9th std science chapter 3 noytes in english version 2020-21

Intext Exercise

Question 1:

In a reaction, 5.3g of sodium carbonate reacted with 6g of ethanoic acid. The products were

2.2g of carbon dioxide, 0.9g water and 8.2g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.

Sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water

Solution 1:

Sodium

 Ethanoi

⎯⎯→

Sodium

 Carbon

 Water

carbonate acid ethanoate dioxide

Na2 CO3 

Given,

CH3COOH

⎯⎯→CH3COONa 

CO2 

H 2O

Mass of sodium carbonate = 5.3 g

Mass of ethanoic acid = 6 g

Mass of sodium ethanoate = 8.2 g Mass of carbon dioxide = 2.2 g Mass of water = 0.9 g

Now, total mass before the reaction = (5.3 + 6) g = 11.3 g

And, total mass after the reaction = (8.2 + 2.2 + 0.9) g = 11.3 g

∴Total mass before the reaction = Total mass after the reaction

Hence, this is in agreement with the 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?

Solution 2:

In water,

H:O (by mass) =1:8

Then,

The mass of oxygen gas required to react completely with 1 g of hydrogen gas = 8 g.

So, the mass of oxygen gas required to react completely with 3 g of hydrogen gas = (8 × 3) g

= 24 g

Question 3:

Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

Solution 3:

The postulate of Dalton’s atomic theory which is based on the law of conservation of mass is: “Atoms are indivisible particles, which can neither be created nor destroyed in a chemical reaction.”

Question 4:

Which postulate of Dalton’s atomic theory can explain the law of definite proportions?

Solution 4:

The postulate of Dalton’s atomic theory which is based on the law of definite proportion is:

“The elements consist of atoms having fixed mass and that the number and kind of atoms of each element in a given compound is fixed.”

Intext Exercise 2

Question 1:

Define atomic mass unit.

Solution 1:

Mass unit equal to exactly one-twelfth  1 

 12th 

the mass of one atom of carbon-12 is called

one atomic mass unit. It is represented by as ‘a.m.u.’ or ‘u’.

Question 2:

Why is it not possible to see an atom with naked eyes?

Solution 2:

The size of an atom is too small which is not possible to see it with naked eyes. To see an atom we need advanced machines.

Intext Exercise 3

Question 1:

Write down the formulae of

(i) sodium oxide

(ii) aluminium chloride

(iii) sodium sulphide

(iv) magnesium hydroxide

Solution 1:

(i) Sodium oxide: Na2O

(ii) Aluminium chloride: AlCl3

(iii) Sodium sulphide: Na2S

(iv) Magnesium hydroxide: Mg(OH)2

Question 2:

Write down the names of compounds represented by the following formulae: (i) Al2(SO4)3

(ii) CaCl2 (iii) K2SO4 (iv) KNO3 (v) CaCO3

Solution 2:

(i) Al2(SO4)3 : Aluminium sulphate

(ii) CaCl2 : Calcium chloride

(iii) K2SO4 :Potassium sulphate

(iv) KNO3 : Potassium nitrate

(v) CaCO3 : Calcium carbonate

Question 3:

What is meant by the term chemical formula?

Solution 3:

The chemical formula of a compound is the symbolic representation of the composition of a compound. The chemical formula of a compound, tells us that the number and kinds of atoms of different elements that constitute the compound.

For example, from the chemical formula H2O of water, we come to know that one oxygen atom and two hydrogen atoms are chemically bonded together to form one molecule of the

compound, water.

Question 4:

How many atoms are present in a

(i) H2S molecule and (ii) PO3 − ion? Solution 4:

(i) In H2S molecule, total 3 atoms are present of which two hydrogen atoms and one sulphur

atom.

(ii) In

atoms.

PO3 − ion, total 5 atoms are present of which one phosphorus atom and four oxygen

Intext Exercise 4

Question 1:

Calculate the molecular masses of H2, O2, Cl2, CO2, CH4, C2H6, C2H4, NH3, CH3OH.

