Includes solved exercises, review questions, MCQ, keyboard questions, and chapter reviews. Class 9 Chemistry Note for kpk mardan boards Class 9 Chemistry Notes Short Q, Long Q Chapter 1.
Chemistry, a subject that has many concepts that begin with a single definition of the smallest particle of the so-called “atom” in our universe. Usually in the eighth grade, we all know that we have to study only one book which combines chemistry, physics, and biology called science. But, as we move into ninth grade, things get more complicated. We got three books instead of just one.
And, if you ask me, alchemy is the most complex (maybe not for you). Then again, ninth-grade chemistry notes are enough. Understanding chemistry means memorizing many concepts. Also, the concepts of any subject that are usually taught in the ninth grade are the basis on which you can develop in that particular subject. Therefore, we are providing you the best chemistry notes for class 9.
Class 9 Chemistry Notes Short Q
Q.1) How many elements are present in each of the following?
a) HF and Hf
b)Co and CO
c) Si and SiO2
d) PoCl2 and POCl3
Answer:
a) HF and Hf
HF: 2 elements H and F with atomic ratio 1:1
Hf: 1 element
b) Co and CO
Co: 1 element
CO: 2 elements C and O with atomic ratio 1:1
c) Si and SiO2
Si: 1 element
SiO2: 2 elements with atomic ratio 1:2
d) PoCl2 and POCl3
PoCl2: 2 elements with atomic ratio 1:2
POCl3: 3 elements with atomic ratio 1:1:3
Q.2) Cm is the chemical symbol for curium, named after the famous scientist Madam Curie. Why wasn’t the symbol C, Cu or Cr used instead?
Answer:
Curium was discovered in 1944, long after discovery of carbon (C), copper (Cu) and chromium (Cr), which were discovered in 3750 BC, 9000 BC and 1797 respectively. The symbols C, Cu and Cr were already used for elements carbon, Copper and chromium respectively. Using again these symbols would lead to confusion. Therefore, its chemical symbol could not be set as that of carbon, copper and chromium and taken as Cm.
Q.3) What is the atomic number of an element? How does it differ from the mass number?
Answer:
Atomic number of an element is equal to the number of protons present in the nucleus of its atoms. It is represented by symbol ‘Z’.
Atomic number = No. of protons
Mass number of an element is equal to the sum of protons and neutrons present in the nucleus of its atoms. It is represented by symbol ‘A’
Mass number = No. of protons + No. of neutrons
A= Z+n
For example nitrogen atom has atomic number 7 and mass number 14, it can be written as 7N14.
Q.4) Students often mix up the following elements. Give the name for each element.
a) Mg and Mn
b) K and P
c) Na and S
d) Cu and Co
Answer:
a) Mg: Magnesium
Mn: Manganese
b) K: Potassium
P: Phosphorus
c) Na: Sodium
S: Sulphur
d) Cu: Copper
Co: Cobalt
Read more: List of Biology Notes For Class 9 All Chapters 1 to 9
Q.5 a) Classify the following molecules as mono-atomic, diatomic, triatomic, and polyatomic molecules.
H2O, N2, S8, He, HCl, CO2, Ar, H2SO4, C6H12O6
Answer:
| Monoatomic | Diatomic | Triatomic | Polyatomic |
| He Ar | N2 HCl | H2O CO2 | S8 H2SO4 C6H12O6 |
Q.5 b) Classify the following molecules as cation, anion, molecular ion, free radical and molecule.
CH+, O-2, CH.3, CO+, CO2, Cl–, Mg+2, CO3-2, O2, Na+, C2H5O-1, H2O, Cl2
Answer:
| Cation | Anion | Molecular ion | Free radical | Molecule |
| Mg+2 Na+ | O-2 Cl– | CH+ CO+ CO3-2 C2H5O-1 | CH.3 | CO2 O2 H2O Cl2 |
Q.6) Calculate the number of moles of butane, C4H10 in 151g of butane (At.Masses C=12 amu and H=1 amu).
Answer:
Given that
Mass of C4H10 = 151 g
Molar mass of C4H10 = 4(12) + 10(1) = 58 g/mol
Moles of C4H10 =?
Formula used
Moles of C4H10 = Mass of C4H10 / Molar mass of C4H10
= 151 / 58 = 2.603 moles
Q.7) What is the mass of 5 moles of ice? (Atomic masses: H=1amu, O=16 amu)
Answer:
Given that
Moles of ice (H2O) = 5 moles
Molar mass of ice (H2O) = 2(1) + 16 = 18 g/mol
Mass of ice (H2O) =?
