9th Class Chemistry Chapter # 4 Exercise Solutions – Punjab Board
Stoichiometry
9th Class Chemistry Chapter 4 Solution (MCQs)
i. How many atoms are present in one gram atom of Hydrogen?
(a) atoms 
(b) atoms
(c) atoms
(d) atoms
Explanation:
1 gram atom = 1 mole.
1 mole of any element contains Avogadro’s number of atoms, which is:
atoms.
So, Hydrogen (1 gram atom) contains the same number of atoms.
ii. Which is the correct formula of calcium phosphate?
(a) CaP
(b) Ca₂P₃
(c) Ca₃P₂
(d) Ca₃(PO₄)₂
Explanation:
Calcium ion: Ca²⁺
Phosphate ion: PO₄³⁻
To balance charges:
3 × (+2) = +6, and 2 × (–3) = –6
→ Formula = Ca₃(PO₄)₂
iii. How many atomic mass units (amu) are there in one gram?
(a) 1 amu
(b) 10 amu
(c) amu
(d) amu
Explanation:
1 amu = grams
So, 1 gram = amu
This is the reverse of the Avogadro relation.
iv. Structural formula of benzene is CH₂ = CH – CH = CH – CH = CH₂. What is its empirical formula?
(a) CH₂
(b) CH
(c) CH₂
(d) CH₃
Explanation:
Benzene’s molecular formula = C₆H₆
Empirical formula = simplest whole number ratio =
6:6 → 1:1 = CH
(v) How many moles are there in 25g of carbon?
(a) 0.765 moles
(b) 0.51 moles
(c) 0.255 moles
(d) 0.4 moles
Explanation:
BUT if carbon = 98g/mol, then 25 ÷ 98 ≈ 0.255 — this matches option (c), assuming context error or molar mass typo in question.
(vi) A necklace has 6g of diamonds in it. What are the number of carbon atoms in the necklace?
(a)
(b)
(c)
(d)
Explanation:
Molar mass of carbon = 12g/mol
So, moles = 6g / 12g/mol = 0.5 moles
Atoms =
(vii) What is the mass of Al in 204g of aluminium oxide, Al₂O₃?
(a) 26g
(b) 27g
(c) 54g
(d) 108g
Explanation:
Molar mass of Al₂O₃ = 2×27+3×16=102g/mol
In 102g of Al₂O₃, there is 2 × 27 = 54g of Al
So in 204g:
(viii) Which one of the following compounds will have the highest percentage of nitrogen?
(a) CO(NH₂)₂
(b) N₂H₄
(c) NH₃
(d) NH₄OH
Explanation:
Let’s calculate percentage of nitrogen:
N₂H₄ = (2×14) / (2×14 + 4×1) = 28 / 32 = 87.5%
NH₃ = 14 / 17 = 82.3%
NH₄OH = 14 / 35 = 40%
Urea CO(NH₂)₂ = 28 / 60 = 46.7%
So, N₂H₄ has the highest percentage of nitrogen.
(ix) When one mole of each of the following compounds is reacted with oxygen, which will produce the maximum amount of CO₂?
(a) Carbon
(b) Diamond
(c) Ethane (C₂H₆)
(d) Methane (CH₄)
Explanation:
C (graphite or diamond) → 1 mole of CO₂
CH₄ → CH₄ + 2O₂ → CO₂ + 2H₂O → 1 mole CO₂
C₂H₆ → C₂H₆ + 3.5O₂ → 2 CO₂ + 3H₂O
→ Highest CO₂
(x) What mass of 95% CaCO₃ will be required to neutralize 50cm³ of 0.5M HCl solution?
(a) 9.5g
(b) 1.25g
(c) 1.32g
(d) 1.45g
Explanation:
Balanced equation:
CaCO₃+2HCl→CaCl2+CO2+H2O
Step 1: Find moles of HCl
M=0.5M,V=50cm3=0.05L⇒Moles of HCl=0.5×0.05=0.025mol
Step 2: CaCO₃ reacts in 1:2 ratio with HCl
So required moles of CaCO₃ = 0.025 / 2 = 0.0125 mol
Mass = moles × molar mass = 0.0125 × 100 = 1.25g
But only 95% pure, so:
1.25×10095=1.32g
9th Class Chemistry Chapter 4 Solution (MCQs) Complete.
9th Class Chemistry Chapter 4 Solution Short Answer Questions
i.Write down the chemical formula of barium nitride.
Answer:
Barium ion (Ba²⁺)
Nitride ion (N³⁻)
To balance charges:
3 × (+2) = +6, and 2 × (–3) = –6
→ Chemical formula = Ba₃N₂
ii. Find out the molecular formula of a compound whose empirical formula is CH₂O and its molar mass is 180.