Solution 1:

Molecular mass of H2 = 2 × Atomic mass of H

= 2 × 1u

= 2 u

Molecular mass of O2 = 2 × Atomic mass of O

= 2 × 16 u

= 32 u

Molecular mass of Cl2 = 2 × Atomic mass of Cl

= 2 × 35.5 u

= 71 u

Molecular mass of CO2 = Atomic mass of C + 2 × Atomic mass of O

= (12 + 2 × 16) u

= 44 u

Molecular mass of CH4 = Atomic mass of C + 4 × Atomic mass of H

= (12 + 4 × 1) u

= 16 u

Molecular mass of C2H6 = 2 × Atomic mass of C + 6 × Atomic mass of H

= (2 × 12 + 6 × 1) u

= 30 u

Molecular mass of C2H4 = 2 × Atomic mass of C + 4 × Atomic mass of H

= (2 × 12 + 4 × 1) u

= 28 u

Molecular mass of NH3 = Atomic mass of N + 3 × Atomic mass of H

= (14 + 3 × 1) u

= 17 u

Molecular mass of CH3OH = Atomic mass of C + 4 × Atomic mass of H + Atomic mass of O

= (12 + 4 × 1 + 16) u

= 32 u

Question 2:

Calculate the formula unit masses of ZnO, Na2O, K2CO3, given atomic masses of Zn

= 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.

Solution 2:

Formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O

= (65 + 16) u

= 81 u

Formula unit mass of Na2O = 2 × Atomic mass of Na + Atomic mass of O

= (2 × 23 + 16) u

= 62 u

Formula unit mass of K2CO3 = 2 × Atomic mass of K + Atomic mass of C + 3 × Atomic mass of O

= (2 × 39 + 12 + 3 × 16) u

= 138 u

Intext Exercise 5

Question 1:

If one mole of carbon atoms weighs 12 gram, what is the mass (in gram) of 1 atom of carbon?

Solution 1:

One mole of carbon atoms weighs 12 g (Given) Thus, the mass of 1 mole of carbon atoms = 12 g

Then, mass of 6.022 1023

number of carbon atoms = 12 g

Therefore, mass of 1 atom of carbon  g

6.022 1023

 1.9927 10−23 g

Question 2:

Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na = 23 u, Fe = 56 u)?

Solution 2:

Atomic mass of Na = 23 u (Given) Then, gram atomic mass of Na = 23 g

Now, 23 g of Na contains = 6.022 1023

number of atoms

6.022 1023

Thus, 100 g of Na contains  100

number of Na atoms

= 2.6182 1024 number of Na atoms

Again, atomic mass of Fe = 56 u(Given) Then, gram atomic mass of Fe = 56 g

Now, 56 g of Fe contains = 6.022 1023

number of atoms

6.022 1023

Thus, 100 g of Fe contains  100

number of Fe atoms

 1.0753 1024

2.6182 1024  1.07531024

number of Fe atoms

Therefore, 100 grams of sodium contain a greater number of atoms than 100 grams of iron.

NCERT 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.

Solution 1:

Mass of boron = 0.096 g (Given) Mass of oxygen = 0.144 g (Given) Mass of sample = 0.24 g (Given)

The percentage of boron by weight in the compound  0.096  100%

0.24

= 40%

And, percentage of oxygen by weight in the compound  0.144  100%

0.24

= 60%

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 combinations will govern your answer?

Solution 2:

Carbon + Oxygen ⎯⎯→ Carbon dioxide

3 g of carbon reacts with 8 g of oxygen to produce 11 g of carbon dioxide.

If 3 g of carbon is burnt in 50 g of oxygen, then 3 g of carbon will react with 8 g of oxygen to form11 g of carbon dioxide.

The remaining (50 –8) = 42 g of oxygen will be left unreacted. The above answer is governed by the law of constant proportions.

Question 3:

What are polyatomic ions? Give examples?

Solution 3:

A polyatomic ion is a group of atoms carrying a charge either positive or negative. For

example, ammonium ion NH4  , hydroxide ion (OH ), carbonate ion , CO

 sulphate ion

SO2 − .

