaO + CO₂
1. How many moles of Calcium carbonate is involved in this reaction?
2. Calculate the gram molecular mass of calcium carbonate involved in this reaction.
3. How many moles of CO₂ are there in this equation?
Answer Key:
CaCO₃ → CaO + CO₂
1. 1 mole of CaCO₃ is involved in this reaction.
2. Gram molecular mass of calcium carbonate
CaCO₃ = (40 + 12 + 3 × 16) = 52 + 48 = 100 g
3. 1 mole of CO₂ is in this equation.

 

IX.  Solve the following problems

1. How many grams are there in the following?
(i) 2 moles of a hydrogen molecule, H₂
(ii) 3 moles of chlorine molecule, Cl₂
(iii) 5 moles of sulphur molecule, S₈
(iv) 4 moles of a phosphorous molecule, P4
Answer Key:
(i) 2 moles of a hydrogen molecule, H₂
Mass of 1 mole of hydrogen molecule = 2 g
Mass of 2 moles of hydrogen molecule = 2 × 2 = 4 g.
(ii) 3 moles of chlorine molecule, Cl₂
Mass of 1 mole of chlorine molecule = 71 g
Mass of 3 moles of chlorine molecules = 71 × 3 = 213 g.
(iii) 5 moles of sulphur molecule, S₈
Mass of 1 mole of sulphur molecule = 32 g
Mass of 5 moles of sulphur molecules = 32 × 5 = 160 g.
(iv) 4 moles of the phosphorous molecule, P₄
Mass of 1 mole of phosphorous molecule = 30.97 g
Mass of 4 moles of phosphorous molecules = 30.97 × 4 = 123.88 g.

Question 2.
Calculate the % of each element in calcium carbonate. (Atomic mass: C – 12, O – 16, Ca – 40)
Answer Key:

Formula to find % of each element

$\mathbf{=}\frac{\mathbf{M}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{e}\mathbf{l}\mathbf{e}\mathbf{m}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{ }\mathbf{i}\mathbf{n}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{c}\mathbf{o}\mathbf{m}\mathbf{p}\mathbf{o}\mathbf{u}\mathbf{n}\mathbf{d}}{\mathbf{M}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{ }\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{c}\mathbf{o}\mathbf{m}\mathbf{p}\mathbf{o}\mathbf{u}\mathbf{n}\mathbf{d}}\mathbf{\times }\mathbf{100}$

$\mathbf{M}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{ }\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{C}\mathbf{a}\mathbf{C}{\mathbf{O}}_{\mathbf{3}}\mathbf{=}\mathbf{100}$

$\mathbf{\%}\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{c}\mathbf{a}\mathbf{=}\frac{\mathbf{M}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{C}\mathbf{a}\mathbf{ }\mathbf{i}\mathbf{n}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{c}\mathbf{o}\mathbf{m}\mathbf{p}\mathbf{o}\mathbf{u}\mathbf{n}\mathbf{d}}{\mathbf{M}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{ }\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{C}\mathbf{a}\mathbf{C}{\mathbf{O}}_{\mathbf{3}}}\mathbf{\times }\mathbf{100}$

$\mathbf{\%}\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{C}\mathbf{=}\frac{\mathbf{M}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{C}\mathbf{ }\mathbf{i}\mathbf{n}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{c}\mathbf{o}\mathbf{m}\mathbf{p}\mathbf{o}\mathbf{u}\mathbf{n}\mathbf{d}}{\mathbf{M}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{ }\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{C}\mathbf{a}\mathbf{C}{\mathbf{O}}_{\mathbf{3}}}\mathbf{\times }\mathbf{100}$

$\mathbf{\%}\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{O}\mathbf{=}\frac{\mathbf{M}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{O}\mathbf{ }\mathbf{i}\mathbf{n}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{c}\mathbf{o}\mathbf{m}\mathbf{p}\mathbf{o}\mathbf{u}\mathbf{n}\mathbf{d}}{\mathbf{M}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{ }\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{C}\mathbf{a}\mathbf{C}{\mathbf{O}}_{\mathbf{3}}}\mathbf{\times }\mathbf{100}$

 

Question 3.
Calculate the % of oxygen in Al2(SO4)3.
(Atomic mass: Al – 27, O – 16, S – 32)
Answer Key:

Formula:

$\mathbf{\%}\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{O}\mathbf{=}\frac{\mathbf{M}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{o}\mathbf{x}\mathbf{i}\mathbf{g}\mathbf{e}\mathbf{n}\mathbf{ }\mathbf{i}\mathbf{n}\mathbf{ }\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{ }\mathbf{c}\mathbf{o}\mathbf{m}\mathbf{p}\mathbf{o}\mathbf{u}\mathbf{n}\mathbf{d}}{\mathbf{M}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{ }\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{A}{\mathbf{l}}_{\mathbf{2}}\mathbf{(}\mathbf{S}{\mathbf{O}}_{\mathbf{4}}{\mathbf{)}}_{\mathbf{3}}}\mathbf{\times }\mathbf{100}$

Molar mass of Al2(SO4)3 = [2(Atomic mass of Al) + 3(Atomic mass of S) + 12(Atomic mass of O)]
= 2(27) + 3(32) + 12(16) = 342 g
% of Oxygen = $\frac{\mathbf{12}\mathbf{\left(}\mathbf{16}\mathbf{\right)}}{\mathbf{342}}\mathbf{\times }\mathbf{100}$ × 100 = 56.14%.

Question 4.

Calculate the % relative abundance of B – 10 and B – 11, if its average atomic mass is 10.804 amu.

Answer Key:

% of relative abundance can be calculated by the formula.

Average atomic mass of the element

= Atomic mass of 1st isotope × abundance of 1st isotope + Atomic mass of 2nd isotope × abundance of 2nd isotope

∴ Average atomic mass of Boron

= Atomic mass of B – 0 × abundance of B -10 + Atomic mass of B – 11 × abundance of B – 11

Let the abundance of B – 10 be ‘x’ and B – 11 be (1 – x)

So, 10.804 = 10 × x + 11 (1 – x)

10.804 = 10x + 11 – 11x

x = 11 – 10.804

x = 0.196

1 -x = 1 – 0.196 = 0.804
Therefore % abundance of B – 10 is 19.6% and B – 11 is 80.4%
[OR]
Let the % of the isotope B – 10 = x

Then the % of the isotope B – 11 = 100 – x

$\mathbf{A}\mathbf{v}\mathbf{e}\mathbf{r}\mathbf{a}\mathbf{g}\mathbf{e}\mathbf{ }\mathbf{a}\mathbf{t}\mathbf{o}\mathbf{m}\mathbf{i}\mathbf{c}\mathbf{ }\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{=}\frac{\mathbf{10}\mathbf{\times }\mathbf{x}\mathbf{+}\mathbf{11}\mathbf{(}\mathbf{100}\mathbf{-}\mathbf{x}\mathbf{)}}{\mathbf{100}}$

$\mathbf{10.804}\mathbf{=}\frac{\mathbf{10}\mathbf{x}\mathbf{+}\mathbf{1100}\mathbf{-}\mathbf{11}\mathbf{x}}{\mathbf{100}}$

$\mathbf{10.804}\mathbf{=}\frac{\mathbf{1100}\mathbf{-}\mathbf{x}}{\mathbf{100}}$

1100 – x = 1080.4

x = 19.6

% abundance of B – 10 = 19.6%

% abundance of B – 11 = 80.4%