KPK 9th Class Chemistry Chapter 1 Fundamental of Chemistry Short Questions Answers
KPK 9th Class Chemistry Chapter 1 Fundamental of Chemistry Short Questions with answers are combined for all 9th class(Matric/ssc) Level students.Here You can prepare all Chemistry Chapter 1 Fundamental of Chemistry short question in unique way and also attempt quiz related to this chapter.Just Click on Short Question and below Answer automatically shown. After each question you can give like/dislike to tell other students how its useful for each.
Class/Subject: 9th Class Chemistry
Chapter Name: Fundamental of Chemistry
Board: All KPK Boards
- Malakand Board 9th Class Chemistry Chapter 1 Fundamental of Chemistry short questions Answer
- Mardan Board 9th Class Chemistry Chapter 1 Fundamental of Chemistry short questions Answer
- Peshawar Board 9th Class Chemistry Chapter 1 Fundamental of Chemistry short questions Answer
- Swat Board 9th Class Chemistry Chapter 1 Fundamental of Chemistry short questions Answer
- Dera Ismail Khan Board 9th Class Chemistry Chapter 1 Fundamental of Chemistry short questions Answer
- Kohat Board 9th Class Chemistry Chapter 1 Fundamental of Chemistry short questions Answer
- Abbottabad Board 9th Class Chemistry Chapter 1 Fundamental of Chemistry short questions Answer
- Bannu Board 9th Class Chemistry Chapter 1 Fundamental of Chemistry short questions Answer
Helpful For:
- All KPK Boards 9th Class Chemistry Annual Examination
- Schools 9th Class Chemistry December Test
- KPK 9th Class Chemistry Test
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KPK 9th Class Chemistry Chapter 1 Fundamental of Chemistry Short Questions Answers
A) HF and H Cl, B) Co and CO, C) Si and SiO2, and D) POCl2 & POCl3
A) I he are two, two elements present in each HF and HCl.
B) Co [cobalt] is a single element and Co (Carbon monoxide) contain two elements.
C) Si (silicon) is a single element and SiO2 contain two elements silicon & oxygen.
D) POCl2 contain 3 elements and POCl3 also contain 3-elements i.e phosphorus, Oxygen and Chlorine.
I he element carbon, Cuprum (copper) and chromium were discovered before curium. Ihe symbol C, was assigned to carbon, Cu to copper and Cr to chromium so when the element Curium was discovered & prepared by Madam Curie, the symbol Cm was assignment /used to it instead of c, Cu, Cr in order to aviode ambiguity.
Atomic Number = No of P or No of E
Element Atomic number (Z)
H 1
C 6
O 8
Na 11
Mass Number: It is the number of proton + neutron present in an atom. It is also called as atomic mass and represented by A.
Element Atomic Number
H 1amu
C 12amu
O 16amu
Na 23amu
Elements Names
Mg & Mn Magnesium & Manganese
K & P Potassium & phosphorus
Na & S Sodium & Sulphur
Cu & Co Copper & Cobalt
Classification of given species are listed below.
| Sr. No. | Given specie | Kind | Sr. No. | Given specie | Kind |
| 1 | CH3+ | Molecular cation | 8 | O2 | Molecule |
| 2 | O-2 | Anion | 9 | Na+ | Cation |
| 3 | CH3 | Free radical | 10 | O2H5O | Complex anion |
| 4 | CO+ | Molecular cation | 11 | H2O | Molecule |
| 5 | Cl– | Anion | 12 | Cl2 | Molecule |
| 6 | Cl+2 | Cation | 13 | CO2 | Molecule |
| 7 | CO3-2 | Complex anion |
Given Data: Mass of Butane
Molar mass of C4H10 = 12 x 4 + 1 x 10 = 58 g/mol.
Required:
n= ?
solution: we know that;
n= Mass in gram/Molar mass
n=151/58 = 2.603 moles
Give Data
Moles of ice = H2O(s)) 5 moles
M. mass of H2O = 18 g/mol.
Required
Mass = ?
Solution: we know that;
n = Mass in gram/Molar mass
On re – arrangement we get
Mass = n x M.mass
Mass = 5 x 18 = 90 gram.
Give Data
Moles of CH4= 6.5 moles
NA = 6.02 x 1023
Isotope Natural abundance (%) Relative atomic mass
7.5 6.0151
92.5 7.0160
Given data:
%age abundance of Li-6=7.5%
%age abundance of Li-7=92.5%
Mass of 6Li= 6.0151 amu
Mass of 7Li= 7.0160 amu
Required : Average atomic mass=?
Solution: We know that:
Average atomic mass
=A mass of 1st isotop x its %age abundance + A mass of 2nd isotopes x its %age abundance/100
Average atomic mass of Li
=A mass of Li-6 x its %age abundance +A.mass of Li-7 x its %age abundance/100
Average atomic mass = 6.01151 x 7.5 + 7.0160 x 92.5/100
Average atomic mass of Li = 6.94 amu
Given Data:
No of PCL3 molecules = 6.68×1023 moles .
M.mass of PCI = 31+35.5=66.5 g/mol.
Required: Mass =?
Solution: We know that ;
No of Particles = mass in grams/M.mass X NA
On re-Arrangement we get
Mass = No of Particles/NA x M.mass
Mass = 6.68 x 1028/6.02 x 1028 x 66.5 = 151.620g
