Chemistry Chapter 4 Short Questions For Class 11 (with Answers)

The most important chemistry chapter 4 short questions for class 11. The 1st-year chapter 4 of chemistry is related to liquids and solids. These questions are for the Punjab Textbook Board and can be used within all of Punjab where this syllabus is taught.

Students are advised to prepare these questions in order to perform the best in the board examination.

Chemistry Chapter 4 Short Questions for Class 11

Q.1: Define Avogadro’s number with examples.
Ans: Avogadro’s number:
“It is the number of atoms, molecules, formula units or ions present in one gram atom of an element, one gram molecule of a substance, one gram formula of an ionic compound or one gram ion of ionic specie”.

Avogadro’s number is denoted by Nₐ and its value is 6.02 × 10²³.
Examples:
1.008g of hydrogen = 1 mole of H = 6.02 × 10²³ atoms of H
18g of water = 1 mole of H₂O = 6.02 × 10²³ molecules of H₂O
58.5g of NaCl = 1 mole of NaCl = 6.02 × 10²³ formula units of NaCl
96 g of SO₄⁻² ion = 1 mole of SO₄⁻² = 6.02 × 10²³ ions of SO₄⁻²

Q.2: Differentiate between mole and Avogadro’s number.
Ans:

Mole	Avogadro’s Number
1) “The mole is the amount of a substance which contains as many elementary entities as there are atoms in 0.012 kg (12 g) of carbon-12.”	1) “The number of entities present in one mole of a substance is a constant number named Avogadro’s Number, i.e. 6.02 × 10²³.”
It is represented by n.	It is represented by Nₐ.
2) 1 Mole of C = 12g 1 Mole of CO₂ = 44g	2) 1 Mole of C = 12g = 6.02×10²³ atoms 1 Mole of CO₂ = 44g = 6.02×10²³ molecules
3) No. of moles of a substance = Mass of substance / Molar mass	3) No. of particles of a substance = (Mass of substance / Molar mass) × Nₐ

Q.3: Calculate the mass in grams of 2.74 moles of KMnO₄.
Ans:
Data:
Mass of KMnO₄ = ?
No. of moles of KMnO₄ = 2.74 moles
Molar mass of KMnO₄ = (39) + (55) + (16 × 4) = 158g mol⁻¹
Using Formula:
No. of moles = mass of KMnO₄ / Molar mass
Mass of KMnO₄ = No. of moles × Molar mass
= 2.74 moles × 158g.mol⁻¹
= 432.92g Ans.

Q.4: Calculate the mass in grams of 2.78 x 10²¹ molecules of CrO₂Cl₂?
Ans:
Data:
No. of molecules of CrO₂Cl₂ = 2.78 × 10²¹ molecules
Molar mass of CrO₂Cl₂ = (52) + (16 × 2) + (35.5 × 2) = 155 g.mol⁻¹
Solution:
No. of molecules = (Mass of CrO₂Cl₂ / Molar Mass) × Nₐ
Mass of CrO₂Cl₂ = (No. of molecules × Molar Mass) / Nₐ
= (2.78 × 10²¹ × 155 g.mol⁻¹) / (6.02 × 10²³)
= 0.7158 g Ans.

Q.5: Calculate the mass in kilogram of 2.6 × 10²⁰ molecules of SO₂?
Ans:
Data:
No. of molecules of SO₂ = 2.6 × 10²⁰
Molar mass of SO₂ = (32) + (16 × 2) = 64g.mol⁻¹
Using Avogadro’s formula:
No. of molecules of SO₂ = (Mass of SO₂ / Molar Mass) × Nₐ
Mass of SO₂ = (No. of molecules × Molar mass) / Nₐ
= (2.6 × 10²⁰ × 64g.mol⁻¹) / (6.02 × 10²³)
= 2.764 × 10⁻²g
To convert it into kilogram,
Mass of SO₂ in kg = (2.764 × 10⁻²) / 1000
= 2.764 × 10⁻⁵ kg. Ans.