Formula applied
Mass of ice = Number of moles of ice x Molar mass of ice
Mass of ice = 5 x 18 = 90 g
Q.8) Calculate the number of molecules in 6.50 mol of CH4.
Answer:
Given that
Moles of CH4 = 6.50 mol
No. of molecules of CH4 =?
Formula applied
No. of molecules of CH4 = Moles of CH4 x Avogadro’s number
= 6.50 x 6.02 x 1023
= 39.13 x 1023 = 3.913 x 1024 moles
Q.9) Calculate the average atomic mass of lithium from the following data:
| Isotope | Natural Abundance (%) | Relative Atomic Mass (amu) | |
| 6Li | 7.5 | 6.0151 | |
| 7Li | 92.5 | 7.016 |
Given that
Isotope I = 6Li = 7.5%
Isotope II = 7Li = 92.5%
Average atomic mass of Li =?
Formula applied
Answer:
Given that
Isotope I = 6Li = 7.5%
Isotope II = 7Li = 92.5%
Average atomic mass of Li =?
Formula applied
Q.10) Calculate the mass of 6.68 x 1023 molecules of PCl3.
Answer:
Given that
Molecules of PCl3 = 6.68 x 1023
Molar mass of PCl3 = 31 + 3(35.5) = 137.5 g/mol
Mass of PCl3 =?
Formula applied
Mass of PCl3 = (Molar mass of PCl3 x Molecules of PCl3 ) / Avogadros’s number
= (137.5 x 6.68 x 1023) / 6.02 x 1023
= 152.574 g
Long Questions Chemistry 9th Notes
Q.1) State and explain with examples:
a) The empirical formula of a compound.
b) The molecular formula of a compound.
Answer:
a) The empirical formula of a compound
Empirical formula of a compound is a chemical formula showing the simplest ratio of atoms or ions in that compound rather than the actual number of atoms or ions of the elements. Empirical formula is not the actual formula of a compound, but gives the correct ratio of elements in the compound.
Examples
Glucose has the actual formula C6H12O6. The simplest ratio of carbon, hydrogen and oxygen atoms in this molecule is 1:2:1. Therefore, the empirical formula of glucose is CH2O.
Benzene has the actual formula C6H6. The simplest ratio of carbon and hydrogen atoms in the molecule is 1:1. Therefore, the empirical formula of benzene is CH.
b) The molecular formula of a compound
Molecular formula of a compound is a chemical formula showing the actual number of atoms or ions of each element present in a molecule of that compound. Molecular formula can be derived from the empirical formula by the following relationship:
Molecular formula = (Empirical formula)n
Where n is 1, 2, 3 and so on.
Examples
Molecular formula of benzene is C6H6.
Molecular formula of hydrogen peroxide is H2O2.
Some compounds have same empirical and molecular formula e.g. hydrochloric acid (HCl), water (H2O), etc.
Q.2) What do you understand by the terms mole and Avogadro’s number. Explain with suitable examples.
Answer:
Mole
Mole, abbreviated mol, is an SI unit used in chemistry to express a definite amount of a substance.
A mole is the amount of pure substance containing the same number of elementary particles as there are atoms in exactly 12 g of carbon-12 (i.e. 6.02 x 1023). Therefore, one mole is equal to 6.02 x 1023 atoms or molecules. The number of atoms or other particles in a mole is the same for all substances.
The mole may be atoms, molecules, ions or other particles. As a matter of fact, mole is the atomic mass, molecular mass or formula mass of a substance expressed in grams
Atomic mass of H expressed as 1 g = 1 mol of H
Molecular mass of H2O expressed as 18 g = 1 mol of H2O
Formula mass of NaCl expressed as 58.5 g = 1 mol of NaCl
The concept of the mole helps in understanding quantitative information about a chemical equation on a macroscopic level. The mole helps to determine the simplest formula of a compound and to calculate the quantities involved in chemical reactions.
Avogadro’s number
Avogadro’s number is a collection of 6.02 x 1023 particles. A mole of a substance represents 6.02 x 1023 atoms, molecules or formula units depending on the nature of the substance. An Italian scientist, Amadeo Avogadro, determined this number and is represented by symbol ‘NA’. In simple words, 6.02 x 1023 particles are equal to one mole.
One mole of C = 12 g of C = 6.02 x 1023 atoms of C
One mole of H2SO4 = 98 g of H2SO4 = 6.02 x 1023 molecules of H2SO4
One mole of NaCl = 58.5 g of NaCl = 6.02 x 1023 formula units of NaCl
Q.3 a) Compare and contrast a mixture and a compound. Give examples of each of them.
