AP 9th Class Physics 3rd Lesson Questions and Answers Atoms and Molecules

AP State Board new syllabus AP Board Solutions Class 9 Physics 3rd Lesson Atoms and Molecules Questions and Answers.

AP 9th Class Physical Science 3rd Lesson Questions and Answers Atoms and Molecules

9th Class Physics 3rd Lesson Atoms and Molecules Questions and Answers (Exercise)

Question 1.
A 0.24 g sample of compound of 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.
Answer:
Weight of O = 0.144 g
Weight of sample =0.24g
% of Oxygen in the sample =

% of Boron = 100 – 60 = 40%

Question 2.
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?
Answer:
3 g of Carbon reacts with 8 g of Oxygen.
So, when 50 g of Oxygen provided 50 – 8 = 42 g left unutilised.
So, law of Chemical combination satisfied.

Question 3.
What are polyatomic ions? Give examples.
Answer:
Polyatomic ions: Two (or) more atoms to form a charged particle is called polyatomic ions. Examples :
Nitrate \(\left(\mathrm{NO}_2^{-}\right) \), Carbonate \(\left(\mathrm{CO}_3^{2-}\right)\) , Sulphate \(\left(\mathrm{SO}_4^{2-}\right)\), Phosphate \(\left(\mathrm{PO}_4^{3-}\right)\)

Question 4.
Write the chemical formulae of the following.
a) Magnesium chloride
b) Calcium oxide
c) Copper nitrate
d) Aluminium chloride
e) Calcium carbonate
Answer:

Question 5.
Give the names of the elements present in the following compounds.
a) Quick lime
b) Hydrogen bromide
c) Baking powder
d) Potassium sulphate
Answer:
Compound Elements — Elements
a) Quick lime (CaO) — Calcium, Oxygen
b) Hydrogen bromide (HBr) — Hydrogen, Bromine
c) Baking powder Sodium, Hydrogen, Tartaric Acid, — (Sodium Hydrogen Tartarate) Carbon & Oxygen
d) Potassium sulphate (K₂SO₄) — Potassium, Sulphur, Oxygen

Question 6.
Calculate the molar mass of the following substances.
a) Ethyne, C₂H₂
b) Sulphur molecule, S₈
c) Phosphorus molecule, P₄ (Atomic mass of phosphorus = 31)
d) Hydrochloric acid, HCl
e) Nitric acid, HNO₃
Answer:
Molar Mass :
a) Ethyne -C₂H₂ :
It contains Carbon & Hydrogen
Mass of C =12 g
Mass of H=1 g
Molecular Mass of C₂H₂=2 ×12+2×1
= 24+2
= 26 g

b) Sulphur – S₈ :
Mass of S = 32 g
Molecular Mass of S₈ = 8 × 32
= 256 g

c) Phosphorus – P₄ :
Mass of Phosphorus P = 31 g
Molecular Mass of
P₄ = 4 × 31 = 124 g

d) Hydrochloric acid – HCl :
It contains Hydrogen and Chlorine atoms.
Mass of H = 1 g
Mass of Cl = 35.5 g
Molecular Mass of HCl
= 1 ×1+1× 35.5
= 1+35.5=36.5g

e) Nitric Acid – HNO₃ :
It contains Hydrogen, Nitrogen & Oxygen.
Mass of H = 1 g; Mass of N=14 g;
Mass of O = 16g
Molecular Mass of HNO₃
= 1 × 1+1× 14+3 × 16
= 1+14+48 = 63 g

9th Class PS 3rd Lesson Questions and Answers (InText)

Question 1.
In a reaction, 5.3 g of sodium carbonate reacted with 6 g of acetic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium acetate. Show that these observations are in agreement with the law of conservation of mass. sodium carbonate + acetic acid → sodium acetate + carbon dioxide + water
Answer:
Mass of Sodium Carbonate + Mass of Acetic acid
= Mass of Carbondioxide + water + Mass of Sodium acetate
5.3 g+6 g = 2.2 g + 0.9g + 8.2 g
11.3 g = 11.3 g
The mass of reactants is equal to the mass of the products, so it proves law of conservation of mass.

