Go To Index
Class IX (Science)
Chemistry I
Target Paper 2024
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:
- Atomic mass
- Molecular formula
- Formula Mass
- Molar Mass
- Mole
- Atomic Number
- Dipole Dipole Interaction
- Hydrogen Bonding
- Covalent Bond
- Brownian movement
- Diffusion
- Electrolyte
- Electrolysis
- Isotopes + Hydrogen
- Electro Affinity
- Transition elements
- Ionization energy
- Diffusion
- Pressure
- Boiling Point
- Intermolecular force
- Empirical formula
- Broglie hypothesis
- Modern Periodic Table
- Dobereiners Law of trial
- Faraday’s 2nd Law of electrolysis
- Boly’s Law with applications
Q.5) Write down Properties of followings:
- Group 1A Active metals
- Group 7A Halogen
- Group 8A Nobel gases
- Neil Bohr’s Atomic Model
- Schrodinger’s Models
- Ionic compound
- Covalent bonds vs Co-ordinate covalent bond
- Saturated Vs Unsaturated
- Oxidation vs Reduction
- Polar compound vs Non polar compound
Q.9) Write down working of following with the help of diagram:
- Dry Cell
- Lead storage Battery
- Electroplating
- Platinum
- Magnesium
- Sodium
- Plasma
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
- Jabir bin Hayyan
- Al-Razi
- Robert Boyle
- Why ionic compound is solid?
- Why salt dissolve in Water?
- Why Gasoline does not dissolve in water?
- Milk is Colloidal Solution.
- Gases diffuses more rapidly.
- Glucose
- Hydrogen peroxide
- Benzene
- Sodium carbonate
- Sodium bicarbonate
- Caustic Soda
- Sulphuric Acid
Q.18) Describe Period in detail
Q.19) Properties of alpha, Beta & Gamma rays
Q.20) Application of electrolytic cell
Q.21) 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
- 100 °C to K
- 150 °F to °C
- 120 °C to °F
9) Write down Electronic configuration of following:
- Carbon
- Calcium
- Magnesium
11) A 600 ml sample of gas is heated from 27°C to 77°C at constant pressure. What would be the final volume.
12) 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?
13) Calculate the molarity of 40 gm of sodium hydroxide to make the solution 500 ml.
SOLUTION OF BALANCED EQUATIONS
Q.13) Balance the chemical equationa) 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 |
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 |
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 |
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 |
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
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 = NA = 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 NA
= 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 = NA = 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.
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.
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.9) Write down Electronic configuration of following C = 6, Cl = 17
Ans: Electronic Configuration Of Carbon (C):
Atomic number of Carbon (C) = 6
Electronic configuration: 1s2, 2s2, 2p4
Electronic Configuration Of Calcium (Ca):
Atomic number of Calcium (Ca) = 20
Electronic configuration: 1s2, 2s2, 2p6, 3s2, 3p6, 4s2
Electronic Configuration Of Magnesium (Mg):
Atomic number of Magnesium (Mg) = 12
Electronic configuration: 1s2, 2s2, 2p6, 3s2
10) Solution of 20 cm3 of alcohol is dissolved in 80 cm3 of water. Calculate the concentration (V/V) of this solution.
Solution:
Data:
- Volume of Solute = 20 cm
- Volume of Solution = 80 cm
- Volume / volume percent (V/V %) = ?
11) A 600 ml sample of gas is heated from 27°C to 77°C at constant pressure. What would be the final volume.
12) 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 H2SO4.13) Calculate the molarity of 40 gm of sodium hydroxide to make the solution 500 ml.
Solution:
Data:
- Mass of Sodium hydroxide = 40 g
- Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol
- Volume of solution = 500 ml
- Molarity (M) = ?
Ans: The molarity of 40 g of Sodium hydroxide to make the solution 500 ml is 2.00 mole/dm3
No comments:
Post a Comment