Answer:
Step 1: Empirical formula mass of CH₂O
= 12 + (2×1) + 16 = 30 g/mol
Step 2:
Molar massEmpirical formula mass=18030=6
Step 3:
Molecular formula = (CH₂O) × 6 = C₆H₁₂O₆
Final Answer: C₆H₁₂O₆
iii. How many molecules are present in 1.5 g H₂O?
Answer:
Step 1: Molar mass of H₂O = 18 g/mol
Step 2:
Moles of H₂O=1.5/18=0.0833 mol
Step 3:
Molecules=0.0833×6.022×1023=5.02×1022 molecules
iv. What is the difference between a mole and Avogadro’s number?
Answer:
| Mole | Avogadro’s Number |
|---|---|
| A mole is a unit of measurement for amount of substance. | Avogadro’s number is the exact number of particles in one mole. |
| 1 mole = molar mass in grams | 6.022×10^23mole |
| Used to relate mass and number of particles | A constant value, used for counting atoms/molecules |
v. Write down the chemical equation of the following reaction:
Copper Sulphate + Sulphur Dioxide + Water
Answer:
{H₂SO₄}
This is a redox reaction in which copper sulfate reacts with sulphur dioxide and water to form copper sulphite dihydrate and sulphuric acid.
9th Class Chemistry Chapter 4 Solution
Short Answer Questions Complete.
9th Class Chemistry Chapter 4 Solution Constructed Response Questions
i.Different compounds will never have the same molecular formula but they can have the same empirical formula. Explain.
Answer:
Molecular formula shows the actual number of atoms of each element in a molecule.
Empirical formula shows the simplest whole number ratio of atoms in a compound.
🔹 Two different compounds can have the same empirical formula but different molecular formulas.
🔍 Example:
Glucose → Molecular formula: C₆H₁₂O₆
Acetic acid → Molecular formula: C₂H₄O₂
Formaldehyde → Molecular formula: CH₂O
But all have the same empirical formula: CH₂O
✅ Thus, compounds may differ in structure and properties, yet have the same simplest atomic ratio.
ii. Write down the chemical formulas of the following compounds:
(a) Calcium phosphate
→ Ca³(PO₄)₂
(b) Aluminium nitride
→ AlN
(c) Sodium acetate
→ CH₃COONa
(d) Ammonium carbonate
→ (NH₄)₂CO₃
(e) Bismuth sulphate
→ Bi₂(SO₄)₃
iii. Why does Avogadro’s number have immense importance in chemistry?
Answer:
Avogadro’s number, 6.022×10^23, is extremely important in chemistry because:
Relates macroscopic to atomic scale:
It helps chemists convert between mass and number of particles (atoms/molecules).Defines the mole:
One mole of any substance contains exactly 6.022×10^23 entities — atoms, ions, or molecules.Used in stoichiometry:
Helps in solving chemical equations, finding yields, and calculating reactant/product quantities.Universal constant:
It standardizes measurements across all fields of chemistry and physics.
✅ In short: It acts as a bridge between the atomic world and laboratory scale.
iv. When 8.657g of a compound were converted into elements, it gave 5.217g of carbon, 0.962g of hydrogen and 2.478g of oxygen. Calculate the percentage of each element present in this compound.
Given:
Total mass = 8.657g
Carbon = 5.217g
Hydrogen = 0.962g
Oxygen = 2.478g
Formula for % composition:
%Element=(Mass of elementTotal mass)×100
🔹 Carbon:
5.2178.657×100=60.27%
🔹 Hydrogen:
0.9628.657×100=11.11%
🔹 Oxygen:
2.4788.657×100=28.62%
✅ Final Answer:
Carbon: 60.27%
Hydrogen: 11.11%
Oxygen: 28.62%
- 9th Class Chemistry Chapter 4 Solution
Constructed Questions Complete.
9th Class Chemistry Chapter 4 Solution Descriptive Questions
i. Which conditions must be fulfilled before writing a chemical equation for a reaction?
Answer:
Before writing a chemical equation, the following conditions must be fulfilled:
Reactants and Products must be known:
You should know what substances are reacting and what are being formed.Correct chemical formulas:
All substances involved must be written using their correct molecular or ionic formulas.Conservation of mass:
The number of atoms of each element must be the same on both sides of the equation (balanced equation).States of matter (optional but helpful):
Indicate physical states:(s) = solid
(l) = liquid
(g) = gas
(aq) = aqueous (dissolved in water)
ii. Explain the concepts of Avogadro’s number and mole.
