Chemistry I - For Class IX (Science) - Target Papers 2026 - By Sir Sajjad Akber Chandio

Go To Index
Class IX (Science)
Chemistry I
Target Paper 2025

Sir Sajjad Akber Chandio

• Target Paper
• Numericals
• Solution Of Balanced Equations
• Numerical's Solution

Target paper 2026

Q.1) Define Chemistry and describes its branches with uses.
Q.2) Define Following terms:

1. Electrolysis
2. Effusion
3. Sheild effect
4. Mole
5. Atomic Number
6. Dipole Dipole Interaction
7. Hydrogen Bonding
8. Diffusion
9. Brownian movement
10. Transition elements
11. Isotopes + isotopes Hydrogen
12. Pressure and Evaporation + factor
13. Electrolyte
14. Chemical reactivity
15. Electro Affinity + trends
16. Ionization energy + trends
17. Intermolecular force BEC(5 estate of matter)
18. Atomic radius + trends
19. Moh scale
20. Tydall Effect

Q.3) State the following Laws with Example:

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

Q.4) Explain discharge tube 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
7. Rutherford atomic model
8. Neil bohr’s atomic model

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

1. Coordinate covalent bond
2. Ionic compound

Q.7) Write down difference between the following

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

Q.8) Explain the corrosion of iron and describe any four methods to prevent metals from it.
Q.9) Write down construction working of following with the help of diagram:

1. Electroplating
2. Lead storage Battery

Q.10: Write down uses of following

1. Platinum
2. Magnesium
3. Sodium
4. Calcium
5. Palsma (4 estate of matter )
6. Adhesive material
7. Glue and epoxy resin

Q.11) Name the three most abundant elements found in air, Earth’s crust, Universe and Human Body.
Q.12) Identifying flame test Cl^- and I^- 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 contribution of scientist
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.

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
Q20) Molecules and its types

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) Practices Electronic configuration KLMN, spdf and dot cross
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.
13) Chlorine is naturally found in two stable isotopes ₁₇Cl³⁵ and ₁₇Cl³⁷ in the proportion of 75% and 25% respectively. Calculate its atomic mass.
14) An atom has one OR five electron in its M shell then what would be its atomic number? Identify that element.

GO TO TOP

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.

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

GO TO TOP

For Stepwise process of Balance Equation CLICK HERE

NUMERICAL'S SOLUTION

1) Calculate number of moles and molecules in 50 g of H₂O.
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.
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?
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.
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
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
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
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
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) 50 °C to K
2) 150 °F to °C
3) 120 °C to °F

1) 50 °C to K
Data:

• Temperature in Celcius (C) = 50 °C
• Temperature in Kelvin (K) = ?

Formula:

K = 273 + °C

Calculation:

K = 273 + 50 = 323 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
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?
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.

Q.11) Define Molarity and calculate Molarity 2 moles in 500 ml of HCl.
Ans: Molarity:
Definition: Molarity is define as the number of moles of solute dissolved in one dm³ of the solution. It is denoted by 'M'
Unit: Molarity is the concentraion unit in which the solute is expressedin gram and amount of solution is expressed in dm³. its unit is mol/dm³.
Formula:
Molarity (M): Number of moles of solute / Volume of solution in dm³

Numerical:
Data:
No.of moles = 2 moles
Volume of solution = 500 ml = 0.5L
Molarity of HCl = ?
Formula:
Molarity (M): Number of moles of solute / Volume of solution in dm³ or Liter

Calculation:
Molarity (M): 2 / 0.5 = 4M
Ans: Molarity of HCL solution is 4M.

Q.13) Calculate the mass of sodium hydroxide in 5M with 300 ml.
Data:
Molaritys = 5M
Volume in ml / cm³ = 300 ml = 0.5L
Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol
Mass of NaOH = ?
Formula:
Molarity (M): Mass of solute/Molar mass of solute X 1000/Volume of solution in cm³ or ml

Calculation:
5 = Mass Of NaOH/40 X 1000/300
Mass of NaOH = 5 x 40 x 300 / 1000 = 60 gm
Ans: Mass of sodium hydroxide in 5M with 300 ml is 60 gm

Q.13) Chlorine is naturally found in two stable isotopes ₁₇Cl³⁵ and ₁₇Cl³⁷ in the proportion of 75% and 25% respectively. Calculate its atomic mass. Data:
Atomic mass of ₁₇Cl³⁵= 35
Atomic mass of ₁₇Cl³⁷= 37
% abundance of ₁₇Cl³⁵ = 75%
% abundance of ₁₇Cl³⁷ = 25%

Formula:
Atomic mass = Mass x % of abundance

Solution:
Atomic mass of ₁₇Cl³⁵ = 35 x 75/100 = 35 x 0.75 = 26.25 a.m.u
Atomic mass of ₁₇Cl³⁷ = 37 x 25/100 = 35 x 0.25 = 9.25 a.m.u
Ans: Atomic of ₁₇Cl³⁵ is 26.25 a.m.u and ₁₇Cl³⁷ 9.25 a.m.u

Q.14) An atom has one OR five electron in its M shell then what would be its atomic number? Identify that element. Data:

• An atom has one electron in its M shell
• An atom has five electron in its M shell
• Atomic number of element with one electron in M-shell =?
• Atomic number with five electron in M-shell =?
• Name of element with one electron in M-shell =?
• Name of element with five electron in M-shell =?

Solution:
a) Element with one electron in M-shell:
Atomic number:
i. It means element has 2 electrons in the K-shell, 8 electrons in the L-shell and 1 electron in the M-shell.
Electronic configuration: 1s², 2s², 2p⁶, 3s¹
ii. Its atomic number of that element is: 2 + 8 + 1 = 11
Name of element with one electron in M-shell = Sodium (Na)
Ans: The elements is Sodium (Na) and its atomic number is 11.

a) Element with five electron in M-shell:
Atomic number:
i. It means element has 2 electrons in the K-shell, 8 electrons in the L-shell and 5 electron in the M-shell.
Electronic configuration: 1s², 2s², 2p⁶, 3s², 3p³,
ii. Its atomic number of that element is: 2 + 8 + 5 = 15
Name of element with one electron in M-shell = Phosphorous (P)
Ans: The elements is Phosphorous (P) and its atomic number is 15.

GO TO TOP