Chemistry I - For Class IX (Science) - Target Papers 2025 - By Sir Sajjad Akber Chandio

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Class IX (Science)
Chemistry I
Target Paper 2025

By Sir Sajjad Akber Chandio
Q.1) Define Chemistry and describes its branches and importance and role of chemistry in society.
Q.2) Define Following terms:

1. Atomic mass
2. Molecular formula
3. Formula Mass
4. Molar Mass
5. Mole
6. Atomic Number
7. Dipole Dipole Interaction
8. Hydrogen Bonding
9. Diffusion
10. Brownian movement
11. Transition elements
12. Isotopes + Hydrogen
13. Pressure + factor
14. Electrolyte
15. Chemical reactivity
16. Electro Affinity
17. Ionization energy
18. Intermolecular force
19. Empirical formula
20. Atomic radius
21. Moh scale

Q.3) State the following Laws with Example:

1. Modern Periodic Table
2. Dobereiner's Law of trial
3. Faraday’s 1^st and 2^nd Law of electrolysis
4. Boyle’s Law with applications
5. Charle’s law with application

Q.4) Describe Rutherford atomic Models With the help of Diagram.
Q.5) Write down Properties of followings:

1. Group 1A Active metals
2. Group 7A Halogen
3. Group 8A Nobel gases
4. Cathode ray tube
5. Schrodinger’s Models
6. DE Broglie hypothesis

Q.6) Define Chemical Bond and explain following with the help of example and characteristics

1. Coordinate covalent bond
2. Covalent bond and types

Q.7) Write down difference between the following

1. Ionic bond ,Covalent bonds and Co-ordinate covalent bond
2. Oxidation vs Reduction
3. Saturated , Unsaturated and supersaturated solution
4. Polar compound vs Non polar compound
5. Metal ,non-metal and metalloids
6. Solution, suspension and colloid
7. Sodium and Iron
8. Period and group

Q.8) Define solubility and describe its factors.
Q.9) Write down construction working of following with the help of diagram:

1. Dry Cell
2. Lead storage Battery
3. Discovery of electron

Q.10: Write down uses of following

1. Platinum
2. Magnesium
3. Sodium
4. Calcium

Q.11) Name the three most abundant elements found in air, Earth’s crust, Universe and Human Body.
Q.12) Define Allotropy and name 4 allotropy of carbon. Write characteristics of Diamond and Graphite.
Q.13) Balance the chemical equation

• a) H₂ + O₂ → H₂O
• b) N₂ + H₂ → NH₃
• c) KClO₃ → KCl + O₂
• d) NH₃ + O₂ → NO + H₂O
• e) NaHCO₃ → Na₂CO₃ + H₂O + CO₂
• f) KI + Cl₂ → KCl + I₂
• g) CaCO₃ + HCl → H₂O + CaCl₂ + CO₂

Q.14) Write down one contribution of following scientist.

1. Jabir bin Hayyan
2. Al-Razi
3. Robert Boyle

Q.15) Scientific reason:

1. Why ionic compound is solid?
2. Why salt dissolve in Water?
3. Why Gasoline does not dissolve in water?
4. Milk is Colloidal Solution.
5. Gases diffuses more rapidly.

Q.16) (a) Write down empirical and molecular formula:

1. Glucose
2. Hydrogen peroxide
3. Benzene

(b) Formulas:

1. Sodium carbonate
2. Sodium bicarbonate
3. Caustic Soda
4. Sulphuric Acid
5. Sugar
6. Glucose

Q.17) Define Alloy and write down uses of Bronze and Brass.
Q.18) Describe Period in detail
Q.19) Application of electrolytic cell and Isotopes
Q.20) Identifying flame test Cl^- and I^-

NUMERICAL

1) Calculate number of moles and molecules in 50 g of H₂O.
2) Calculate the number of atoms present 9.2 gm of Ca
3) 20 gm of salt is dissolved in 500 cm³ of a solution. Calculate the molarity of that solution.
4) The 700 cm³ of a gas is enclosed in a container under a pressure of 650 mm of Hg. if the volume is reduced to 350 cm³, what will be the pressure then?
5) A sample of hydrogen gas has a volume of 350 cm³ at 40 °C. If a gas is allowed to expand up to 700 cm³ at constant pressure. Find out its final temperature.
6) Find out the number of Proton, Neutron and Electron is present in the following elements?