Question 2.
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 ?
Answer:
Ratio of H : O by mass in water is :-
Hydrogen: Oxygen =1: 8
Then 3 g of hydrogen : ? g of oxygen
∴ 1: 8 = 3 : x
x=8 × 3 ⇒ x = 24 g
∴ 24 g of oxygen gas would be required to react completely with 3 g of hydrogen gas.

Question 3.
Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Answer:
(ii) – Postulate
Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction.

Question 4.
Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Answer:
(v) – Postulate
Atoms combine in the ratio of small whole numbers to form compounds.

Question 5.
Define the atomic mass unit.
Answer:
Atomic mass unit : The atomic mass unit is equal to \(\frac{1}{12}\)th the mass of an atom of carbon-12.

Question 6.
Why is it not possible to see an atom with naked eyes?
Answer:
Atom is too small to see with naked eye.

Question 7.
Write down the formulae of
i) Sodium Oxide
ii) Aluminium Chloride
iii) Sodium Sulphide,
iv) Magnesium Hydroxide.
Answer:

Question 8.
Write down the names of compounds represented by the following formulae :
i) Al₂(SO₄)₃
ii) CaCl₂
iii) K₂SO₄
iv) KNO₃
v) CaCO₃
Answer:
i) Al₂(SO₄)₃ – Aluminium Sulp̊hate
ii) CaCl₂ – Calcium Chloride
iii) K₂SO₄ – Potassium Sulphate
iv) KNO₃ – Potassium Nitrate
v) CaCO₃ – Calcium Carbonate

Question 9.
What is meant by the term chemical formula?
Answer:
Chemical formulae is a symbolic representation of atom of elements present in a compound. Ex: CH₄ – Methane.

Question 10.
How many atoms are present in a
(i) H₂S molecule and
(ii) \(\mathrm{PO}_4{ }^{3-}\) ion ?
Answer:
Atoms are present in
i) H₂S₃ atoms,
ii) \(\mathrm{PO}_4{ }^{3-}\) – 5 atoms

Question 11.
Calculate the molecular masses of H₂, O₂, Cl₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, CH₃OH.
Answer:
Molecular Mass
i) H₂ – Hydrogen gas :
Atomic mass of H = 1 g
Molecular mass of H₂= 2 × 1=2 g

ii) O₂ – Oxygen gas :
Atomic mass of O=16 g
Molecular mass of O₂= 2 × 16 = 32 g

iii) Cl₂ – Chlorine gas :
Atomic mass of Cl = 35.5 g
Molecular mass of Cl₂ = 2 × 35.5 = 71 g

iv) CO₂ – Carbondioxide :
It contains Carbon & Oxygen atoms.
Atomic mass of C = 12 g
Atomic mass of O = 16 g
Molecular mass of CO₂
=1 × 12+2 × 16
=12+32=44 g

v) CH₄ – Methane :
It contains Carbon & Hydrogen atoms.
Atomic mass of C=12g
Atomic mass of H=1g
Molecular mass of CH₄
= 1 × 12+4 ×1
= 12+4=16g

vi) C₂H₆ – Ethane :
It contains Carbon  Hydrogen atoms.
Atomic mass of C = 12g
Atomic mass of H = 1g
Molecular mass of C₂H₆
= 2 × 12+1 ×6
= 24+6=30g

vii) C₂H₄ – Ethene :
It contains Carbon & Hydrogen atoms.
Atomic mass of C=12g
Atomic mass of H=1g
Molecular mass of C₂H₄
= 2 × 12+4×1
= 24+4=28 g

viii) NH₃ – Ammonia :
It contains Nitrogen and Hydrogen atoms:
Atomic mass of N=14g
Atomic mass of H=1
Molecular mass of NH₃
= 1 × 14+3 × 1
= 14+3=17g

ix) CH₃OH – Methanol :
It contains Carbon, Hydrogen & Oxygen atoms.
Atomic mass of C=12g
Atomic mass of H=1g
Atomic mass of O=16 g
Molecular Mass of CH₃OH
= 1 × 12+1 × 16+4 × 1
= 12+16+4=32g