 - 2 −

Question 4:

Write the chemical formulae of the following: (a) Magnesium chloride

(b) Calcium oxide

(d) Aluminium chloride (e) Calcium carbonate Solution 4:

(a) Magnesium chloride: MgCl2

(b) Calcium oxide: CaO

Question 5:

Give the names of the elements present in the following compounds: (a) Quick lime

(b) Hydrogen bromide

(d) Potassium sulphate

Solution 5:

Compound Chemical formula Elements present

Quick lime CaO Calcium, oxygen

Hydrogen bromide HBr Hydrogen, bromine

Baking powder NaHCO3 Sodium, hydrogen, carbon, oxygen

Potassium sulphate K2SO4 Potassium, sulphur, oxygen

Question 6:

Calculate the molar mass of the following substances: (a) Ethyne, C2H2

(b) Sulphur molecule, S8

(e) Nitric acid, HNO3

Solution 6:

(a) Molar mass of ethyne, C2H2 = 2 × 12 + 2 × 1 = 28 g/mol

(b) Molar mass of sulphur molecule, S8 = 8 × 32 = 256 g/mol

(c) Molar mass of phosphorus molecule, P4 = 4 × 31 = 124 g/mol (d) Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5 g/mol (e) Molar mass of nitric acid, HNO3 = 1 + 14 + 3 × 16 = 63 g/mol

Question 7:

What is the mass of:

(a) 1 mole of nitrogen atoms?

(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)? (c) 10 moles of sodium sulphite (Na2SO3)?

Solution 7:

(a) The mass of 1 mole of N- atoms = 14 g.

(b) The mass of 4 moles of Al-atoms = (4 × 27) g [Atomic mass of Al= 27 u]

= 108 g

= 10 × [2 × 23 + 32 + 3 × 16] g (Atomic mass of Na = 23 u, Atomic mass of S=32 u Atomic mass of O = 16 u)

= 10 × 126 g = 1260 g

Question 8:

Convert into mole.

(a) 12 g of oxygen gas

(b) 20 g of water

Solution 8:

(a) 32 g of oxygen gas = 1 mole

Then, 12 g of oxygen gas = 12

moles

= 0.375 mole

(b) 18 g of water (H2O) = 1 mole

Then, 20 g of water (H2O) =

moles = 1.11 moles (approx.)

Then, 22 g of carbon dioxide (CO ) = 22

mole = 0.5 moles

Question 9:

What is the mass of:

(a) 0.2 mole of oxygen atoms?

(b) 0.5 mole of water molecules?

Solution 9:

(a) Mass of 1 mole of oxygen atoms = 16 g

Then, mass of 0.2 mole of oxygen atoms = 0.2 × 16g = 3.2 g

(b) Mass of 1 mole of water molecules (H2O) = 18 g

Then, mass of 0.5 mole of water molecules (H2O) = 0.5 × 18 g = 9 g

Question 10:

Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.

Solution 10:

1 mole of solid sulphur (S8) = 8 × 32 g = 256 g

i.e., 256 g of solid sulphur contains = 6.022 × 1023 molecules

6.022 1023

Then, 16 g of solid sulphur contains

 16

256

molecules

= 3.76 × 1022 molecules

Question 11:

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)

Solution 11:

Given, atomic mass of Al = 27 u, atomic mass of O = 16 u

1 mole of aluminium oxide (Al2O3) = (2 × 27 + 3 × 16) = 102 g i.e., 102 g of Al2O3 = 6.022 × 1023 molecules of Al2O3

6.022 1023

Then, 0.051 g of Al2O3 contains =

102

 0.051 molecules of Al2O3

= 3.011 × 1020 molecules of Al2O3

The number of aluminium ions (Al3+) present in 1 molecule of aluminium oxide is 2. Therefore, the number of aluminium ions (Al3+) present in 3.011 × 1020 molecules (0.051 g) of aluminium oxide (Al2O3) = 2 × 3.011 × 1

Thanks for reading 9th std science chapter 3 noytes in english version 2020-21

Tuesday, 4 May 2021

9th std science chapter 3 noytes in english version 2020-21

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