Q.6: Calculate the moles of Cl atoms in 0.822 g of C₂H₄Cl₂?
Ans:
Data:
Mass of C₂H₄Cl₂ = 0.822g
Molar mass of C₂H₄Cl₂ = (12 × 2) + (1 × 4) + (35.5 × 2) = 99g.mol⁻¹
Step I: Calculate the No. of Moles of C₂H₄Cl₂.
No. of moles of C₂H₄Cl₂ = Mass of C₂H₄Cl₂ / Molar Mass
= 0.822g / 99g.mol⁻¹
= 8.30 × 10⁻³ moles
Step II:
Calculate the No. of moles of Cl atoms.
C₂H₄Cl₂ : Cl
1 mole : 2mole
8.30 × 10⁻³ : 2 × 8.3 × 10⁻³
= 0.0166 moles of Cl atom Ans.

Q.7: Calculate the number of molecules in 10 gram of ice?
Ans:
Data:
Given mass of ice (water) = 10.0g
Molar mass of water = 18g mol⁻¹
Solution:
Number of molecules of H₂O = (Mass of water in gram / Molar mass of water in g mol⁻¹) × Avogadro’s number
= (10 / 18g mol⁻¹) × 6.02 × 10²³
Number of molecules of water = 0.55 × 6.02 × 10²³
= 3.31 × 10²³ molecules of H₂O

Q.8: Calculate the number of molecules in 3.6 gram of water?
Ans:
Data:
Given mass of ice (water) = 3.6 g
Molar mass of water = 18g mol⁻¹
Solution:
Number of molecules of H₂O = (Mass of water in gram / Molar mass of water in g mol⁻¹) × Avogadro’s number
= (3.6 / 18g mol⁻¹) × 6.02 × 10²³
Number of molecules of water = 0.2 × 6.02 × 10²³
= 1.204 × 10²³ molecules of H₂O

Q.9: One mole of H₂O has two moles of bonds, three moles of atoms, ten moles of electrons and twenty-eight moles of total fundamental particles.
Ans:
Water molecule is formed when two atoms of hydrogen combine with one atom of oxygen as.
¹⁶₈O = 8p + 8e + 8n
¹₁H = 1p + 1e

According to Avogadro’s principle, 1 mole of water H₂O contains 6.02×10²³ molecules.
i) Bonds: 1 molecule of water contain 2 bonds.
H₂O molecules : bonds
1 : 2
6.02×10²³ : 2 × 6.02×10²³
: 1.204 x 10²⁴ (2 moles) Ans.
ii) Atoms: 1 molecule of water contain 3 atoms.
H₂O molecules : atoms
1 : 3
6.02×10²³ : 3 × 6.02×10²³
: 1.806 x 10²⁴ (3 moles) Ans.
iii) Electrons: 1 molecule of water contain 10 electrons. (8e⁻ of O + 2e⁻ of H)
H₂O molecules : electrons
1 : 10
6.02×10²³ : 10 × 6.02×10²³
: 6.02 × 10²⁴ (10 moles) Ans.
iv) Fundamental particles:
(1 molecule of water contains 28 electrons. [O (8p + 8e⁻ + 8n) + 2H (2p + 2e⁻)])
H₂O molecules: Fundamental particles
1 : 28
6.02×10²³ : 28 × 6.02×10²³
: 16.8 x 10²⁴ (28 mole) Ans.

Q.10: One mole of H₂SO₄ should completely react with two moles of NaOH. How does Avogadro’s number help to explain it?
Ans:
In a neutralization reaction one mole of H⁺ ions (i.e. 6.02×10²³) from an acid combines with one mole of OH⁻¹ (i.e. 6.02×10²³) ions from a base to form one mole of water as.
H⁺(aq) + OH⁻¹(aq) ———> H₂O(l) ……………………(1)
One mole of H₂SO₄ ionizes in water to produce 2 moles (2 x 6.02×10²³) of H⁺ ions as.
H₂SO₄(aq) ⇌ 2H⁺(aq) + SO₄²⁻(aq)
While, one mole of NaOH ionize to give one mole (6.02×10²³) of OH⁻¹ ions.
NaOH(aq) ⇌ Na⁺¹(aq) + OH⁻¹(aq)
Now according to Equation 1, to neutralize two moles (2 x 6.02×10²³) of H⁺ ions of H₂SO₄, we need two moles (2×6.02×10²³) of NaOH.