Question 12.
Calculate the formula unit masses of ZnO, Na₂O, K₂CO₃, given atomic masses of Zn=65u
Na=23u, K=39u, C =12u and O=16u.
Answer:
Formula Unit Mass :
i) ZnO – Zinc Oxide :
It contains Zinc & Oxygen atoms.
Atomic mass of Zn=65 u (or) Da
Atomic mass of O=16 u (or) Da
Formula unit mass of ZnO
= 1×65+1 × 16
= 65+16
= 81 Da (or) u

ii) Na₂O – Sodium Oxide :
It contains Sodium and Oxygen atoms.
Atomic mass of Na=23 u (or) Da
Atomic mass of O=16 u (or) Da
Formula unit mass of Na₂O
=2 × 23+1 × 16 = 46+16
= 62u (or) Da

iii) K₂CO₃ – Potassium Carbonate :
It contains Potassium, Carbon & Oxygen atoms.
Atomic mass of K = 39u (or) Da
Atomic mass of C=12u (or) Da
Atomic mass of O=16u (or) Da
Formula unit mass of K₂CO₃
= 2 × 39+1 × 12+3 ×16
= 78+12+48
= 138u (or) Da

Examples:

Question 1.
a) Calculate the relative molecular mass of water (H₂O)
b) Calculate the molecular mass of HNO₃.
Answer:
a) Atomic mass of hydrogen =1u, oxygen =16u
So, the molecular mass of water, which contains two atoms of hydrogen and one atom of oxygen is =2 ×1+1 × 16=18 u
b) The molecular mass of HNO₃= the atomic mass of H
+ the atomic mass of N+3 × the atomic mass of O
=1+14+48=63 u

Question 2.
Calculate the formula unit mass of CaCl₂.
Answer:
Atomic mass of Ca+(2 × atomic mass of Cl) = 40+2 × 35.5=40+71=111 u

9th Class Physical Science Chapter 3 Questions and Answers (Lab Activities)

(Page No. 52) Activity 3.1

Question 1.
Write an activity to prove the law of conservation of mass.
Answer:
Aim : To verify the law of conservation of mass.
Apparatus required: Conical flask, small ignition tube, cork, thread, chemicals, weighing machine.

Procedure :
1. Take one of the following sets, X and Y of chemicals –

2. Prepare separately a 5% solution of any one pair of substances listed under X and Y each in 10 ml in water.
3. Take a little amount of solution of Y in a conical flask and some solution of X in an ignition tube.
4. Hang the ignition tube in the flask carefully; see that the solutions do not get mixed. Put a cork on the flask (see Figure.)

Ignition tube containing solution of X, dipped in a conical flask containing solution of Y
5. Weigh the flask with its contents carefully.
6. Now tilt and swirl the flask, so that the solutions X and Y get mixed.
7. Weigh again.

Observations :
1. What happens in the reaction flask ?
Answer: X & Y undergoes a chemical change.

2. Do you think that a chemical reaction has taken place?
Answer: Yes. A chemical reaction has taken place.

3. Why should we put a cork on the mouth of the flask?
Answer: Cork is put on the mouth of the flask in order to avoid spillage of chemical substances.

4. Does the mass of the flask and its contents change ?
Answer: Mass of the flask will not change, but contents will change as they undergo a chemical change.

(Page No. 62) Activity 3.2

Question 2.
How do you calculate the ratio by number of atoms for water?
– Refer to Table 1 for ratio by mass of atoms present in molecules and Table 2 for atomic masses of elements. Find the ratio by number of the atoms of elements in the molecules of compounds given in Table 1.

Table 1: Molecules of some compounds

Compound	Combining Elements	Ratio by Mass
Water (H2O)	Hydrogen, Oxygen	1:8
Ammonia (NH3)	Nitrogen, Hydrogen	14:3
Carbon dioxide (CO2)	Carbon, Oxygen	3:8

The ratio by number of atoms for a water molecule can be found as follows:
For example:

Element	Ratio by mass	Atomic mass (u)	Mass ratio/ atomic mass	Simplest ratio
H	1	1		2
0	8	16	\(\frac{8}{16}=\frac{1}{2}\)	1

Thus, the ratio by number of atoms for water is H : O = 2:1.
Answer:
1. Ratio of Number of atoms of Water – H₂O

2. Ratio of Number of atoms of Ammonia – NH₃


3. Ratio of Number of atoms of Carbondioxide – CO₂