• Fe₂₆⁵⁶


• O₈¹⁷


• U₉₂²³⁵


• C₆¹⁴

7) Convert

1. 50 °C to K
2. 150 °F to °C
3. 120 °C to °F

8) How would you prepare 500 cm³ of 0.20 M NaOH from a stock solution of 1.5M NaOH
9) Write down Electronic configuration of following: KLMN
10) A sample of sulphuric acid has molarity 20 M. How many cm³ of solution should you use to prepare 500 cm³ of 0.5 M H₂SO₄?
11) calculate Molarity 2 moles in 500 ml of HCl.
12) Calculate the mass of sodium hydroxide in 5M with 300 ml.

SOLUTION OF BALANCED EQUATIONS

Q.13) Balance the chemical equation
a) H₂ + O₂ → H₂O
Step no.1: Write correct formula of all reactants on left side and product on right side of an equation.

H₂ + O₂ ⟶ H₂O

Step no.2: Check the number of atoms on each side.

Reactants	Products
H = 2	H = 2
O = 2	O = 1

Oxygen is unbalanced.

Step no.3: (Balance Oxygen on product side)
Now multiply the formula of Water (H₂O) with co-efficient 2 on product side.

H₂ + O₂ ⟶ 2H₂O

Reactants	Products
H = 2	H = 2 x 2 = 4
O = 2	O = 1 x 2 = 2

Step no.4: (BalanceHydrogen on the reactant side)
Now again check and balance the equation by placing 2 in front of Hydrogen (H₂) on reactant side.

2H₂ + O₂ ⟶ 2H₂O

Reactants	Products
H = 2 x 2 = 4	H = 4
O =  2	O = 2

Now the chemical equation is balanced.

2H₂ + O₂ ⟶ 2H₂O

b) N₂ + H₂ → NH₃
N₂ + 3H₂ → 2NH₃ (Equation is Balanced)

c) KClO₃ → KCl + O₂
Step no.1: Write correct formula of all reactants on left side and product on right side of an equation.

KClO₃ → KCl + O₂

Step no.2: Check the number of atoms on each side.

Reactants	Products
K = 1	K = 1
Cl = 1	Cl = 1
O = 3	O = 2

Oxygen is not balance.

Step no.3: (Balance Oxygen on reactant side first)
Now multiply the formula of Potassium chlorate (KClO₃) with co-efficient 2 on reactant side

2KClO₃ → KCl + O₂

Reactants	Products
K = 1 x 2 = 2	K = 1
Cl = 1 x 2 = 2	Cl = 1
O = 2 x 3 = 6	O = 2

Step no.4: 
Now multiply the formula of Potassium chloride (KCl) with coefficient 2 on the product side to balance the Potassium and Chlorine and Multiply Oxygen (O₂) atoms with coefficient 3 to balance the Oxygen.

2KClO₃ → 2KCl₂ + 3O₂

Reactants	Products
K = 1 x 2 = 2	K = 1 x 2 = 2
Cl = 1 x 2 = 2	Cl = 1 x 2 = 2
O = 2 x 3 = 6	O = 2 x 3 = 6

Now the chemical equation is balanced.

2KClO₃ → 2KCl + 3O₂

d) NH₃ + O₂ → NO + H₂O
4NH₃ + 5O₂ → 4NO + 6H₂O (Equation is Balanced)

e) NaHCO₃ → Na₂CO₃ + H₂O + CO₂
2NaHCO₃ → Na₂CO₃ + H₂O + CO₂ (Equation is Balanced)


f) KI + Cl₂ → KCl + I₂
Step no.1: Write correct formula of all reactants on left side and product on right side of an equation.

KI + Cl₂ → KCl + I₂

Step no.2: Check the number of atoms on each side.

Reactants	Products
K = 1	K = 1
Cl = 2	Cl = 1
I = 1	I = 2

Chlorine and Iodine is not balanced is not balance.

Step no.3: (Balance Iodine on reactant side first)
Now multiply the formula of Potassium iodide (KI) with co-efficient 2 on reactant side

2KI + Cl₂ → KCl + I₂

Reactants	Products
K = 1 x 2 = 2	K = 1
Cl = 2	Cl = 1
I = 1 x 2 = 2	I = 2

Step no.4: 
Now multiply the formula of Potassium chloride (KCl) with coefficient 2 on the product side to balance the Potassium and Chlorine.