Q.11: Which laws are to be obeyed during stoichiometric calculations.
Ans:
The following two laws are obeyed during stoichiometric calculations.
(1) Law of Conservation of mass:
“Mass can neither be created nor destroyed during a chemical reaction, but it only changes from one form to another form”.
(2) Law of Definite proportions:
“This law states that a pure chemical compound always contains same elements chemically combined together in a fixed ratio by mass.

Q.12: How many moles of CO₂ can be produced from burning one mole of octane.
Ans:
Octane burns in air as follows:
2 C₈H₁₈(l) + 25 O₂(g) ———> 16 CO₂(g) + 18 H₂O(l)
According to the balanced chemical equation,
Octane : CO₂
2 mole : 16 mole
1 : 16/2
1 : 8 moles of H₂O Ans. (Note: Although the text says H2O here, it refers to CO2 based on the question and calculation)

Q.13: What are the steps involved in the identification of a limiting reactant?
Ans: Steps for the identification of limiting reactant:
To identify a limiting reactant, the following three steps are performed.
i. Calculate the number of moles from the given amounts of reactants.
ii. Find out the number of moles of product with the help of a balanced chemical equation.
iii. Identify the reactant which produces the least amount of product as limiting reactant and the other as an excess reactant.

Q.14: What is the importance of limiting reactant in industry?
Ans: Importance of limiting reactant in industry:
Frequently, a large amount of inexpensive reactant is supplied because of the following reasons:
a. To ensure that whole of the mass of the expensive reactant is completely converted to the desired product.
b. To produce maximum amount of product.
c. To increase the rate of reaction.

Q.15: How concept of limiting reactant is applied in smothering of fire?
Ans: Smothering of fire:
Fire is a combustion reaction in which fuel and oxygen, O₂, combine, usually at high temperatures, to form water and carbon dioxide. Once the fire has started, it is self-supporting. An effective way to quench a fire is smothering, which reduces the amount of available oxygen below the level needed to support combustion. In other words, smothering decreases the amount of the excess reactant. Foams, inert gas, and CO₂ are effective substances for smothering.

Q.16: How can we calculate the efficiency of a chemical reaction? / Define percentage yield. Give its formula.
Ans:
The efficiency of a chemical reaction is calculated in the term of % age yield.
Percentage yield:
“Percentage yield is defined as: it is the ratio of the actual yield to the theoretical yield multiplied by 100”.
% age yield = (Actual yield / Theoretical yield) × 100

% age yield show the efficiency of a chemical reaction.

Q.17: Write two important aspects of stoichiometry for synthesis of drugs?
Ans: Importance of Stoichiometry in drug synthesis:

Stoichiometry ensures the accuracy of drug synthesis. Any deviation can result in incomplete reaction or contamination with un-reacted reactants or by-products.
Stoichiometry allows chemists to precisely control chemical reactions to produce drugs, to ensure its efficiency, effectiveness and safe use.

Q.18: Give any three points for the importance of stoichiometry in medicine?
Ans: Importance of Stoichiometry in Medicine:
Stoichiometry is very important in the field of medicine and is used to:

In the preparation of antibiotics, the stoichiometry ensures that each dose matches the active ingredient and target bacteria.
To determine the cholesterol level in the blood of patients. Cholesterol is a form of fat that is not all bad. However, cholesterol can have harmful effects.
To determine the glucose level in the blood of diabetic patient. Use of insulin relies on the stoichiometry to precise control of blood sugar levels.

Q.19: What is the importance of stoichiometry in determination of correct dose for patients?
Ans: Importance of correct dose of medicine:
To determine the amount and number of drugs to give a dosage to a patient. The medicine has no effect when given in small amounts and can cause toxic state or death when given in large amounts.
Example:
Paracetamol is used as a pain killer and to decrease fever. An overdose may result a blood thinning, organ damage and severe liver damage.