Answer:
| Compound | Mixture |
| i) It is formed by the chemical combination of atoms of the elements. | It is formed by the physical combination or mixing up of the substances. |
| ii) The constituents lose their original properties. | The constituents retain their properties. |
| iii) A compound always has fixed composition by mass. | Mixtures do not have a fixed composition by mass. |
| iv) The components of the compound cannot be separated by physical methods. | The components of the mixture can be separated by physical methods. |
| v) Every compound is represented by its chemical formula. | It consists of two or more components and does not have any chemical formula. |
| vi) A compound has a homogeneous composition. | A mixture may be homogeneous or heterogeneous in composition. |
| vii) A compound has a sharp and fixed melting point. | A mixture does not have sharp and fixed melting points. |
| Examples: Water (H2O), sodium chloride (NaCl), etc. | Examples: Air, soil, soda water, etc. |
Q.3 b) How will you classify molecules? Support your answer with at least two examples of each.
Answer:
On the basis of the nature of atoms present in the molecules, they may be classified into two groups:
Homoatomic molecules
The molecule made up of atoms of the same element is called homoatomic molecule. They are also called homonuclear molecules. For example, oxygen (O2), hydrogen (H2), chlorine (Cl2), sulphur (S8), etc.
Heteroatomic molecule
The molecule made up of atoms of the different elements is called heteroatomic molecule. They are also called heteronuclear molecules. For example, carbon dioxide (CO2), water (H2O), hydrochloric acid (HCl),etc.
Q.4 a) What is the molecular mass of a compound? How will you differentiate it from formula mass?
Answer:
Molecular mass is the sum of atomic masses of all the atoms present in one molecule of a molecular compound. For example, molecular mass of H2O is 18 amu and that of CO2 is 44 amu.
Molecular mass of H2O = 2(1) + 16 = 18 amu
Molecular mass of CO2 = 12 + 2(16) = 44 amu
Formula mass is for ionic solids, which do not have molecules. It is based on the empirical formula of the compound. Therefore, formula mass is the sum of atomic masses of all the atoms in one formula unit of an ionic compound. For example, the formula mass of NaCl is 58.5 amu and that of K2SO4 is 174 amu.
Formula mass of NaCl = 23 + 35.5 = 58.5 amu
Formula mass of K2SO4 = 2(39) + 32 + 4(16) = 174 amu
Q.4 b) Calculate the molecular mass or formulae mass, as the case may be of the following compounds in amu.
i) Benzene(C6H6)
ii) Ethane gas (C2H6)
iii) Aluminum Chloride, AlCl3
iv) Iron Oxide, Fe2O3
Answer:
i) Benzene, C6H6
Molecular mass of C6H6 = 6(12) + 6(1) = 78 amu
ii) Ethane gas, C2H6
Molecular mass of C2H6 = 2(12) + 6(1) = 30 amu
iii) Aluminium chloride AlCl3
Formula mass of AlCl3 = 27 + 3(35.5) = 133.5 amu
iv) Iron oxide Fe2O3
Formula mass of Fe2O3 = 2(56) + 3(16) = 160 amu
Q.5 a) Find out the number of protons, electrons and neutrons in the following elements.
11Na23, 14Ag107, 26Fe56, 82Pb207, 18Ar40, 92U238
Answer:
a) 11Na23
No. of protons = 11
No. of electrons = 11
No. of neutrons = 23 – 11 = 12
b) 14Ag107
No. of protons = 14
No. of electrons = 14
No. of neutrons = 107 – 14 = 93
c) 26Fe56
No. of protons = 26
No. of electrons = 26
No. of neutrons = 56 – 26 = 30
d) 82Pb207
No. of protons = 82
No. of electrons = 82
No. of neutrons = 207 – 82 = 125
e) 18Ar40
No. of protons = 18
No. of electrons = 18
No. of neutrons = 40 – 18 = 22
f) 92U238
No. of protons = 92
No. of electrons = 92
No. of neutrons = 238 – 92 = 146
Q.5 b) Complete the following table:
| Symbol | Atomic No. | Number of Protons | Number of electrons | |
| a) | K | |||
| b) | 8 | |||
| c) | 15 | 15 | ||
| d) | 20 | |||
| e) | Cl |
Answer:
| Symbol | Atomic No. | Number of Protons | Number of electrons | |
| a) | K | 19 | 19 | 19 |
| b) | O | 8 | 8 | 8 |
| c) | P | 15 | 15 | 15 |
| d) | Ca | 20 | 20 | 20 |
| e) | Cl | 17 | 17 | 17 |
Read more: List of Physics Notes For Class 9 Chapters 1 to 9