2KI + Cl₂ → 2KCl + I₂

Reactants	Products
K = 2	K = 1 x 2 = 2
Cl = 2	Cl = 1 x 2 = 2
I = 2	I = 2

Now the chemical equation is balanced.

2KI + Cl₂ → 2KCl + I₂

g) CaCO₃ + HCl → H₂O + CaCl₂ + CO₂
Step no.1: Write correct formula of all reactants on left side and product on right side of an equation.

CaCO₃ + HCl → H₂O + CaCl₂ + CO₂

Step no.2: Check the number of atoms on each side.

Reactants	Products
Ca = 1	Ca = 1
C = 1	C = 1
Cl = 1	Cl = 2
H = 1	H = 2
O = 3	O = 2 + 1 = 3

Chlorine and Hydrogen is not balance.

Step no.3: (Balance either Hydrogen or Chlorine on reactant side first)
Now multiply the formula of Hydrochloric acid (HCl) with co-efficient 2 on reactant side

CaCO₃ + 2HCl → H₂O + CaCl₂ + CO₂

Reactants	Products
Ca = 1	Ca = 1
C = 1	C = 1
Cl = 1 x 2 = 2	Cl = 2
H = 1 x 2 = 2	H = 2
O = 3	O = 3

Now the chemical equation is balanced.

CaCO₃ + 2HCl → H₂O + CaCl₂ + CO₂

For Stepwise process of Balance Equation CLICK HERE

NUMERICAL'S SOLUTION

1) Calculate number of moles and molecules in 50 g of H₂O.
Solution:
Data:

• Number of moles of H₂O = ?
• number of molecules of H₂O = ?
• Given mass of of H₂O = 50 g
• Avogadro’s Number = N_A = 6.02 x 10²³

Calculation:
Molecular mass of Water H₂O = (2 x 1) + (1 x 16) = 2 + 16 = 18 a.m.u

Formula:

Number of molecules = Number of moles x N_A
= 2.78 x 6.02 x 10²³
= 16.736 x 10²³ molecules of H₂O Ans.
Ans: There are 2.78 moles and 16.736 x 10²³ molecules in 50 g of H₂O .


2) Calculate the number of atoms present in 9.2 gm of Calcium (Ca).
Data:
Number of atoms of Ca = ?
Avogadro’s Number = N_A = 6.02 x10²³
Solution:
Atomic mass of Calcium (Ca) = 40
1 g atomic weight of Calcium = 40 gm
40 g of Calcium contains =6.02 x 10²³ atoms of Calcium
Formula:

Ans: 1.384 x 1023 atoms are present in 9.2 gm of Ca.

3) 20 gm of salt is dissolved in 500 cm³ of a solution. Calculate the molarity of that solution.
Solution:
Data:

• Mass of solute = 20 g
• Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
• Volume of solution = 500 cm³
• Molarity (M) = ?

Ans: The molarity of that solution is 0.683 mole / dm³.




4) The 700 cm³ of a gas is enclosed in a container under a pressure of 650 mm of Hg. if the volume is reduced to 350 cm³, what will be the pressure then?
Solution:
Data:

• V₁ = 700 cm³
• P₁ = 650 mm
• V₂ = 350 cm³
• P₂ = ?

Formula:

P₁ V₁ = P₂ V₂

5) A sample of hydrogen gas has a volume of 350 cm³ at 40 °C. If a gas is allowed to expand up to 700 cm³ at constant pressure. Find out its final temperature.
Solution:
Data:

• T₁ = 40 °C = 40+273 K = 313 K
• V₁= 350 cm³
• V₂= 700 cm³
• T₂ = ?

6) Find out the number of Proton, Neutron and Electron is present in the following elements?

• Fe₂₆⁵⁶


• O₈¹⁷


• U₉₂²³⁵


• C₆¹⁴

i) Fe₂₆⁵⁶
Solution:
Data:

• Atomic number = Z = 26
• Atomic mass = A = 56
• Number of protons = ?
• Number of electrons = ?
• Number of neutrons = ?

