9th science

Atoms Molecules

NCERT Solution

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: Percentage of boron in given compound `= (0.096/0.24) xx 100 = 40%`
Percentage of oxygen in given compound `= (0.144/0.24) xx 100 = 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 combination will govern your answer?

Answer: We know that atoms of different elements combine in definite proportions to make a particular compound. Hence, 3 g of carbon will always give 11 g of carbon dioxide.


Question 3: What are polyatomic ions? Give examples.

Answer: An ion with more than one atom is called a polyatomic ion. Examples: Sulphate (SO42-), Nitrate (NO32-)

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: (a) MgCl2, (b) CaO, (c) CuNO3, (d) AlCl3, (e) CaCO3

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: (a) Calcium, oxygen, (b) Hydrogen, bromine, (c) sodium, hydrogen, carbon, oxygen, (d) Potassium, sulphur, oxygen


Question 6: Calculate the molar mass of the following substances.

(a) Ethyne, C2H2

Answer: `2 xx 12 + 2 xx 1 = 26` u

(b) Sulphur molecule, S8

Answer: `8 xx 32 = 256` u

(c) Phosphorus molecule, P4 (Atomic mass of phosphorus = 31)

Answer: `4 xx 31 = 124`

(d) Hydrochloric acid, HCl

Answer: `1 xx 1 + 1 xx 35.5 = 36.5` u

(e) Nitric acid, HNO3

Answer: `1 xx 1 + 1 xx 14 + 3 xx 16 = 63` u


Question 7: What is the mass of:

(a) 1 mole of nitrogen atoms?

Answer: Atomic mass of nitrogen = 14 u
Hence, mass of 1 mole of nitrogen atoms = 14 g

(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?

Answer: Atomic mass of aluminium = 27 u
Hence, mass of 4 moles of aluminium atoms `= 4 xx 27 = 108`

(c) 10 moles of sodium sulphite (Na2SO3)?

Answer: Molecular mass of sodium sulphite
`= 2 xx 23 + 1 xx 32 + 3 xx 16 = 46 + 32 + 48 = 126 u`
Hence, mass of 10 mole of sodium sulphite `= 10 xx 126 = 1260 g`


Question 8: Convert into mole.

(a) 12 g of oxygen gas

Answer: Mass of 1 mole of oxygen = 32 g
Hence, number of mole in 12 g oxygen `= 12/32 = 0.375 M`

(b) 20 g of water

Answer: Molecular mass of water `= 2 xx 1 + 1 xx 16 = 18 u`
So, mass of 1 mole of water = 18 g
Hence, number of mole in 20 g of water `= 20/18 = 1.11 M`

(c) 22 g of carbon dioxide.

Answer: Molecular mass of carbon dioxide `= 1 xx 12 + 2 xx 16 = 44 u`
So, mass of 1 mole of carbon dioxide = 44 g
Hence, number of mole in 22 g of carbon dioxide `= 22/44 = 0.5 M`

Question 9: What is the mass of

(a) 0.2 mole of oxygen atoms?

Answer: Mass of 1 mole oxygen atoms = 16 g
Hence, mass of 0.2 mole of oxygen atoms `= 0.2 xx 16 = 3.2 g`

(b) 0.5 mole of water molecules?

Answer: Mass of 1 mole of water = 18 g
Hence, mass of 0.5 mole of water `= 0.5 xx 18 = 9 g`


Quesiton 10: Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.

Answer: Mass of 1 mole of sulphur `= 8 xx 32 = 256 g`
Hence, number of mole in 16 g of sulphur `= 16/256 = 0.625 M`

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)

Answer: Formula of aluminium oxide = Al2O3
Mass of aluminium atoms in aluminium oxide `= 2 xx 27 = 54 u`
Molecular mass of aluminium oxide `= 2 xx 27 + 3 xx 16 = 54 + 48 = 102 u`

Here; 2 mole aluminium ions are present in 102 g aluminium oxide
Hence, number of aluminium ions in 0.051 g aluminium oxide
`= (2/102) xx 0.051 = 2/2000` mole
`= (6.022 xx 10^23)/1000 = 6.022 xx 10^20`



Introduction

Atomic Mass

Existence of Atoms

Mole Concept

In Text Solution

NCERT Solution