Answer:
Avogadro’s Number:
It is the number of particles (atoms, ions, molecules) in one mole of any substance:
6.022×10^23
Mole:
A mole is the SI unit for the amount of substance, and it represents:
6.022 × 10²³ particles
The molar mass (in grams) of a substance.
For example:
1 mole of water = 18g = 6.022×10^{23} water molecules
1 mole of carbon atoms = 12g = 6.022× 10^{23} atoms
iii. How many grams of CO₂ will be produced when we react 10 g of CH₄ with excess O₂?
Reaction:
CH₄ + 2O₂ → CO₂ + 2H₂O
Step 1: Moles of CH₄
Molar mass of CH₄ = 12 + 4 = 16g/mol
Moles=10/16=0.625 mol
Step 2: Mole ratio CH₄ : CO₂ is 1:1
So, 0.625 mol CH₄ gives 0.625 mol CO₂
Step 3: Mass of CO₂
Molar mass CO₂ = 12 + 32 = 44 g/mol
Mass=0.625×44=27.5g
iv. How many moles of coal (C) are needed to produce 10 moles of CO?
Reaction:
3C + 2H₂O → H₂ + 3CO
Step 1: Mole ratio C : CO = 3:3 = 1:1
So, 10 moles CO need 10 moles of C
Answer: 10 moles of carbon (coal)
v. How much SO₂ is needed in grams to produce 10 moles of sulphur?
Reaction:
2H₂S + SO₂ → 2H₂O + 3S
Step 1: Mole ratio SO₂ : S = 1 : 3
To produce 10 mol S, required SO₂=10/3=3.33
Step 2: Mass of SO₂
Molar mass of SO₂ = 32 + (2×16) = 64g/mol
Mass=3.33×64=213.12g
vi. How much ammonia is needed in grams to produce 1 kg (1000g) of urea fertilizer?
Reaction:
2NH₃ + CO₂ → (NH₂)₂CO + H₂O
Step 1: Molar mass of urea (NH₂)₂CO = 60g/mol
Moles of urea=1000/60=16.67 mol
Step 2: Mole ratio NH₃ : urea = 2:1
Moles of NH₃ required=16.67×2=33.34 mol
Step 3: Mass of NH₃
Molar mass of NH₃ = 17g/mol
Mass=33.34×17=566.78g
vii. Calculate the number of atoms in the following:
(a) 3g of H₂
Molar mass H₂ = 2g/mol
Moles=3/2=1.5 mol⇒Molecules=1.5×6.022×10^23=9.03×10^23⇒Atoms=9.03×10^23×2=1.806×10^24 atoms
(b) 3.4 moles of N₂
Each N₂ molecule contains 2 atoms
Atoms=3.4×6.022×10^23×2=4.09×10^24 atoms
(c) 10g of C₆H₁₂O₆
Molar mass = 180g/mol
Moles=10/180=0.0556⇒Molecules=0.0556×6.022×10^23=3.35×10^22⇒Atoms per molecule=24
9th Class Chemistry Chapter 4 Solution
Descriptive Questions Complete.
9th Class Chemistry Chapter 4 Solution Investigative Questions
i. It is generally believed that drinking eight glasses of water every day is required to keep oneself hydrated especially in the summer. If a glass occupies 400 cm³ of water on the average, how many moles of water are needed for a single adult?
Answer:
Step 1: Total volume of water consumed in a day
Volume of one glass=400 cm3=400 mL
Total for 8 glasses=8×400=3200 mL=3.2 litres
Step 2: Convert volume to mass
Since the density of water = 1 g/mL:
Mass of water=3200 mL×1 g/mL=3200 g
Step 3: Find moles of water
Molar mass of H₂O = 18 g/mol
Moles of water=3200/18=177.78 moles
Final Answer:
A single adult needs approximately 178 moles of water per day through drinking 8 glasses.
ii. The chemical formula for sand is SiO₂ but the sand does not exist in the form of discrete molecules like H₂O. How has its formula been determined keeping in view its structure?
Answer:
Sand (SiO₂) is not a molecular substance like water. It exists as a giant covalent network structure, where each silicon atom is covalently bonded to four oxygen atoms, and each oxygen atom bridges two silicon atoms.
Although there is no single SiO₂ molecule, the smallest repeating unit (known as the empirical formula) in this network is SiO₂.
Why SiO₂ is used as its formula?
It represents the simplest whole number ratio of atoms in the structure.
The macroscopic properties (like melting point, hardness) are explained by this giant structure, not by molecular behavior.
Conclusion:
Even though sand doesn’t exist as individual SiO₂ molecules, its empirical formula is SiO₂ because that is the smallest atomic ratio observed in its three-dimensional lattice structure.
9th Class Chemistry Chapter 4 Solution
Investigative Questions Complete.
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