Formula:
Atomic number= Z = Number of proton in nucleus = Total number of electron around nucleus
Neutron (N) = Mass number (A) - Atomic number (Z)

Calculation:
Therefore,
Number of protons is Z = 26
Number of electrons is Z = 26
Number of neutrons = A - Z = 56 - 26 = 30

ii) O₈¹⁷
Solution:
Data:

• Atomic number = Z = 8
• Atomic mass = A = 17
• Number of protons = ?
• Number of electrons = ?
• Number of neutrons = ?

Formula:
Atomic number= Z = Number of proton in nucleus = Total number of electron around nucleus
Neutron (N) = Mass number (A) - Atomic number (Z)

Calculation:
Therefore,
Number of protons is Z = 8
Number of electrons is Z = 8
Number of neutrons = A - Z = 17 - 8 = 9

iii) U₉₂²³⁵
Solution:
Data:

• Atomic number = Z = 92
• Atomic mass = A = 235
• Number of protons = ?
• Number of electrons = ?
• Number of neutrons = ?

Formula:
Atomic number= Z = Number of proton in nucleus = Total number of electron around nucleus
Neutron (N) = Mass number (A) - Atomic number (Z)

Calculation:
Therefore,
Number of protons is Z = 92
Number of electrons is Z = 92
Number of neutrons = A - Z = 235 - 92 = 143

iv) C₆¹⁴
Solution:
Data:

• Atomic number = Z = 6
• Atomic mass = A = 14
• Number of protons = ?
• Number of electrons = ?
• Number of neutrons = ?

Formula:
Atomic number= Z = Number of proton in nucleus = Total number of electron around nucleus
Neutron (N) = Mass number (A) - Atomic number (Z)

Calculation:
Therefore,
Number of protons is Z = 6
Number of electrons is Z = 6
Number of neutrons = A - Z = 14 - 6 = 8

7) Convert
1) 100 °C to K
2) 150 °F to °C
3) 120 °C to °F
Solution
1) 100 °C to K
Data:

• Temperature in Celcius (C) = 100 °C
• Temperature in Kelvin (K) = ?

Formula:

K = 273 + °C

Calculation:

K = 273 + 100 = 373 K Ans.

2) 150 °F to °C
Data:

• Temperature in Farenheit (F) = 150 °F
• Temperature in Celcius (C) = ?

Formula:

C = (F - 32) x ⁵/₉

Calculation:

C = (150 - 32) x ⁵/₉

C = (118) x ⁵/₉

C = ^{118 x 5}/₉ = ⁵⁹⁰/₉

C = 65.56 °C Ans.

3) 120 °C to °F
Data:

• Temperature in Celcius (C) = 120 °C
• Temperature in Farenheit (F) = ?

Formula:

F = (C x ⁹/₅) + 32

Calculation:

F = (^{C x 9}/₅) + 32

F = (^{120 x 9}/₅) + 32

F = (¹⁰⁸⁰/₅) + 32 = 216 + 32

F = 248 °C Ans.

8) How would you prepare 500 cm³ of 0.20 M NaOH from a stock solution of 1.5M NaOH
Solution:
Data:

• M₁ = 1.5 M NaOH
• M₂ = 0.2 M NaOH
• V₂ = 500 cm³
• V₁ = ?

Calculation:
Formula:

Concentrated Soluiotn = Dilute Solution

M₁ V₁ = M₂ V₂

1.5 x V₁ = 0.20 x 500

= 66.67 cm³

Ans: Take 66.67 cm3 of concentrated NaOH and placed in a measuring flask and dilute it by adding water up to the mark. It will become 0.2M solution of NaOH.

10) A sample of sulphuric acid has molarity 20 M. How many cm³ of solution should you use to prepare 500 cm³ of 0.5 M H₂SO₄?
Solution:
Data:

• M₁ = Molarity of given H₂SO₄ = 20 M
• M₂ = Molarity of required H₂SO₄ = 0.5 M
• V₁ = volume of concentrated solution needs to be diluted =? 1
• V₂ = Volume required in H₂SO₄ on = 500 cm³

Formula:

M₁ V₁ = M₂ V₂

Calculation:

V₁ = 12.5 cm³

Ans: 12.5 cm³ of 20 M is used to make 500 cm³ aqueous solution to form 0.5 M H₂SO₄.